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Tags: matematika informatika imtihonga tayyorgarlik
Year: 2017
Text
ABITURIYENT KUTUBXONASI
TEST TOPSHIRIQLARI
YECHIMI
MATEMATIKA
va INFORMATIKA
2016
Toshkent
«Spectrum Media Group»
2017
1-variant
1. Oltin va durdan yasalgan
bezakning oglrligi 3 misqol, narxi
24 dinor. Agar 1 misqol oltin 5 dinor,
1 misqol dur 15 dinor tursa, bezakda
qancha misqol oltin bor?
Yechish:
x - oltin, у - dur.
ix + y = 3
[5x + 15y = 24=*
(5x + 5y = 15
=>[5x + 15y = 24
-10y = -9
у = 0,9, x = 2,1
2,1 misqol oltin.
Javob: 2,1.
2. ctg2nx- л/з^ 2x - 5x2 = 0 ildizlari
yig’indisini toping.
Yechish:
Aniqlanish sohasi:
(sin 2лх Ф 0
\3-2x-5x2 >0^
[2лх*яп
[5x2 +2x-3<0^
n
ХФ —
I 2 =>
(x + 1)(5x-3)<0
хф— ,neZ
=> 2
-1 <x<0,6
---G —G------G----e----•-
-1 -0,5 0 o,5 0,6
Tenglamani yechamiz.
1) ctg2xx = 0, cos2nx = 0,
2xx = — + лп,х = — + —, n GZ
2 4 2
2)1}3-2х-5х2 =0
3 - 2x- 5X2 = О, 5X2 + 2x- 3 = 0
x =-1, x = 0,6
Aniqlanish sohasiga tegishli
yechimlarni topamiz.
1)x=-+-,nCZ
4 2
n = Oda x = —
4
n =-1 da x = ~—
4
n = -2 da x = —0,75
2) x = 0,6, x = -1 aniqlanish sohasiga
tegishlisi x = 0,6.
Demak, tenglamaning ildizlari:
x = 0,25,x = -0,25, x = -0,75,
x = 0,6
Ildizlari yig ‘indisi:
0,25 + (-0,25) + (-0,75) + 0,6 = -0,15.
Javob: -0,15.
fx + 2y = 5
3. { tenglamalar
[x +y + 5 = 4xy
sistemasining yechimlaridan iborat
nuqtalarni tutashtiruvchi kesmaning
uzunligini toping.
Yechish:
Birinchi tenglamadan x ni topib
ikkinchi tenglamaga qo'yamiz.
(x = 5-2y
[(5 - 2y)* 2 + у + 5 = 4(5-2y) -у
25 - 20y + 4У2 + y + 5 = 20y - бу2
12/ -39y + 30 = 0
4/ -13y+10 = 0
У1=2,
5
xi=5-2y1=5-4=1 (1;2)
c л — 5 5 ,5 5 i
x2 = 5-2y2 = 5—=-
y 2 2
A(1;2), B(^;^)
Ava В nuqtalarni tutashtiruvchi
kesmaning uzunligi
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Yechimlar. Matematika va informatika 2017
1-variant
4. a-i, az.as ketma-ketlikda
ixtiyoriy uchta ketma-ket hadining
уig'indisi 40 ga teng. Agar ketma-
ketlikning uchinchi hadi 9 ga teng
bo'lsa, birinchi va sakkizinchi
hadlarining yig'indisi nechaga teng?
Yechish:
a? + аг + аз = аз + a-» + as = ... =
= as + ay + as = 40
аз = 9
ai + аг + аз - 40, ai + аг - 31
аг + аз + Э4- 40, a2 + Э4 = 31
аз + в4 + as = 40, а^ + as = 31
ai + ав = а4 + as- 31.
Javob: 31.
5. е1”2* + x"1* = 2e4 tenglamani yeching.
Yechish:
e1"2' + x"“ = 2e4
(elnx)lnx + x"“ = 2e4
elnx = x bolganligi uchun
+ jnx _ 2e4
2xnx = 2e4, 4nx = e4 tenglikning ikkala
qismini logarifmlaymiz.
!nXnx = Ine4 ln2x = 4, Inx = ±2
1) Inx = 2, x = e2
2) Inx = -2, x = e~2.
Javob: e2 va -l, .
e
6. Tenglamani yeching:
4x-2 + j2x-5 +
+ yJx + 2 + 3j2x-5 = 7 42 .
Yechish:
j2x-5 = a belgilash kiritib yechamiz.
2x - 5 = a2, 2x = a2 + 5
a2 +5
x =-----
2
a2+5
------2 + a +
2
J^^-+2+3a=7x/2
N 2
1 a2 + 5-4 + 2a
2
a2 +5 + 4 + 6a _? ft
N 2
la2 + 2a+ 1 la2 + 6a+ 9 rr
N 2 + V 2 -
l(a+1)2 ! l(a + 3)2 _7д
V 2 V 2
V2 V2
a + 1 + a + 3 -7-42 -^2
2a+4= 14, 2a = 10,a = 5
42x-5=5, 2x-5 = 25
2x = 30, x = 15.
Javob: 15.
7. (x2 + Зх + З)^ + x + 3) = 3xz
tenglama haqiqiy ildizlari yig'indisi
modulini toping.
Yechish:
x2 + 3 = a belgilash kiritamiz.
(x2 + 3x + Sjix2 + x + 3) = 3x2
(a + 3x)(a + x) = Зх2
a2 + 4xa + Зх2 = Зх2
a2 + 4xa = 0, a(a + 4x) = 0
a = 0, a = -4x
x2 + 3 = 0va)4 + 3= -4x
14 + 3 = 0 tenglama haqiqiy ildizga ega
emas.
x2 + 4x + 3 = 0, xi + x2 = —4
|xr + x2\ = |-4| = 4.
Javob: 4.
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Yechimlar. Matematika va informatika 2017
1-variant
8. Silindr o'qiga parallel qilib
o'tkazilgan tekislik uning asosidan <p
kattalikdagi yoyni ajratadi. Silindr o'qining
uzunligi h bo'lib, undan kesuvchi
tekislikkacha bo'lgan masofa a ga teng.
Kesim yuzini toping.
Yechish:
OO7 = H,
Z AOiB - (p. OiC = a,
AAi = BB1 = h.
Skesim — AB'AAj
AAOjB dan = tg—
O,C 2
AC = OiCtg— = a-tg—
2 2
AB = 2OiC
Skesim ~ 2atg^ h.
Javob: 2atg-h.
2
9. Iog3(5 - x2) л/3-2х-х2 = 0
tenglama ildizlari yig'indisini toping.
Yechish:
Aniqlanish sohasini topamiz.
^5-x2>0 \x2-5<0
\3-2x-x2 >0 \x2 +2x-3<0
f(x-V5)(x + V5)<0
[(x - 1)(x + 3) < 0
1) (x~45 )(x+45 ) <0
(-45; 45)
2)(x-1)(x + 3)<.0
-3 -1
3 -J5 1 J5
xC(-45; 1]
Tenglamani yechamiz.
1одз(5 -x2) = Ova
3-2x-4 = 0
5-x2 = 3°
x2 = 4, x = ±2
x = -3, x = 1
Aniqlanish sohasiga x = -2, x = 1
tegishli. Ildizlari yig'indisi
-2+1 = -1.
Javob: -1.
10. Hisoblang:
log6(V2-V3 +4г + 4з ).
Yechish:
^2-43 + ^2 + 43 = x ifoda qiymatini
topamiz.
(^2-43 + ^2 + 43 )2=4
2 -43 + 2^(2 - 43)(2 + 43) +
+ 2+4з =x2
4 + 244-3
^ = 4+ 2,>? = 6,х=4б
1оде(^2-4з +42 + 43 ) =
- 1
- 1од64б - 1одв62 =—.
Javob: —.
2
11. lx2 - 5ах| = 15а tenglama yagona
yechimga еда boladigan a ning
barcha qiymatlari yig'indisini toping.
Yechish:
lx2 - 5ax| - 15a tenglamada 15a>0,
a>0bolishikerak. a = 0bolganda
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Yechimlar. Matematika va informatika 2017
1-variant
esa tenglama yagona yechimga ega
bo'ladi, chunk)
lx2 - 50-x| = 150 yoki lx2) = 0, x = 0.
Javob: a = 0.
12. Tenglama ildizlari yig'indisini
toping |log4(|x| + 3)| = |x|.
Yechish:
Aniqlanish sohasi
|x| + 3 > 0, x C R.
flogXI XI +3) = x
(log4(jx|+3) = -x
J|x|+3 = 4X (x = 1
[I x | +3 = 4~x \x = -1
x = ±1 - tenglama ildizlari.
Ildizlari yig'indisi 1 + (-1) -0.
Javob: 0.
13. To‘g‘ri burchakli uchburchakning
bir burchagi 38° ga teng bo'lsa, to‘g‘ri
burchak uchidan tushirilgan bissektrisa
va balandlik orasidagi burchakni toping.
Yechish:
A E D В
Z C = 90°, Z A = 38°
CD - balandlik, CE - bissektrisa
Z ECD ni topamiz.
CE- bissektrisa, Z ACE = Z ECB = 45°
CD - balandik, Z ACD = 90° - 38° = 52°
Z ECD-Z ACD - Z ACE =
= 52° - 45° = 7°.
Javob: 7°.
14. Tenglama ildizlari yig'indisini
2 ^3
toping: arcsin(x - 5x) = arccos( ——)
Yechish:
.. I | it 5it
1) arccosi-------= л- — = —
i 2 , 6 6
2) arcsinfx2 -5x) = —
6
. . ? c 5it 1
sinarcsinlZ - 5x) =—, sin— = —
6 6 2
x2 - 5x =—, x2 - 5x- 0
2 2
D>0, ildizlari yig'indisi x-i + хг = 5.
Javob: 5.
15. V4x + 13-V3x + 12=-Vx + 1
tenglamaning ildizlari yig'indisini toping.
Yechish:
Tenglamaning aniqlanish sohasi
13
4x + 13>0 x~ 4
3x + 12>0=><x>-4 =>x>-7
x + 1>0 x>-1
V4X + 13 =V3x+12-VxT7
tenglikning ikkala qismini kvadratga
ko'taramiz.
4x + 13 = 3x + 12-
- 2 7(3x + 72)(x + 1) + x + 1
2yj(3x + 12)(x + 1) = 0
3x+12 = 0,x=-4
x + 1 = 0, x = -1
x = -4 aniqlanish sohasiga tegishli
emas.
x =-1 tenglama ildizi.
Javob: -1.
16. /(x) = (sinx)C0S* bo'lsa, f’( —)
6
ni toping.
Yechish:
Daraja ko'rsatkichli funksiya bolganligi
uchun
у = (sinx)C0SX ning ikkala qismini
logarifmlaymiz.
у = (sinx)cosx, Iny = ln(sinx)C0SX
Iny = cosxln(sinx)
Hosilani topamiz.
(Iny)' = (cosxlnsinx)'
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Yechimlar. Matematika va informatika 2017
1-variant
y' ... cosx
— = - sm x In sin x + cos x-
у sinx
2
y' = y(-sinxlnsinx + cos x )
sinx
У = (sinx) '(-smx'lnsmx +---------)
sinx
Hosilaning — dagi qiymatini topamiz.
6
a = 0, b < 0, x + c # 0, x # -с, c > 0
Demak, a - 0, b <0, c> 0.
A) a2 + be = be < 0- to‘g‘ri
B) be-a > 0, be < 0 bolishi kerak-
noto‘g‘ri.
C) ac = O- to‘g‘ri
D) a4-bc = -be >0, be <0- to‘g‘ri.
Javob: be - a > 0, be < 0 bolishi
kerak - noto'g'ri.
18. Agar у > x > 0 bo'lsa,
2 x3 X4 x5
ус +—T + —-- + ... ni soddalashtiring.
У У У
Yechish:
у >x > 0
x2 + —у + —5- + -^- + - geometrik
progressiya.
x
y> x> 0, q =— < 1 cheksiz
У
kamayuvehi geometrik progressiya.
x3
e .. b, У4 X\
1-q /'I у)
у
17. Rasmda у = a +..- funksiya
x+c
grafigi tasvirlangan. Quyidagilardan
qaysi biri noto‘g‘ri?
Yechish:
b
у = a +---
x + c
x3 у x3
у4 y-x y3(y-x)
Yig'indini hisoblaymiz:
x3 2 x3
+ y3(y-x) X+y4-xy3'
Javob: x2+—--------.
у - xy
19. у = 4coszx + sinzx funksiya
butun qiymatlari yig'indisini toping.
Yechish:
1) sin2x + cos2x = 1
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Yechimlar. Matematika va informatika 2017
1-variant
2) 4cos2x + sin2x = 3cos2x +
+ cos2x + sin2x = 3cos2x + 1
3) 0 < cos2x < 1 bo'lganligi sababli
у = 3cos2x + 1
1 <y<4
[1; 4] butun yechimlariyig'indisi:
1+ 2 + 3 + 4= 10.
Javob: 10.
20. Teng yonli trapetsiyaning
asoslari 4 va 12 bo'lsa, unga ichki
chizilgan doira yuzini toping.
Yechish:
ABCD - teng yonli trapetsiya,
AB = CD, BC = 4,AD = 12
1) BN = BK = NC = CM = 2
KA = AP = PD = DM = 6
2) AB = BK + KA=2 + 6 = 8
3) AE = AP-EP = 6-2 = 4
4) BE2=AB2-AE2 = 8?-42 =
= 64-16 = 48
BE =443
5) R = OP = &L=i^l=243
2 2
6)S = nf4 = (243 )4 = 12п.
Javob: 12л.
21. Tenglamaning eng kichik
yechimini toping (x + 1)'°Sg<x+1’ = 2.
Yechish:
Aniqlanish sohasi: x + 1 > 0, x > -1
Tenglikning ikkala qismini ikki asosga
ko'ra logarifmlaymiz.
log2(x + 1)^(x+,)=log22
log2(x + 1)log2(x + 1)=1
log24(x + 1) = 1
log2(x + 1) =±1
1) log2(x + 1) = 1, x+ 1 =2, x= 1
1 1
2) log2(x + 1) =-1, x + 1 , x = -—
1
Tenglamaning yechimlari - — va 1.
Eng kichik yechim -0,5.
Javob: -0,5.
22. (3x - 8) log5(7x - 2X2 - 5) = 0
tenglama ildizlari ko'paytmasini toping.
Yechish:
Aniqlanish sohasini topamiz.
7x-2x2-5>0
2x2-7x + 5<0
к
2(x- 1)(x-^)<0, 1 <x<2,5
Tenglamani yechamiz.
1) 3x - 8 = 0, x =^= 2^ aniqlanish
sohasiga tegishli emas.
2) log5(7x - 2X2 - 5) = 0
7x-24-5 = 1
2x2 — 7x+6 = 0
2(x-j)(x-2) = 0
x = 1,5, x = 2 aniqlanish sohasiga tegishli.
1,5- 2 = 3 ildizlar ko'paytmasi.
Javob: 3.
23. Uchburchakning uchlari to‘g‘ri
burchakli dekart koordinatalar
sistemasida quyidagicha berilgan:
A(0; 0), B(-1; -2), C(-2; 0).
Uchburchak yuzini toping.
Yechish:
A(0; 0), B(-1; -2), C(-2; 0)
ABC - teng yonli uchburchak.
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Yechimlar. Matematika va informatika 2017
1-variant
AC- asos, AC = 2, BD- balandlik,
BD = 2
AC-BD 2-2
° две - 2 ~ 2 ~Z
Javob: 2.
24. (52x+7 - 25) -V4-x2 = 0
tenglama ildizlari yig'indisini toping.
Yechish:
Aniqlanish sohasi
4-x2 ^.0, x2 -4<0,
(x - 2)(x + 2)<0,-2^xi2
Tenglamani yechamiz:
1)52x+7 -25 = 0
52x+7 = 25, 52x+7 = 52
2x + 7 = 2, 2x=-5, x = -2,5
2)4-x? = 0,x? = 4,x = ±2
x = -2,5 yechim aniqlanish sohasiga
tegishli etnas.
x = ±2 tenglama ildizlari.
Ildizlariyig'indisi: 2 + (-2) = 0.
Javob: 0.
25. /(x) = e';''2 3'4 bo'lsa, /'(8) ni
toping.
Yechish:
1) (ex)' = ex, (e^b)' = (kx + b)'-ekx+b
(e'C^j = е4^,-^хг-Зх-4)
2x-7 = аг,2х = a2 + 7
a + 5 + a + 3 = 5- ^/2 • V2
2a + 8 = 10, 2a = 2, a = 1
>l2x-7 =1, 2x-7 = 1, 2x = 8, x = 4.
Javob: 4.
27. Tenglama nechta butun
yechimga ega? V2x-x2 = 2-ln|x — 7|
^Jx2 3x 4 2x 3
2у/хг-Зх-4
2) f(8) = в'*2 334
2-8-3
2л182-3-8-4
13 _13ee
2-6~ 12
. u 13e6
Javob: ----
12
26. x + Q + 5л/2х — 7 +
+ л/х + 1 + Зл/2х^7 = 5>/2
Yechish:
V2x - 7 = a belgilash kiritamiz.
Yechish:
Aniqlanish sohasi:
(2x-x2>0 ixs-2x<0
[|x-1\>0 \x>1,x<1
fx(x-2)<0 f0<x<2
[x<1,x>1 [x<1,x>1
——
0 12
[0; 1) U (1; 2]
Aniqlanish sohasiga tegishli butun sonlar
tenglamaning yechimi bolishimumkin.
x = 0, x = 2
1)x = 0da y/2 0-0 = 2ln\0 - 1\
0 = 2ln1, 0 = 0
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Yechimlar. Matematika va informatika 2017
1-variant
2)x = 2da y/2-2-2* 2 = 2ln\2 - 1\
0 = 2ln1, 0 = 0
Demak, x = 0va x = 2 tenglamaning
yechimi.
Javob: 2 ta butun yechim.
28. (tgjrx + 1)л/2 + Зх-2х2 = 0
tenglama ildizlari yig'indisini toping.
Yechish:
Aniqlanish sohasi:
[costtx * 0
[г+зх-гх^о^
Л
1Г.ХФ-1-7ГП
2
2x2-3x-2<0
1
Х& — + П
2
4 ( n
2^x+-J(x-2)<0
( 1
x*- + n,n<EZ
2
<x<2
I 2
n=O,xjt-
2
4 o.3
n = 1, хФ—
2
n = -1, x/-—
2
2) j2 + 3x-2x2 = 0,
> 1
2x2-3x-2 = 0,x=--,x = 2
2
1 . .
x=~— tenglama yechimi emas.
Tenglama ildizlari yig'indisi:
-0,25 + 0,75+ 1,75 + 2 = 4,25.
Javob: 4,25.
29. J2x3dx ni hisoblang.
0
Yechish:
f x"dx = —— + С да asosan
J n + 1
f2x3dx = 2fx3dx =
0 0
= 2-—1
4 0
x4 1 =1_
2 0~ 2
. u 1
Javob: -.
2
30. To'rtburchakli muntazam prizma
diagonali d ga teng bo'lib, yon yoq
tekisligi bilan a burchak tashkil etadi.
Prizma hajmini toping.
Yechish:
ABCDA1B1C1D1 - muntazam
to‘rtbdrchakli prizma.
B1D = d, Z B1DC1 = a,
V= ?
Y — Sasos'hi, Sasos ~ 3
CiD yon yoq diagonali.
_1 1 i 2
2 2 2
Tenglamani yechamiz.
1) tgrcx + 1 = 0, tgxx = -1,
лх = -—+лп,х=-— + п, aniqlanish
sohasiga tegishli yechimla'ri:
1
n = 0 da x= —
4
n = 1 da x = 0,75
n = 2 da x = 1,75
1) ABiDCi dan asos tomonini topamiz.
BA .
—L-L = sin a ,
B,D
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Yechimlar. Matematika va informatika 2017
1-variant
chunki ABiC-iD to'g'ri burchakli.
B1C1 = BiD sina yoki a = d-sina
2) Sasos = a2 = cfsirfa
3) H = BjB ni topamiz. &B,BD to‘g‘ri
burchakli.
B,B = B,D2 - BD2
BD - kvadrat diagonal!
BD = aj2
H = yjd2-(ay[2)2 = у/d2-2a2 =
= yld2 -2d2s'ma =
= dy/l -2sm2 a = dy/cos2a
4) V = Sasos-H = cf-sin2a-d-
yjcos2a - d3sin2a Vcos2a .
Javob: d3sin2a Vcos2a .
31. MS Excel 2003 dasturida «=МОПРЕД» funksiyasi bu:
Yechish:
MS Excel dasturida МОПРЕД funksiyasi matrisa determinantini hisoblaydi.
=MOnPE£(massiv).
massivda albatta sonli kvadrat massiv bolishi lozim. Kvadrat deganda
massivda ustunlar soni satrlar soniga teng bolishi kerak.
Masalan:
(2 ЗА
4 5 matrisa beril9an-
Bu matrisa 2 ta ustun va 2 ta satrdan iborat. Demak, kvadrat matrisa. Bu
matrisaning determinantini matematik yo‘l bilan ishlaymiz.
det(A) = 2-5-3-4=10-12 = -2.
Demak, A matrisa determinanti-2. Bu misolni MS Excelda yozish uchun A1
yacheykaga 2 ni, B1 yacheykaga 3 ni A2 yacheykaga 4 ni, B2 да 5 ni kiritib A3
yacheykaga quyidagi formulani kiritamiz.
=МОПРЕД(А1 :B2)
natijada A3 yacheykada -2 hosil bo'ladi.
Javob: maydonda joylashgan determinantning qiymati.
32. Tenglik o'rinli bo‘lishi uchun sonlarning asosi qanday bo'lishi kerak?
24003(X)=26010(xr2004(x)
Yechish:
2004(X) + 24003(X) = 26010(X).
7 lik sanoq sistemasi qo‘shish jadvali.
+ 1 3 4 5 6
1 3 5 6
1 1 2 4 6 10
2 2 3 5 6 10 11
3- -3- —4— -6- -4i ) 11 12
-4- -4- —ё—\ 10 11 12 13
5 5 6 to 11 12 13 14
6 6 10 11 12 13 14 15
Qo‘shish bosqichi
1) Raqamlarni oxiridan boshlab qo'shamiz.
4 + 3 = 10 0 yoziladi 1 yodda.
0 + 0 = 0 1 yodda bolgani uchun 0+1 = 1
0 + 0 = 0
2 + 4 = 6
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Yechimlar. Matematika va informatika 2017 1-variant
2004 + 24003 = 26010.
+ 2004(7)
24003,7,
26010(7)
x - yettilik sanoq sistemasi.
Javob: yettilik.
33. HTMLtiiidagi hujjatda <PRE>...</PRE> teglarining vazifasi:
Yechish:
HTML tilida probellarni saqlash maqsadida <pre>.. </pre> tegi ishlatiladi. Bu
teg yordamida nafaqat probellar, balki matnning oldindan formatlangan holatini,
harxil surishlar bajariigan bo'lsa, ulaming holatini saqlab qolish uchun ham ishlatiladi.
Javob: probellarni o‘z o’rnida saqlaydi.
34. Faqat axborot ko'rinishlari berilgan javobni aniqlang.
Yechish:
Axborot - inson hayotida muhim omil bo'lib, axborot orqali inson o'z
tasawurini boyitib boradi, atrof-muhitdagiyangiliklardan xabardor bo'lib boradi.
Axborotning uzlukli va uzluksiz turlari mavjud.
Uzlukli, ya’nidiskret axborotlarga: rasm, tasvir, grafik, jadval, matn va boshqalar.
Uzluksiz axborotga, ya’ni analog axborotga: ob-xavo, vaqt, harorat, tovush
va boshqalar.
Javob: matn, grafik, tovush.
35. Hisoblang va javobini 16 lik sanoq sistemada ifodalang:
1001011(2)*4(io) + 234(8).
Yechish:
410-X2.
4to - 100(2).
1001011(2)-100(2) = 100101100(2).
234(8) -< 010011100(2). (1-chi jadval) (ilovaga qarang).
100101100(2) + 010011100(2).
100101100
0100111-00
111001000(2) 000111001000- 1C816.
1 C 8
Javob: 1C8.
36. Ushbu o'zgarmaslar ichidan to‘g‘ri yozilganini aniqlang:
Yechish:
O'zgarmas son suzuvchi vergulni o'z ichiga olmasligi kerak.
Masalan, 0,5 suzuvchi vergulni, chunki 0,5 = 5-10~1 - 50-10~2.
Shuning uchun o'zgarmas 92624E-2 = 92624-10~2.
Javob: 92624E-2.
12
Yechimlar. Matematika va informatika 2017
2-variant
2-variant
. log,(5x-6) „ . .
1. 72....—— = 2 tenglama
Vid-3x
ildizlari yig'indisini toping.
Yechish:
Aniqlanish sohasi:
[5x-6>0 J5x>6
[10 - 3x > 0 |-3x > -10
' 6
X > 5 6 10
=>< => — < x < —
10 5 3
x<—
I 3
Aniqlanish sohasiga tegishli butun
sonlar x = 2, x = 3.
1)x = 2da 10ёз(5-2-6) =
J10-3-2
_ log34 _ log34
V? 2
2)x = 3da 1овз(5'3~£> =
•J10-3-3
-^.logj9.2
Demak, x-3 tenglama yechimi.
Tenglama ildizlariyig'indisi 3.
Javob: 3.
2. Piramidaning asosi teng yonli
uchburchak bo'lib, asosidagi burchak a.
Piramida asosidagi har bir ikkiyoqli
burchak <p = 90° - a. Piramida yon sirti
S bo'lsa, to'la sirtini toping.
Yechish:
DABC - uchburchakli piramida.
ABC- tengyonli uchburchak. AB = AC
о
c
AABC-AACB = a
Z AED = <p = 90°-a
Syon — S
1) sasos ~ S cos<p -
= S-cos(90° - a) = S-sina
2) Sto‘la — Casos + Syon “ C’COStp + S —
= S(1 + cos<p) = S(1 + sina) =
= 2Scos2(45° /
Javob: 2Scos2(45° ~^)-
3. Uchburchakli muntazam
prizmaga shar ichki chizilgan, bu shar
prizmaning uchala yog'iga va ikkala
asosiga urinadi. Shar sirtining prizma
to'la sirtiga nisbatini toping.
Yechish:
ABCA1B1C1 - muntazam uchburchakli
prizma.
S - shar sirti. S = 4nR?
Si - prizma to'la sirti.
13
Yechimlar. Matematika va informatika 2017
2-variant
KEN- teng tomonli uchburchak prizma
asosiga parallel tekislik. Bu tekislik
shar kesimida KEN uchburchakka ichki
chizilgan shaming DFQ katta doirasini
hosil qiladi. AEOD da OD = R,
Z DEO = 30°, ED = R43 , EN = 2R^3 ,
Syon = 3aH = 3-2R^3 -2R = 12R*43
агУз (2Rj3)2V3 г к
asos 4 4 'J
StO'ia = 2-ЗЕ?4з + 12R2 л/з = ISR2^
S:Si = 4nR2:18R2y/3 = 2л; 9^3
S 2л
s,~ э4з'
i. 2?r
Javob: —.
9V3
4. 3х + 2х1°ег3 -x = 0 tenglama
ildizlari ko'paytmasini toping.
Yechish:
3х + 2Х'°®23 x = 0
3х + 210823' x = 0
a^b = b
3х + 3х x = 0, ?(1 + x) = 0
3х # 0, 1 + х = 0, х = -1
х = —1 - tenglama ildizi. Ildizlari
ko'paytmasi-1 да teng.
Javob: -1.
[xy = 2е
5. { sistema nechta
[log2x = y-1
yechimga ega?
Yechish:
Birinchi tenglamaning ikki qismini
2 asosga ko'ra logarifmlaymiz.
log2xy = fog22e
ylog2x = 6, log2x =—
У
log2x = y- 1,
6 2
- = y- 1,y2-y-6 = 0
У
Tenglama ildizlari у = -2 va у = 3.
1) у = -2 da log2x = -2-1, log2x = -3
2) у = 3da log2x = 3-1, log2x = 2
x = 22 = 4, (4; 3)
2 ta yechim.
Javob: 2.
6. у = f (x) ftinksiya maksimum
nuqtasi xo bo'lsin. U holda qaysi
tengsizlik xo ning qandaydir atrofidagi
barcha x lar uchun o'rinli?
Yechish:
У = f(x)’ xo 6 D(y) maksimum nuqta
uchun shunday r> > 0 mavjudki,
x C(x0- o; xo + o) oraliqda f(xQ) > f(x)
bo'ladi.
Javob: f (x) < /(xo).
7. ABC uchburchakning yuzi S, bir
tomoni AC = b va Z CAB = a. Shu
uchburchakni AB tomon atrofida
aylantirishdan hosil bo'lgan jism
hajmini toping.
Yechish:
ABC uchburchakni AB tomon atrofida
aylantirishdan ikkita konus hosil bo'ladi.
1 1
V = -ttDC2 AD + -лОС2 DB =
3 3
= -DC2(AD + DB) = -DC2 AB
3 3
_ DC AB . .
S =-----, DC = b sina
2
DCAB = 2S
14
Yechimlar. Matematika va informatika 2017
2-variant
VDC-AB-DC =— -2Sb-
3 3
2xSb sin а
sina = ----.
3
, , 2iSbsina
Javob:-------.
3
8. To'gTi parallelepiped asosi
rombdan iborat bo'lib, uning diagonal
kesimlarining yuzalari Si va S2 da teng.
Parallelepipedning yon sirtini toping.
Yechish:
Asosi romb.
Si = df H, S2 = d2H
Byon “ 4a'H
S, d,-H = d,
7 S2 d2 H d2
ctg\ 5x + — | = —^L = ->/3
I 3) 4з
5x + — = arcctg(-y/3) + лп,п e Z
Л Л
5х + — -л + лп
3 6
Л Л
5х = л-----+ лп
6 3
. . Л ЛП
Javob: — + —
10 5
10. у = 7sin5x + 5sin7x funksiyaning
hosilasini toping.
Yechish:
1) (sin(kx + b))' = kcos(kx + b)
(sin5x)' = 5cos5x
(sin7x)' = 7cos7x
2) y’ - 7(sin5x)'+ 5(sin7x)' = 35cos5x +
+ 35cos7x = 35(cos5x + cos7x)
3) cos5x + cos7x yig'indidan
ko'paytmaga o‘tamiz.
c „ 5x + 7x
cos5x + cos7x = 2cos----------
2
2)~~4 a
/df+df_1 !s2 S2 _
V 4 2yH2 H2
Jsf+s2
2H
S =4- H = 2Jsf + S:
уол 2H ' '
Javob: 2^S2 + S2 .
5x-7x „ „
cos------= 2cosoxcosx
2
4) y' = 35-2cos6xcosx = 70cosxcos6x.
Javob: 70cosx cos6x.
9. Tenglamani yeching:
V3ctg^5x + -|^ = -3.
Yechish:
J~3ctg f 5x + = —3
11. 2^ = cos(x2 - 9) tenglama
ildizlari yig'indisini toping.
Yechish:
1) Aniqlanish sohasi:
x-3>0,x>3
2) -1 < cosfx2 - 9) < 1
0 < 2^ < 1, 2^ < 2°
Vx-3 s 0, x < 3
3) x^3va x<3 da x = 3 ekanligi kelib
chiqadi. x = 3 tenglama yechimi.
2^3 = cos(32 - 9)
2? = cost)
1 = 1
15
Yechimlar. Matematika va informatika 2017
2-variant
Demak, x = 3 tenglama ildizi. Ildizlari
yig'indisi 3.
Javob: 3.
12.11 ga karrali uch xonali natural
sonlar nechta?
Yechish:
Eng katta uch xonali son 999, eng
katta ikki xonali son 99. 11 ga karrali
sonlarni topamiz.
11 ga karrali uch xonali sonlar 81 ta.
Javob: 81.
13. /(x) = e* logxe bo'lsa, /’(e) = ?
Yechish:
1) logxe = Inx
2) f(x) = ex-lnx ko'paytmadan hosila
olamiz.
f(x) - exlnx + ex — = ex (lnx +—)
x x
3) f’(e) = ee(lne +-) = ee-(1 +-) =
e e
= e + e .
Javob: ee + ee1.
14. A va В to'plamlarning barcha
umumiy elementlaridan tuzilgan
to'plam qanday nomlanadi?
Yechish:
A va В to'plamlarning kesishmasi deb
ulaming barcha umumiy elementlaridan
tuzilgan to'plamga aytiladi.
Javob: A va В to'plamlarning
kesishmasi.
15. Agar barcha x, у lar uchun
x3 + 4xzy + axy2 + 3xy - bxcy + 7xy2 +
+ dxy + y2 = x3 + y2 ayniyat bajarilsa,
|a + b + c|(a - b - d) ni toping, (c > 1)
Yechish:
x3 + 4x2y + axy2 + 3xy - bxcy + Zxy2 +
+ dxy + y2 = x3 + y2
4x2y + axy2 + 3xy - bx°y +
+ Yxy2 + dxy - 0
4i?y - Ьх"у + axy2 + Txy2 +
+ 3xy + dxy = 0
c = 2
Ayniyat bo'lganligi uchun
icy 4 - b = 0 b = 4
xy2a+7 = 0a=-7
xy3 + d = 0d = -3
a = ~7, b = 4, d = -3, c = 2 da
|a + b + c\(a-b-d) =
= |-7 + 4 + 2|f-7- 4 + 3) = = -8.
Javob: -8.
16. Tenglama nechta yechimga ega?
(1 - 2sin2x)log7(18 + x - 4Х2) = 0
Yechish:
Aniqlanish sohasi:
18 + x- 4x2 > 0
4X2 -x - 18 < 0
4(x + 2)(x-2,25)<0
-2 < x < 2,25
1) 1 - 2sin2x = 0, sin2x =— ,
X = ±— + яП
4
n = Oda x- aniqlanish sohasiga
tegishli yechimlar.
2) log7(18 + x-4^)=0
18+X-4X2 = 7°
4X2 - x - 17 = 0
8
aniqlanish sohasiga
tegishli yechimlar.
Demak, tenglama 4 ta yechimga ega.
Javob: 4 ta.
17. To'g'ri prizmaning asosi to'g'ri
burchakli uchburchak bo'lib, uning
gipotenuzasi c va bir o'tkir burchagi a.
Ostki asosining gipotenuzasi va ustki
mi
16
Yechimlar. Matematika va informatika 2017
2-varian
asosidagi to'g'ri burchakning uchi orqali o'tkazilgan tekislik bilan p burchak tashkil etadi. Prizmadan tekislik kesib ajratgan uch burchakli piramidaning hajmini toping. Yechish: ABCDA1B1C1D1 - uchburchakli to‘g‘ri prizma. ABC, AiBiC1 uchburchaklar to‘g‘ri burchakli. AB = A1B1 - c Z ВАС = a, Z C1DC = p A, £ _____ в, \ y* CD gipotenuzaga tushirilgan balandlik CD=AC^-. AB AC = ABcosa, CB = ABsina AB2 sin a cos a AB = ABsinacosa = csina-cosa Q AB CD _ C2sinacosa 0 — 2 2 H = CC1 = CD-tgfl, V = ^S-H = C2 sin a cosa -Csin a cos a . „ 2-3 W‘ C3(sinacosa)2 . „ 6 ,9/,= = £^2a.w . . C3 sin2 2a . . Javob: —— tgp . 18. Bir savat olma 20000 so‘m, bir savat nok 30000 so'm, bir savat olxo'ri 40000 so'm turadi. Sakkiz savat mevaning bahosi 230000 so'm bo'lsa, ulardan eng ko'pi bilan nechtasida olxo'ri bo'lishi mumkin? Yechish: 1 savat olma 20000 so'm 1 savat nok 30000 so'm 1 savat olx'ori 40000 so'm 8 savat meva 230000 som x savat olma, у savat nok, z savat olxo'ri x + у + z = 8 20000x + ЗООООу + 40000z = 230000 2x + 3y + 4z = 23 Tanlash usuli bilan yechamiz. (x + y + z=8 |2x + 3y + 4z = 23 f2x + 2y + 2z=16 |2x + 3y + 4z = 23 Ikkinchi tenglamadan birinchi tenglamani ayirsak у + 2z = 7. Bundan у = 1 bo'lsa z = 3, у = 3 bo'lsa z = 2, y = 5 bo'lsa z - 1 bo'ladi. Demak, olxo'ridan eng ko'pi bilan uch savat olish mumkin. Javob: 3. 19. Tetraedrning qirrasi a ga teng. Shu tetraedrning bir qirrasi orqali o'tib, uning qarshisidagi qirrani 2:1 nisbatda bo'luvchi tekislik bilan kesilgan. Hosil bo'lgan kesimning yuzini toping. Yechish: DABC - tetraedr, AB - a BE:EA = 2:1. Kesim DEC uchburchak. Kesim yuzi S. D S'7 1) DEC uchburchak teng yonli, chunki ZAEC=ZAED AC = AD, Z CAE =Z DAE = 60° Bundan EC = ED. 2) ADEC da EN balandlik o'tkazamiz. ~aEN 0 — 2
17
Yechimlar. Matematika va informatika 2017
2-variant
3) EN ni topamiz. ЛАСЕ da
EC2 =AC2+ AE2 - 2ACAEcos60° =
2 , a2 2a2 1 7 a2
= a +— ------=----
9 3 2 9
ЛЕЫС dan EN = -jEC2 -NC2 =
_ 17a2 a2 a Jl9
9 4 6
Q _a-EN _a2y/l9
2 ~ 12
у = kx + b
(-1 = -6k + b
[1 = -2k + b
-2 = -4k
к =—da 1 =-2- + b
2 2
b = 2
Javob: ----
12
20. Tenglama ildizlari ko'paytmasini
toping: 731og2(-x) - log2 y/x2 = 0.
Yechish:
Aniqlanish sohasi:
J-x>0 (x<0
[31og2(-x)>0=>'[-x>21’
fX<0 . 4
< =>x<-1
[x<-1
V?=|x|
x < -1 bolganligi sababli yfx2 = | x |= -x
731og2(-x) = log2(-x) ikkala qismini
kvadratga kolaramiz.
3log2(-x) = log22(-x)
log2(-x)(log2(-x) -3) = 0
1) log2(-x) = 0,-x = 2°, -x = 1, x = -1
2) log2(-x) = 3,-x = 23,-x = 8, x = -8
x = -8 ya x = -1 aniqlanish sohasiga
tegishli.' Ildizlari ko‘paytmasi: -8(-1) = 8.
Javob:8.
21. A(-6; -1) nuqta C(n; 5) va
B(-2; 1) nuqtalardan o'tuvchi to‘g‘ri
chiziqda yotsa, n ning qiymati
nechaga teng bo'ladi? '
Yechish:
A (-6; -1) va В (-2; 1) nuqtalardan
oluvchi to‘g‘ri chiziq tenglamasini
tuzamiz.
C(n; 5) nuqta
x
у =— + 2 to‘g‘ri chiziqqa
tegishli, shuning uchun
5 = — + 2, — = 3,n = 6.
2 2
Javob: 6.
22. Yig'indining oxirgi raqamini
toping: 2O142015 + 2O152014.
Yechish:
4 ning ixtiyoriy juft darajasi 6 bilan, toq
darajasi 4 bilan tugaydi.
2014го15 = ...4
5 ninqjxtiyoriy darajasi 5 bilan tugaydi.
Demak, 2O142015 + 2O152014 =
= ...4 + ...5= ...9.
Yiglndining oxirgi raqami 9 bilan tugaydi.
Javob: 9.
23. a = -4 bo'lsa,
a+1
J (ln(sin2 3x + cos2 3x) + 1)dx aniq
a
integralni hisoblang.
Yechish:
1) sin23x + cos23x = 1
2) In1 = 0
a+1 a+1
3) J (0 + 1)dx = j 1dx = x
a a
a + 1
a
= a+1-a = 1.
Javob: 1.
18
Yechimlar. Matematika va informatika 2017
2-variar
24. Diagonallarining soni
tomonlarining sonidan 1,5 marta ko‘p
bo'lgan qavariq muntazam
ko'pburchakning barcha ichki burchaklari
va bitta tashqi burchagi yig'indisini toping.
Yechish:
D - diagonallar soni, n - tomonlar soni.
D=n(n-3)
2
— = n^n~3\n-3 = 3,n = 6
2 2
Muntazam oltiburchak ekan. Ichki
burchaklari yig'indisi
180°(6-2) = 720°.
, 360’ 360’ M . . .
а =----= —— = 60 bitta tashqi
burchagi.
720° + 60° = 780°.
Javob: 780°.
25. (V3 + 2V2 )sinx +
+ (7з-2>/2 )sinx = — .
3
Yechish:
73 + 2V2 J3-2V2 = V9-8 = 1 -
o'zaro teskari sonlar.
(^3 + 242 )sinx = a
1 10
a +— = —
a 3
3a2 -10a + 3 = 0
1
a=—,a=3
3
• „ / r————— \ sin x
/ I ~smx 1 I 2 A
V3 + 2V2 = J(7 + V2)
= (V2+7)S,nX
,)(J2 *<)''=!
1
sinx — log ^ -=-10g^+J3 <-1
J
sinx < -1, 0
2}(V2+7)S,nX =3
sinx = log %; 3 > 1, sinx > 1, 0.
Javob: 0.
26. Arifmetik progressiya n-hadi
n-2 * n
an =-----ga teng. Progressiyanmg
5
ayirmasini toping.
Yechish:
d = a? — ai =
= X lzZk_l = _0,2.
5 < 5 ) 5
Javob: -0,2.
27. |3 -VTT5 | > --6-8
tengsizlikning butun yechimlari nechta?
Yechish:
Shunga asosan
□ l—c x~8
6
О I----E
3-yJx + 5 <—--
6
[ 6л/x + 5 < 26 - x
[6>/x + 5 > 10+ x
1) 6jx + 5 <26-x
26-x>0
36(x + 5)<676 - 52x + x2
x>-5
x<26
x2-88x + 496>0
'5 44-12-Л0 26 44+12Vl6
[-5;44-12>li0)
19
Yechimlar. Matematika va informatika 2017
2-variant
2) 6jx + 5 > 10 + x
x + 5>0
10 + x>0
36{x + 5) >100 + 20x + x* 1 2 3
Jx>-5
\x2-16x-80<0
-s го ^_4; 20j
Tengsizlikning yechimi:
[-5; -4) U (-4; 20)
Tengsizlikning butun yechimi 24 ta.
Javob: 24.
28. sin(^ + (x +1)2) = x2 + 2x + 2
tenglama ildizlari yig’indisini toping.
Yechish:
1) Keltirish formulasiga asosan:
sin(^ + (x + ))2) = cos(x+ f)2
2)^ + 2x + 2 = (x+1)2 + 1
3) -1 < cos(x + 1)2 < 1
-1 < (x + 1)2 + 1 < 1
-2<(x + 1)2<0
a) (x + 1)2 s-2, xCR
b) (x + 1)2 < 0, x + 1 - 0, x =-1
x = -1 tenglama ildizi, chunki
sin(| + (-1 + 1)2) = (-1/ + 2(-1) + 2
sin — = 1 - 2 + 2, 1 = 1
2
Tenglama ildizlariyig'indisi-1 да teng.
Javob: -1.
29. Uchburchakli muntazam prizma
yon yog’ining diagonali asos tekisligi
bilan <p burchak tashkil etadi. Prizma
yon sirti S bo’lsa, uning hajmini toping.
Yechish:
Syon = S
ABCA1B1C1 - muntazam uchburchakli
prizma. AtC- yon yoq diagonali.
Z ACA1 = <p, AA1 = H
AB=BC=AC=a
PI
tg</> = ~, H = a-tg<p
a
S - yon sirtiyuzi. S = За-H
u S S
H= —, — = atg<p
За 3a
a2 =—— , a= I——
3tg<p \ 3tgg>
S
3tg<p
V=Sasos-H=^-^--^-
4 3a
а^з-s _43s Fs~_ sVs
12 ~ 12 ytg<p~ 12jt^'
Javob:
sVs
12^/tg^
30. Katta diagonali d va o’tkir
burchagi <p bo’lgan romb, o’tkir
burchagining uchi orqali katta
diagonaliga perpendikulyar qilib
o’tkazilgan o’q atrofida aylantirildi.
Hosil bo’lgan jism hajmini toping.
Yechish:
ABCD - romb. AC = d, Z ВАС = <p.
Aylanish jismining hajmi AKBC va
ANDC trapetsiyalarning aylanishidan
hosil bo’lgan ikkita teng kesik konus
hajmlarining yig’indisidan AKB va AND
uchburchaklarning aylanishidan hosil
bo’lgan ikkita teng konus hajmlari
yig’indisining ayirmasiga teng. AC = d,
20
Yechimlar. Matematika va informatika 2017
2-variant
я-ВО d2} _ Drx
-------------- itCs-BO
3 4 )
AAOBdan BO = —tg^-
2 w 2
nd3tg—
v =-----
2
xd3tg -
Javob:------2..
2
31. Brauzer so’zining ma’nosi:
Yechish:
Brauzer- bu shunday dasturki, tarmoqdagi axborotlamiko’rish yokiizlashni
ta’minlovchi dastur.
Shundan brauzer ma’nosiko’rinishni ta’minlash va ko’rsatish.
Javob: ko'rinishni ta’minlash, ko’rsatish.
32. A1 = -8, B1 = 9, B2 = 3 bo’lsin. Quyidagi formula natijasi -78 ga teng
bo'lishi uchun A2 katakka kiritilishi kerak bo'lgan qiymatni aniqlang.
=ЕСЛИ(И(А1+В2<А2*В1;А1*А2<>0);А1*В2-В1-А2;А1*В1-В2+А2)
Yechish:
A1B2-B1 -A2 = -78.
-8-3- 9- A2 = —78.
-24-9+78=A2.
A2 = 45 bunday javob yo’q.
A1B1 -B2+A2=-78.
—8-9—3+A2 = -78.
-72-3 + A2 = —78.
A2 = —78 + 75 = -3.
Javob: -3.
33. IP(lnternet Protocol) manzil nima?
Yechish:
IP - bu manzil bo’lib, ingliz tilida Internet Protocol Address, ya’ni lokal
tarmoqqa yoki internet tarmog’iga ulangan kompyuterning unikal, ya’ni
qaytarilmaydigan identifikatoridir.
Identifikator deganda manzil nomi tushunilib, manzil nomi odatda protokolning
IPv4 versitasida 4 bayt yoki 32 bit uzunlikdagi ikkilik sanoq sistemasi bilan beriladi.
Protokolning IPv6 turida esa IP manzil 16 bayt yoki 128 bit uzunlikka ega.
Javob: protokollar.
34. ... - kompyuterga dasturiy ta’minotni o’rnatish jarayonidir.
Yechish:
Installyatsiya (angl. installation) - o’rnatish, joylashtirish, ma’nosini anglatadi.
Aynan installyatsiya - bu kompyuterga dasturiy ta'minot o’rnatishdir.
Javob: installyatsiya.
21
Yechimlar. Matematika va informatika 2017 3-variant
35. Hisoblang va javobini 16 lik sanoq sistemada ifodalang:
16(ю)*4(8) + 254(8).
Yechish:
1610-Хг.
1612
161812
©8142
@41212
©x®
©
16ю — 10000(2)-
4(8) —* 100(2).
254(8) -* 0101101100(2) (1-chijadval bo'yicha).
10000(2)-100(2) — 1000000(2).
1000000
010101100
011101100(2)
011101100(2)-^ x16.
11101100 -> EC16. (1 -jadval bo ‘yicha)
E C
Javob: EC.
36. Sonning butun qismi qaysi standart funksiya yordamida ifodalanadi?
Yechish:
Sonning butun qismi INT(x) kabi belgilanib, integer (sonning butun qismi)
kabi belgilanadi.
Javob: INT(X).
3-variant
1. Tenglama nechta yechimga ega? 12 - 7x + x2 = 4(x-3)Vx Yechish: Aniqlanish sohasi x>0. 12 — 7x + x2 = x2 - 7x + 12 = = (x-3)(x-4) (x-3)(x-4) = 4(x-3)Jx (x-3)(x-4-4jx ) = 0 (x-3)(x-4y/x-4)=0 x-3 = 0, x = 3 x-4-fx -4 = 0 tenglama ildizlarini topamiz. jx = a a2 -4a-4 = 0 2 2 1) 4x = 2 + 2^2 x = (2 + 2y/2)2 2) y[x = 2-2y/2 yechim bo'la olmaydi, chunk!2-2^2 <0. Tenglama ildizlari x = 3 va x = (2 + 2y/2)2. Tenglama 2 ta yechimga ega. Javob: 2. 2. /(x) = ex - x bo'lsa, ni /(2) toping.
22
Yechimlar. Matematika va informatika 2017
Yechish:
f (x) = ex-x
f(4) = e< 4 = (e2)2 _ = (e2 - 2)(e2 + 2)
f(2)=e2-2
y(4) (e2-2)(e2 + 2)
7®------------------e +2'
Javob: e2 + 2.
3. Tenglamani yeching:
0,(3)cos2x - 0,(6)sinx = 0,(1) - 0,(3)sin2x.
Yechish:
Davriy kasrdan oddiy kasrga otamiz.
0,(3) =- = -
9 3
0,(6)=— = —
9 3
0,(1) =-
9
1 2 2 . 11-2
—cos x—smx =--------sin x
3 3 9 3
3cos2x - 6sinx = 1 - 3sin2x
3cos2x + 3sin2x- 6sinx-1=0
3(cos2x + sin2x) - 6sinx -1=0
3 - Osinx —1 = 0
6sinx — 2 = 0, sinx=—=— = 0,(3)
3 9
sinx = 0,(3)
x = (~1)narcsin0,(3) + тгп, n C Z.
Javob: (-1)narcsin0,(3) + лп.
4. a = 30°, a = (tga)‘9“, b = (tga)ct9“,
c = (ctga)‘9“ bo’lsa, quyidagilardan
qaysi bin o’rinli?
Yechish:
а = 30° da
1
a = (tg30’/930'=(4=T=(^F
b = (tg3O°f930' =[^=J =(73)^
c = (ctg30°)ta30° = (>/з)л
of
а = (7з)Ль = рз)Лс = (.
Ko‘rsatkichli funksiyada asos 1 dan kaua
bo‘lsa, darajasi kattasi katta boladi.
c>a > b.
Javob: c > a > b.
5. Kvadratlarining ayirmasi 55 ga
teng bo’lgan barcha natural sonlar
juftligini toping.
Yechish:
x2 - y2 = 55
Qisqa ko'paytirish formulasiga asosan
^-Уг = (х-у)(х + у)
55 ni ko'paytiruvchilarga ajratamiz.
55 = 1-55
55 = 5-11
x2 -y2 = 55 tenglama
(x-y=1 (x-y = 5
1 va <
(x + y = 55 |x + y = 55
tenglamalar sistemasiga teng kuchli
boladi.
x,yCN
(x—y=1
1 , x = 28, у = 27
|x + y = 55 7
(28; 27)
\x-y = 5
2 ) < 7 , x = 8, у = 3
[x + y = 55
(8; 3).
Javob: (8; 3) va (28; 27).
6. /(x) = cos( у tg^-) bo’lsa, /'(1)
4 4
ni toping.
Yechish:
f'(x) = cos —tg—-
I 4
.2
. I 71 л 7UX\
= -sm —tg— •
4 } л2 2
4 7 4 cos —
4
f'(1) =-sm -tg—
\4 4 )
23
Yechimlar. Matematika va informatika 2017
3-variant
2 2
7Г . 7C 7C
-------------------------- -Sin 7-
16cos —------------16 —
4 2
42 x* 1 2 3 4 * * _ л24?
2 8 16 '
. . it2 42
Javob:---------.
16
7. x = 15 bo'lsa,
lg(x2 + 3x - 4)
-log(J[_1)(x + 4)
ifodaning qiymatini toping.
Yechish:
Ifodani soddalashtirib olamiz.
Iga ,
= log6 a shunga asosan
Igb
lg(x2 + 3x-4)
—-------------- - log . (x + 4) =
lg(x-1)
= log K,(x2 + 3x - 4) - logx_, (x + 4) =
=|оёх<
x2 + 3x-4
x + 4
, (x + 4)(x-7)
-'°8-' =
Demak, ifodaning qiymati 1 да teng.
x = 15 da ham ifodaning qiymati 1 boladi.
Javob:1.
8. 6х = 0,25 bo'lsa,
a/49x-10-7х +25 + 7х + 2,5 ifodaning
qiymatini toping.
Yechish:
1) 449х-10-7х + 25 =
= 7(7x-5)2 =|7x-5|
2) 6х = 0,25 da x < 0 ekanligi kelib
chiqadi.
3) x <0 bo'lganligi uchun
17х- 5| =5- 7х
4) ^49х-10-7x +25 + 7x + 2,5-
= 5- 7x + 7x + 2,5 = 7,5.
Javob: 7,5.
9. To'g'ri burchakli parallelepiped
asosining tomonlari a va b ga teng.
Parallelepiped diagonali yon qirrasi
bilan q> burchak tashkil qilsa, uning
hajmini toping.
Yechish:
a, b — tomonlari
AB = CD = a, AD = BC = b
BD - diagonal
DCi - yon qirra
A D
Z B1DD1 = q>
AB1D1D to'g'ri burchakli
Bp, = ^B.C2 +C.D2 = 4a2+b2
H = D1D 4'- - = 4 a2 +b2ctg<p
tg<p
Sasos “ Q"b
X/ = Sasos H = ab 4a2 +b2 ctg<p.
Javob: ab Va2 +b2 ctg<p .
( 3 1A
10. Agar A -3—; 5— va
I 4 2J
B(-0,8; -1,4) nuqtalar berilgan bo'lsa,
AB kesma o'rtasining koordinatasini
toping.
Yechish:
Kesma o'rtasining koordinatasi
А(хь yi), B(x2; у?), С(х3; y3).
, _У1+У2
3 n
C nuqta kesma o'rtasi.
24
Yechimlar. Matematika va informatika 2017
3-variant
-3- °’8 _зу5^о,8
2 ~ 2
11
= -2,275= -2——
40
13. у > 0 bo'lsin. To'rtburchakning
uchlari to'g'ri burchakli dekart
koordinatalar sistemasida quyidagicha
berilgan: A(0; 0), B(0; y), C(5; y) va
D(7; 0). To'rtburchak diagonallarining
o'rtalari orasidagi masofani toping.
Yechish:
у > 0, A(0; 0), B(0; y), C(5; y), D(7; 0),
EF=?
(11 A
C -2—; 2,05 .
\ 40 )
( 11
Javob: -2—; 2,05
I 40
11. у = 2cos2x + cos4x
funksiyaning hosilasini toping.
Yechish:
1) (cos(ax + b))' = -asin(ax + b)
(cos2x)' = -2sin2x
(cos4x)' = ~4sin4x
2) y' = 2(cos2x)' + (cos4x)' =
= -4sin2x - 4sin4x = -4(sin2x + sin4x)
3) sin2x + sin4x yig'indidan
ko'paytmaga otamiz.
4y4-
sin4x + sin2x = 2sin------
2
4x-2x „ . -
•cos---------- 2sin3xcosx
2
4) y' = -4-2sin3xcosx = -8sin3xcosx.
Javob: -8sin3xcosx.
12. {x|x e N, -5 < x < 5} to'plamni
nechta usul bilan ikkita kesishmaydigan
qism-to‘plamlarga ajratish mumkin?
Yechish:
Ikkita kesishmaydigan qism to'plamga
o‘n xil usul bilan ajratish mumkin.
A={1, 2, 3, 4}-4 ta element.
1) {1} va {2} 2) {1} va {3}
3) {1} va {4} 4) {1} va {2; 3}
5) {1} va {2; 4} 6) {1} va {3; 4}
7) {1; 2} va {3; 4} 8) {1; 2} va {3}
9) {1; 2} va {4} 10) {1; 2; 3} va {4}.
Javob: 10.
1) AD - asos, a = 7, BD- asos, b -5
2) EM - ACD uchburchak o'rta chizig'i.
AD _ r
EM =----= 3,5
2
3) EN ABC uchburchak o'rta chizig'i.
EN = — = 1,5
2
EN = FM
4) EF = EM-EN = 3,5-1,5 = 2.
Javob: 2.
14. p = logi,43 bo'lsa,
72P _ g2p
—z--------------x- ifodaning
72p+2-7p-5₽+52₽
qiymatini toping.
Yechish:
p = logics da
72p -52p
—---------------— ifodaning
72p + 2-7p-5P +52p
qiymatini topamiz.
Soddalashtirib olamiz.
72p-52p _(7P-5P)(7P + 5P) _
(7P+5P)2” (7P+5P)2
m₽ j 5 у
_ 7P-5P _ tsj Ы = 1,4P-1
~7P+5P~(7~\P <5y ~1,4P+1
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Yechimlar. Matematika va informatika 2017
3-variant
p = logi,43 da
1,4^,4Э-4 3-4 _2 _1
1,4l°61-43 + l'~3+1 ~~4~~2
Javob: 0,5.
15- To'g'ri parallelepiped asosining
tomonlari a va b, o'tkir burchagi a.
Asosining katta diagonali parallelepiped
kichik diagonaliga teng. Parallelepiped
hajmini toping.
AB = CD = a, AD = BC = b
AC = BiD
AC2 = AD2 + CD2 + 2ADCDcosa
BD2=AB2 + AD2- 2ABAD cosa
H2 = BiD2 -BD? = 4ABAD
H - 2 jab cos a
^asos — dbsifltt
V = Sesos H = absina- 2y/ab =
= 2sina yjfabj cosa .
Javob: 2sina7(ab)3 cosa.
16. a = 7"l + V2 bo'lsa, З''г'"та2а,1ок°’
ifodaning qiymatini toping.
Yechish:
logo,га + log5a ning qiymatini topamiz.
0,2 = 5~1, logo,za = -log5a
logo,га + log5a = -log# + log# = 0
Demak, 3^^^^ =3^=3»=1
Ifodaning a =-Jl + y/2 dagiqiymati
ham 1 да teng.
Javob: 1.
17. Ig(lgx) + lg(lgx3 - 2) = 0
tenglamani yeching.
Yechish:
Aniqlanish sohasi:
Jlgx>0 fx>1
(lgx3-2>0=> (igx3 >2^
x>1
S , => X
|x3 > 100
t/iod
Tenglamani yechamiz.
Ig(lgx) + lg(lgx3 - 2) = 0
lg(lgx (3lgx - 2)) = 0
lgx(3lgx - 2) = 1(f, Igx = a
a(3a - 2) = 1
3a2-2a - 1 =0, a = 1, a=-—
3
1) a = 1 da Igx = 1, x = 10
1 1
2)a = -- dalgx=~,
3 3
-1 1
x=10 3 = ~
</7o
x= 10 aniqlanish sohasiga tegishli,
1
x =—f= aniqlanish sohasiga tegishli
j10
emas.
Javob: 10.
18. f—j= hisoblang.
J xVInx
Yechish:
Hy
1) f — = In | x | +C dan foydalanib
1 x
yechamiz.
f dx = rd(lnx) (Inx)"^
xVta7 Vinx -- + 1
2
= ^J^^ + C = 2(lnx)3 +С = 2>/Ых+С.
2
Javob: 2>/inx + C .
26
Yechimlar. Matematika va informatika 2017
3-variant
19. Radius! R bo’lgan sharga kesik
konus ichki chizilgan. Kesik konusning
asoslari shardan o‘q kesimidagi yoylari
a va p ga teng ikkita segment kesadi.
Kesik konus yon sirtini toping.
Yechish:
ABCD teng yonli trapetsiya kesik konus
o’q kesimi. Z AOB= a, Z DOC = ft
R2 = DF = Rsin—
2
ZAOD = ——= 180’-
2 2
l = AD = 2Rcos^~
4
Syon = ttI(Ri + R?) =
= 2xFfcos---— (sin — + sin —) =
4 2 2
= 2xF?cos . 2sin ^JL cos (C—JL =
4 4 4
O2 a + fi a~P
= 2n:R sin--—cos--—.
2 4
. . „ . a + j§ а,- в
Javob. 2ttR sin---cos----.
2 4
20. Uchburchakli muntazam prizmada
yon yog‘i diagonali bilan ikkinchi yon
yog‘i orasidagi burchak 30°. Asos
qirrasi a. Prizma yon sirtini toping.
Yechish:
ABCA1B1C1 - muntazam uchburchakli
prizma. ABC va A1B1C1 - muntazam
uchburchaklar.
AB = a, AB1 - yon yoq diagonali.
AB1 diagonal bilan BB1C1C yon yoq
orasidagi burchak 30° ga teng.
1) A nuqta BC tomon o'rtasi D nuqtaga
proyeksiyalanadi. Proyeksiya BrD
boladi.
2) Z AB,D = 30°. AB1BD dan
н = bb, =7b,d2 -BD2
BD=—
2
3) B1D niADBi to’g'ri burchakli
uchburchakdan topamiz.
Z ABfD = 30°, Z ADB1 = 90°,
Z B1AD = 60°.
AD _ B,D g _ AD sin 60°
sin 30° sin 60° 1 sin 30°
ал/З Уз
ДО- —, В,О = —Ц—?- = —
2 1 1 2
Syon = Р-Н = 3a-aV2 =3a2^2 .
Javob: 3a2y[2.
21. у > 0 bolsin. To'rtburchakning
uchlari to'g’ri burchakli dekart
koordinatalar sistemasida quyidagicha
berilgan: A(1; 0), B(1; y), C<—1; y) va
D(-3; 0). To’rtburchak diagonallarining
o’rtalari orasidagi masofani toping.
Yechish:
1) у > 0, A(1; 0), B(1; y), C(-1; y),
D(-3;0),EF=?
2) E nuqta - BD kesma o’rtasi,
F nuqta -AC kesma o’rtasi
27
Yechimlar. Matematika va informatika 2017
3-variant
2) f[ — = f[0;
{.22) { 2)
3) EF kesma uzunligi:
EF=(-1-0)* 1 2+(---} =1.
V I? 2)
Javob: 1.
22. Agar |x - 10| + a tenglama
bitta yechimga ega bo'lsa, a ning
qiymatini toping.
Yechish:
|x - 10\ + a tenglama bitta
yechimga ega bolishi uchun tenglikning
o‘ng tomoni nolga teng bolishi kerak.
Chunki |x - a| = 0 tenglama doimo
bitta yechimga ega boladi.
x = a
Shuning uchun \x - 10\ - 0 va^ + a = 0.
Bundan x = 10 va a =-—= = -5.
2 2
Javob: a = -5.
23 = <28
[lg(x + y) = lg40-lg(x-y)
tenglamalar sistemasini
qanoatlantiruvchi x va у noma’lumlarning
qiymatlari yig'indisini toping.
Yechish:
Birinchi tenglamadan:
X у I
210,21O =128*
x+y 7
2 10 —2/
x + y 7 ..
----— = — m topamiz.
10 x
Ikkinchi tenglamadan:
!g(x + y) + lg(x -y)= lg40
lg/x + y)(x-y) =lg40,
x -y2 = 40 ni topamiz.
x + y 7 f 2
---- = — x2 + xy = 70
\ 10 x ,
X2-y2 =40 V-y=40
x(x + y) _ 70
(x+yXx-y) 40’
4x=7x- 7у
7y = 3x bundan x = 7,y = 3 ekanligini
topamiz.
x+y-7 + 3=10.
Javob: 10.
24. /(x) = 18sin4xsin5x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko'paytmadan yig'indiga otamiz.
1
sin4x sin5x =— (cos(4x - 5x) -
1
- cos(4x + 5x)) =— (cosx-cos9x)
2) f(x) = 18 — (cosx-cos9x) =
- 9cosx - 9cos9x
Boshlang‘ich funksiyasini topamiz.
F(x) = 9sinx - sin9x + C.
Javob: 9sinx - sin9x + C.
25. Uchburchakning b va c tomonlari
va ular orasidagi burchak a berilgan.
Shu uchburchak a burchagining uchi
orqali uchburchak tashqarisidan o'tgan
hamda b va c tomonlar bilan bir xil
burchak hosil qiluvchi o'q atrofida
aylantirildi. Hosil bo'lgan jism hajmini
toping.
Yechish:
Aylanish jismining hajmi V - OOiBC
trapetsiyaning aylanishidan hosil
bolgan kesik konus hajmidan AOiB
28
Yechimlar. Matematika va informatika 2017
3-variant
va AOC uchburchaklar aylanishidan
hosil bo’lgan 2 ta konus hajmlari
yig’indisidan ayrilganiga teng.
Z BAO1 = Z CAO bo’lgani uchun
Z BAOi = Z CAO = 90°,
ZO1BA=ZOCA=~.
2
H = bsin —, R = bcos—
2 2
,. а а
h = csin—, r = ccos-
2 2
V (H + hJfR2 + Rr + r2) -
Я I ti-^2 Я t_2 7Г . CC 2 (X
- — HFr hr =— sin — cos —
3 3 3 2 2
((b + c)(b2 + be + c2) -b3- c3) =
а
я . а 2— .
= — sin — cos 2 (b + c) =
= — smacos— (b + c)-bc.
6 2
Javob: — sinacos —(b + c)bc.
6 2
__ x2 4-6x4-21, . .
26. у =--------- funksiyaning
11 + 6x + x
qiymatlar sohasiga tegishli barcha
butun sonlar yig'indisini toping.
Yechish:
x2 + fix 4- 21
у = ——_—— qiymatlar sohasini
topamiz.
x2+6x + 21 . 10
у =---------= 7 -I---------=
x2 +6X + 11 x2 + 6x + 11
. 10
— 1 4----7--
(x + 3)2+2
Утах ~ 1 4"- — 1 4- 5 — 6
2
yC(1;6]-2, 3, 4, 5, 6
2 + 3+4 + 5 + 6 = 20.
Javob: 20.
27. Asosining radiusi R va
balandligi bilan yasovchisi orasidagi
burchak 30° bo'lgan konus ichiga
chizilgan shar konusning asosiga va
yon sirtiga urinadi. Konusning shar
ustidagi qismining hajmini toping.
Yechish:
Konus o‘q kesimi teng yonli uchburchak.
V hajm ECK konus hajmidan EFK shar
segment hajmining ayrilganiga teng.
AD = DB = R, Z ACD = 30°
V EN2 CN - ttNF2(OENF)
AAOD dan r = OD = OF = ADtg/3
J3=ZDAO=ZDAC =^ = 30°
2 2
AD
r = ADtg30° =~^
AOEN dan Z OEN = 30°, chunk! OEN
va ECN burchaklarning tomonlari
o’zaro perpendikulyar.
х/з
EN = OEcos30° = r- —
2
ON = OEsin30° =-
2
NF=OF-ON = r--=-
2 2
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Yechimlar. Matematika va informatika 2017
3-variant
NC = NE-ctg30° =43 r'^=y
it 3 , 3r itr2 ( 1 r\
3 4 2 4 3 2)
_ Зг37г 5r3Tt _ itr3
~~~8 24~~6~
л f R V _ irR3
в\7з; ~ 1в4з'
l u л/?3
Javob:-----t= .
18>/3
28. /(x) =——— funksiya uchun
4
/~1(0) ni hisoblang.
Yechish:
(x - 312
f(x) =---funksiyaga teskari
4
funksiyani topamiz.
4f(x)=(x-3)2
yj4f(x) = x-3, x = 3 + J4f(x)
Г1(х)=3+44х
Г1(0)=3+44^0=3.
Javob: 3.
29. а = 2^3 + 4б bo'lsa, Z4108’7'08498
ifodaning qiymatini toping.
Yechish:
a = 2^3+ 45 da ning
qiymatini topamiz.
4loga7 log4sa ni soddalashtirib olamiz.
1
4loga7-log49a = 4-- loga7-log7a =
7,-*loge7 log49a _j2 =4Q
Ifodaning a = 2^3 + 45 dagiqiymati
ham 49 ga teng.
Javob: 49.
30. [0; 300] kesmada 3 ga
bolinganda qoldiq 1 ga, 4 ga bo'linganda
qoldiq 2 ga, 5 ga bolinganda qoldiq
3 ga va 6 ga bolinganda qoldiq 4 ga
teng boladigan natural sonlar nechta?
Yechish:
a - son.
a sonini3 ga bo'lganda 1 qoldiq.
a = 3n + 1
4 ga bo'lganda 2 qoldiq.
a = 4k + 2
5 ga bo'lganda 3 qoldiq.
a = 5m + 3
6 ga bo'lganda 4 qoldiq.
a = 6c + 4
3n + 1 = 4, 3(n + 1) = a + 2
4k + 2 = a, 4(k + 1) = a + 2
5m + 3 = a, 5(m + 1) = a + 2
6c + 4 = 6(c + 1) = a + 2
Demak, a + 2 soni 3, 4, 5, 6 sonlariga
karrali son. 3, 4, 5, 6 sonlarining
EKUKni topamiz.
3-4-5 = 60, bundan a + 2 = 60, a - 58.
[0; 300] kesmada 58 ga karrali natural
son 5 ta, bu!ar58, 116, 174, 232, 290.
Javob: 5 ta.
31. Ikkilik sanoq sistemasidagi 11001112 sonini o'nlik sanoq sistemasiga
o'tkazing:
Yechish:
65 4 3210
1100111w-Xw
7-2® + f-25 + fl-24 + 0-23 + 1? + 1-21 + 1-2P = 64 +32+0 + 0 + 4 + 2+1 = 10310.
Javob: 103.
30
Yechimlar. Matematika va informatika 2017 3-variant
32. Paskal dasturlash tilida berilgan ushbu ifodaning qiymatini toping.
trunc(sqrt(abs(trunc(4,5)-sqrt(400)*round(1,5)))).
Yechish:
round(1,5) = 1 (eng yaqin butun songacha yaxlitlash)
SQRT(400) =y[400 = 20
trunc(4,5) = 4 (sonning butun qismi)
abs(4-20-1) = abs(-16) = 16
SQRT(16) = Jl6-4
trunc(4) = 4.
Javob:4.
33. HTML tilida Web sahifaga yugurikli chiziq joylashtirish uchun qaysi teg
ishlatiladi?
Yechish:
HTML - tilida veb sahifa yaratish uchun teglar ishlatiladi.
<Marquee> </Marquee> teglari yugurukli chiziq o'rnatish uchun ishlatiladi.
Javob: <Marquee> va </Marquee>.
34. Web brauzerda matnning ko’rinishi quyidagicha bolishi uchun uning HTML
kodi qanday bolishi kerak? Kvadrat tenglama ax* + bx + c = 0 ko’rinishida bo’ladi.
Yechish:
Kvadrat tenglama ax2 + bx + c = 0 ko‘rinIshida bo'ladi.
<i> kursiv yozuv matni </f>
<stong> qalin yozuv matni </strong>
<sup> sonni daraja ko'rinishida yozish </sup>
<p> <i> kvadrat tenglama
<strong>ax<sup>2</sup>+bx+c=</strong> ko'rinishida bo'ladi. </i></p>.
Javob: <p></> Kvadrat tenglama
<strong>ax<sup>2</sup>+bx+c=0</strong> ko’rinishida bo’ladi. </i></p>.
35. HTML tilidagi web-sahifada ta’riflash ro’yxatini hosil qilish uchun qanday
teg ishlatiladi?
Yechish:
HTMLga ro'yxat yaratish uchun
<UL> raqamlanmagan ro'yxat
<OL> raqamlangan ro'yxat
<DL> ta’riflash ro'yxatini tashkil etish.
Javob: <DL>.
36. TCP protokolining ish tamoyili nimalardan iborat?
Yechish:
TCP (transmission control protocol) - asosiy protokollardan biridir. Bu
protokolyordamida ma’lumotlarnimanzilga uzatadi, ya’nimarshrutlaydi.
Javob: ma’lumotlarni marshrutlaydi.
31
Yechimlar. Matematika va informatika 2017
4-variant
4-variant
1 1 + 1 1
1 1 + 1 1 1 1 + 1 f
--1 1 1 1
x 2 x 2 x 2 x 2
= — tenglamani yeching.
36
Yechish:
2)—-— = —1— = 2x
A l 2 + x 2 + x
x 2 2x
1 2 + x
’ 2x 2x ~ 4X
2 + x 2+x
4) 1 - 4x - 2x
} 2 + x , 2 + x ~ 4 + 2x~ 2 + x
----1-----
4x 4x
5)JA = LL
2 + x 36
x + -2
72x = 2х + хг
72x- 2x — x2 = 0
X2 - 70x = 0, xtO
x=70.
Javob: 70.
2. m = 9 bo'lsa,
+ V2 • (LLm -1)
m + 2-42m
ifodaning qiymatini toping.
Yechish:
aA + bLb=[L~a]3 + {Lb]3 =
= (Va + Lb^(La^ -Lab+ LtL^ =
= (La + Lb^(a - Lab + bj formuladan
foydalanib yechamiz.
7_ m^+2L2 (Я3 + №3
m+ 2- L2m m - L2m + 2
(Lm + L2j(m - LLm + 2^
m-L2m + 2
= Lm+L2
2. Lm +L2 + L2(L2m -1) =
= Lm +L2 + 2Lm -L2 = 3Lm
3. m = 9 bolsa 3Lm = 3L9 = 3-3 = 9.
Javob: 9.
3. Teng yonli uchburchakka ichki
chizilgan aylana yon tomonini urinish
nuqtasida uchidan boshlab hisoblaganda
8 va 4 ga teng kesmalarga ajratsa,
uchburchakning perimetrini toping.
Yechish:
ABC - teng yonli uchburchak.
AB = BC, AD = 8, DB = 4, P = 2
Bitta nuqtadan otkazilgan urinmalar
tengligidan
AD=AE=8
DB = BK = КС = CE = 4
P=AD+DB+BK+KC+CE+AE=
= 16+16 = 32.
Javob: 32.
4. Agar a = 31° va p = 270° bo'lsa,
sina-sin(P - a) + sin I 2 J ni hisoblang.
Yechish:
sina-sin(fi - a) + sin \ 2 ) ni
soddalashtiramiz.
32
Yechimlar. Matematika va informatika 2017
4-variant
1
1) sina-sin(fl ~a)=— (cos(2a ~fl)~ cosfl
2)sin2[^-a} = 1-^^^-^
1 1
-~~cos(2a-fl)
,(fl A
sinasinffl - a) + sin I ~ - a I =
1 11
= — cos(2a - fl) cosfl +— -
1 11
~ cos(2a - fl) = ~ cosfl +-^ =
1 11
= --cos270° +-=-.
2 2 2
i
Javob: —.
2
5. (a, 3, b) arifmetik progressiya,
(a, 2, b) geometrik progressiya bo'lsa,
a2 + b2 ni toping.
Yechish:
(a; 3; b) arifmetik progressiya, (a; 2; b)
geometrik progressiya.
. -r ... , a + b _
Arifmetik progressiyada = 3.
Geometrik progressiyada ab = 2?.
Bundan a + b = 6, a b = 4.
a2 + b2=(a + b)2 - 2ab = 62-2-4 =
= 36-8 = 28.
Javob: 28.
6. /(x) = arccos(cosx) bo'lsa, f'( )
ni toping.
Yechish:
1) arccos(cosx) = x, x 6 [0; я].
2) f(x) = x, f(x) = 1,
Javob: 1.
7. f(x) = -—ifOda berilgan.
x-4
f (x) funksiyaga teskari funksiyani toping.
Yechish:
f(x)(x-4) = 4-2f(x)
f (x)(x - 4) + 2f (x) = 4
f(x)(x-2)=4
4
f(x) =---funksiyaga teskari
x-2
funksiyani topamiz.
4 4
f(x) =---о- х-2 = У7^’
x-2 /(x)
4 _ 4 4 „ 4 + 2x
Tw2'f (x,=r2“"
. . 4 + 2x
Javob: -----
8. To'g'ri burchakli uchburchakda
to'g'ri burchagining bissektrisasi
gipotenuzani a va b ga teng kesmalarga
bo'ladi. Uchburchak yuzini toping.
Yechish:
lc - bissektrisa
„ _ a1 b, a^_ £
2 ’ b, b’
ai = a x, bi = b-x
c2 = a-t2 + b2
(a + b)2 = (ax)2 + (bx)2
„2 _ (a + b)2
a2 + b2
a b x2 _ ab(a + b)2
2 ~ 2(a2+b2)'
Javob:
ab(a + b)2
2(a2+b2) ’
33
Yechimlar. Matematika va informatika 2017
4-variant
.-0.5
_ ic к <1 25'1 + log, a 1
9. a = 16 bo Isa, — --——
5' -log4a
ifodaning qiymatini toping.
Yechish:
a = 16 da
25 1 + log, a' 3 .
--------—-------ifodaninq
5’-log, a a05
qiymatini topamiz.
Soddalashtirib olamiz.
5~2-log2a 3д0.5 =
5 ’ -log2,a
5~2-log, а 3д0,5
5~1~^og2a
a = 16 da
^^16-3^16 =
5-1-MOg216
3
a-0-5 * * *
5--T4 — ~2
2 5
1
= - + 2-12 = 0,2-10 = -9,8.
5
Javob: -9,8.
10. Vx'°83^ > 3 tengsizlikning
yechimi bolmaydigan eng katta va
eng kichik tub sonlar yig'indisini toping.
Yechish:
V?®® > 3 ikkala qisminikubga
oshiramiz.
xto83x5 >33 *, x^S 27
(x’°«x)5 >27, x'°MX >273
log3x“'J' >log339, log2x>9
(log3x - 3)(1одзх + 3) > 0
log3x < —3, /оозх > 3
x < З'3, х > 3?
1
x <— va x> 27
27
1
(-«=;-^)U(27;<x>)
Aniqlanish sohasi: x > 0, demak,
tengsizlikning yechimi
(0;-^) U (27; v).
Tengsizlikning yechimi bolmaydigan
1
27
1 г 1 1
; 27 . —; 27 oraliqdagi
eng kichik tub son 2, eng katta tub
son 23, ularning yig'indisi 2 + 23 = 25.
Javob: 25.
oraliq
11. 639 sonini 2:3:4 kabi nisbatda
bo'ling.
Yechish:
639 sonini 2:3:4 kabi nisbatda bo'lamiz.
2x + 3x + 4x = 639
9x = 639, x=71
2x = 2-71 = 142
3x = 3-71 = 213
4x = 4-71 = 284.
Javob: 142, 213, 284.
12. Aylanaga teng yonli trapetsiya
tashqi chizilgan. Bu aylananing radiusi
trapetsiyaga tashqi chizilgan aylana
radiusidan marta kichik. Trapetsiya
asosidagi burchakni toping.
Yechish:
ABCD - teng yonli trapetsiya.
n 2r
R:r=46 , sina = —
c
S a+b+c+c
r=—, p=------------
P 2
Trapetsiyaga aylana ichki chizilgan,
2c = a + b, p = 2c
34
Yechimlar. Matematika va informatika 2017
4-variant
S = y/a-b-c-c = Cyfab bundan
Trapetsiyaga tashqi chizilgan aylana
radiusi
R = -^yf(ab+424ac+bc)2 =
2c2y]ab + c2 с Г c2'
-----r=—, R = —J1 +—
4cyab 2 у ab
4R2 л с2 6-4r2
—г~ = 1 +—г,
с 4г
4г2 1
bundan—-- = -
с2 2
c2
c2 ' ' 4r2
2r 1
-fi -1 Л 0 1 J5
(-42 ;-1) U (42 ;<x>)
Tenglamani yechamiz.
- 2x) = log/3 ,(3x2 + 2x)
х3 — 2х = Зхг + 2х
x3 — 4x—3x2 = 0
x(x2-3x-4) =0,
x = 0,x=-1,x = 4
x = 0, x = -1 aniqlanish sohasiga
tegishli emas.
x = 4 aniqlanish sohasiga tegishli.
1 ta yechim.
Javob: 1.
sina = ——, а = 45° =—
2 4
Javob: —.
4
13. logx2 i(x3-2x) =
= log х2 (Зх2 + 2х) tenglama nechta
yechimga ega?
Yechish:
Aniqlanish sohasini topamiz.
x2-1>0 (x - 1)(x +1) > 0
X2-1 ^1 x2 ф 2
x3-2x>0 x(x2 - 2) >0
3x2 + 2x>0 x(3x + 2)>0
x<-1,x>1
x*±^2
x(x-V2)(x +V2)>0
2
x<-—,x>0
3
о &.
(-42 ;0)U(y/2 ;«>)
Yechimlarni umumlashtiramiz.
14. C nuqta - AB kesmaning o'rtasi.
AC va BC kesmalarda mos ravishda
M, N nuqtalar shunday olinganki,
AM:MC = CN:NB munosabat bajariladi.
Agar AB kesma uzunligi 36 ga teng
bo'lsa, MN kesma uzunligini toping.
Yechish:
AB - kesma, AB = 36
AC = CB = 18
AM:MC = CN:NB
M C N
A 8
Agar M nuqta AC kesma o'rtasi, N nuqta
CB kesma o'rtasi bo'lsa, и holda
AM = MC = CN = NB = 9 bo'ladi,
bundan esa MN = MC + CN = 9 + 9=18
ekanligi kelib chiqadi.
Javob: 18.
15.log* (9-x2)-2log (9-x2)-8<0
5 5
tengsizlikning eng katta butun
yechimini toping.
Yechish:
Aniqlanish sohasi:
9 —x2>0, x2 —9<0,
(x - 3)(x + 3)<0,-3<x<3
log 1 (9 - x2) = a belgilaymiz.
5
35
Yechimlar. Matematika va informatika 2017
4-variant
a2 - 2a - 8 < 0
(a + 2)(a— 4) <0, -2<a<4
- 2 < log 1 (9 - x2) < 4
5
- 2 < -logs(9 -x2)<4
- 4 < Iog5(9 -x2) <2
cr4 s n xz2 c-z
1
625
-76<x2<9-—-
625
0^8™,0*x<l^
625 N 625
Tengsizlikning eng katta butun
yechimi 2.
Javob: 2.
16. To'g'ri to'rtburchakning
perimetri 30 sm, uning ikki qo'shni
tomonlariga chiziigan kvadratlar
yuzalarining yig'indisi 137 sm2. To'g'ri
to'rtburchakning yuzini toping.
Yechish:
P = 30, Si + S2 = 137
P = 2(a + b)~ 30
a + b = 15
S, = a2, S2 = b2, a2 + b2= 137
S = ab, a2 + b2 = 137
(a + b)2 - 2ab = 137
2ab = (a + b)2 - 137
. (a + b)2-137
2
. 152-137 225-137 ,
2 2
Demak, S = ab = 44.
Javob: 44.
17. у = log5(sin2x + cos2x)
1
funksiyaning x = — nuqtadagi ikkinchi
tartibli hosilasining qiymatini toping.
Yechish:
1) sin2x + cos2x - 1
2) log51 = 0
3) у = 0, y'=0
У" = 0, y" = 0.
Javob: 0.
18. Besh burchakli muntazam
piramida asosi yuzi S, yon sirti Q ga
teng. Piramida yon yog'ining asos
tekisligiga og'ish burchagini toping.
Yechish:
Sasos = S, Syon = Q
Натта yon yoqlari asos tekisligi bilan
bir xil tp burchak hosil qiluvchi piramida
tola sirti quyidagicha:
2Sasos-cos2^
sto.(a=----------1
cosip
Shundan foydalanib yechamiz.
Sto'la ~~ Sasos + Syon = S + Q
fS + Q)cos(p = 2S-cos2^
Scostp + Qcos<p = 2S- -
Scos<p + Qcostp = S + Scosp
S S
cos<p =—, <p = arccos—.
. u S
Javob: arccos —.
Q
19. Asosi a ga, yon tomoni b ga
teng bo'lgan teng yonli uchburchakning
yon tomoniga tushirilgan mediana
uzunligini toping.
Yechish:
a - asosi, b - yon tomoni, ть - yon
tomonga tushirilgan mediana.
36
Yechimlar. Matematika va informatika 2017
4-variant
mh = -42a2 + 2b2-b2 = -J2a2 + b2 .
b 2 2
Javob: ~^2a2 + b2 .
2
I V / 3
20 J x \У ~ > sistemadan x + у
|x-y = 21
ni toping.
Yechish:
Ikkinchi tenglamani qisqa ko‘paytirish
formulasidan foydalanib
ko'paytuvchilarga ajratamiz.
x-y = 21,
(4х~4у )(4х + 4у ) =21
4х-4у = 3 bolganligi uchun
3-(4^ + 44)=21
\4x~4y =3
[Vx + y/y = 7
24x=10
4x = 5, y[y = 2
x = 25, у = 4 (25; 4)
x + у = 25 + 4 = 29.
Javob: 29.
21. To'g'ri burchakli uchburchakning
gipotenuzasini unga ichki chizilgan
aylananing urinish nuqtasi a va b
kesmalarga boladi. Agar biror kateti c ga
teng bo'lsa, uchburchak yuzini toping.
Yechish:
c - katet, AB = c, CK= a, KA = b,
S=?
СВ2 = CA2 - AB2 Pifagorteoremasiga
ko'ra:
CB = 7(a + b)2 -c2
Demak, uchburchak yuzi:
CB AB _ Cyj(a + b)2-44
о —------—--------------.
2 2
cV(a + b)2 - c2
Javob: - '----------.
2
22. x = 2 - V3 bo'lsa, +-1
x
ifodaning qiymatini toping.
Yechish:
alo8eb = b formuladan foydalanamiz.
yj _^10g2X2 „^S^2* __
_ 222logz5t — 2log2X — X
2) x + - = 2-43+—1~^ =
x 2-43
=2-4з+-----^4^—== = 2-43
(2 + 43)(2-43)
t_£±4-.2_^+£l^=
22-(43)2 4-3
= 2-43 + 2 + 43=4 .
Javob: 4.
23. O'zaro teng bo'lmagan x va
у sonlari x2 - 36x = y2 - 36y tenglikni
qanoatlantirsa, x + у ni toping.
Yechish:
x#y,
x2 - 36x = y2 - 36y
x2-36x-y2 + 36y = 0
x2-y2-36(x-y)^0
(x-y)(x + y)-36(x-y) = 0
(x-y)(x +y-36) =0
1) x-y = 0, x=ymasala shartida x^y.
2)x+y-36 = 0, x + y = 36.
Javob: 36.
37
Yechimlar. Matematika va informatika 2017
4-variant
ifodani n = 32 dagi qiymatini toping.
Yechish:
1) п12-з4з = (п°-у-(УзУ =
n° 8 + 3 - JЗп° 8 n0-8+3--j3n08
(n04-4з)(п°а -4зп°* + з)
пО8-^Зп°-8 +3
= п04-4з
2) n04 -л/3-л/3(73п°-8-7) =
= n0,4 - J3 -3^ + V3 = n°4 - 3n°’4 =
= -2n°’4
3) n = 32 da -2n04 = -2nr° = -2n5 =
= -2 325 = -2 2 = -2-22 = -8.
Javob: -8.
25. у = ex lnx bolsa, y" - y' ni toping.
Yechish:
1) Ko'paytmadan hosila olamiz.
1
y’ = (e Inx)' = exlnx + ex- — =
X/, ? !
= e (Inx +—)
x
2) y" = (ex(lnx +-))' = ex(!nx +-) +
X X
X/1 1 . X// 2 1 .
+ e (---?) = e (lnx +---)
XX XX
3)y"-y' = ex(lnx + 2—L)-
X X
„ ’ 1 , 2 1
- e (Inx +—) = e (Inx +-- - Inx -
X XX
1 11 f
x xx v 7
Javob: ex I x2 I.
26. Teng yonli trapetsiyaning
diagonal! yon tomoniga perpendikulyar.
Yon tomoni b katta asosi bilan
a burchak tashkil qiladi. Trapetsiyaning
katta asosi atrofida aylantirishdan
hosil bolgan jism sirtini toping.
Yechish:
ABCD - teng yonli trapetsiya,
BD - diagonal!± AD, AD = b, Z DAK = a.
AB katta asosi atrofida aylantirsak,
DD1C1C sitindr va DAD-i, CBC1
konuslar hosil bo'ladi. Hosil bo'lgan
jismning sirti: S = 2itRH + 2rrRb
R = bsina, H = KN = AB-AK- NB
AK = NB = bcosa
AB =
AD
cos a
b
cos a
H = —-----2b cos a
cos a
S = 2tcR(H + b) = 2irb-sina-
( fa .
•------2bcosa + D
Vcosa
2jrb* 1 2 sin a - 2 ,
----------(1 _ 2cos a + cosa) =
cos a
= 27tb2tga(cosa - cos2a) =
= 2xb2tga-2sin— sin —.
2 2
л t.2. . a . 3a
Javob: 4nb tga sin — • sin —.
27. Katetlari yiglndisi I ga, to‘g‘ri
burchagi uchidan tushirilgan balandligi
h ga teng bolsa, uning yuzini toping.
Yechish:
a, b - katetlar
a+b=l, h = hc- gipotenuzaga
tushirilgan balandlik,
S = ?
38
Yechimlar. Matematika va informatika 2017
4-variant
„ ch a-b , .
S =---=-----, ab = ch
2 2
c* 1 2 = a2 + b2 = (a + b)2 - 2ab = i2 - 2ch
c2 + 2ch - У2 = 0 kvadrat tenglamadan
c ning qiymatini topamiz.
-2h + 44h2+4l2
c -----------------
2
2
S = ^ = ±(4lf^-h).
Javob: + l2 -h^.
28. Hisoblang:
V3j2-4 V34 + 24V2 t/324 .
Yechish:
(a +• 4b)2 = a2 + 2a4b + b formuladan
foydalanib yechamiz.
1) 34+ 2442 =^34 + 2-4-342 =
-4118 + 2-342-4 + 16 =
= ^(з42+4)2 =у1з42+4
2) 4з42-4-^з42 + 4-4182 =
= у1(з42-4)(з42+4)-18 =
= у/((342)2-42)-18 =
= 4(18-16)-18 = 42-18 = 436 = 6.
Javob: 6.
29. Bir sonning 25% ortig'i shu
sonning 20% kamidan necha foiz ko‘p?
Yechish:
x- son
1)x-100%
Xi - 125% x-i = 1,25x
2)x- 100%
X2 - 80% X2 = 0,8x
1,25x — y
0,8x — 100%
1,25x-100% 1250%
V =------------=-------= 7 00, ZOto
7 0,8x 8
156,25°% - 100°% = 56,25°%.
Javob: 56,25%.
30. To'g’ri burchakli trapetsiyaning
diagonali yon tomoniga teng. Agar uning
balandligi 2 ga va yon tomoni 7 ga teng
bo’lsa, trapetsiya o’rta chizig’ining
uzunligini toping.
Yechish:
a, b - trapetsiya asoslari. c - yon
tomoni. d - diagonali. d-c-7, h=2
m~o‘rta chizig'i
b
a
a + b
2
b2 = c2-h2 = 72-2г =45
b =445=345
3b 3-345 . к
m = — =-----= 4,545 .
2 2
Javob: 4,54b .
31. Sakkizlik sanoq sistemasida 54,218 va 13,23s sonlarining ko’paytmasini
toping.
Yechish:
54,21 w 13,23(8) = ?
54,21(8) = 101100,010001(2).
13,23(8) = 1011,010011(2).
39
Yechimlar. Matematika va informatika 2017
4-variant
101100,010001
x 1011,010011
101100010001
101100010001
000000000000
000000000000
101100010001
000000000000
101100010001
101100010001
000000000000
101100010001
111110100,000100000011
111110100,000100000011(2) = 764,0403(8) (1-chijadval bo'yicha).
Javob: 764,0403.
32. 221ю, 101Ю, 142ю butun sonlarning barchasini yozish mumkin bolgan
eng kichik asosli sanoq sistemasida shu sonlar raqamlari yig'indisini aniqlang.
Yechish:
221io-X2. 221w = 11011101, .
V /
raqamlar yig ‘ indisi 6
22112
2_ liiOI2_
2 10_Г55|2_
2 10 4_l27|2_
ф 10 152_fl3|2
"@14 7121612
ф 6 фб[3[2
ф <@Дф
ф
101ю-Хг. 10110= 11001012 .
raqamlar yig'indisi 4
10112
’10 l50j2_
7D 4 j25i2_
10 2_M2i2
10 5 121612
© 4 "@6|312
Ф @2©
Ф
142ю-Хг. 142w = 10001110, .
raqamlar yig'indisi 4
14212
"14 |71T2
2 6j35l2_
2 11 2_IT7I2
@10 15 161812
ф)14фб|412
Ф @4(212
©2©
©
6+4 + 4=14.
Javob: 14.
40
Yechimlar. Matematika va informatika 2017 4-variant
33. Xab(huB) qanday qurilma?
Yechish:
Xab qurilma yordamida kompyuterlarning boshqa kompyuterlar bilan telefon
tarmogl orqali axborot almashishi qurilmasidir.
Javob: kompyuterning boshqa kompyuterlar bilan telefon tarmog'i orqali
axborot almashinuv qurilmasi.
34. 2000 Kbayt axborot necha bitga teng?
Yechish:
1 bayt - 8 bit
1 Kbayt - 2ю bayt - 1024 bayt
1 Kbayt — 210- 8 bit
2000 Kbayt = 21°-8-2000 bit = 210-^-2-1tf bit = 210-2 (2-10)3 bit = 210+1-20? =
= 211-203 bit.
Javob: 23o-2O11.
35. Ikkilik sanoq sistemasida amallami bajarib, natijani o'nlik sanoq
sistemasida aniqlang: 1001 + 110011.
Yechish:
1001 + 110011 =Xio.
1001
110011
111100(2)
54321 0
111100(2)-x10=?
1-25 + 1-24 + 1-23 + 1-22 + 0 + 0 = 32 + 16 +8+4+0= 60(10).
Javob: 60.
36. Sakkizlik sanoq sistemasida 42644s va 122418 sonlarining yig'indisini
hisoblang.
Yechish:
42644b + 122418 = ?
Qo‘shish uchun avval 1-jadval bo‘yicha ikkilikka otib, keyin raqamlarni
qo‘shib chiqamiz.
42644(8) = 100010110100100(2).
12241(8) = 1010010100001(2).
100010110100100
+ 1010010100001
101101001000101(3)
Endi bu sonni triadalarga ajratamiz.
101101001000101(2) = 55105(8) (1-chijadval bo‘yicha).
Javob: 55105.
41
Yechimlar. Matematika va informatika 2017
5-variant
5-variant
1. x = 2,125 bolsa, л/4х2-5(4х-5) +
+2^'9 + x(x + 6) ifodaning qiymatini
toping.
Yechish:
a2 ± 2ab + b2=(a± b)2
1) д/4x2 -5(4x- 5) +
+2^/9 + x(x + 6) = yj4x2 -20x + 25 +
+2>/9 + 6x + x2 = y](2x-5)2 +
+2^(x + 3)2 = \2x - 5| + 2(x + 3)
2) x = 2,125 da\2x-5\ = 5-2x bo'ladi.
5 - 2x + 2(x + 3) = 5 - 2x + 2x + 6 - 11.
Javob: 11.
2. Konus ichiga balandligi konus
asosining radiusiga teng silindr chizilgan.
Siiindr tola sirtining konus asosining
yuziga nisbati 3:2 kabi. Konus o‘qi bilan
yasovchisi orasidagi burchak topilsin.
Yechish:
A О = OB = R- konus radiusi.
AO = 00 =H- silindr balandligi.
OK = ON = r silindr radiusi.
Silindr tola sirti:
Sfo'fa = 2m2 + 2rtrR
Masala shartiga ko'ra
2ЛГ2 + 2nrR =-nR2
2
r2 + Rr = 0 kvadrat tenglamani
yechamiz.
R 3n
— va r = —R
2 2
R
— tenglama ildizi bo'ladi.
8
АСОВ va AENB o'xshash.
ZOCB = ZNEB = <p
R
, NB R-r 2 1
tQ<p — —-----— — — —
EN R R 2
1
<p-arctg—.
Javob: arctgi.
3. To'g'ri burchakli uchburchakning
to‘g‘ri burchagi uchidan tushirilgan
balandlik h ga, katetlarining
gipotenuzadagi proyeksiyalarining
ayirmasi I ga teng. Uchburchak yuzini
toping.
Yechish:
h - gipotenuzaga tushirilgan balandlik,
x, у - katetlarning gipotenuzadagi
proyeksiyalari.
x-y = l, S=?
a
„ _ ab _ ch
О —---—----
2 2
ft =7Х У , xy = h2, c = x + у
42
Yechimlar. Matematika va informatika 2017
5-variant
(x-y)2 = f, 4 + y2 - 2xy = f
J2 ,/ _ 2 ^2
? +2h2 + 2h2 = c,
c = -Jl2 + 4h2
s=EJL=^ylp +4if.
2 2
Javob:—V/2 + 4h2 .
2
4. yjx + 1-4>Jx-3 +
+7% +1 + 4>Jx-3 ifodaning
x = 3,185 dagi qiymatini toping.
Yechish:
Soddalashtirib olamiz:
1) -J x +1 - 4-Jx~-~3 =
= ^x-3-4-Jx-3 + 4 -
= ^(Vx-3)2 -2 -Vx-3 -2 + 2" =
= ^х-3-2^ =|Vx-3-2|
2) |Vx-3 - 2| ifoda x = 3,185 da
|-Jx-3 -2| = 2- Vx-3 bo‘ladi.
3) 2-4x^3 + 2 + 4x^3=4
Ifodaning qiymati 4 ga teng.
Javob: 4.
5. x = 2,61 bo'lsa,
(3-x)-1-7(x-3)2(x + 1).
Yechish:
1) (3-x)_J^(x-3)2(x + 1) =
=———-\x—3\--J x+ 1
3-x 1 1
2) x = 2,61 da |x - 3| = 3 - x ga teng
bo'ladi.
3)---—(3-x)-Vx+7 = Vx+7
4) x = 2,61 da
Vx + 1 = ^2,61 + 1 = Дб? = 1,9.
Javob: 1,9.
6. = (0,25)z~x tenglamani
yeching.
Yechish:
256 = 2s, 0,25 = 2~2
~ = (2-2 , 2&x~3 = 2”4+2/
5 - x = -4 + 2x
9 = 3x, x = 3.
Javob: 3.
7. Agar uchburchakning medianalari
12, 15 va 21 sm ga teng bo'lsa, uning
yuzini toping.
Yechish:
ma = 12, ть = 15, mc = 21 uchburchak
medianalari. Uchburchak yuzini
medianalari orqali hisoblaymiz.
1)m= т.+ть + тв = 12 + 15 + 21 = 24
2 2
2)S = --jmfrn - 777a)(m - mb)(m - mc) =
= ^24(24-12y24-15)(24 - 21) =
= -J24-12-9-3 = -Jl2-2-12<F3 =
3 3
= -12-342^3 = 48^6 .
3
Javob: 48 4б .
8. Tetraedrning qirrasi b.
Qirralardan birinchi o'rtasidan bir-biri
bilan kesishmaydigan ikkita qirrasiga
parallel qilib tekislik o'tkazilgan. Hosil
bo'lgan kesimning yuzini toping.
Yechish:
DABC - tetraedr.
AB = b. Kesuvchi tekislik AB qirraning
o‘rtasi F nuqta orqali AC va BD
qirralarga parallel qilib o'tkazilgan.
43
Yechimlar. Matematika va informatika 2017
5-variant
FN kesma ABC uchburchakning o'rta
chizigl.
FN = -AC=-
2 2
N nuqta BC qirra o‘rtasi.
NL = —BD = —
2 2
NL\\BD
L nuqta CD qirra o‘rtasi.
KL\\AC
К nuqta AD qirra o'rtasi.
Demak, FNLK - romb. BD qirra BDE
tekisligida yotadi. E nuqta AC qirra
o‘rtasi. Bu tekislik AC qirraga
perpendikulyar. BD1AC KF\\BD va
FA/ЦДС, bundan FK1. FN. Demak,
FNLK kvadrat ekan. Sk = f—1 = —.
4
. U b2
Javob: —.
. 4
л 75-2^6 . u. U1
9. —r- r-—r=—7=- ni hisoblang.
(4/3 + V2)«/3 - ^/2)
Yechish:
(Уа->)(^а + №)=Уа-уЪ
1) V5-2V6 = y/5-2y[3-42 =
= ^7з-42)2 =43-42
л/5-2>/б = ^-42
\</3+</2)(</3-</2) 43^2 ‘
Javob: 1.
10. Si + S2 + S3 = ? Bu yerda
О - tashqi chizilgan aylana markazi.
Yechish:
О - tashqi chizilgan aylana markazi.
Si + S? + S3 = ?
S = 100, a2 =10, a - 10
a
$ _ 2 + a a _ За a _ 3a2
1 ~^2
a a
Sz=2^ = ^_
2 2 8
„ _S_a2
3 2 2
S7 + S2 + S3 =— +— + — = a2 = 100.
8 8 2
Javob: 100.
11. a = 4,125 bo'lsa,Va-4Va^4 +
+Va + 4>/a-4 ifodaning qiymatini
toping.
Yechish:
Soddalashtirib olamiz.
1) y]a-4y/a^ =
= Va-4 -4^a-4+4 =
= ^a-4j2-2ja-4-2 + 22^
= ^Va-4-2') =|Va-4-2|
2) a =4,125 da
|Va-4 -2| = 2- Ja-4 ga teng bo'ladi.
3) 2-Ja-4 +Ja + 4 +2 = 4.
Javob: 4.
44
Yechimlar. Matematika va informatika 2017
5-variant
12. ^7-yl4x + 2+7 = 2
tenglamaning ildizlari quyidagi
oraliqlardan qaysi biriga tegishli?
Yechish:
Aniqlanish sohasi
x>-2
Tenglikning ikkala qismini kvadratga
ko'taramiz.
7-^4x + 2 + 7 =4
'IJx + 2+7 = 3, yJx + 2+7^9
Jx + 2 = 2, x + 2 = 4, x = 2
Tenglamaning ildizi 2 C [2; 4) oraliqqa
tegishli.
Javob: [2; 4).
13. f(x) = ln(ex- xex) bolsa, /'(2) ni
toping.
Yechish:
1) f'(x) = (ln(ex-xex))’ =
_ e* - e* - xex _ -xex
ex-xex ex(1-x)
-x _ x
1-x x-1
2
2)V(2)^~j = 2-
Javob:2.
14. Soddalashtiring:
I T
(18+8>/2)-(4-V2 )2 + 4 J20— .
Yechish:
(18 + 842 ) (4-42 )2 + j =
= (18 + 8 42 ) (16 - 8 42 + 2) +
+ 4-^ = (18 + B42 )(18-842 ) +
+ 4 ^ = 2 (9 + 442 )-2-(9-442 ) +
+ 18 = 4-(92 - (442 )2) + 18 =
= 4(81 - 32) + 18 = 4-49 + 18 = 214.
Javob: 214.
15. a = 1 bo'lsa,
a+1
j (ln(sin* 2 * 2x + cos2 2x) + 1)dx aniq
a
integralni hisoblang.
Yechish:
1) sin2x + cos2x = 1
2) In1 = 0
a+J a+I a + 1
3) f (0 + 1)dx = \1dx = x
a a a
= a + 1 -a = 1.
Javob: 1.
16. Qarang: 1-variant 22-savol
(8-bet).
1
cos—
17. f—~dx hisoblang.
J x
Yechish:
1
COS— 7 Z-J \ 7
f—--dx =-[cos—dl — =-sin— + C.
x x \x) x
1
Javob: -sin—+ C.
x
18. Ifodani soddalashtiring:
cos44a + sin8a - sin44a.
Yechish:
1) cos44а - sin44а = (cos24а - sin24а)-
(cos24a + sin2 4a) = cos8a
2) cos8a + sin8a = cos8a +
z \ 8« + — — 8a
+ cos --8a = 2cos-------------
<2 J 2
8a- — + 8a
2 n x:
cos-----------= 2cos—
2 4
•cosI 8a - — I - 2- cos(8a - 45°) =
= 42 cos(8a - 45°).
Javob: 42 cos(8a - 45°).
45
Yechimlar. Matematika va informatika 2017
5-variant
19. Asosi m va asosidagi burchagi
a bo'lgan teng yonli uchburchakning
perimetrini toping.
Yechish:
m
m - asos, а - asosidagi burchak.
P = 2a +m ni topamiz.
m
~2 rn
— = cosa, a =------
a 2cosa
2 cos a
_ ( 1 _ m(1 + cos a)
l.cosrz ) cosa
л • 2
2m cos — 2/n • sm a cos —
2 =2 =
cosa cos a sin a
2
cos —
= 2mtga-------2.—
2 sin - cos —
2 2
Javob: mtga ctg^.
^mtga-ctg-.
20. /(x) = 40sin3xcos7x uchun
boshlang'ich funksiyani toping.
Yechish:
f(x) = 40sin3xcos7x
1) Ko‘paytmadan yig'indiga o'tamiz.
1
sin3xcos7x =— (sin(3x + 7x) +
1
+ sin(3x- 7x)) =— (sinlOx + sin(-4x)) =
1
=— (sin 10x - sin4x)
1
2) f(x) - 40- — (sin 10x - sin4x) =
= 20sin10x - 20sin4x
Boshlang'ich funksiyasini topamiz.
F(x) = -2cos10x + 5cos4x + C =
= 5cos4x - 2cos10x + C.
Javob: 5cos4x - 2cos10x + C.
21. Agar Z ACE = 70°, Z EDF = 30°
va CnB yoyi 110° bo'lsa, Z CAF ning
qiymati topilsin.
Yechish:
Z ACE = 70°, Z EDF = 30°
CnByoy110°, ZCAF=?
ODLAF
1) Z COB = 110°
CO = OB = R,
Z OCB = Z OBC = 35°
2) CO = OE = R,
Z OCE =Z OEC = 35°
Z COE = 110°
3) EO = OD = R,
Z ODE = Z DEO = Z EOD = 60°
4) Z BOD = 360° — 110° —
- 110°-60° = 80°
5) ABOD to'rtburchak.
Z ABO = 180° - 35° = 145°
Z BAD = 360° - 145° - 80° - 90° = 45°
Z CAF =Z BAD = 45°.
Javob: 45°.
22. ABCD trapetsiya AB yon
tomonining o'rtasidagi E nuqtadan
CD tomonga parallel qilib AD katta
asos bilan G nuqtada uchrashguncha
to'g'ri chiziq o'tkazilgan. Agar
AG = 5 dm Va GD = 2,5 m bo'lsa,
trapetsiyaning asoslarini (m) toping.
Yechish:
AG = 5 dm = 0,5 m
GD = 2,5m
46
Yechimlar. Matematika va informatika 2017
5-variant
EG parallel CD ga.
EK - o‘rta chiziq.
EK=GD = 2,5
AD = a = 0,5m+2,5m = 3m katta asos.
EK=^,
2
b=2EK-a- 2-2,5-3=5-3= 2m
a = 3 m, b = 2 m.
Javob: 3; 2.
23. Prizmaning qirralari soni 66 ga
teng. Uning yoqlari sonini toping.
Yechish:
Prizma qirralari soni 66 ta bo'lsa,
yoqlari soni 66:3 = 22 ta.
Javob: 22.
tenglamani yeching.
Yechish:
Aniqlanish sohasi:
2jx- xiO
Jx (2 -Jx)± 0, x^O, 4, x>0
Umumiy maxrajga keltirib yechamiz.
2 1 _ 4
2-y/x 2 yfx(2-4x)
4jx +Jx (2 -yfx ) = 8
4\lx + 2y/x -8 = 0
x - б4х + 8 = 0 tenglama ildizlarini
topamiz.
jx = 2, 4x= 4
Bunda x = 4 aniqlanish sohasiga
tegishli emas.
x= 16 tenglama ildizi.
Javob: 16.
b,+b.+b, 7
25. ——-—- = ~ geometnk
b. + b2 6
progressiya maxrajini toping.
Yechish:
b1+b1q + b1q2 _7
b, + b,g 6
b,(7 + q + q2) 7
b,(1 + g) 6
6 + 6q + 6q2 = 7 + 7q
6q2-q - 1 = 0 tenglama ildizlari.
1 1
1 1
Javob: —va —.
3 2
26. Parallelepipedning asoslari
tomoni 4 ga teng kvadratlardan, barcha
yon yoqlari romblardan iborat. Yuqori
asosining uchlaridan biri ostki asosining
barcha uchlaridan baravar uzoqlikda
joylashgan. Parallelepipedning hajmini
toping.
Yechish:
ABCDA1B1C1D1 - parallelepiped
ABCD va A1B1C1D1 - kvadrat,
AA1BB1 - romb.
81
AB=AAi = a = 4
AiB — AiA — AjC — A1D — a — 4
AB = 4, AC = JAB2+BC2 = aj2
a
= 42
J2
V = Sasos-H = a2 H = 42- 2J2 = 32J2 .
Javob: 32^2.
47
Yechimlar. Matematika va informatika 2017
5-variant
27. Hisobiang:
"1 Q Q
log2 (-^— + + --- + 1,375).
0,(4) 0,(6) 0,(8)
Yechish:
3 3 3 . .
log?(-----------*------1,375) nt
0,(4) 0,(6) 0,(8)
hisoblaymiz.
3 3 3
Awal------+-----+------+ 1,375 ni
0,(4) 0,(6) 0,(8)
hisoblaymiz.
3(1:-+ 1:-+1:~)+ 1,375 =
9 9 9
79 9 9>
= 3- - + - + -1+1,375 =
(4 6 8)
(1 1 1\ y?
= 3-9- - + - + - + 1,375 = 27— +
{4 6 8) 24
4 9-13 .3 .120 .c
+ 1,375 =----+ 1— =1---= 16
8 8 8
log216 = 4.
Javob: 4.
28. a = 1 bo'lsa, V2a-7a2 +2
y]2a + 4a2 +2 ifodaning qiymatini
toping.
Yechish:
2a — y]a2 + 2 д/2а + Va2 + 2 =
= ^2a-Va2 + 2^2a + Va2 + 2j =
= )(2a)2 -(Ja2 +2) =
= ^4a2-a2-2 =^3a2-2
a = 1 da
y)3-12-2=^3-2=4l =1.
Javob:1.
Yechish:
ABCDA1B1C1D1 - kub
Sto'/a — 6a ~ 36, a = 6
a — у/б
EN kesma AA1 va BC ayqash qirralari
o'rtasi orasidagi masofa.
en2 - ae2 + an2
AE = °=^
2 2
AN2 = AB2 + BN2
Javob: 3.
30. Agar teng yonli uchburchakning
asosi b ga, yon tomoniga tushirilgan
balandligi h ga teng bo'lsa, uning
yuzini toping.
Yechish:
b - asosi, h - yon tomonga tushirilgan
balandlik.
29. Tola sirti 36 sm2 bo'lgan
kubning ayqash qirralari o'rtalari
orasidagi masofani toping.
2
AB = BC = a, DC -x, BD = a-x
48
Yechimlar. Matematika va informatika 2017
5-variant
Uchburchak ADC va ADB to‘g‘ri
burchakli.
h2 = b2 - x2, h2 = a2 - (a - x)2
x2 = b2 - h2, x = ~Jb2-h2
h2 = (a-a + x)(a + a-x) = x-(2a - x)
. h2 +x2 h2 +b2 - h2
2 a =---= —
x Jb^h2
b2 b2
—===, a =—;==
JT^h2 24b^7f
a h _ b2h
~~2~~Ub2 -h2'
i. b2h
Javob: —r==
4^b2-h2
31. Ma’lumotlar ombori bilan Bilimlar omborining o’rtasidagi asosiy farq nimada?
Yechish:
Ma’lumotlar ombori - bu harxil turga mansub, kompyuter xotirasida doimo
saqlanuvchi, ma’lum bir soha holatiniaks ettiruvchi, foydalanuvchi uchun
mo’ljallangan ma’lumotlar jamlanmasi.
Bilimlar ombori- maxsus ma’lumotlar ombori bo’lib, и bilimlar ustida amallar
bajarish uchun mo’ljallangan. Bilimlar ombori ma’lumotlar ombori javob
berolmaydigan savollarga javob berib, intellektual va tekshiruv tizimining asosiy
qismi hisoblanadi.
Javob: bilimlar ombori axborotni yetishmayotgan faktlar bilan to'ldirib borish
imkoniyatlariga ega.
32. A1 = -4, A2 = -1, B1 = 8, B2 = 5 bo’lsin. Natijasi 9 ga teng bo’ladigan
formulani aniqlang.
Yechish:
A1 = -4; A2 = -1; B1 = 8; B2 = 5.
Natija = 9.
=MAKC funksiyasi eng katta sonni aniqlaydi.
Masalan: MAKC(4;5) —> natija 5 bo’ladi.
=MAKC(ABS(A 1 )+B2;A2+B1);
=MAKC(ABS(-4)+5;-1+8)
=MAKC(4+5;7)
=MAKC(9;7)=9.
Javob: =MAKC(ABS(A1)+B2;A2+B1).
33. Elektron jadvaldagi murojaat (ssilka) - bu:
Yechish:
Elektron jadval yacheykasiga formula yozilganda bu formula boshqa
yacheykaga nusxalanganda bu formula va undagi yacheykadagi qiymat
o'zgarishi nazarda tutiladi.
Javob: yacheykadagi formula nusxalanganda undagi bor ma’lumotlar o’zgaradi.
34. Qanday dasturlar majmuasi kompyuterning va kompyuter tarmoqlarining
ishini ta’minlaydi?
Yechish:
Kompyuterda ishlash uchun mo’ljallangan dasturlar bir qancha turga bo’linadi.
Foydalanuvchi uchun mo’ljallangan, uning ish faoliyatiniyengillashtiruvchi
49
Yechimlar. Matematika va informatika 2017
5-variant
dasturlar. Bu dasturlardan mutaxassis bolmagan foydalanuvchi bemalol
foydalana oladi.
Yana bir dastur turlari, dasturlar yaratish uchun moljallangan. Bunda
foydalanuvchi ma’lum o‘quv va tajribaga ega bolishi kerak. Bunday dasturlar
yaratish vositasi hisoblanadi.
Foydalanuvchi va kompyuter o‘rtasidagi muloqotni o‘rnatish uchun maxsus
tizim (sistema) dasturlari mavjud. Bu dasturlar foydalanuvchi buyruqlarini
mashina tushunadigan kodlarga o'chirib, amallarni bajarib, yana foydalanuvchi
tushunadigan ko‘rinishda javobni qaytaradi. Ular kompyuter va kompyuter
tarmog'i ta ’minlovchi qurilmalar hisoblanadi.
Javob: tizimli dastursiz ta’minot.
35. Ikkilik sanoq sistemasida amallarni bajaring:
1111011,011 + 1,01 — (1-25 + 1-22 + 1-21).
Yechish:
1111011,011 + 1,01 -(1-32 + 1-4 + 2) = ?
32 + 4 + 2 = 38ю ->• X2.
38|2_
~2_fl9|2
18 181912
18 (D8WI2
© ©4I2I2
©2©
©
3810- 100110(2).
1111011,011
+ 1,01
1111100,101
"100110,000
1010110,101(2)
1111100,101(2)- 100110(2) = 1010110,101(2).
Javob: 1010110,101.
36. O‘n oltilik sanoq sistemasidagi 7A,84i6 sonini o'nlik sanoq sistemasida
ifodalang.
Yechish:
7A, 84 - sonini o‘n oltilik sanoq sistemasidan o'nlik sanoq sistemasiga otish
uchun quyidagi amal bajariladi.
7A, 84 = 7-161 + A-1ff, 8-16~1 + 4-1&2.
A o'rniga 10 to'g'ri keladi.
7-16+10-1,
± + Л~ = 122^
16 16-6 64
Ф 122,515625;.
Javob: 122,515625.
50
Yechimlar. Matematika va informatika 2017
6-variar
6-variant
1. Uch burchakli muntazam piramida
asosiga perpendikulyar va asosining
ikki tomonini teng ikkiga bo'luvchi tekislik
bilan kesilgan. Dastlabki piramida
asosining tomoni a ga va asosidagi
ikki yoqli burchak a ga teng. Tekislik
kesib ajratgan piramida hajmini toping.
Yechish:
DABC - muntazam uchburchakli
piramida.
AB=AC=BC=a
AAPD=а
1) ABC asosga perpendikulyar
asosining AB va AC tomonlarini teng
ikkiga bo'luvchi kesim KNE
uchburchak. KNE tekislik ABC
tekislikka perpendikulyar. KF]]DO.
2) KANE piramidaning asosi ANE
uchburchak. A ANE ning yuzi A ABC
.. 1 .. . o a2V3
yuzmmg— qismiga teng. Sme
4 76
AFK va AOD uchburchaklar o'xshash.
KF balandlikni OD orqali ifodalaymiz.
1 1
AF =— AO, chunkiAF = — AP
4 2
AO=~AP, KF=—OD
3 4
OD kesmani ADOE dan topamiz.
°£ = tga, OD = OP tga,
chj3 З\[3 .
OP = r =----, OD =------tga
6 6
izr- 3 a>/3 . а^З
KF ---------tga =-----tqa
4 6 8
и 1 1 a2q3 aq.
3 3 16 8
. . a3tga
Javob: ——.
128
tga =----
128
2. JVx2 -2x + 1 dx hisoblang.
0
Yechish:
J Vx2 -2x + 1 dx =
0
2 ------- 2
= J V(x - 1)zdx = x -11 dx
0 0
2
J| x -11 dx ni hisoblaymiz.
0
1-1
8 = 28л = 2—= 1
2
J| x -11 dx = 1.
0
Javob: 1.
3. Ifodani soddalashtiring:
^Ц2^Ц=+(о,(5))^
Yechish:
51
Yechimlar. Matematika va informatika 2017
6-variant
[3x-12y=27
4. (x; y) sonlar jufti-!^- g
tenglamalar sistemasini qanoatlantirsa,
x + у ning qiymatini toping.
Yechish:
Birinchi tenglamani ko‘paytuvchilarga
ajratamlz.
3x-12y = 27, x-4y- 9,
(4x ~ 2y[y )(4x + 2y/y ) = 9
( Jx - 2 Jy )’9 = 9
4x ~2y[y = 1
Jx-2jy=1
4x + 2y[y = 9
2\[x=10
4x = 5, \[y = 2
x = 25, у = 4 (25; 4)
x + у = 25 + 4 - 29.
Javob: 29.
_ 1 + cos3a + cos2« + cosa ., ,
5. -----------------.----ifodani
2cos a + cosa-1
soddalashtiring.
Yechish:
.. , n n «+ В a-f>
1) cosa + cosp = 2cos cos—
„ n a + 3a
cosa + cos3a = 2cos------
2
a-3a „ „
cos----- = 2cos2a-cosa
2
2) 1 + cos2a = cos2a + sin2 a +
+ cos2a — sin2a = 2cos2a
Kasrning surati: 1 + cos3a + cos2a +
+ cosa = 1 + cos2a + cos3a + cosa =
= 2cos2acosa + 2cosza =
= 2cosa(cos2a + cosa)
Kasrning maxraji: 2cos2a + cosa - 1 =
= 2cos2a + cosa - cos2a - sin2a =
= cosza - sin2a + cosa = cos2a + cosa
_ , 2cosa(cos2« + cosot) „
Demak, -------------------- = 2cosot.
cos 2a + cosa
Javob: 2cosa.
6. f(x + 2) = e* * * x-g(x2 + 1) va g(1) = 5
bo’lsa, f '(2) ni toping.
Yechish:
g(1) = 5 ekanligi sababli
f(x + 2) = exg(x2 + 1)dax = 0 deb
olamiz.
f(2)=e°g(1) = 1-5 = 5
f(2) = 5.
Javob: 5.
7. Tenglama nechta yechimga ega?
л/5-х2 n
tgx = 0
X
Yechish:
Aniqlanish sohasini topamiz:
-45 <x<45
x^O
X Ф — + Ttn
52
Yechimlar. Matematika va informatika 2017
6-variant
,, n=0,x*-
2
, It
n = -1, xt-
2
Tenglamani yechamiz.
1) ^5~x =o, д/5-х2 =0
X
x2 -5 = 0, x2 = 5, x = ±45
2) tgx = 0, sinx = 0, x = m
n = 0, x = 0 aniqlanish sohasiga
tegishli emas.
Tenglamaning yechimi:
x = ±45
Tenglama 2 ta yechimga ega.
Javob: 2 ta.
8. Tenglama nechta yechimga ega?
(1---V-)-V9^T2 = 0
sin X
Yechish:
Aniqlanish sohasini topamiz.
fsin x ф 0 f x ф Tin lx * яп
\9-x2 >0^\x2-9<0{-3<x<3
-------•-----e •---
-3 0-3
[-3; 0) U (0; 3]
Tenglamani yechamiz.
1)1----= 0 , sin2x = 1,
sin x
sinx = ±1, x = — + itn, n CZ,
2
_ Я
n = 0, x = —
2
. я
n = -1, x=----
2
2) 49-x2 = 0,9-X2 = 0,
x2 = 9, x = ±3
Tenglamaning yechimlari:
x =—, x=~—, x = 3, x = -3, demak,
2 2
tenglama 4 ta yechimga ega.
Javob: 4 ta.
9. Teng yonli uchburchakning yon
tomoni 20 ga, asosi 24 ga teng. Bu
uchburchak medianalarining kesishish
nuqtasi bilan bissektrisalarining kesishish
nuqtasi orasidagi masofani toping.
Yechish:
AB = BC = 20, AC = 24
AN, CM, BD - mediana
AK, CE, BD - bissektrisa
О - medianalar kesishgan nuqta,
Oi - bissektrisalar kesishgan nuqta.
BD2=AB2-\—\ = 2(4 - Т4 =
I 2 J
= 400- 144 = 256
BD = 16
Mediana BD = BO + OD =
= 2OD + OD = 3OD
Bissektrisa OiD = r=—- =
P
„ 24 -16
OiO = 0,0-00=6-—=-.
3 3
. c. 2
Javob: —.
3
53
echimlar. Matematika va informatika 2017
6-variant
10. Asoslari 13 va 17 ga teng
jo'lgan teng yonli trapetsiyaning
diagonallari o'zaro perpendikulyar.
Trapetsiyaning yuzini toping.
Yechish:
a = 17, b = 13, ch±d2
S = ?
AD = a, BC = b
3OC va AOD uchburchaklar teng yonli
to‘g‘ri burchakli uchburchaklar.
AO = OD = y, BO = OC = x.
В c
/X/
/ гк v \
A’-----'-------*D
x2+x2 = 62 x2= —
2
h,= — = — = -
’ b 2b 2
. y2 a2 a
h, = — = — = —
2 a 2a 2
. . , a+b
h - hi + h2----
2
_ a+b , (a + b^ (
S=-----h= ----- = -
2 I 2 M
= 152 =225.
Javob: 225.
2 J
11. O'zaro teng bo'lmagan x va у
sonlari x2 r- 6x = y2 - 16y tenglikni
qanoatlantirsa, x + у ni toping.
Yechish:
xty
x2 - 16x = y2 - 16y
x2 - 16Х-У2 + 16y = 0
x2 -y2- 16(x-y) = 0
(x-y)(x+y)-16(x-y) = 0
(x-y)(x + y— 16) = 0
1)x-y = 0, x = у masala shartida x / y.
2) x + y- 16 = 0, x + у = 16.
Javob: 16.
12. Iog2(2Vx~+5 + 5) +
+ log015(-x - 0,5) = 1 tenglamaning
butun yechimlari nechta?
Yechish:
Aniqlanish sohasi:
jx + 5>0 Jx>-5
|-x-0,5>0 [x<-0,5
-5 ix <-0,5
log2(2 л/х + 5 + 5) - log2(-x -0,5) = 1
log2(2 Jx + 5 + 5) = tog22 + tog2(-x - 0,5)
log2(2y/x + 5 +5) = log2(-2x- 1)
2jx + 5 + 5 = -2x- 1
2 Vx + 5 = -2x - 6
Jx + 5 = —x — 3, —x - 3 S 0,
x<-3
x + 5 = (-x-3)2,
x + 5 = x2 + 6x+ 9
x2 + 5x + 4-0,
x=-1, x = -4
-5^x<-0,5vax<-3 bolganligi
sababli x =-4 tenglama ildizi bo'ladi.
Demak, tenglama 1 ta yechimga ega.
Javob: 1 ta.
13. Agar /(x) = axz + bx3 - 2
funksiya uchun /(-3) = -2 shart
bajarilsa, f (3) qiymatni toping.
Yechish:
f(x) = ax7 + bx3- 2 funksiya uchun
f(-3)=-2
-2 = (~3)7a + b(-3)2 - 2
(-3)7a + b(-3)3 -2 + 2 = 0
(~3)7a + b(-3)3 = 0
(~3)4a = -b, b = -81a
f(x) = ax7 —81 ax3 -2
f(3) = 37-a-81a-3l -2 =
= 37a -37 a-2=-2
f(3)=-2.
Javob: -2.
14. Tenglama nechta yechimga ega?
(—----1)-V4^x2'= о
sinx
54
Yechimlar. Matematika va informatika 2017
6-variant
Yechish:
Aniqlanish sohasini topamiz.
[sinx^O (x^nn (х + лп
{4-x2>0(x2-4<0[-2<x<2
n = 0 da x / О
х^лп
2хг +17x + 8 <0
х + лп
'2 0 2
[-2; 0) U (0; 2]
Tenglamani yechamiz.
1) ——1 = 0, sinx = 1,
sinX
2;V4-x2 =0, x2-4 = 0,
x2 - 4, x = ±2
Aniqlanish sohasiga tegishli yechimlar
x =—, x = 2, x =-2.
2
Tenglama 3 ta yechimga ega.
Javob: 3 ta.
15. ABC uchburchakning A burchagi
2a ga teng va AC = b, AB = c bo'lsa,
A burchak bissektrisasini toping.
Yechish:
Z A = 2а, AC - b, AB = c
AD - bissektrisa
2a
2bc- cos—
. _ 9 2bc-cosa
AD =--------=-----------
b + c b + c
. . 2bc cosa
Javob: ---------.
16. Tenglama nechta yechimga ega?
(3 - ctg2x) V-8-17x-2x2 = 0
Yechish:
Aniqlanish sohasi:
fsinx^O
\-8-17х-2х2>0^
х*лп,[-8;-^]
п =-1, х^-л
п = -2, xt -2л
-8 -2тт -П 1
~2
[-8; -2л) U (2л; -л) U (-л; -~]
Tenglamani yechamiz.
1)3- ctgzx = 0, ctg'x = 3,
ctgx = ±43
6
. FZ 5л
ctgx = -yj3 , x =— + лп
5л л
n =-1 da x =---, x =—
6 6
„ , 11л 7л
п =-2 da x=----, x =----
6 6
~ . 17л 13л
п = 3 da x =---, x =----
6 6
л 5л 7л 11л
Х=---, X =-----, X =-, X =----,
6 6 6 6
х=-1-!~- lar aniqlanish sohasiga
tegishli yechimlar.
2) у!-8-17х-2х2 =0
-2X2 -17x-8 = 0
x=-8,x=-~
2
Demak, tenglama 7 ta yechimga ega.
Javob: 7 ta.
17. Geometrik progressiyada
— = 2— bo'lsa, maxrajini toping.
b5 b13
55
Yechimlar. Matematika va informatika 2017
6-variant
Yechish:
bn = br<f~1
bn = bn-rq
formutalardan foydalanamiz.
b3 _ 2
b3q2 b10-q3
^=2~,q^0,q = 2.
q q
Javob: 2.
18. a > 0 ning qanday qiymatlarida
a 3
tengsizlik o'rinli. J e*dx > -
-a 2
Yechish:
1) | exdx = e* + C ga asosan
a
J exdx = ex
-a
& a -a
-e -e
-a
2) ea-e-a>^,ea=y
уЛЛ>0,^у-2>0
У 2 у
2(y + 0,5)(y-2) ? c
У
-0,5 0 2
-0,5 <y<0, у >2
-0,5 < ea < 0 tengsizlik yechimga ega
emas.
ea > 2, lnea > In2, a > ln2
(In2; co).
Javob; (In2; co).
19. Yuzi Q ga, diagonallarining
nisbati m.n bolgan rombning tomonini
toping.
Yechish:
S = Q, di:d2 = m:n,
a = ?
_ , . _ d, • d,
Romb yuzi Q = 1 - -
di = mx, d2 = nx
mnx2 2 2Q [2Q
Q =------, x2 =—, x = .—
2 mn Nmn
, f2Q , [2Q
di=m. — ,d2 = n —
V mn У mn
2 i 2 .2 -Г d3
4a = di + d2 , a ----2-
2
1 l2Qm2 2Qn2
a = —.---+-----=
2 v mn mn
l2Q(m2 + n2) _ Iq m2 + n2
V 4mn N 2mn
i u
Javob: ,Q------.
V 2mn
20. Teng yonli uchburchak perimetri
2p ga, asosidagi burchak a ga teng.
Shu uchburchakka ichki chizilgan
aylana radiusini toping.
Yechish:
ABC - teng yonli uchburchak. P = 2p,
а - asosidagi burchak. r = ?
P = 2a +b = 2p
2p = 2a + 2acosa
p =a(1 +cosa)
a = ^P^ = ~rL._
7 + cosa o,, „2 o.
2 cos -
2
_ a-b . 2a2cosasina
S = sm a =---------------
2 2
_ a2 sin2«
2
56
Yechimlar. Matematika va informatika 2017
6-variant
2 a2 sin 2а
2 _ asin2a _ psin2a
4acos2 — 4 cos2” Seos'*—
2 2 2
. . psin2«
Javob: —-------.
8cos4-
2
21. у = 3sin2x + 3sin2( - x)
funksiyaning qiymatlar sohasiga
nechta butun son tegishli?
Yechish:
Keltirish formulasiga asosan
у = 3sin2x + 3cos2x = 3(sin2x + cos2x) - 3
Funksiya qiymatlar sohasiga 1 ta
butun son tegishli.
Javob: 1 ta.
л/З
22. — radiusli sferaga muntazam
to'rtburchakli piramida ichki chizilgan.
Uchidagi yassi burchak 45° ga teng
bo'lsa, piramida yon sirtining yuzini
toping.
Yechish:
SABCD - muntazam to‘rtburchakli
piramida. AO = R. О - sfera markazi.
SOi = H- piramida balandligi.
Z ASB = 45°.
R=~, AB = a, а = 45°
2
Ph
Syon = , P = 4a, ha- apofema.
ha = SE, Syon = 2a ha
R2 = (H - R)2 +— bundan
o a2+2H2
l\ — ———————
4H
16R2sin2 — cos«
=---------------= SFfsina-cosa =
. a
tg2
= 4R2sin2a
R= — , a = 45°
2
57
Yechimlar. Matematika va informatika 2017
6-variant
3
Syon = 4 — sin90° = 3.
4
Javob:3.
23. Tengsizlikning eng katta butun
va eng kichik musbat butun yechimlar
yig'indisini toping:
3 -x2 + 5x - 6 < x2 + 5x-32x - 2-32x+1 .
Yechish:
З2* / - 5x - 6 < x2 + 5x-32x - г-З2^
З2*/- бх-З2* + 2-^-3 + 5x- 6-х2 S О
З^х2- 5x + 6) - (x2 —5x + 6) < 0
(x2-5x + 6)(32x - 1) <0 tengsizlik
quyidagi tengsizliklar sistemasiga teng
kuchli.
, (x2-5x + 6<0
1) 1
[32x-1>0
|x2 -5x+6>0
2) 1
[32x-1<0
, fx2-5x+6<0 (2<x<3
1) I =>5
|32* > 1 [2x>0
\2<x<3 „
=>2<x<3
[x>0
lx2-5x + 6>0 \x<2,x>3
2) i => •<
|32x<7 |x<0
x<0
(-^;0]U[2;3]
Eng katta musbat butun yechim 3.
Eng kichik musbat yechim 2.
3 + 2 = 5.
Javob: 5.
25. Uchburchakning bir tomoni
21 sm, qolgan ikki tomoni 3:8 nisbatda
bo'lib, ular orasidagi burchak 60° ga
teng bo'lsa, uning perimetrini toping.
Yechish:
a = 21, b:c = 3:8,
а = 60°, P = ?
Kosinuslar teoremasidan:
a2 = b2 + c2- 2bc cosa
b = Зх, c = 8x
212 = 9x2 + 64X2 - 48x2-cos60°
212 - 73X2 - 48X2 —
2
49X2 = 212, 7x = 21,x = 3
b=3-3 = 9,c = 8-3 = 24
P = a+ b + c = 21 + 9 + 24 = 54.
Javob: 54.
26. arccos < arccosx tengsizlikni
yeching.
Yechish:
Aniqlanish sohasi:
-1<-<1 (-2<x<2
-1<x<1 L
-1 <x< 1
у = arccosx funksiya kamayuvchi,
shuning uchun arccos— < arccosx.
24 21+10823 — 3 !°845 + 5,OS43 — 1,og23
ifodaning qiymatini toping.
Yechish:
formulalardan foydalanib yechamiz.
2^+bg23 __glo$45 +£]а&43 _ylog23 _
= 21 2,Og23 — 3*°W5 4. 5 — 7 —
= 2-3-1 = 5.
Javob: 5.
— - x > 0, > 0, x < 0
2 2
Aniqlanish sohasi bilan birgalikda
x C [-1; 0) bo'ladi.
Javob: [-1; 0).
58
Yechimlar. Matematika va informatika 2017
6-variant
27. To'g'ri burchakli uchburchakka
ichki va tashqi chizilgan aylanalarning
radiustari mos ravishda 2 sm va
5 sm ga teng. Agar aylanalar markazi
orasidagi masofa 5 ga teng bo'lsa,
uchburchakning katetlarini toping.
Yechish:
r = 2, R = 5, OO, = 5
c = 10, a + b = 14
(a + b)2 = 142, a2 +b2 + 2ab = 142
2ab = 142 - 102 = 24-4
ab = 48
(a + b = 14
[ab = 48
a - 8, b - 6 yoki a = 6, b = 8.
Javob: 6 va 8.
28. Tenglama nechta butun yechimga
ega? 8 '^х>]4-х + log4(x - 2) = 8.
Yechish:
Aniqlanish sohasi:
[x>0
fx>0
4-x>0^>
x <4^> 2 <x< 4
x-2>0 [x>2
Tanlash yo‘li bilan yechamiz.
Aniqlanish sohasiga tegishli butun son
tenglama yechimi bolishi mumkin.
x = 3, x = 4
1) x = 3da
8'™3 44-3+^„(3-2) =
= 8-1 + log41 =8 + 0 = 8
2) x = 4 da
8i"f3iy4-4+\o<li4(4-2') =
7
= 8’°83'’-0-log42 = --^* 8
Demak, tenglamaning yechimi
x = 3. Tenglama 1 ta butun yechimga
ega.
Javob: 1 ta.
29. Tenglama nechta yechimga ega?
Iog3(9 - x^-tgx = 0
Yechish:
Aniqlanish sohasi:
9-x2 >0
cosXrO
x2-9<0
X
Х*— + ХП
2
(-3<x<3
x* — + xn,neZ
1) tgx = 0, x = xn, n 6 Z
2) fc>g3(9-4) = 0,9-4 = 43,
X2 = 8, x = ±242
n = 0 dax = x-0 = 0 aniqlanish
sohasiga tegishli. Tenglama 3 ta
yechimga ega, bular x = 0,x= +242 .
Javob: 3 ta.
30. Ushbu /(x) = —
x +x-2
funksiyaning boshlang'ich funksiyasini
toping.
Yechish:
Boshlang'ich funksiyani topish amaliga
integrallash deyiladi, demak:
F(x) = f f(x)dx = J -?-xt7--dx =
J Jx +x-2
f d(x2 + x - 2) 2
= I —Ц-------- = lnx+x-2+C =
J x2 + x-2
= ln(\x — 7||х + 2У + C.
Javob: ln(|x — 1|-|x + 2|) + C.
59
Yechimlar. Matematika va informatika 2017 6-variant
31. CELLSPACING tegining vazifasi:
Yechish:
CELLSPACING tegi yacheykalar atrofidagi ramkani butun jadval ramkasidan
ajratish (qalinligini belgilash) imkonini beradi.
Masalan:
<Table border cellspacing=1>
Yacheyka 1 Yacheyka 2
3 4
Ramka qalinligi 1.
<Table border cellspacing=6>
Ramka qalinligi 6.
Javob: jadval yacheykalar! ramkalarining qalinligini belgilab beradi.
32. Paskal. Quyidagi dasturning ekrandagi natijasini aniqlang.
var a, b: integer; s: real;
Begin a:=-2; s:=1; for b:=+10 to 6 do s:=s*a*b;
writein (s:4:2); end.
Yechish:
Bu dastur xatolik beradi, chunki for a:=b to c do sikl operatorida b<c amal
bajarilishi kerak. Bizning misolda aksi b>c. Shunga dastur xatolik beradi.
Javob: Xatolik haqida xabar chiqadi.
33. Qaysi qatorda elektron jadval katagi to‘g‘ri belgilangan?
Yechish:
Yacheyka nomida avval ustun nomi lotin alifbosi (А, В, C, D, E, F, ...) so‘ng
satr raqami ko'rsatiladi.
D23 - ya’ni D inchi ustun va 23-chi satr kesishmasida joylashgan yacheyka
manzili.
Javob: D23.
34. Qanday teg yordamida HTML hujjatlarida hujjatning bir joydan boshqa
joyida o'tish yoki boshqa hujjatga o'tish mumkin?
Yechish:
HTML da murojaat tegi <A> teg hisoblanib, uning asosiy atributi href
hisoblanadi.
href ga murojaat manzili ko'rsatiladi.
Masalan:
<A href="murojaat manzili"> sichqonni bosganda shu manzilga otish matni
</A>.
Javob: <A>.
60
Yechimlar. Matematika va informatika 2017
7-vahant
35. Ikkilik sanoq sistemasida arifmetik amallarni bajaring:
110,01+11,0101=...
Yechish:
110,01 + 11,0101 = ?
110,01
+ 11,0101
1001,1001(2)
Javob: 1001,1001.
36. Protokol - bu:
Yechish:
Kompyuterlararo ma’lumot uzatish qonun-qoidalar asosida bajariladi. Bunday
qonun-qoidaiar toplamiprotokollar deyiiadi. Protocol yordamida
kompyuterlararo aloqa o'rnatish, ma’lumotlar qabul qilish, uzatishni ta’minlaydi.
Javob: kompyuterlar orasida aloqa o‘matilishida, ma’lumotlar qabul qilishi va
uzatilishida foydalanadigan signallar to’plamidir.
7-variant
1. /(x) = 8cosxsin3x uchun
boshlang’ich funksiyani toping.
Yechish:
1) Ko‘paytmadan yig'indiga o'tamiz.
1
sin3xcosx (sin(3x + x) +
1
+ sin(3x -x))=— (sin4x + sin2x)
1
2) f (x) - (sin4x + sin2x) =
= 4sin4x + 4sin2x
3) F(x) = -cos4x - 2cos2x + C.
Javob: -cos4x - 2cos2x + C.
2. Tenglamani yeching:
7x + 2 +V2x + 3 +
+ ^/x + 6 + 3 V2x + 3 = 11V2 .
Yechish:
^2x + 3 = a belgilash kiritamiz.
2x + 3 = a2, 2x = a2- 3,
+J-—- + 6 + 3a=11yf2
N 2
la2-3 + 4 + 2a
N 2 +
V 2
la2 +2a + 1 [a2 + 6a + 9
N 2 N 2
V2 42
a + 1 +a + 3=1l42-42
2a+4 = 22, 2a = 18,a = 9
42x + 3=9, 2x + 3 = 81
2x = 78, x = 39.
Javob: 39.
3. у = 5sin6x - 3sin10x
funksiyaning hosilasini toping.
Yechish:
1) (sin(ax + b))' = a cos(ax + b)
(sin6x)' = 6cos6x
(sin10x)'= 10cos10x
61
Yechimlar. Matematika va informatika 2017
7-variant
y' = (5sin6x - 3sin10x)' =
= 30cos6x - 30cos10x =
= 30(cos6x - coslOx)
2) cos6x - coslOx ayirmadan
ko'paytmaga o'tamiz.
cos6x - coslOx = -2sin ——
2
. 6x + 10x „ . . „ . . o
sin-------= -2sm(-2x)-sm8x =
= 2sin2xsin8x
3) y' = 30(cos6x - coslOx) =
= 60sin2x-sin8x.
Javob: 60sin2xsin8x.
4. ABC o'tkir burchakli uchburchak
berilgan. Uchburchakning BC tomonini
C uchidan boshlab hisoblanganda 2:3
nisbatda bo'luvchi AN to'g'ri chiziq
o'tkazilgan. Agar ABN uchburchakning
yuzi 15 ga teng bo'lsa, ABC
uchburchakning yuzini toping.
Yechish:
CN:BN = 2:3, Sabn = 15, SABC = ?
„ AB-BN . .
SABN ~ 2 ’Sin^
„ ABBC . .
sAbc^—2-------
BN = Зх, BC = 5x
AB3x . o
sm.fi
_3
SABC ^5x.sin/? 5
2
c _ 5SAbn _ 5-15 _ _
ABC 3 3
Javob: 25.
5. Qarang: 1-variant 6-savol
(4-bet).
6. Uchburchakli piraniidaning ikki
yon yogi teng yonli to'g'ri burchakli
uchburchaklar bo'lib, ularning
gipotenuzalari b va bu gipotenuzalar
o'zaro a burchak hosil qiladi. Piramida
hajmini toping.
Yechish:
DABC - uchburchakli piramida. ADC
va BDC - teng yonli to'g'ri burchakli
uchburchaklar. AD = DB = DC,
AB = BC = b. Sasos = -b'sina.
2 ВАС = 90°-—
2
BC = 2Rsin 9a --
I 2
OC=R=—^—
2cos—
2
1 1 .г .
-----b sine •
3 2
b !-------
------л/cosct
2cos—
2
1 ,3 . a I----
= —b sm—.Vcosa .
6 2
. . 1 • a I---
Javob: — b sin—Vcos« .
6 2
7. 0,29 - (((-0,23) - (-0,06) + 0,37) -
- ((-0,47) - (-0,37))) ni hisoblang.
Yechish:
0,29 - (((-0,23) - (-0,06) +0,37)-
- ((-0,47) - (—0,37))) = 0,29-
62
Yechimlar. Matematika va informatika 2017
7-variant
- (-0,23 + 0,06 + 0,37-(-0,47 + 0,37)) =
= 0,29 + 0,23 - 0,06-0,37-0,47 +
+ 0,37 = 0,46 -0,47 = -0,01.
Javob: -0,01.
8. Uchburchakning tomonlari a = 13,
b = 14, c = 15. Bulardan ikkitasi (a va b)
markazi uchinchi tomonida yotgan
doiraga urinma bo'ladi. Doira radiusini
toping.
Yechish:
a = BC = 13, b = CA = 14
c=AB=15, OE = OF=R
Sabc - Sboc + Saoc
„ 14R
^boc ~ 2 ’ _ 2
„ _27R _a + b + c
^ABC — 2 ' P ~ 2
S№C = ,j21(21 -15)(21 -14)121 -13) = 84
2™=84,R^M = 56=62
2 27 9 9
2
Javob: 6—.
9
9. Tenglama nechta yechimga ega?
(tg2x - 1) Vl0x-x2-21 = 0
Yechish:
Aniqlanish sohasi:
cosx ф 0
10x-x2-21>0
Х*-+ЯП
2
x2 -Юх + 21 <0
Я
х* — + яп
=> 2
(x-3)tx-7)<0
n = 1 dax* —
2
я
x* — + nn,nsZ
2
3 3l 7
~2
[3;3X)U(3^;7]
Tenglamani yechamiz.
1) tg2x- 1=0, tg2x = 1, tgx = ±1
Я ЯП
x = — + —
4 2
n , 5я
п = 2 da x = —
4
„ . 7я
n = 3dax =—
4
5я 7я . . .
x =—, x=— aniqlanish sohasiga
4 4
tegishli yechimlar.
2) yll0x-x2-21 = 0,
X2 - 10x + 21 =0, x = 3, x =7
Demak, tenglama 4 ta yechimga ega.
Javob: 4 ta.
10. Nomanfiy x, у sonlar uchun
a = va b = 2jxy bo'lsin. Qaysi
tengsizlik har doim oTinli?
Yechish:
Ma’iumki, x> 0, у > 0 bo'lsa,
x + y> 2y/xy tengsizlik doimo o'rinli.
Shunga asosan yechamiz.
a = 2x +^ y, b = 2jxy
2x + — у s 2^2x -^y > 2y[xy bundan
a>b ekanligi kelib chiqadi.
Javob: a > b.
x-z = -1
11. \x + y = 2 tenglamalar
y-z = -5
sistemasini qanoatlantiruvchi x, у va
z sonlarining o‘rta arifmetik qiymatini
toping.
63
Yechimlar. Matematika va informatika 2017
7-variant
Yechish:
x-z = -1
x + y = 2 tenglamalar sistemasining
у-z =-5
birinchi va ikkinchi tenglamalar
ayirmasini topamiz.
x-z~-1
x + y = 2 yoki у + z- 3
-z-y = -3
fy + z = 3
Endi < sistemadan yvaz ni
[y-z--5
topamiz.
y = -1, z = 4.
Shuningdek x ni topamiz.
x = 2-y = 2-(-1) = 2 + 1 = 3
(3;-1;4)
x + y + z_ 3-1 + 4_£_2
3 ~ 3 ~3~
Javob: 2.
12. Uchburchak ABC da Z A = 75°,
2hc = AB bolsa, C burchak kattaligini
toping.
Yechish:
Z A = 75°, 2hc =AB, ZC=?
ДАВС dan------ikki.
sin(75° + a) cos a
tenglikdan - “- = cosa
2sin 75°
2sin75°cosa = sin(15° + a)
2sin30°cos(45° + a) = s in (a - 75°)
45° + a + a - 75° = 90°, a = 60°
ZC=15° + 60° = 75°.
Javob: 75°.
13. (x2 + 11X + 11)-
(x2 + x + 11) = Их2 tenglama eng
kichik ildizi eng kattasidan qancha
kichik?
Yechish:
x2 + 11 = a belgilash kiritamiz.
(x2 + 11 + 11x)(x2 + 11 + x) =
= (a + 11x)(a + x) = a2 + 12xa + Их2
a2 + 12xa + 11x2 = Их2
a2 + 12xa = 0, a(a + 12x) = 0
a = 0, a = -12x
x2 + 11 = 0 tenglama haqiqiy ildizga
ega emas.
x + 11 = -12x
x2 + 12x + 11 = 0 tenglama ildizlari-11
va -1.
-1-(-11)=-1 + 11 = 10.
Javob: 10.
14. cos0,2x - V3 sin0,2x < 0
tengsizlikni yeching.
Yechish:
1
Tengsizlikning ikkala qismini — ga
ko‘paytiramiz.
1 Гз
— cos0,2x ~ — sin0,2x < 0
2 2
cos- cos0,2x- sin — sin0,2x < 0
3 3
cos| 0,2x + — I < 0
I 3)
— + 2тт<0,2х + — < — + 2пп, nCZ
2 3 2
n „ x 7it „
— + 2itn < — < — + 2лп
6 5 6
— + 10лп<х<~+10лП, nBZ.
6 6
Javob: | — +10я-п;—+10xn |,
16 6 )
nCZ.
15. у = 4sin6x + 3sin8x
funksiyaning hosilasini toping.
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Yechimlar. Matematika va informatika 2017
7-variant
Yechish:
1) (sin(ax + b))' = acos(ax + b)
(sin6x)' = 6cos6x
(sin8x)' = 8cos8x
y' = 4(sin6x)' + 3(sin8x)' =
= 24cos6x + 24cos8x =
= 24(cos6x + cos8x)
2) cos6x + cos8x yig'indidan
ko‘paytmaga o'tamiz.
cos6x + cos8x = 2cos + —
2
6x-8x „ v
cos-------= 2cos7xcosx
2
3) y' = 24 2cosx cos7x =
= 48cosx cos7x.
Javob: 48cosxcos7x.
16. Katta katetni diametr qilib, unga
yarim aylana tashqi chizilgan. Kichik
kateti 30 sm, to'g'ri burchak uchini
yarim aylana gipotenuzani kesgan
nuqta bilan tutashtiruvchi vatar 24 sm,
yarim aylana uzunligini toping.
Yechish:
в
AB = 30, AD = 24
ABD uchburchakdan
BD2 = BA2-AD2
BD =у1з02 -242 = 754-6 = 18
BC - kesuvchi urinma.
BCBD = BA2
BC = ^ = 3-°L = 50
BD 18
AC2 = ВСг-ВА2 = 50?— 30? = 40?
AC = 40
o AC . 2nR D
2 2
Javob: 20л.
17. Agar J(x) = 2х-2x bo'lsa, f'(x) > 0
tengsizlikni yeching.
Yechish:
1)(t)' = tln2
2) Ko'paytmadan hosila olamiz.
f'(x) = (t-2x)' = 2?!n2-2x + t-2 =
= t-2(xln2 + 1)
3) 2x-2(xln2 + 1)>0
t - doimo musbat
1
xln2 + 1 >0, xln2 >~1, x>-----
ln2
. 1 Ine .
x >-----=-------= -log?e
ln2 ln2
(-1одге; <*>).
Javob: (-log2e; co).
18. Qarang: 2-variant 4-savol
(14-bet).
19. {x|x 6 N, 6<x4 40} to'plamning
nechta qism-to'plamlari mavjud?
Yechish:
6 £ X* 2 < 40, x e N
4б<х<-Дб A={3, 4, 5, 6}
A to'plam 4 ta elementdan iborat.
1.0 c: A
3. {3} c A
5. {5} c A
7. {3; 4} c A
9. {3; 6} c A
11. {4; 6} c A
13. {3; 4; 5} c A
15. {3; 5; 6} c A
A to‘plamning qis
2.A<zA
4. {4} c A
6. {6} c A
8. {3; 5} c A
10. {4; 5} c A
12. {5; 6} c A
14. {3; 4; 6} g A
16. {4; 5; 6} c A
to'plamlari 16 ta.
Javob: 16.
20. (-12):((+3) + (-15)):(-5) ni
hisoblang.
Yechish:
(-12):((+3) + (-15)):(-5) =
= -12:(3- 15):(-5)=-12:(-12):(-5) =
= 1:(-5) =—0,2.
Javob: -0,2.
21. a < 0 ning qanday qiymatlarida
0
j(3“гх - 2-3'x)dx >0 tengsizlik o'rinli
bo'ladi?
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Yechimlar. Matematika va informatika 2017
7-variant
Yechish:
0_
a
3~2a
1) \axdx = — + C
J Ina
]\3 2x -2-3 X)dx =
a
Г 32' 2-3~,A
21n3 ln3 J
13°
= + 2-3° + ----23a) =
ln3 2 2
1 1 3~2a
= ——(—- + 2 + —----2-3°) =
ln3 2 2
= —1— (3 + 3 2a -2-2-31)
2in3
2
2)^-—(3 + 3~2a-4-3a}>0
21n3V 1
ЗГ2а-4-ЗГа + 3>0
(3^ - 1)(3Ta -3)>0
3~a<1, ^>3
a) 3^ < 1, 3~a < 3°, -a < 0, a > 0
Masala shartida a <0, a>0 yechim
emas.
-1
(-°°;-7J.
Javob: (-oo; -1].
22. Qarang: 2-variant 16-savol
(16-bet).
23. Geometrik progressiya n-hadi
bn =---5n+2 ga teng. Progressiyaning
4
maxrajini toping.
Yechish:
Javob: 5.
24. Tenglamani yeching:
Vx+3+47x^1+
+ >/x + 8 + 6>/x-1 =17.
Yechish:
To 7a kvadratga keltirib yechamiz.
1. \jx + 3 + 4^x-1 =
= Vx-7 + 4Vx-7+4 =
= yj(y/x-1 +2^ = -Jx-1 + 2
2. s]x + 8 + 6y/x-1 =
=^x-1+6^x-1+9 =
= 7(Vx-1 + 3)3 = Jx~1 + 3
з. +2+Vx^7+3=17
2-Jx^l = 12, 4x^1 = 6
x- 1=36,x=37.
Javob: 37.
25. — = 2>/2 geometrik
progressiya maxrajini toping.
Yechish:
Geometrik progressiyada
btf-T) „ _b,(g3-7)
6 q-1 3 q-1
S6:S3 -^(Q6 -1) . b^q3 -1) _
q-1 q-1
b,(q3 -7)(q3 +7) q-1
q-1 b,(q3-1)
= q3 + 7
q3 + 7 =242 , q3 = 242~ 1,
q=^242-1.
Javob’: ^242-1 .
26. To'g'ri burchakli
uchburchakning gipotenuzasi c, o'tkir
burchaklaridan biri a ga teng. Shu
uchburchakka ichki chizilgan aylana
radiusini toping.
Yechish:
c - gipotenuza, a - o'tkir burchak.
a = 2 CAD, AO - bissektrisa
66
Yechimlar. Matematika va informatika 2017
7-variant
Z.ABO = 1-(90,‘-a)=45°-~
AOD va BOD uchburchaklardan
AD = OD ctg^,
DB = OD ctg(45° )
c=AB=AD+DB= OD(ctg^ +
+ ctg(45° - ^))
c
. О. ..Lr. а
ctg~ + ctgl45 --
ctg~ + ctg>45 - —
cos- cos 45’-“
__2+ I------22 =
sin^ sin) 45-° I
2 I 2
( a а
sin 45 — + —
V 2 2?
ct . f .г-, а
sin —sm 45---------
2 I 2
sin 45°
. а . f a
sin-sn 45----------
2 I 2
72
2 sin- sin j 45° - —
2 \ 2
2
2sin—sin| 45° - —
2 [ 2
c • 2sin -sin 45-------
2 V 2
•J~2
= c42 sin—sinf 45° -—I
2 I 2
Javob: cV2sin—sin| 45°- —
2 I 2
27. ABC uchburchakda AB = 10,
BC = 17, AC = 21 sm. Uchburchak
ichidan olingan N nuqtadan AC va BC
tomonlarigacha bo'lgan masofaiar
2 va 4 sm bo'lsa, N nuqtadan AB
tomongacha bo'igan masofani toping.
Yechish:
AB = 10, BC = 17,
AC = 21, ND = 2, NE = 4
S& = ^24 (24 -10^24^17\24 -21) =
= ^24-14-7-3=84
Sa = Sr + S? + S3
Si - AANC yuzi
s AC±D = 2^1 = 21
1 2 2
S2 ~ ABNC yuzi
CB^E = 1^4=34
2 2 2
S3 - AANB yuzi
o AB -NK 10 -NK
S, =-----=-----= 5NK
3 2 2
84 = 21 + 34 + 5NK
2Q
NK=— = 5,8.
5
Javob: 5,8.
67
Yechimlar. Matematika va informatika 2017 7-variant
28. Tenglama nechta yechimga ega?
(2cosx- 1) л/9-5х-4х2 = 0
Yechish:
Aniqlanish sohasi:
9-5х-4хг> 0,4k2 + 5x-9<0
о
4(x- 1)(x+-)<0
4
9 „ _
— < x<1
4
Tenglamani yechamiz.
1
1) 2cosx —1=0, cosx =— ,
2
x = ± — + 2itn,n e Z
3
n = 0 da x =~ aniqlanish sohasiga
tegishli.
2) -]9-5x-4x2 =0
9
4)C + 5x-9 = 0, x= 1, x = —
4
Tenglama 3 ta yechimga ega, bular
n. . 9
X =---, X= 1, X =-.
3 4
Javob: 3 ta.
29. у =
2>
x-- (x + 0,5)
(3-x)(x + 2)
funksiyaning aniqlanish sohasiga
tegishli bo'lgan barcha butun sonlar
ko'paytmasini toping.
Yechish:
9 A
x- — (x + 0,5)
---—--------> 0 yoki
(3-x)(x + 2) У
(X-0,4)(x + 0,5)<0
^(x-3)(x + 2)~~“
Oraliqlar usuli bilan yechamiz.
(-2;-0,5] U [0,4; 3)
Aniqlanish sohasiga tegishli butun
sonlar-1, 1, 2.
Ildizlari ko‘paytmasi-1-1-2 =-2.
Javob: -2.
30. To'g'ri burchakli uchburchakka
yarim aylana ichki chizilgan, uning
diametri gipotenuzada yotadi, markazi
esa gipotenuzani 15 sm va 20 sm li
kesmalarga bo'ladi. Yarim aylananing
katetlar bilan urinish nuqtalari
orasidagi yoyning uzunligini toping.
Yechish:
ABC - to‘g‘ri burchakli uchburchak.
ODBE- to'rtburchakning B, Dva E
burchaklari to'g'ri burchak va DO = OE
bo'lgani uchun bu to'rtburchak kvadrat
bo'ladi.
DE yoy butun aylana uzunligining
to'rtdan biriga teng. ADO va OEC
uchburchaklar teng.
в
AD _OE
АО ОС
AD = ^AO2-OD2 = Jl52 -R2
J152-R2 _ R
------,— — f~\ — / £
15 20
12
r - — = 3 I = 2nr = 2-3n = 6л.
4
Javob: 6л.
31. HTML tilidagi hujjatda bgcolor=FFFFFF atributining vazifasini aniqlang:
Yechish:
bgcolor=FFFFFF oq rangli bildirib HTML tilida orqa fonga rang berish tegi
hisoblanadi. Bundan tashqari, rangni quyidagicha ham bersa bo'ladi.
68
Yechimlar. Matematika va informatika 2017
7-variant
bgcolor="green” yashil rang foni
bgcolor="white" fon rangi oq va xokazo.
Javob: oq fonni aniqlaydi.
32,22012ю, 122O2io butun sonlarning barchasini yozish mumkin bo'lgan eng
kichik asosli sanoq sistemasida shu sonlar raqamlarining yig'indisini hisoblang.
Yechish:
2201210-*X2
22012|
~2
2
_______
1110061
~10 I
10
012 10
12 I
©J .
2_
7
6
i 15
06 14
6 10 ,
©10 15
3
2 11
ф 10
12 l687|
17 6
16 i
15 i
14 _
Л) 6 3 10 54_[Fl2
i_ Г34312
8 2_ |171I2_
8 14 16_l85[2_
7 14 11 8_|42I2
ф 2 ф 4 2 2 11012
ф ф 2"фЮ|5[2
© TQ54I2I2
ф2ф
©
X = 10101011111100(2). Bu sonlar raqamlar yig'indisi 10 ga teng.
1220212
"12_ I6IOIT2
2 6 1305012___
2 10 2 1152512
02 10 10 14_ 176212
’ 2 (T)10 12 ~6_ 138112
© 5 12 16 2 i19OI2_
4 5 16 18 18 I95I2_
10 4 2 18 10 8_l47|2_
10 ф 2 'll) 10 15 4J23I2_
© © ©14 72_Hll2
©'6 3 101512
ф 2 ©4|2I2
Ф Ф2©
©
X = 10111110101010(2). Bu sonlarraqamlaryig'indisi9 ga teng.
10 + 9= 19.
Javob: 19.
33. Elektron jadvalda B4:E12 kataklar bloki nechta katakni o'z ichiga oladi?
Yechish:
B4:E12.
Bu yerda В ustunning nomi 4 satr bo'yicha joylashuvi. Ya’ni ustun 4 satr
kesishmasida B4 yacheyka joylashgan.
69
Yechimlar. Matematika va informatika 2017
8-variant
Jami 4 ta ustun
Yacheykalarini qamrab olgan.
Demak: B4:E12 diapazonda 9-4= 36 ta yacheyka joylashgan.
Javob: 36.
34. Quyidagilardan qaysi biri axborot ko’rinishi hisoblanadi?
Yechish:
Axborot ko'rinishlari: uzlukli va uzluksiz axborot ko'rinishlari mavjud.
Javob: grafikli, tovushli.
35. Internetdagi ma’lumotlarni uzatish qoidalari ... deyiladi.
Yechish:
Internet xizmatlaridan foydalanuvchilar uchun ma’lumotlarni uzatish tartibini
belgilovchi yagona qoidalar majmuyi belgilangan. Bunday qoidalar majmuyi
PROTOKOLLAR deb ataladi. Masalan, http, ftp va boshqa protokollar.
Javob: protokollar.
36. HTML tilidagi hujjatda <Framest COLS= “25%, 70%”>teglari nimani
anglatadi:
Yechish:
Brauzer oynasini oyna ostilarga bo'lish uchun <Frame> tegi ishlatiladi, uning
atributlari COLS, ROWS.
COLS - ekranni ustunlarga bo'lish.
ROWS - brauzerni oynasini satriarga bo'ladi.
<Framest COLS="25%,70%> ekranni 2 ta ustunga ajratadi.
Javob: brauzer ekranini ikkita ustunga bo’ladi.
8-variant
1. Tenglamani yeching:
a/V2x-7 +1 = 5-V2x-7 .
Yechish:
Aniqlanish sohasi:
5 - 1/2X-7 > 0, Zj2x-7 < 5
2x-7< 125, x < 66
ll2x-7 = a belgilash kiritamiz.
Ja + 1 = 5-a tenglikning ikki qismini
kvadratga ko'taramiz.
(y/a+T)2 = (5-a)2
a + 1 = 25- 10a + a2
a2 - 11a~+ 24 = 0 tenglama yechimi
a = 3, a = 8.
1) V2x-7 = 3, 2x - 7 = 27
2x = 34,x=17
2) ll2x-7 = 8, 2x-7=512
2x = 519, x = 259,5
x = 259,5 aniqlanish sohasiga tegishli
emas. Demak, tenglama yechimi
x=17.
Javob: 17.
70
Yechimlar. Matematika va informatika 2017
8-variant
2. J(a2+(4-4a)x + 4x3)dx<12
1
tengsizlikni qanoatlantiruvchi a ning
barcha qiymatlarini toping.
Yechish:
1) [(a2 + (4-4a)x + 4x3)dx =
1
Y2 \2
= (a2x-4(1-a)~ + x4)l =
2 _
1 ~
= (a2x - (2 - 2a)x2 + x4)
= 2a2 - 4(2-2a) + 16-a2 +
+ (2-2a) - 1 = 2a2-8 + 8a + 16-a2 +
+ 2- 2a - 1 = a2 + 6a + 9 = (a + 3)2
2) (a + 3)2 < 0, a + 3 = 0, a = -3.
Javob: {-3}.
3. P nuqta ABCD to'g'ri to'rtburchak
ichidagi nuqta. Agar |PC| = -73 ,
|PB| = V2 , |PD| = V6 bo'lsa, |PA| = ?
toping.
Yechish:
Uchburchak BPC da PE - balandlik,
PE = h, CB = a, PB = x
h2 = (43 )2 -(a- x)2
h2 = (V2 /-x2
3 - (a - x)2 = 2 - x2
Uchburchak APD da PK - balandlik.
(4б )2 - (a - x)2 = c2 - x2
6-(a-x)2 = c2 -x2
(6-(a-x)2 = c2 -x2
|3-(a-x)2 =2-x2
3 = c2-2
c2 = 5, c=45 , PA = c =45 .
Javob: 45 .
4. Tenglama nechta butun yechimga
ega? (5 - x2) 2 (cos(tt3x) + 1) = 0
Yechish:
Aniqlanish sohasi:
5 - x2 >0, x2 - 5 <0,
-45<x<45
Tenglamani yechamiz.
cos(?r • 3х) + 1 _q
45-х2
cos(n-37) + 1= 0
cos(ir37) = —1, л-3* = л + 2лп, n C Z
3х = 1 +2n
n = 0da37 = 1,x = 0
n = 1 da 37 = 3,x = 1
n = 4da^ = 9, x = 2
x = 0, x= 1, x = 2 aniqlanish sohasiga
tegishli yechimlar. Tenglama 3 ta
butun yechimga ega.
Javob: 3 ta.
5. Qarang: 1-variant 12-savol
(6-bet).
6. Asosi a ga, yon tomoni b ga teng
bo'lgan teng yonli uchburchakning yon
tomoniga tushirilgan balandlik
uzunligini toping.
Yechish:
a - asosi, b - yon tomoni,
hb - yon tomonga tushirilgan balandlik.
a
hb2 -a2-x2, hb2 = b2-(b-x)2
71
Yechimlar. Matematika va informatika 2017
8-variant
a2
b
a^x^tf-jb-x)2
a2 - b2 = x2 - (b - x)2
(x - b + x)(x + b-x) =a2 -b2
(2x-b)-b = a2- b2
„ . a2-b2
2x-b =------,
b
„ a2-b2
2x =----
b
a2
x = —
2b
. 2 2 & x.
"• -a ’^=a'
>2 4b2 -a2
4b2
hb =-^--j4b2-a2 .
2b
Javob: -^--^4Ь2-a2 .
2b
7. Trapetsiya katta asosidagi
burchaklari 20° va 70°, asoslari
o’rtalarini tutashtiruvchi kesma uzunligi
2 ga, o’rta chizig’i 4 ga teng bo’lsa,
asoslarini toping.
Yechish:
AD, BC-asoslari, EK-o'rta chiziq.
MN - asoslari o'rtalarini tutashtiruvchi
kesma.
А д, |j D, D
MN = 2, Z BAN = 70°,
Z CDN = 20°, EK = 4
AN = ND, BM = MC, AD = a,BC = b
BM=AAi= —
2
А^=^-, ZA1MD1 = 90°
2
AjDi - gipotenuza, MN - mediana.
MN A,D, = 2MN = 4
2
A1D1 = a-b = 4
a + b . а
-----= 4, a + b = 8
2
(a + b = 8 \a = 6
[a-b=4[b = 2'
Javob: 2 va 6.
8. Qarang: 3-variant 29-savol
(30-bet).
9. Iog381 + 1од39 + 1одзЗ + ... ni
hisoblang.
Yechish:
ai = 1одз81 = 4,
a2 - log39 = 2,
a3 = log33 = 1
a2 2 1
q =-?- = - = —
a, 4 2
Cheksiz kamayuvchi geometrik
progressiya. Shuning uchun
2 2
Javob: 8.
1 1
4—=----+... 4, - ......:.. ni
Jf+yjlO 42011+42014
hisoblang.
Yechish:
Kasr maxrajini irratsionallikdan
qutqaramiz.
1
+ 42011 +42014 ”
44—4i
~ (44+4i)(44-4i)
72
Yechimlar. Matematika va informatika 2017
8-variant
•J7--J4
+ (V7+V4)(V7-V4) +
>Ji0->/7 ।
+ (л/7О+77)(л/7О-л/7) '" +
________>/2014--J 2011_________
(>/2014 + >/2011) • (>/2014 - >/2011)
>/4-1 -J7--J4 >/l0->/7
—------1--------1----------h ... +
4-1 7-4 10-7
>/2014->/2011 _y[4-1
+ 2014-2011 3 +
+>/7->[4+>/lO->/7 +...+
+>/2014 - >/2011) = ^(>/2014 -1).
Javob: .
3
11. Qarang: 1-variant 26-savol
(9-bet).
12. Boshlang'ich funksiyani toping:
/(x) = sin I — + 5 I.
Yechish:
,, . . (x -'l
f(x) = sin — + 5
\4 )
F(x) = -4cos + 5^ + C.
Javob: F(x) = -4cos^ + + C.
13. f(x) = ln(sin(x + 1)) bo'lsa,
f(^p)ni toping.
Yechish:
1)f'(x)=(ln(sin(x+1)))- =
cos(x +1) . . ..
= —;—±=Ctg(x+1)
sin(x +1)
= ctg]--1 + 1I = ctg— = 0.
Javob: 0.
14. x2 + 10x + y2+ 16y = 32
tenglama bilan berilgan aylana
chegaralab turgan soha yuzini toping.
Yechish:
x2 + 10x + y2 + 16y = 32
x2 + 10x + y2 + 16y- 32 = 0
(x + 5)2-25 + (y + 8)2 - 64 - 32 = 0
(x + 5)2 + (y + 8)2 = 121
1^ = 121
8 = ^= 121л:.
Javob: 121л.
15. Agar tg2a = - va a C (0; )
6 2
bo'lsa, cos2a - sin2a ni hisoblang.
Yechish:
cos2а - sin2a ni topamiz.
cos2a - sin2a = cos2a, cos2a ni tga
orqali ifodalaymiz.
1
cos2a 1~^2а- 6_6_5
T^~^T-6+i~7
6 6
. U 5
Javob: -
7
16. Qarang: 1 -variant 25-savol
(9-bet).
17. ABC uchburchakda Z C = 90°,
AC =2л/3 , BK = 1, CK uchburchak
balandligi bo'lsa, Z ACK ni toping.
Yechish:
Z C = 90°, AC =2>/3
BK= 1
ЛСКА va ABKC o'xshash.
, BK 1
tga =---=---
CK CK
73
Yechimlar. Matematika va informatika 2017
8-variant
sin a =
CK CK
AB ~ 2j3
1
tga sina =—== bundan a = 30°
243
Z ACK = 90°-a = 90° - 30° = 60°.
Javob: 60°.
18. ABCD to'rtburchakning har bir
tomoni chizmada ko'rsatilgandek o‘z
uzunligiga teng uzunlikda davom
ettirilgan. Agar A'B'C’D' to'rtburchak
yuzasi 5 ga teng bo'lsa, ABCD
to'rtburchak yuzasini toping.
A'B'C'D' ixtiyoriy to'rtburchak berilgan.
Biz A'B'C'D' kvadrat bo'lgan hoi uchun
yechamiz. Sab'ctt = 5, Sabcd = ?
A'B'C'D'- kvadrat.
AA' = AD = DD' = CD = CC =
= BC = BB' = AB = x
A'B' = a
AA'AB' to'g'riburchakli.
(A’BJ2 = (AB)2 + (A'A)2
a2 = (2x)2 + x2 = бх2
a = xV5
Sabcd = x2
Sa'B'cd' = 4-Saab + Sabcd =
= 4. ?2L1X + x2 = 4x2 + x2 = 5x2
2
5 = 5x2, x2 = 1, x = 1
Sabcd - 1.
Javob:1.
19. Iog2x + log2Vx + log2 Vx + ... +
+ log227x + ... = 10 tenglamada x ni
toping.
Yechish:
log2x + log24x + log2t/x + = Ю
1 1
log2x + log2x2 +log2x‘' + ...= 10
1 1
log2x +—log2x + —log2x + ...= 10
1 1
log2 x(1+ -+- + ...) = 10
1 1
1 + ^ + ~^— _ cheksiz kamayuvchi
geometrik progressiya.
bi = 1, b2=-, q=-
2 2
S = -^~ = ^— = 2
2
2 log2x = 10, log2x = 5
x = 25 = 32.
Javob: 32.
20.1 + {x} = cos(3x) tenglamani
yeching, bunda {x} - x ning kasr qismi.
Yechish:
-1 < cos3x 5 1 bo'lganligi sababli
-1 <1 + {x}< 1
-2<{x}s0
{x} x ning kasr qismi.
0 S{x} < 1 shuning uchun x = 0.
Javob: 0.
21. To'g'ri to'rtburchakning
tomonlari 1:3 nisbatda. To'g'ri
to'rtburchak yuzining unga tashqi
chizilgan doira yuziga nisbatini toping.
Yechish:
BC.AB = 1:3, BC = x,
AB = 3x, Sabcd:Su
74
Yechimlar. Matematika va informatika 2017
8-variant
1) Sabcd — AB-BC = Зх2
2) =тгАОг
AC2 = AB2 + BC2 = (3x)2 + x2=10x2
AC=xJid, A0 = ~ = ^~
2 2
10х2л 5х2л
d 4 2
3) SABCD:Sd = 3^-.~^ =
3x2-2 _ 6
5х2л 5л
. u 6
Javob: —.
5л
„ (x-2)(x + 1)2(x-3)
(x + 2)2(x-4)
tengsizlikning eng katta manfiy butun
yechimining eng kichik musbat butun
yechimiga nisbatini toping.
Yechish:
(x-2)(x + 1)2(x-3) . ... .
--------———-----— > 0 tengsizltkm
(x + 2) (x-4)
oraliqlar usulida yechamiz.
x*-2,x*4
Son o‘qida suratning ildizlarini va
maxrajning uzulish nuqtalarini
belgilaymiz va har bir oraliqda
kasrning ishorasini aniqlaymiz.
x C [2; 3] U (4; oo)U{~1}
Tengsizlikning eng kichik musbat
butun yechimi 2, eng katta manfiy
butun yechimi-1.
Yechimlar nisbatini topamiz.
1_
2 '
u 1
Javob: —
2
23. у = Iog3(sin22x + cos22x)
1
funksiyaning x = — nuqtadagi ikkinchi
tartibli hosilasining qiymatini toping.
Yechish:
1) sin22x + cos22x = 1
2)y = logs1 = 0
3) y'= (0/, y" = 0.
Javob: 0.
24. 8 - V?6-Vx = ifre + jx
tenglamani yeching.
Yechish:
^76 +Vx + ^76-4x = 8 tenglikning
ikkala qismini kubga oshiramiz.
^76 + Jx + V76-Vx J* = 83
76 +4x + 3^(76+ 4x)2
$76-4х + 3V(76-Vx)2
^76 + y/x + 76~Jx = 512
152 + 3^(76 + Vx)(76-Vx)-
• ^76 + 4x + V76-Vxj = 512
3^762 -x -8 = 512-152
^762 -x = 360:24
7& - x = 153
x = 762- 153 = 5776 - 3375 = 2401.
Javob:2401.
25. Qarang: 3-variant 1-savol
(22-bet).
26. Uchburchakning uchlari to'g'ri
burchakli dekart koordinatalar
sistemasida quyidagicha berilgan:
A(0; 0), B(1; -2), C(1; 0). Uchburchak
yuzini toping.
Yechish:
A(0; 0), B(1;-2), C(1; 0),S=?
75
Yechimlar. Matematika va informatika 2017
8-variant
-АСОВ 1-2
2 2
Javob: 1.
- 7 3 7
= log , 22 =--------logj 2 - — .
23 3 2 2 2
27. Hisoblang:
Yechish:
Soddalashtiramiz.
1) 1 + ^ 1~^5
44+47 4з~4~5
1 + 47 45-1
— —. T'=T 4 . й=-
44 + 47 у13-45
_ (47+1)4444 ,
44+47 -^4-47
! (45-1)4з+45
4з-45-4з + 45
4(47 + 1)(47 + 1)(4-47)
416-7
t 4(45-1x45-1X3 + 45)
49^5
44/7 +1X347 -~3)
=---------------h
3
! 44/5-1X245 + 2) _
43(47 + 1X47-1)
=-------1 4-
3
! 42(45-1~\X'/5+1) _
_43-(7-1) , 72(5-7).
3 2
3 2
2)loge242 =log232-25 =
28. Qarang: 1-variant 27-savol
(9-bet).
29. 5-in2<10«, = 4x2-12x+ 10
tenglama ildizlari yiglndisini toping.
Yechish:
1) 4)4 - 12x +10 = 4(>?-3) + 70
2) 0 <: sin2(10rcx) < 1
^=1,51 = 5, 5~1 =4
5
1 2
* < g-sin (10kX) <
5 ~
0,2<4x2- 12x+10<. 1
1) 4X2 - 12x + 9,8>0,xGR
2) 4)4 — 12x + 9 0
(2x-3)2<0, 2x-3 = 0
x = 1,5.
Javob: 1,5.
30. Teng yonli uchburchakning
uchidagi burchagi a ga teng. Shu
uchburchakka ichki va tashqi chizilgan
doira radiuslari nisbatini toping.
Yechish:
ABC - uchburchak teng yonli.
AB = BC, A ABC = а
Ог - tashqi aylana markazi, Oi - ichki
aylana markazi.
в
1) ЕВОг uchburchakda
76
Yechimlar. Matematika va informatika 2017
8-variant
1 AR
BE =- AB, R = 02B =
2 2cos—
2
2) Uchburchak ADOi da
Z DAOi =- ZDAB =
2
= L (go°-—) = 45°- —
2 ’ 2 4
r = OiD=ADtg(45°--)
4
3) ABD uchburchakda
AD = AB-sin— bo'lgani uchun
R:r=-AB—:ADtg(45° --) =
2cos—
2
AB__________1_______
2cos~ ABsin—tg( 45°-—У
2 2 { 4)
cfg^45’-^
sin a
cfg|45"--l
Javob: —-------~.
sin a
31. Qarang: 1-variant 33-savol (12-bet).
32. A1 = —4, A2 = 0, B1 = 9, B2 = 4 bolsin. Natijasi 5 ga teng bo'ladigan
formulani aniqlang.
Yechish:
A1 = -4; A2 = 0; B1 = 9; B2 = 4, natija = 5.
ЕСЛИ(А1*В2<0; A2+5; B2+9);
Bu yerda ЕСЛИ mantiqiy funksiya. Bu funksiyaning sintaksisi quyidagicha.
=ЕСЛИ(А;В;О);
A - mantiqiy ifoda; Agar bu mantiqiy ifoda rost bo'lsa, В amalga oshiriladi.
Agar A - mantiqiy ifoda yolg'on bo'lsa, D bo'ladi.
A1 = -4; A2 = 0; B1 = 9; B2 = 4.
=ЕСЛИ((-4)*4<0; 0+5; 4+9).
Bizning misolda mantiqiy ifoda o'mida (-4)-4 < 0
-16 <0bu mantiqiy ifoda rost qiymat qabul qiladi. Shuning uchun
A2 + 5 = 0 + 5 = 5 amalga oshiriladi.
Javob: =ЕСЛИ(А1‘В2<0;А2+5;В2+9).
33. user_name@inbox.ru elektron pochta manzili berilgan. Bu yerda
foydalanuvchi nomini aniqlang.
Yechish:
user_name@inbox. ru
Elektron pochta manzili foydalanuvchi nomi user_name bo'ladi.
Javob: user_name.
34. Buyruq fayllari kengaytmasini aniqlang.
Yechish:
.pas - pascal dasturi fayli.
.bas - beysik dastur fayllari.
.bat - buyruq fayllari kengaytmasi.
Javob: bat.
77
Yechimlar. Matematika va informatika 2017 9-variant
35. Microsoft Excel dasturida quyidagi formulaning natijasini toping:
=Срзнач(Корень(9); 6; Длстр(3,1415))
Yechish:
=СРЗНАЧ(КОРЕНЬ(9);6;ДЛСТР(3,1415)).
СРЗНАЧ funksiyaning vazifalari qavs ichidagi sonlarning o'rta arifmetik
qiymatini hisoblaydi.
Masalan:
=CP3HA4(4;2;3) bu funksiyaning qiymati 3 bo'ladi, chunki o'rta arifmetik
qiymat + + = 3 kabi hisoblanadi.
КОРЕНЬ funksiyasi qavs ichidagi sonni ikkinchi darajali iidiz ostidan
chiqaradi.
Masalan:
=KOPEHb(49)=7 chunki /49=7.
ДЛСТР funksiyasi satr uzunligi, ya’ni qavs ichidagi belgilar sonini beradi.
Masalan:
ДЛСТР(3,1415)=6 ga teng.
Chunki qavs ichida vergulni hisobga olib, 6 ta belgidan iborat.
=СРЗНАЧ(КОРЕНЬ(9);6;ДЛСТР(3,1415))=5; chunki:
=KOPEHb(9)=3.
=ДЛСТР(3,1415)=6.
=CP3HA4(3;6;6;)=3/~6. = 1J_ = 5.
Javob: 5.
36. HTML tilidagi hujjatda Aniqlik ro'yxatlari qaysi teg bilan boshlanadi:
Yechish:
Aniqlik ro'yhatini tashkil etish teglari.
<DL> tegi bo'lib, и yopiluvchi teg hisoblanadi.
<DL>
<LH> Ro'yxat nomi </LH>
<DT> Terminlar nomi
<DD> Termin uchun aniqlik ta’rifi.
</DL>. ,
Javob: <DL>.
9-variant
1. Qarang: 1 -variant 24-savol
(9-bet).
2. To'g'ri prizmaning asosida to'g'ri
burchakli ABC uchburchak. Z C = 90°
Z A = 60°, AC = b. Prizmada
AB gipotenuza orqali o'tuvchi yon
yog'ining diagonali AC katet orqali
о tuvchi yon yoq bilan 30 burchak
hosil qiladi. Prizma hajmini toping.
Yechish:
ABCA1B1C1 to'g'ri burchakli
uchburchakli prizma.
ZC = 90°, ZA = 60°,
AC = b
78
Yechimlar. Matematika va informatika 2017
9-vahant
1) ABi diagonalning AAiCiC yoqdagi
proyeksiyasiACj bo'ladi. Z B1AC1 = 30°
2) Prizma balandligi
CC1 = ^AC2-AC2
AC1 ni B1AC1 uchburchakdan topamiz.
Z B1C1A - to'g'ri burchakli.
tg30 =-~-,
AC1
AC1 = BiCrctg30°
BfCf = BC = ACtg60° = -j3 b
AC1 = 3b
Н = СС^ yl(3b)2-b2 = 2b/2
S =AC BC _b2-4з
asos 2 2
h2J3 r- r-
V = Sasos-H =—=- 2bj2 = Ь34б .
Javob: b3V6 .
3. Agar
ni
1 a 4 К .1
— + — = — bo Isa,
2a 1,5 3
0,53 a2 ., .
-v+—ni toping.
a2 4,5
Yechish:
1 a 4 0,53 a2
— + — = — bolsa, ——+----------
2a 1,5 3 a2 4,5
hisoblaymiz.
1 a 4 . . .
— + — = — tenglamadan a m
2a 1,5 3
topamiz. Umumiy maxrajga keltiramiz.
a 4=0
3 + 4a2 = 8a
4a2 - 8a + 3 = 0
1 3
a=—,a = —
2 2
0,53 a2 .... ..
~- +-----ning qiymatini topamiz.
a 4,5
nt.. 0,53 0,52
a = 0,5 da--5- +---= 0,5 +
0,5 4,5
_1_-1 I.-!®-®
+ 9-2 ~ 2 + 18 ~ 18 ~ 9
... 0,53 1,52
a = 1,5 da —= + -'— =
1,52 4,5
„ JO,5"\ . 1,5\ . c 1
= 0,5 --- +1,5 ---- =0,5—i-
<1,5) <4,5) 9
, c 1 1 1 3 1 1 15
3 2 9 2 3 18 2 9
. U 5
Javob: —.
9
4. Parallelepipedning asoslari
tomoni 6 ga teng kvadratlardan, barcha
yon yoqlari romblardan iborat. Yuqori
asosining uchlaridan bin ostki asosining
barcha uchlaridan baravar uzoqlikda
joylashgan. Parallelepipedning hajmini
toping.
Yechish:
ABCDAjB^iDi - parallelepiped,
ABCD, A1B1C1D1 - kvadrat.
Asosi kvadrat, yon yoqlari - romb.
AB = AAi=a = 6
Sasos ~ a = 6^ = 36
V = Sasos'H
H = A1O =^AA2~^AO2
2 2
79
Yechimlar. Matematika va informatika 2017
9-variant
h. ‘ .3ji
\ 2 J 42 42
V = 36- 342 = 10842 .
Javob: IO8V2.
x2 + 6 x + 21
5. у =---------funksiyaning eng
11 + бх + х
kichik butun qiymatini toping.
Yechish:
-x? +6x + 21 10 _
x2+6x + 11 x2 + 6x + 11
„ 10
— 1 4---------
(X + 34 + 2
10
ymax=1 +^=1 + 5 = 6
Qiymatlar sohasi (1; 6]. Eng kichik
butun qiymati 2.
Javob: 2.
6. Qarang: 1-variant 16-savol
(6-bet).
7. Qarang: 1-variant 8-savol
(5-bet).
8. Konus o‘q kesimi uchidagi
burchak 2<p ga, o‘q kesimi yuzi esa Q ga
teng bo'lsa, konus hajmini toping.
Yechish:
1 1
^=j-SaxsH = ~rrR2H
^=tg<p, R = tg<p-H
rl
\tgv
V=^--tg2<p-~- I^ = ~q4Q^
3 tgg>\tgq) 3
Javob: —Q4Qtgp
3
9. Qarang: 2-variant 11-savol
(15-bet).
10. Radiusi 8 ga teng bo'lgan
aylanaga ichki chizilgan muntazam
oltiburchakning perimetrini toping.
Yechish:
R = 8, P = ?
R =--—
2»^
6
a
71
2
= a
Z ASB = 2<p
S,esm = H'^ = HR = Q
R = a, P = 6a = 6-8 = 48.
Javob: 48.
.. „ 2-8x-x . .... .
11. 3 -----5-----tengsizhkni
x -1
qanoatlantiruvchi eng katta butun
manfiy sonni toping.
Yechish:
3 > —Ox—x gfxfcfaiggirtirfo oiamiz.
x2-1
з-г-8 *,х-х\о
X -1
80
Yechimlar. Matematika va informatika 2017
9-variani
3x2-3-2 + 8x + x2
x2 -1
4x2 + 8x- 5 . л .
-------------> 0 kasrning surat va
x -1
maxrajini ko'paytuvchilarga ajratamiz:
/ 5Y
4 x + — x—
I 2 2)
—------—-----> 0 son o‘qida
(x-1)(x + 1)
suratning ildizlarini va maxrajning
uzilish nuqtalarini belgilaymiz va har bir
oraliqda kasrning ishorasini aniqlaymiz.
(-oo; -2,57 U (-1; 0,5] U (1; oo )
Eng katta butun manfiy son -3.
Javob: -3.
AC = AB = 4, BC = 4V2
ACAD va ABAE- to'g'ri burchakli.
AE = EC, AD = DB
CD2 = AC? + AD? = 42 + 2? = 20,
CD =245
BE2 = AB2+AE2 = 42 + 22 = 20,
BE=2y[5
CO = -CD = ^^,
3 3
12. Agar barcha x, у lar uchun
x3 + 4x2y + axy2 + 3xy - bxcy + 7xy2 +
+ dxy + y2 = x3 + y2 ayniyat bajarilsa,
a-b + c + dni toping, (c > 1)
Yechish:
Ayniyat bo'lganligi uchun o'xshash
darajalar oldidagi koeffitsiyentlarni
tenglashtirib topamiz.
c = 2
x2/4- b = 0, b = 4
xy2 a + 7 = 0, a= -7
xy 3 + d = 0, d = -3
a=-7,b = 4,
c = 2, d = -3 da
a -b + c + d = -7-4 + 2-3 = -12.
Javob: -12.
13. Uchburchakning uchlari to'g'ri
burchakli dekart koordinatalar
sistemasida quyidagicha berilgan:
A(1; 0), B(5; 0), C(1; 4). O'tkir
burchaklar medianalari orasidagi
o'tmas burchak kosinusini toping.
Yechish:
1) A(1; 0), B(5; 0), C(1; 4), cosa = ?
2) BE, ED - medianalar. ABC-to'g'ri
burchakli.
BO=-BE=^-
3 3
Kosinuslar teoremasidan Z COB ni
topamiz.
CO2 + BO2-BC2
cosa =-----------
2CO-BO
z r-\2 / Г~\2
,2
cosa =
2 4'^ 4^
3 3
80+80-32-9 _ 10-18 ___8_ __4
2-80 ~ 10 10~ 5
4
Javob: —.
5
14. Qarang: 1-variant 21-savol
(8-bet).
15. Qarang: 3-variant 16-savol
(26-bet).
16. Qarang: 2-variant 29-savol
(20-bet).
17. Qarang: 3-variant 9-savol
(24-bet).
81
Yechimlar. Matematika va informatika 2017
9-variant
18. Z BED = 60°, Z EDP = 15°,
Z BPD = ?
Yechish:
Z BED = 60°, Z EDP = 15°, Z BPD = ?
Z BED = Z AEC = 60°
Z BEA =Z1 DEC = 120°
Z ECD = 180° - 120° - 15° = 45°
Bundan BD yoy 90° ekanligi kelib chiqadi.
Z BED = 60° ligidan AC yoy 30°
ekanligi kelib chiqadi.
Z BPD = 30°.
Javob: 30°.
19. Aylana kvadratning ikki qo'shni
tomoniga urinadi, qolgan ikki tomonini
2 va 23 bo'lgan kesmalarga ajratadi.
Aylana radiusini toping.
Yechish:
PE = FB = 2,
AB = BC = 23 + 2 = 25
OF = R, AOPF da Z OPF = 90°
OP = OE - PE = R - 2, AE = R
PF = AB-AE = 25-R
OF2 = OP? + PF2
R2 = (R — 2)2 + (25 - R)2
R? = — 4R + 4 + 625-50R + R2
R2 -54R + 629 = 0
R=17,R = 37
R = 37yechim bo'la olmaydi, chunki
AB> R bolishi kerak.
Javob: 17.
20. Hisoblang:
3 2 _9___4_ 27_________8_
7 5 + 49 25+ 243 125
Yechish:
3 2 9___4_ 27______8_ +
7 5 + 49 25 + 243 125
3 9 27 _2_±____8__
~7 + 49 + 243+" 5 25 125
3 9 27 (2_ 4 8
~7+T9+143 + " [5 + 25 + 125+~"
3
q = 7
3
3 9 27
1) — +— +---+ ...-cheksiz
7 49 243
kamayuvchi geometrik progressiya.
a 3 A 9
b. = b, =—,
’ 7 2 49
3
s=-^-=-^=-
1-q 3 4
7
_. 2 4 8 . ..
2) — + — +--+... - cheksiz
5 25 125
kamayuvchi geometrik progressiya.
. 2 . 4 2
b^5’b2 = 25’Q-5
2
1-q -f-l з
5
_ ,32 9-8 1
Demak,----=----= — .
43 12 12
. u 1
Javob: — .
12
21. Agar barcha x, у lar uchun
x3 + 4x^ + axy2 + 3xy - bxcy +
+ 7ХУ2 + dxy + y2 = x3 + y2 ayniyat
bajarilsa, b - d ni toping, (c > 1)
Yechish:
Ayniyat bo'lganligi uchun o'xshash
darajalar oldidagi koeffitsiyentlardan
foydalanib yechamiz. c = 2
rcy oldidagi koeffitsiyent
4-b = 0,b=4
xy2 oldidagi koeffitsiyent
a + 7 = 0, a =-7
xy oldidagi koeffitsiyent
82
Yechimlar. Matematika va informatika 2017
9-variar
3 + d = 0,d = -3
a =-7, b = 4, c = 2, d = -3
b-d = 4-(-3) = 4 + 3=7.
Javob: 7.
22. |3-VT+5| >^-
tengsizlikning manfiy butun yechimlari
nechta?
Yechish:
Aniqlanish sohasi:
x + 5> 0, x>-5
3-Vx + 5>^^
6
L л—? 8-x
6
[7S — б4х + 5 < 8 - x
Ых + 5 <26-x
|бл/х + 5 > 10 +x
1) 6y[x + 5 <26-x
(26-x>0
\36x+ 180 < 676 - 52x + x2
\x2 -88x + 496>0
-5 <x <44-12^10
[-5; 44 - 12s/Td)
2) 6>/x + 5 > 10+ x
(10 + x>0
\36x + 180>100 + 20x + x2
]x2-16x-80<0
Manfiy butun yechimlari -5, -4, -3,
-2, -1.
Javob: 5 ta.
23. (-3; 4) nuqtaning absissa,
ordinata o'qlariga va koordinata
boshiga nisbatan simmetrik bo'lgan
nuqtalarini tutashtirishdan hosil bo'lgan
uchburchakning balandligini toping.
Yechish:
1) A(-3; 4), h = ?
2) Uchburchak BDC - to'g'ri burchakli.
BC - gipotenuza. BD, CD - katetlar.
A, .4 - / I . / I /1 I 1
.3 1 1 ta 1 < 1
С 7Д d
DE = h,BD = 8,CD = 6
BC2 = BD2 + CD2 = 82 + 62 = 1(4
h_BD-CD_8-6_48_18
BC 10 10 ’
Javob: 4,8.
y2V2 + 3 . u. U1
24. —.....ni hisoblang.
VV2 + 1
Yechish:
(Va+Vb) = a+ b + 2>[ab
^2^2+3 =^3 + 2-42-1 =
= 42 + 2-42-1 + 1 = ^(42 + T)2 = 442 + 1
4242 +3 _442+1 _1
442+1 442 + 1
Javob:1.
25. ABC uchburchakning yuzi
30 sm2. AC tomonidan olingan D nuqta
uni AD:DC = 2:3 nisbatda bo'ladi.
BC tomonga tushirilgan DE
perpendikulyar uzunligi 3 sm bo'lsa,
BC ni toping.
Yechish:
S = 30,
AD:DC = 2:3
DE = 3,
BC ni topamiz.
83
Yechimlar. Matematika va informatika 2017
9-variant
S - AK. AK balandlik tushiramiz.
2
AK parallel DE ga
ДАКС va ADEC o'xshash.
AK DE DE-АС 3-5x c
AC DC DC 3x
BC=^- = ^ = 12.
AK 5
Javob: 12.
26. | a + b | = 8, | a - b | = 4 va
| a | = 3 bo'lsa, | b | ni toping.
Yechish:
|a + b| = 8, |a-b|=4, |a| = 3, |6| = ?
1)\a + b\2 = 82
a2 + 2ab + b2 =64
2)\a~b\2 = 42
a2 -2ab + b2 =16
a2 + 2ab + b2 =64
3K.2 - -2
a -2ab + b =16
2a + 2b2 = 80
a2 + b2 =40, | a | = 3
Ьг = 40-3? = 40-9 = 31
\b\=43i.
Javob: .
27. ABCD parallelogramda AB = 37,
diagonallari kesishish nuqtasidan
AD tomonga tushirilgan perpendikulyar
uni AE = 26 va ED = 14 bo'lgan
kesmalarga ajratadi. Parallelogramm
yuzini toping.
Yechish:
AB = 37,
AE = 26,
ED = 14,
S= ?
А К E D
DEL AD,
Sabd-BK-AD, OE = ^~
2
Uchburchak BKD va OED o'xshash.
B~ =K~, KD = KE + ED
OE ED
BK KE+ED KE + 14
BK~ ED ’ 14
2
KE = 14
AK = AE - KE = 26-14 = 12
BK2 = AB2 - AK2 = 372 - 122 =
= (37- 12)(37 + 12)
BK = >/25-49 = 5-7 = 35
AD = AE +ED = 40
Sabcd = ADBK = 40-35 = 1400.
Javob: 1400.
28. a va b sonlarini 5 ga bo'lganda
qoldiq mos ravishda 4 va 2 ga teng
bo'ldi. a b ko'paytmani 5 ga
bo'lgandagi qoldiqni toping.
Yechish:
a = 5h + 4, b = 5m + 2
ab = (5n + 4)(5m + 2) = 25nm + 10n +
+ 20m + 8 = 5(5nm + 2n + 4m) + 8
5(5nm + 2n + 4m) ifoda 5 ga qoldiqsiz
bo'linadi.
r = 3, qoldiq 3 ga teng.
Javob: 3.
29. a uzunlikdagi vatar silindr
asosida л yoyni tortib turadi. Agar
silindrning balandligi H bo'lsa, uning
to'la sirtini toping.
Yechish:
AB - vatar, AB = a, H. a vatar л yoyni
tortib tursa, bu vatar diametr bo'ladi.
84
Yechimlar. Matematika va informatika 2017
9-variant
a
Stoia ~ Sasos + Syon, Sto‘la ~ TrR' + 2ttRH
R=—
2
, x2 Z x
Sfo'/a = л| — | + 2л- — H = ал[ — + H |.
{2l 2 I 2 )
i q \
Javob: ал — + H .
l2 )
30. 2sin2x + 3sinx > 0 tengsizlikning
[0; 2л] kesmadagi yechimlari to'plamini
toping.
Yechish:
sinx(2sinx + 3) >0
-1 < sinx < 1 bo'lganligi sababli
2sinx + 3 doimo musbat, shuning uchun
[0; 2 л] oraliqdagi yechim (0; л).
Javob: (0; л).
31. Qarang: 3-variant 33-savol (31-bet).
32. Paskal. Takrorlanishlar sonini aniqlang:
For k:=round(1,5) to trunc(sqr(2)) do a:=5; end.
Yechish:
k: = round(1,5)=^ k: = 1.
trunc(sqr(2)) = trunc(22) = 4.
fork: = 1 to 4 do.
a: = 5;
end
k: = 1.
k: = 2.
k: = 3.
k: = 4. Jami 4 ta marta sikl bajariladi.
Javob: 4.
33. Qarang: 1-variant 36-savol (12-bet).
34. Faqat brauzerlar berilgan qatorni ko'rsating.
Yechish:
Brauzer dasturlar Opera, Adwiper, qolgan dasturlar: AutoCad- grafiklar,
tasviryasash dasturi, Iternet Explorer- brauzer, MySQL - ma’lumotlar omboridir.
Javob: Opera, Adwiper.
35. Ketma-ket portning nomi keltirilgan qatorni toping?
Yechish:
Ketma-ket port Com porti bo'lib communications port (последовательный
порт). Com portnining bunday atalishiga sabab shuki, bu port orqali yuborilgan
ma’lumot ketma-ket bir bitdan yuboriladi.
Javob: COM.
36. Qarang: 2-variant 33-savol (21 -bet).
85
achimlar. Matematika va informatika 2017
10-variant
10-variant
1. sin3x + >/з cos3x s 0
rrigonometrik tengsizlikni yeching.
Yechish:
1
Tengsizlikning ikkala qismini — ga
2
ko'paytiramiz.
1 -J3
— sin3x + — cos3x > 0
2 2
sin3x-cos— + sin — cos3x > 0
3 3
sin 3x + — SO
I 3
2лп < 3x +— s л + 2лп
3
-- + 2лп < Зх + 2лп
3 3
л 2лп , , 2л 2лп „
—+—sxi—+—, n e z.
9 3 9 3
, . \ л 2лп 2л 2лп
Javob: — +-----; — +---
9 3 9 3
2. Qarang: 5-variant 29-savol
(48-bet).
3. lx2 - 5ax| = 15a tenglama ikkita
haqiqiy yechimga ega boladigan a ning
natural qiymatlari yig'indisini toping.
Yechish:
lx2-5ax| = 15atenglamada a>0
bolishi kerak.
Grafik usulida yechamiz.
у = lx2 - 5ax|, у = 15a
Tenglama ikkita haqiqiy yechimga ega
bo'lishi uchun 15a > yo
у = lx2 - 5ax| da xo = 2,5a,
yo = 6,25a2
15a > 6,25a2
6,25a2 - 15a <0
a(6,25a - 15) < 0
0 <a <2,4 naturalyechimlaria = 1,
a = 2. a ning natural qiymatlari yig'indisi
1+2 = 3.
Javob: 3.
4. у = 2sin(2x + —) funksiya nechta
4
natural qiymatlarni qabul qiladi?
Yechish:
-1 <sin 2x + — I < 1
I 4)
-2 < 2sin\ 2x + -|s 2
I 4 J
Qiymatlar sohasi [-2; 2]. Qiymatlar
sohasida 2 ta natural son bor, bular
1 va 2.
Javob: 2 ta.
5. Qarang: 3-variant 15-savol
(26-bet).
6. Uchburchak ABC da AC asosiga
DE parallel to'g'ri chiziq o'tkazilgan
Sabc = Sdec = 2 bo'lsa, DE:AC ni
toping.
Yechish:
SABC = &, SDBE = 2, DE:AC = ?
Uchburchak ABC ga uchburchak DBC
S 7 АСУ*
o'xshash, bundan — yoki
SMC lOfJ
5МС fDEV
$abc I AC 7
86
Yechimlar. Matematika va informatika 2017
10-variant
2 fDE? 1 DE
8"UcJ ’ 2 ~ AC
DE:AC = 1:2.
Javob: 1:2.
У Д log8 27 _ у logg 3 + 3 logg 7 ___ 1 tog7 8
ifodaning qiymatini toping.
Yechish:
1) ato8ab = b,
2) a'°et,c = с'"е!>3
Is
3) log b‘=-logab,
a n
4) 1n = 1
j^Z3 33 _ ylogs 3 + 7I0SS 3 _ у _
_ ^log? 3 _ у _ 2* 1 2 loS2 3 _ 1 — 2loS2 32 _ у _
= 32-1 = 9-1 = 8.
Javob: 8.
8. О yarim aylana markazi
Z DBC = 45°, Z AOD = 30°.
|BC| = 1 sm bo'lsa, |DC| ni toping.
Yechish:
9. Qarang: 1 -variant 30-savol
(10-bet).
10. Ushbu f(x) =—--------
x + 3x + 2
funksiyaning boshlang'ich funksiyasini
toping.
Yechish:
Boshlang'ich funksiyani topish amaliga
integrallash deyiladi.
F(x) = jf(x)dx =
- [ 2x + 3 dx_rd(x2 + 3x + 2) _
x2+3x + 2 x2 + 3x + 2
= ln\x* + Зх + 2| + C =
= Infix + 1|-|x + 2I) + C.
Javob: ln(|x + 1 | |x + 2|) + C.
11. /(x) = 72sin7x-sin11x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko'paytmadan yig'indiga otamiz.
sin7x sin11x (cos(7x- 11x) -
1
- cos(7x + 11x)) =— (cos4x - cos18x)
2
A О °
Z DBC = 45°, Z AOD = 30°, |BC| = 1,
DC = ?
Z AOD = 30°, AD yoy 30°,
Z DBC = 45°, DC yoy 90°
BC yoy 180° - (30° + 90°) = 60°
Bundan ZCDB = - °- = 30
2
Sinuslar teoremasiga ko ra:
sin 45° _ DC
sin 30° BC
42
2 DC
1 1 ’
2
DC =42.
Javob:42 .
2) f(x) = 72'^ (cos4x- cos18x) =
= 36cos4x - 36cos18x
F(x) = 9sin4x - 2sin18x + C.
Javob: 9sin4x - 2sin18x + C.
7x
12. у = logo 25--funksiyaning
x2+49
eng kichik qiymatini toping.
Yechish:
Eng kichik qiymatini topish uchun
funksiyadan hosila olamiz.
( 7x / _ 7(x2 + 49)-7x-2x _
^x2 + 49j x2+49
_7-49-7x2
x2+49
7-49-7x? = 0,-4 =49,x = +7
87
Yechimlar. Matematika va informatika 2017
10-variant
Aniqlanish sohasiga ko'ra x> 0, x = 7.
7-7 . 49
У og025 7г+4д 0&2.49
= log,^ = log2.2 2-'=i
У min ““ •
15. a = 5 bo'lsa,
a+1
J (ln(sin2 3x + cos2 3x) + 1)dx aniq
a
integralni hisoblang.
Yechish:
1) sin23x + cos23x = 1
2) In1 = 0
a+1 a+1
3) J (0 + 1)dx = j 1dx = x
a a
= a + 1 -a = 1.
13. x3 - 32 -4xVx = 0 tenglamaning
haqiqiy ildizlari ko'paytmasini toping.
Yechish:
xVx = 7x-x2 = (Vx)3
x3 - 4(Jx )3 - 32 = 0,
(4x)3 = a
a2-4a-32 = 0
a =-4, a = 8
(Л)3=-4
4x = -</4 tenglama haqiqiy ildizga
ega emas.
(Jx)3=8, Jx=2,x = 4.
Javob: 4.
fy2+xy = 12
14. < tenglamalar
|x +xy = 4
sistemasini yeching.
Yechish:
(y2+xy = 12 (y(x + y)=12
[x2+xy=4 |x(x + y) = 4
y(x + y) 12
x(x + y) 4
-- = 3,y=3x
x
X2 + xy = 4, x2 + Зз^ = 4
4X2 = 4, x2 = 1, x = ±1
x = ±1 day = ±3
(1; 3) va (-1; -3).
Javob: (1; 3)va(-1;-3).
Javob:1.
16. J
n
x-y = —
6 sistemani yeching.
sinx=2
cosy
Yechish:
Я It
X-y= у = X- —
У 6 6
sinx = 2cos x —
I 6
sinx = 2COSXCOS— + 2sinxsin—
6 6
sinx = cosx + sinx,
43 cosx = 0, cosx = 0, x = — + 7tn
2
7Г К 7Г
у = — + яп — = — + яп,п е Z
2 6 3
f— + Ttn-,— + тгл |, п С Z.
<2 3 J
Javob'. ( — + яп; — + яп ), n С Z.
17. у = /(х) funksiya D to'plamda
quyidan chegaralangan bo'lsin. U holda
qaysi munosabat ixtiyoriy x 6 D uchun
o'rinli?
Yechish:
у = f(x) funksiya D to'plamda quyidan
chegaralangan bo'lsin. Uholda xGD
biror к haqiqiy soni uchun f(x) > к
munosabat o'rinli.
88
Yechimlar. Matematika va informatika 2017 10-variant
у ~ f(x) funksiyaning o'zgarish sohasidagi har qanday qiymati uchun shunday o'zgarmas chekli A soni uchun f(x) > A bo'lsa, f(x) quyidan chegaralangan deyiladi. Javob: biror К haqiqiy soni uchun /(x) > K. 21. О - aylana markazi. Z OBA = 10°, Z OCA = 30° bo'lsa, Z ВАС ni toping. A / /°\ X. ( / °\ I bv23c
18. Qarang: 1-variant 28-savol (10-bet). Yechish: Z OBA = 10°, Z OCA = 30°, Z ВАС = ?
ЛЛ OX 5|O854~I°S2516 ... ,. 19. (0,2)2 ni hisoblang. Yechish: а=х +y Z ВАС = Z OBA + zi OCA Z ВАС = 10° + 30° = 40°.
1 —log5 4 _l°g2s I6 ni soddalashtiramiz. ~tog5 22- log 2 2“ = 2 3 1 4 = —•21og52-—log52 = 2 2 = log52 - 2log52 = -logs2 (0,2y‘°es2 = (5-y°852 = 5‘°es2 = 2 . Javob: 2. Javob: 40°. 22. Qarang: 2-variant 8-savol (15-bet). 23. (х-2)л/х2-2x-12 = 6x-12 tenglama ildizlari yig'indisini toping. Yechish: Aniqlanish sohasini topamiz. x2 -2x — 12 >0 2 + yj4 + 48 2 + ^52 X1,2 = = - =
20. Avtomobil haydovchisi birinchi soatda yo'lning yarmini, ikkinchi 1 soatda qolgan yo'lning — qismini, uchinchi soatda qolgan 56 km masofani bosib o'tdi. Haydovchi uch soatda jami qancha (km) yo'l bosib o'tgan? Yechish: 2 2 2 xC(—®; 1 -4l3]U[1 +4i3; oo) Tenglamani yechamiz.
x-km yo'l. X XX 1-soatda — km, x-—= — qolgan yo'l 2 2 2 2-soatda — km 6 3-soatda 56 km 3 soatda —+ — + 56 = x 2 6 x- — - — = 56x, 6x- 3x-x = 56-6 2 6 2x = 56-6, x = 168. Javob: 168. (x-2)~Jx2-2x-12 -6(x-2) = 0 (x-2)(Jx2-2x-12 -6)=0 x-2 = 0, x = 2 yjx2-2x-12 -6 = 0 yjx2-2x-12 =6 x2 -2x-12 = 36 x2-2x-48 = 0 x = 8, x =-6 x = 2 aniqlanish sohasiga tegishli emas. Tenglamaning ildizlari x = 8 va x = -6. Ildizlaryig'indisi 8 + (-6) = 2. Javob: 2.
89
Yechimlar. Matematika va informatika 2017
10-variant
24. Uchburchak muntazam kesik
piramidaning yon qirralari asos
tekisligi bilan a burchak tashkil etadi.
Ostki asosining tomoni a ga, ustki
asosining tomoni b ga teng bo'lsa,
kesik piramida hajmini toping.
Yechish:
ABCA1B1C1 uchburchakli kesik
piramida. ABC, A1B1C1 - asoslari.
Z A1AE = a, AB = BC = AC = a
Kesik piramida hajmi:
v=^(S,+S2 + VvsT)
Si va S2 ABC va AiBiC1
uchburchaklarning yuzalari.
Of ~o —------------
1 4 2 4
H = AiK. AAA1K da 4AtAK = a.
AK=AO-A1Oi
AO - ABC uchburchakka,
A1O1 - A1B1C1 uchburchakka tashqi
chizilgan aylanalarning radiuslari.
AO = ~, AIO1=-^
Л 1 /3
... a-b ,, a-b .
AK — —гЬ="", H — —• fact
Va Va
1 a-b a2j3 i
V =----T^tga-(-----+ -
3 V3 4
4
(a-b)tga
aVa
43.2 ,2 (a3-b3)tga
----(a2 +b +ab) = -----
4 12
. . (a -b )tga
Javob: ---------2—
12
25. _f(x) = tgxsinxctgxcosx bo'lsa,
/'(—) ni toping.
2
Yechish:
1) tgx ctgx = 1
2) f(x) = sinx cosx tgx ctgx =
1 о
= sinxcosx =— sin2x
2
(1 \ 1
3) f'(x) =l —sin2x I = — 2cos2x =
=cos2x
4) f' I — I = cos2 — = cosit = -1.
' \2) 2
Javob: -1.
26. Muntazam uchburchak tomoni a.
Uchburchak tomonlari o'rtalari
tutashtirilib, muntazam uchburchaklar
hosil qilindi. Ichma-ich joylashgan
uchburchaklar yuzalari yig'indisini toping.
Yechish:
a - muntazam ABC uchburchak tomoni.
Muntazam uchburchak yuzi:
4
A1B1C1 muntazam uchburchak tomoni
a . o a" Vi
5.yuz,S,= —
A2B2C2 muntazam uchburchak tomoni
. a2^3
yuzl s2=~^r-
a
4
S3 = .. - cheksiz kamayuvchi geometrik
progressiya.
90
Yechimlar. Matematika va informatika 2017
10-variant
b, =------, b, =------,
’ 4 2 16
q=b2:b-i = t
4
S + Si + S2 + S3 ... yig'indini
hisoblaymiz.
S' = S + St + S2 + S3+ ... =
28. To'g'ri parallelepiped
diagonallari 9 sm vaV33 sm, asosining
perimetri 18 sm, yon qirrasi 4 sm.
Parallelepiped tola sirtini toping.
Yechish:
Di = 9, D2 =433 ,P=18, H = 4, S^a = ?
1-q
3
4___4
1-1= 2
4 4
Ichma-ich joylashgan uchburchak
yuzalari yig'indisi —-— ga teng.
Javob:—
3
27. Aylanadagi bir nuqtadan
uzunliklari 10 va 12 bolgan vatarlar
o'tkazilgan. Kichik vatar o'rtasidan
katta vatargacha bolgan masofa 4 ga
teng bo'lsa, aylana radiusini toping.
Yechish:
AB = 10, AC = 12, DE = 4,
DEI AC, AD = DB
Uchburchak AKB da DE- o'rta chiziq.
в
BK = 2DE = 8
AE = 3, AK = 6
Bundan BC2 = BK2 + КС2
BC2 = 8? + 62 = 102, BC = 10
ДАВС teng yonli: AB = BC.
AB-AC-BC AB2 _ 102
~ 4 AC BK 2BK~ 2-8
2
100 _ „
=----= 6,25 .
16
Javob: 6,25.
P = 2(a +°b)
1) ДСАА1 to'g'ri burchakli
A,C = 9, AA1 = 4
AC2 = A-iC2 - AA,2
AC=492-42 = 4б5
2) AB^BD to'g'riburchakli
B1B = 4, B/D-433
BD? = B1D2 - B-1B2
BD = д/Сл/ЗЗ)2-^ = V77
3) d2 + d22 = 2(a2 + b2)
di =AC, d2 = BD
(465 )2 + (417 )2 = 2(a2 + b2)
82 = 2(a2 + b2)
[a2+b2=41 a = 5
[a + b = 9 b=4
yoki a = 4, b = 5
4) Sto‘la = 2Sasos + Syon
Syon = PH = 18-4=72
Sasos — a- b-sina
_42 +52-(4l7)2 24 3
2-4-5 ~ 40~ 5
sina = 4l -cos2 a = —
4
Sasos = 4-5 — =16
5
Sto/a =2-16 + 72= 104.
Javob: 104.
91
Yechimlar. Matematika va informatika 2017
10-variant
29. cos2x = sin2x tenglamani yeching.
Yechish:
(45 + 2((4б -2) = 5-4 = 1 -o'zaro
teskari sonlar.
„ 1-cos2x
2) cos2x =-------
2
2cos2x = 1 - cos2x, 3cos2x = 1
1 1
cos2x , 2x = ±arccos— + 2irn,
3 3
1 1
x = ±— arccos— + 7tn.
2 3
1 1
Javob: ± — arccos — + лп, n C Z..
2 3
30. (75+2 Г1 = (л/5-2)^
tenglama ildizlari yig'indisini toping.
Yechish:
(45 +2)x~1 =(45-2)^
= (45- 2)~1
(45 - 2)~1<x~1> = (45 - 2),+/ asoslari
bir xH.
( 1 A
(x-1)-\——+l\ = 0
\x+1 )
x + 2
(x-1)-l±±=0
x + 1
x = 1, x = -2, x^-1
Ildizlari yig'indisi 1 + (-2) = -1.
Javob: -1.
31. Excel 2003 dasturida necha turdagi diagramma tuzish mumkin?
Yechish:
Excel 2003 dasturida jam! 14 ta turdagi diagramma tuzish mumkin bo'lib, ular:
Tekislida (на плоскости):
1) гистограмма (gistogramma)
2) графический (grafikli)
3) круговая (aylanali)
4) линейчатая (chiziqli)
5) лепестковая
6) точечная
7) смешанная
8) кольцевая
9) с областями
Fazoda (в пространстве):
1) с областями
2) поверхность
3) круговая
4) график
5) гистограмма.
Javob: 14.
32. 331 ю, 320ю, ЮОю, 102ю butun sonlarning barchasini yozish mumkin
bo'lgan eng kichik asosli sanoq sistemasida shu sonlar yig'indisini aniqlang.
92
Yechimlar. Matematika va informatika 2017
10-variant
Yechish:
> 33ho-X2. X = 101001011,.
raqamlar yig ‘indisi 5
33112
~2_ Ii65|2_
13 16J82I2
12 58_I41I2_
11 ‘4 2 4 |2O(2_
10 ф 2 tD'2 HOI2
ф @ ’©10(512
©4(212
®2ф
©
32010 -> X2. 32010 = 101000000,
raqamlar yig ‘ indisi 2
320I2
2 (16O|2_
12 16 |80[2_
12 “(0)8 (4OI2_
~(d) ©4 I2012
©2 I10I2
'©10(512
©4[2I2
©2®
©
1ОО1О-^Хг. 100ю= 11001002 .
raqamlar yig'indisi 3
10012
10 (50l2_
© 4_(25I2_
10 2_(12I2
10 5 12I6|2_
© 4 «2
Ф @2®
Ф
102ю->Х2. 102ю= 11001102
raqamlar yig ‘ indisi 4
10212
10 Г51Т2
2 4j25:2
2 11 2IT2I2
©10 5‘l2(6l2_
Ф 4 ©613|2
ф ©2ф
Ф
5 + 2 + 3 + 4=14.
Javob: 14.
93
Yechimlar. Matematika va informatika 2017 11-variant
33. “Axborot tizimi" nima?
Yechish:
Tizimning o'zi maqsad yo'lida bir vaqtning o‘zida ham yaxshi, ham o'zaro
bog'langan tarzda faoliyat ko'rsatadigan bir nechta turdagi elementlar
majmuasidir.
Axborot tizimi esa maqsadga erishish uchun axborotni shakliga va
mazmuniga ko'ra turlarga ajratish, saqiash, izlash, qayta ishlash prinsiplari,
qayta ishiashda qo'llaniladigan usullar, shaxslar hamda vositalarning o'zaro
bog'langan majmuasidir.
Javob: belgilangan maqsadga erishish uchun axborotlarni uzatish, qayta
ishlash va saqlashda qo'llaniladigan usullar, shaxslar va vositalarning o'zaro
bog'langan majmuasi.
34. To'g' ri tenglikni ko'rsating:
Yechish:
1 Kbit- 1000 bit.
Ikkilik axborot miqdorini o'lchovchi birlik.
Javob: 1 Kbit = 1000 bit.
35. MS Excel 2003 dasturida berilgan =Если(Степень(3;4)>80;
Сцепить(“АИо”;“Ьиз”); MAKC(15;30;4)) formulaning natijasini aniqlang.
Yechish:
=ЕСЛИ(СТЕПЕНЬ(3;4)>80;СЦЕПИТЬ("АМо";”Ьи8'));МАКС(15;30;4);
Bu yerda agar СТЕПЕНЬ(3;4) ya’ni 34 darajasi 80 dan katta bo'lsa,
CUEriHTb("Avto";''bus'1) funksiyasi bajariladi. Aks holda адагЗ4 darajasi
80 dan katta bo'lmasa, и holda
MAKC(15;30;4);
34 = 81; 81 > 80 bo'lgani uchun CLJEnHTb("Avto";"bus") bajariladi, ya’ni bu
funksiya vazifasi "Avto" va "bus" so'zlarini qo'shib, biryaxlit matn ko'rinishiga
keladi "Avtobus".
Javob: Avtobus.
36. Qarang: 4-variant 33-savol (41-bet).
11-variant
1. Qarang: 2-variant 2-savol
(13-bet).
2. Rombning perimetri 2p,
diagonallarining yig'indisi m bo'lsa,
uning yuzini toping.
Yechish:
P = 2p, di + d2 = m, S = ?
n „ d, d}
Romb yuzi S
di + d2 +2did2 - m
2dtd2 = m2 - 4a2,
,, m2-4a2
did2 =---——
2
Rombperimetri4a = 2p, a
d2 + d22 = 4a2
94
Yechimlar. Matematika va informatika 2017
11-variant
g _ did2 _m2 -p2
2 4
. m2 -p2
Javob: -----—.
4
а(4з~1)
4
3. Tomoni a ga teng bo'lgan teng
tomonli uchburchak ichiga
uchburchakning tomonlariga va
bir-biriga urinuvchi uchta teng doira
joylashtirilgan. O'zaro urinuvchi
doiralar yoylaridan hosil bo'lgan egri
chiziqli uchburchakning yuzini toping.
Yechish:
ABC - muntazam uchburchak,
a - uchburchak tomoni. O1O2O3 -
bir-biriga urinuvchi uchta aylana
markazlarini tutashtirishdan hosil
bo'lgan uchburchak. Smnl egri chiziqli
uchburchak yuzini topishimiz kerak.
O1O2O3 uchburchak yuzidan uchta
O1ML, O2LN, O3NM sektoriaming yuzini
ayirsak, LMN yuzi hosil bo'ladi.
O1O2O3 uchburchak tomoni
O1L = LO2 = r bo'lganligi sababli 2r ga
teng.
AO1 60° li A burchak bissektrisasi
bo'lgani uchun Z O1AD = 30°.
tg30°=~,
AD =
tg30°
= r43
AD = EC = г4з , DE - O1O2 = 2r
a=AD + DE + EC = 2AD + DE =
= 2r43 + 2r = 2r(1 + 43)
2r= a _а(4з~1)
43 + 1 2
_а243(43-1)2
16
Uchta sektorning umumiy yuzi:
q □ тег2 60° _ nr2 _
'J'^sektor ~ —~~ — —
360 2
_тга2(4з~1)2 ла2(4-2у[3)
2-16 ~ 32 ~
_na2(2-43)
16
Slmn = S OjQjOj — 3Sseklor =
_-а-243(4-243) ла2(2-43) _
~ 16 16 ~
= —((2-43)-243-л(2-43)) =
16
_а2(2-4з\243-тг)
16
. . а2(2-73)(273-лг)
Javob: —-----------------.
16
4. Qarang: 9-variant 2-savol
(78-bet).
5. Qarang: 2-variant 17-savol
(16-bet).
6. Iog‘x-log25^- + 9log2^|<
8 x
< 4log25 x tengsizlikning yechimi
bo'lgan eng kichik natural sonni toping.
Yechish:
Aniqlanish sohasi x> 0.
x3 32
logj x - log2 , —- + 91og2 — < 4 log2 x
2 8 x 2
x3 32
log3 x - log2 — + 9 log2 < 4 log2 x
95
Yechimlar. Matematika va informatika 2017
11-variant
( $ A 2
log2x- + 9(log2 32-
к & /
-log2 x* 1 2)<41og2x
log* x - (log2 x3 - log2 8) +
+ 9(5 - 2 log2 x) < 4 log2 x
loq2x = a
а- (За - 3)2 + 9(5 - 2a) < 4a2
a4-9a2 + 18a-9+ 45- 18a-4a2 <0
a4 - 13a2 + 36<0
(a2 - 4)(a2 - 9) < 0
(a + 2)(a - 2)(a + 3)(a -3)<0
-3-2 2 3
-3 <a <-2,
2<a<3
1)-3 <a < -2, —3 < log2x < -2,
1 1
— <x< —
8 4
2) 2 < a < 3, 2 < log2x < 3,
Tengsizlikning yechimi
(1 1\
va (4; 8).
{8 4 J '
Tengsizlikning yechimibo'lgan eng
kichik natural son 5.
Javob: 5.
7. sin3x + (1 - sinx) cos2x =
= 2tg— + 2tg3— + 2tgs * — + 2tg7 8 — + ...
a8 8 8 8
tenglama [0; 2л] kesmada nechta
yechimga ega?
Yechish:
1) 2tg^ + 2tg3^ + 2tgs^ + 2tg7^ + ...=
О О О O*
г\/х It j. 3 Я- x 5 i 7 % \
= 2(tg— + tg3 ~ + tg5- + tg7- +
О О О О
geometrik progressiya.
0<tg—<1
8
bi=tg~, b2=tg — ,q = tg -
ООО
. 7Г 7Г
. tg— 2tg
s= bi 8 8 _
2(i-tg2^\
о \ О J
7 . „ Л 1 t Л 1
=-tg2-=—tg- = -
2 8 2 4 2
я 1
2) sin x + (1 - sinx) cos2x = 2 —
2
sin3x + cos2x - sinxcos2x -1=0
sin3x + 1 - 2sin2x - sinx( 1 - 2sin2x) -
-1 = 0
sin3x - 2sin2x - sinx + 2sin3x = 0
3sin3x - 2sin2x - sinx = 0
sinx(3sin2x - 2sinx - 1) = 0
3) sinx = 0, x = тгп, [0; 2я]
х = лп
n = 0, x = 0
n = 1, X-Tt
n = 2, x = 2n
4) 3sin2x - 2sinx -1 = 0
sinx = 1, sinx = —
3
a) sinx = 1, x = — + 2лп
2
n = 0 da x=^ C[0; 2л]
b) sinx =~,x = (~1)пЧarcsin— + лп
1
n = 0 da x= -arcsin — C [0; 2л]
1
n = 1 da x = arcsin — + лС [0; 2л]
1
n =2 da x = -arcsin — + 2л G [0; 2л]
Demak, tenglamaning yechimlari x = 0,
~ л . 1 .
x = л, x = 2л, x =—, x = arcsin — + лп,
2 3
1
x = -arcsin — + 2л
3
Tenglama 6 ta yechimga ega.
Javob: 6.
96
Yechimlar. Matematika va informatika 2017
11-variant
8. Qarang: 4-variant 25-savol
(38-bet).
9. Sayyoh moljallangan yo'lning
67% ni o'tgach hisoblab ko'rsa,
yo'lning to'rtdan uch qismini o'tish
uchun yana 9 km yurishi kerak ekan.
Moljallangan yo'l necha km bo'lgan?
Yechish:
x-yo‘l
67% x = 0,67x bosib olilgan yo‘l.
3x
—— = 0,67x + 9
4
3x = 2,68x + 36
0,32x = 36
x=^2=112,5
32
Moljallangan yo‘1112,5 km.
Javob: 112,5.
10. /(x) = 16sin3xsin5x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko'paytmadan yig'indiga otamiz.
sin3xsin5x (cos(3x - 5x)) -
1
- cos(3x + 5x)) =— (cos2x - cos8x)
1
2) f(x) = 16 — (cos2x - cos8x) =
= 8cos2x - 8cos8x
3) F(x) = 4sin2x - sin8x + C.
Javob: 9sin4x - 2sin18x + C.
11. Qarang: 6-variant 1-savol
(51-bet).
12. (x2 + 23x + 23ХХ2 + x + 23) = 23Х2
tenglama haqiqiy ildizlari yig'indisini
toping.
Yechish:
x2 + 23 = a belgilash kiritamiz.
(x2 + 23 + 23x>(/ + 23 + x) = 23/
(a + 23x)(a + x) = 23/
аг + 24xa + 23X2 = 23X2
a2 + 24xa = 0, a(a + 24x) = 0
a = 0, a - -24x
x2 + 23 = 0 tenglama haqiqiy ildizga
ega emas.
x2 + 23 =-24x
x2 + 24x + 23 = 0 tenglama ildizlari
xi = -23, x2=—1.
Xi + X2- -24.
Javob: 24.
13. Qarang: 4-variant 6-savol
(33-bet).
14. [50; 150] kesmada 3 ga
bo'linganda qoldiq 1 ga, 4 ga
bo'linganda qoldiq 2 ga, 5 ga
bo'linganda qoldiq 3 ga va 6 ga
bo'linganda qoldiq 4 ga teng
bo'ladigan natural sonlar nechta?
Yechish:
[50; 150], a - son
a = 3n + 1,a + 2 = 3(n + 1)
a = 4k +2, a + 2 = 4(k+1)
a = 5m + 3, a + 2 = 5(m + 1)
a = 6c + 4, a + 2 = 6(c + 1)
a + 2 soni 3, 4, 5, 6 sonlariga karrali son.
3, 4, 5, 6 sonlarining EKUKni topamiz.
3-4-5 = 60, demak, a + 2 = 60, bundan
a = 58 ekanligi kelib chiqadi.
[50; 150] kesmada 58 ga karrali son,
58 va 116.
2 ta.
Javob: 2.
15. Ifodaning eng kichik qiymatini
1 2
toping: — cos4a + sin 2a.
Yechish:
.. . 2_ 1-cos4a 1 1 .
1) Sin 2a =-------=----cos 4a
2 2 2
1 11
2) — cos4a + — - — cos4a =
8 2 2
1 3
= — cos4a
2 8
3) -1 < cos4a < 1
97
Yechimlar. Matematika va informatika 2017
11-variant
4) у =— cos4a
2 8
_1 _3 ._1 _3_1
Ymin 2 8 ' 2 8 8
_ 1 3 , „,1 3 _7
Утах 2 8 2 8 8
Ifodaning eng kichik qiymati
-= 0,125.
8
Javob: 0,125.
16. Qarang: 1 -variant 2-savol
(3-bet).
17. Teng yonli uchburchakning
asosiga va yon tomoniga tushirilgan
balandliklari 10 va 12 sm bolsa, asos
uzunligini toping.
Yechish:
ha = 10, hb=12,a = ?
uchun AOi:OiO2 = О2Оз:ОзВ tenglik
o'rinli. S2 soha yuzini ifodalaydigan
son shu soha perimetrini ifodalaydigan
sondan 50% ga katta bo'lsa, katta
aylana uzunligini toping.
Yechish:
AOi:OiC>2 = O2O3.O3B
AB - diametr, AO1 = O1B = R
O1O2 ~ O1O3 ~ r
- = — , R = 2r
r r
_ лР2 лг2 лгг ttR2
2 2 2 + 2 2
p =^. + ^ + ?^- = 7t(R + 2r)=27[R
2 2 2 2
S2 - 150%
P2 - 100%
2S2 = 3P2, 2- — = 3-2ttR
2
R = 6
C = 2xR = 26n = 12n.
Javob: 12л.
19. Aylana ABCD trapetsiyaning B,
C va D uchlaridan o'tib AB tomoniga
В nuqtada urinadi. Asoslari a va b
bo'lsa, BD diagonalini toping.
a-10 = 12-b, a = 1,2b
Ла2 = б2-[|] ^Ь2-(0,6)2 =
= b2 - 0,36b2 = 0,64b2
ha - 0,8b, b =-^- = —h = —-10 = 12,5
0,8 4 4
. 6 6 25 ...
a = 1,2b =--12,5=------= 15.
5 5 2
Javob: 15.
Rasmda AB katta aylana diametri,
Oi katta aylana markazi, O2 va O3
kichik aylana markazlari bo'lib, ular
Yechish:
BC = a, AD = b
1
1) ZABD=-BED
2
Z ABD = Z BCD
2) Z ADB = Z DBC chunki BC\\AD,
BD - kesuvchi
3) AABD va ADCB o'xshash.
BD2 = AD BC
BD BC
BD =JAD-BC, BD =4adj .
Javob: yja-b .
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11-variant
20. Diagonallarining soni tomonlari
sonidan 2 marta ko'p bolgan
ko'pburchakning ichki burchaklari
yig'indisi topilsin.
Yechish:
D - diagonallar soni, n - tomonlar soni.
D = 2n
D_n(n-3)
2
_ n(n-3) „ . ,
2n = —------n-3 = 4, n = 7
2
Ichki burchaklari yig'indisi:
180°(7—2) = 180°-5 = 900°.
Javob: 900°.
21. к ning qanday qiymatlarida
1
----= 1 - к tenglama manfiy
x + 1
yechimga ega?
Yechish:
— = 1-k
x + 1
1 = x + 1 - kx- к
x(1 -k) = k
к
x =----tenglama manfiy yechimga
1-k
fa
ega bolishi uchun---< 0 bolishi kerak.
к к
< 0 yoki ——>0
1-k к-1
к С (-оо; 0) U (1; со) yoki к <0, к> 1.
Javob: к < 0, к > 1.
n „ . x-9x
cosx - cos9x - -2sin —--
2
. x + 9x . c
sin------ = 2sm4xsm5x
2
3) y' = 9-2sin4x sin5x =
= 18sin4xsin5x.
Javob: 18sin4xsin5x.
23. Rasmda у = a funksiya
x + c
grafigi tasviriangan. Quyidagilardan
qaysi biri doim o'rinli?
Yechish:
b
у = a +----
x + c
a > 0, b < 0, c> 0 bolgani uchun
cb-a <0 doimo to'g'ri.
Javob: C.
24. i +2----~ < о tenasizlikni
x-4
yeching.
Yechish:
Aniqlanish sohasi:
22. у = 9sinx - sin9x funksiyaning
hosilasini toping.
Yechish:
1) (sin(ax + b))’ = acos(ax + b)
(sinx)' = cosx, (sin9x)'= 9cos9x
y' = 9(sinx)' - (sin9x)' = 9cosx - 9cos9x -
= 9(cosx - cos9x)
2) cosx - cos9x ayirmadan
ko'paytmaga otamiz.
x-4 + y/3x-2 n . . ... .. .
-------------< 0 tengsizhk quyidagi
x-4
tengsizliklar sistemasiga teng kuchli.
y. ix-4 + y[3x-2 <0
[x-4>0
_. fx - 4 + J3x-2 > 0
\x-4<0
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Yechimlar. Matematika va informatika 2017
11-variant
. \^3x-2 <4-x
7. 1
|x-4>0
4 - x > 0 bo'lganda 1-tengsizlik
yechimga ega bo'ladi.
[4-x>0
< yechimga ega emas.
[x-4>0
\-j3x-2 >4-x
|x-4<0
\3x~2>16-8x + x2
|x<4
\x2-11x + 18<0 i2<x<9
(x< 4 [x <4
=> 2 < x <4.
Javob: 2 < x < 4.
25. Hisoblang: (4>/3 + 8)
.ъ 3-2^3 V3-2 ч
.(73(73-2) + -^- + -^-- + ...):
:(73 + 1).
Yechish:
bi =4з(4з-2)
, 3 - 2J3
b2-ir
qJ^^:j3(j3-2} =
b,. J3
з-2\[з
~ Тз-Тз-сТз-Л)""
43{4з~2) 1
4з-4з(4з-2) 4з
Cheksiz kamayuvchi geometrik
progressiya, chunki q <1.
4з(4з -2) _з(4з -2)
V3
2) (4j3 + 8)3^-~2):(43+Г) =
= 4(75 + 2).«^> ’
3-1
= 6(7з + 2)(43 - 2) = 6(3 -4) = -6.
Javob: -6.
26. Poyezd 4 minutda 8 kilometr
masofani, motosikl 6 minutda
8 kilometr masofani bosib o'tdi.
Motosiklchining tezligi poyezd
tezligining necha foizini tashkil etadi?
Yechish:
1 S
4 minut = — soat. v = — . Poyezd
15 t
4 minutda 8 km.
„ ,c km km . ...
v1 = 8-15---= 120------ motosikl
soat soat
6 minutda 8 km.
1
6 minut =— soat.
10
D2 -,840-km- =80-km
soat soat
120- 100%
80-x
80-100 200 .c2
120 3 3
Motosiklchining tezligi poyezd
2
tezligining 66 — % ini tashkil etadi.
3
2
Javob: 66-%.
3
27. Murakkab n sonining 1 dan
katta eng kichik bo'luvchisi m bo'lsin.
U holda:
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Yechimlar. Matematika va informatika 2017
11-variant
Yechish:
n - murakkab son. Masalan, n = 6
bolsin. 6 ning natural boluvchilari 1,
2, 3, 6. 1 dan katta eng kichik
bo'luvchisi m = 2.
m <4n , chunki2 <^6 .
Yoki n = 4 bo‘lsa, 4 ning natural
boluvchilari 1, 2, 4. 1 dan katta eng
kichik bo'luvchisi m = 2.
m =Jn chunki 2=^4
Demak, m<4n .
Javob: m Jn .
28. Qarang: 3-variant 20-savol
(27-bet).
29. Qarang: 2-variant 13-savol
(16-bet).
30. 210-28 + 26-24 + 2z- 1
ifodani 9 ga bo'lgandagi qoldiqni
toping.
Yechish:
210 _ 2® + 2® - 24 + 22 - 1 =
= 2®(22 - 1) + 2V22 -1)+(2г-1) =
= (?- 1)(28 + z + 1) =
= (4- 1)(256 + 16 + 1) = 3-273 =
= 3-3-91=9-91
Demak, ifoda 9 ga qoldiqsiz bo'linadi.
Qoldiq r = 0 bo'ladi.
Javob: 0.
31. Tashqi qurilmalarni boshqarish elektron sxemalar- bu:
Yechish:
drayverlar yordamida tashqi qurilmalarni boshqarish mumkin. Odatda
drayverlar Operatsion tizim bilan birgalikda o'rnatiladi. Lekin ba’zi qurilmalar
uchun (printer, videokarta) uchun maxsus drayverlar taqdim qilinadi.
Javob: drayverlar.
32. A1 = -4, B1 = 7, B2 = 3 bolsin. Quyidagi formula natijasi -18 ga teng
bolishi uchun A2 katakka kiritilishi kerak bo'lgan qiymatni aniqlang.
=ЕСЛИ(И(А1 +B2<A2*B1 ;A1*A2<0);A1*B2-15+A2;A1*B1 +5-A2).
Yechish:
A1 = -4; B1 = 7; B2 = 3.A2 = ?
=ЕСЛИ(И(А 1 +B2<A2*B1;A 1 *A2<0);A 1 *B2-15+A2;A 1 *B 1+5-A2).
Bu yerda ЕСЛИ mantiqiy funksiya.
ЕСЛИ(И(А;В);С;В).
Bu yerda A va В mantiqiy ifodalar CvaD- matematik amallar bo'lib xizmat
qiladi.
Agar A va В mantiqiy amallar bir vaqtning o'zida bajarilsa, и holda C ifoda
bajariladi. Aks holda, agar A va В mantiqiy ifodalar bajarilmasa, D ifoda
bajariladi.
Bizning misolimizda:
=ЕСЛИ(И(-4+3<А2*7;(-4)*7;(-4)*А2<0);-4*3-15+А2;-4*7+5-А2);
Formula natijasi-18 bolishi uchun
-4-3 - 15+A2 = —18 yoki-4- 7 + 5-A2= -18.
-12-15 + A2 = -18 yoki-28 + 5-A2 =-18.
A2 =-18 + 27 = 9 yokiA2 = 23-28= -5.
Demak, A2 = 9 yoki A2 = -5 bolishi mumkin. Lekin A2 musbat bolishi kerak.
A2 = 9.
Javob: 9.
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Yechimlar. Matematika va informatika 2017
12-variant
33. 5 lik sanoq sistemasida 3+4 nechaga teng?
Yechish:
1-usul: 5 lik sanoq sistemasida qo'shish jadvali. 3(s> + 4(s> = 12(5)-
+ 0 1
0 0 1
1 1 2
2 2 3
3 3 4
4~ 4 ~TU
2 4
2 4
3 4 10
4 ip 11
10 11 12
-44- -42 13
2-usul: Awal qo'shiluvchi sonlarni 10 lik sanoq sistemaga olkazib, ularni
qo'shib, keyin 5 lik sanoq sistemaga o'tkazamiz.
4(5) -* 4(ю)
3(5) 3(10)
3+4=7w
Xs
7 io - 12(5).
Javob: 12.
34. Qanday dastur Operatsion sistema (tizim)ni faoilashtiradi?
Yechish:
BIOS - mikroprogrammalar jamlanmasi bo'lib, bular yordamida tizimni
blokning ayrim tashkiliy qismlarini sozlashga va operatsion tizimni
faollashtirishga xizmat qiladi.
Javob: BIOS.
35. MS Excel 2003 dasturida berilgan =ЗНАК(0)+МИН(15;16;17)
formulaning natijasini aniqlang.
Yechish:
3HAK(0) 0 chunki qavs ichida turgan son 0 ga teng.
МИН(15;16;17) = 15.
15 + 0 = 15.
Javob: 15.
36. Qarang: 3-variant 36-savol (31-bet).
12-variant
1. Qarang: 3-variant 7-savol
(23-bet).
2. Trapetsiyaning parallel tomonlari
a va b ga teng. Shu tomonlarga
parallel va trapetsiya yuzini teng ikkiga
bolgan kesmaning uzunligini toping.
Yechish:
a, b - asoslari.
Si=S2=~. EK=?
2
ABCD trapetsiya va EBCK
trapetsiyalar o'xshash
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Yechimlar. Matematika va informatika 2017
12-variant
c + b a + c
1) Si - s2, =-y Л
h, = ^-h2
1 c + b 2
2) S = S, + S2
a+b , c+b . a+c .
---h =------h. +----h2
2 2 1 2 2
(a + b)(hi + h2) = (c + b)hi +(a + c)h2
hi(a + b - c - b) = (a + c - a - b)h2
hi(a -c) - (c- b)h2
. -c~b +
h1 h2
a-c
a+c c-b
3) ~-h2 =---h2
c + b a-c
2 Z Z к2
a — с = c — b
„2 2 ,2 2 a2 +b2
2c = a + b , c2 =---
2
a2+b2
c =.-----.
N 2
Javob;
\ 2
3. Rasmda berilgan yarim doirada
shtrixlangan soha yuzi Sadc ning
qiymati topilsin. Bu yerda
ДО = ОВ = 2, AD=DC=BC.
о в
Yechish:
AD = DC = BC = 60° bo'lganligi uchun
&ADC da
ZA = 30°, Z C = 30°, ZD= 120°
AO = ОС = AD = DC = 2
c AD • DC *
SAOC =--------Sm120 =
2 2
4. Hisoblang:
Iog2532-log27625logi2881.
Yechish:
log „ bk = — log b dan foydalanib
3 n
yechamiz.
log26 32 log27 625 log,28 81 =
= log62 25-log33 54 log2?34 =
= ;|jyloe521oe351og23 =
_40 lg2 lg5 \g3 = 40 19
~ 21lg5 lg3 lg2~ 21 21
, к J9
Javob: 1—.
21
5. Daraxtdagi beshta shoxning har
birida qo'nib turgan qushlar soni
baravar. Har bir shoxdan 2 tadan qush
uchib ketganda, ilgari 3 ta shoxda
nechta qush bo'lsa, hamma shoxda
shuncha qush qoldi. Har bir shoxda
nechtadan qush bo'lgan?
Yechish:
x - qushlar soni. 5 ta shohda 5x ta qush.
5(x-2) = 3x, 5x- 10 = 3x
2x=10,x = 5
Har bir shohda 5 tadan qush bo'lgan.
Javob: 5 tadan.
®sin(Inx) , ,,
6. I—i---^dx hisoblang.
1 x
Yechish:
?sin(lnx) ,
|—-------dx =
1 x
e G
= J sin(ln x)d(ln x)=- cos(ln x) 1 =
= -cos(lne) + cos(lnl) = -cos 1 + cosO =
= -cos1 + 1 = 1- cos1.
Javob: 1 - cos1.
7. Qarang: 7-variant 6-savol
(62-bet).
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12-variant
8. Qarang: 5-variant 2-savol
(42-bet).
9. Qarang: 4-variant 2-savol
(32-bet).
10. 0,2 ^+4-5= 5^5 tenglamani
yeching.
Yechish:
Bir xil asosga keltirib yechamiz.
0,2=54 5^5 = 51-52 =5U~2 = 51‘5
^-(xZ+7x+4,5) = ^1,5
-fx2 +7x + 4,5) = 1,5
4 + 7x + 4,5 = -1,5
x2 + 7x +6 = 0
x = -1, x = -6.
Javob: -1 va -6.
11. 0‘zaro teng bo'lmagan x va у
sonlari x2 - 26x = y2 - 26y tenglikni
qanoatlantirsa, x + у ni toping.
Yechish:
x # y,
x2 -26x = y2- 26y
x2 - 26x-y2 + 26y = 0
x2 -y2 - 26(x -y) = 0
(x-y)(x + y) - 26(x-y) = 0
(x-y)(x + y-26) = 0
1) x-y = 0, x = у masala shartida x # y.
2) x + y- 26 = 0, x + у = 26.
Javob: 26.
12. a va b natural sonlarning
umumiy •bo'luvchilari soni 3 ga teng
bo'lsa, a + 3b va b sonlarning umumiy
bo'luvchilari nechta?
Yechish:
a va b ning umumiy bo'luvchilari soni
3 ga teng. Tanlash yo'li bilan yechamiz.
Masalan: a = 22, b = 22-3 bu
sonlarning umumiy bo'luvchilari soni 3
bo'lganda
EKUB(a; b) =2*, т =2+1. U holda
a + 3b =4 + 36 = 40,
a=4,b=12.
EKUB(3a + b;b)= 4 = 2?, bu
sonlarning umumiy bo'luvchilari soni
т = 2+1=3да teng. Yoki a = 32-5,
b = 32 sonlar uchun ham xuddi
shunday bo'ladi.
Javob: 3.
13. AB kesma К aylananing
diametri bo'lsin. L aylana К aylanaga
hamda AB to'g'ri chiziqqa К
aylananing markazida urinadi; M
aylana К va L aylanaga hamda AB
to'g'ri chiziqqa urinadi (chizmaga
qarang). Agar M aylana radiusi 2,5 ga
teng bo'lsa, L aylana radiusini toping.
Yechish:
r = 2,5 M- aylana radiusi, L - aylana
radiusi - R.
D AB AK KB
F\ =s- =---=-----
4 2 2
OK=R
O,D = CK = r
KO-i = 2R-r, OOi = R + r
OiC = x, OC = R-r
OiOC va OiKC uchburchaklardan:
((R + 2,5)2 =(R-2,5)2+x2
[(2R-2,5)2 = 2,52 + x2
bundan
R = 5.
Javob: 5.
14. Qarang: 2-variant 19-savoi
(17-bet).
15. a = 9,75 bo'lsa, л/а + бл/а^Э +
+4a- 6Va-9 ifodaning qiymatini
toping.
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Yechimlar. Matematika va informatika 2017 12-variant
Yechish: Uchburchakning aylanadan
1) ^a-6ja^9 = =: -Jз — 9 — 6 J 3 — 9 + 9 = = ^>/a^9y-2ja^9-3 + 9 = = ^(Va-9-З)2 =|Va-9-3| tashqaridagi qismining yuzini toping. Yechish: c мЛ8к / / A, \ I / \ I
2) a = 9,75 da | ja^9 - 3| = = З-л/а-9 да teng bo'ladi. 3) 3+Ja-9 + 3-y/a-9 = 6. Javob:6. nV A D \_jMl A в . 1) ABC - muntazam uchburchak. OE=~,AB=~ 3 3
16. Qarang: 2-variant 3-savol (13-bet). 2) ODE - to'g'ri burchakli uchburchakda OD - katet.
17. Qarang: 5-variant 8-savol (43-bet). OD ichki chizilgan aylana radiusi. OD = ^- = OE^~
18. 0,2cosZx- 25~™!x < 4 (125Г°'5 tengsizlikni yeching. Yechish: 6 2 Bundan Z DEO = 60° Xuddi shunday Z KFO = 60°. 3) Z EBF = 60°,
CM w о 2 ®. ii Й r-|<N ? Л Г " V <4 Й £ ii lo| 8 сл *l XJ-i v § > II + * I ю ”, VJ 1 s. II ION T- V x Sig й~ X й % » » О 8 • - о о cn~ FT o' o' EO || BF, OF || BE, OEBF - romb. OE = EB = BF = OF =- 3 4) Z EOF = 60°, EOF sektoryuzi _ Tt-OE2-60’ _7t ГаУ _ a2 seklor~ 360° 6 ^3j 54 mm AJ 9 2 18
~— + 2лп <2х<~ + 2тсп 3 3 7C 7Г ™ + 7tn <x < — + 7tn,n eZ . 6 6 S — 3‘(Sromb ~ Ssektor) — _ „(a243 a2A _ „ 2 -7r) _ »5 — s?a — 18 54) 54
Javob: + тгп: — + тип . n e Z . 1 6 6 ) _а2(343-л) 18
19. Muntazam uchburchakning tomoni a. lining markazidan^ radius bilan aylana ichki chizilgan. . . а2(ЗУЗ-тг) Javob: —i -. 18 20. Qarang: 6-variant 14-savol (54-bet).
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12-variant
21. Rasmda у = a +—— funksiya
x + c
grafigi tasvirlangan. Quyidagilardan
qaysi bin noto‘g‘ri?
Yechish:
у = a + —grafikdan a <0, b <0,
x + c
c> 0 ni topamiz.
A) b5 -a4 <0- to'g'ri
B) a + be2 <0- to'g'ri
C) ab + c> 0- to'g'ri
D) a3-b3 <0-noto'g'ri, chunkia <0,
b <0 bo'lganda a3br > 0 bo'lishi kerak
edi.
Javob: a3b3 < 0 - noto'g'ri, chunki
a < 0, b < 0 bo'lganda a3b3 > 0 bo'lishi
kerak edi.
„ sin10a + sin6a + sin2a
22. ----------------------ni
cos 10a + cos 6a + cos 2a
soddalashtiring.
Yechish:
.. . .-on- « + P а-В
1) sma + sin fl = 2sm—— cos—^-
sin 10a + sin 2a =
„ . 10a + 2a 10a —2a
= 2sin——— cos---------=
2 2
= 2sin6acos4a
a-fl
2) cosa + cosfl = 2cos cos
2
cos10a + cos2a =
r. 10a + 2a 10a-2a
= 2cos------cos---------=
2 2
- 2cos6acos4a
sin 10a + sin 2a + sin 6a _
cos 10a + cos 2a + cos 6a
_ 2 sin 6a cos 4a + sin 6a _
2 cos 6a cos 4a + cos 6a
_ sin6a(2cos4a + 7) _ sin 6a _ ^Qa
cos 6a(2 cos 4a+ 1) cos 6a
Javob: tg6a.
23. 2arctg(x2 - 5x + 7)
tenglamani yeching.
Yechish:
1) arctgfx2 - 5x+ 7) = —
4
2) tgarctg(x2 - 5x + 7) = tg~ , tg?- = 1
4 4
хг-5х+7=1, хг-5х + 6 = 0
x = 2, x = 3.
Javob: 2 va 3.
24.
1 +
a b 2 bo'lsa, b-?
2 3 = 4
_a + b~ 5
Yechish:
Birinchi tenglamani 2 ga ko'paytiramiz.
2 8 2
_ a +b 2
2 1_4
,a + b~ 5
8 3 4
-----= 1 —
b b 5
5 1 .
— = —, b 25.
b 5
Javob: 25.
25. |DP| = |PB|, |PA| = 8 sm,
|PC| = 4,5 sm bo'lsa, |AB| ni toping.
Yechish:
|DP| = |PB|, |R4| = 8, |PC| = 4,5, |ДВ| = ?
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Yechimlar. Matematika va informatika 2017
12-variant
AC va DB kesishuvchi vatarlar.
APPC = DP-PB
8-4,5 = (DP)2, DP = 6
ДАРВ to'g'ri burchakli
AB2 ~ АР2 + PB2
AB2 = в2 + в2 = 102, AB = 10.
Javob: 10.
26. Yo'lovchi birinchi soatda yo'lning
— qismini, ikkinchi soatda qolgan
5
1
yo'lning —qismini, uchinchi soatda esa
3
qolgan yo'lning yarmini yurgach,
manzilgacha 4 km masofa qoldi.
Yo'lovchi jami qancha (km) yo'l yurishi
kerak?
Yechish:
x-km yo'l
1-soatda — km, x = —qolgan yo‘l
5 5 5
„ 4x 4x
2-soatda----= —,
5-3 15
4x 4x 8x . „,
-------= — qolgan yo lda
5 15 15
_ , , 8x 4x
3-soatda----= —
15-2 15
x 4x 4x .
--1----)----4 — )(
5 15 15
3x+4x+4x .
------------+ 4 = x
15
11x . 15x-11x .
x —— = 4 , --------= 4 , x = 15.
15 15
Javob: 15.
Z О = 60°, AB = a, AC = b,OA-?
a+p = 60°
AB . o AC
sina =--, smp =----
AO AO
sina AB . . .
-----=--, bsma = asmp
sin p AC
bsina = asin(60° - a)
. . fV3 1 . "l
bsina = a- -cosa---sina
2 2
fa V3
sina — + b = — acosa
{2)2
. -J3a
tga =——,
a + 2b
1
1 + tg2a= .
1 -sm a
1 - sin2a =-1—r-
1+ tg2a
. 2 4 1 tg2a
Sin a = 1 -— =—^—z-
1 + tg a 1 + tga
tga 43а
sina = -. = -;======= =
s/1 + tg2a yj(a + 2b)2 + 3a2
__ 43a _ \[3a
4 4a2 +4ab + 4b2 2^a2 + ab + b2
AB 2a>ja2 +ab + b2
AO =----= ——f=--------=
sina V3a
= —у)3(а2 +ab + b2).
3
Javob: ^3(a2 +ab + b2).
27. 60° li burchak ichida uning
tomonlaridan a va b masofada turgan
nuqtadan burchakning uchigacha
bo'lgan masofani toping.
Yechish:
28. arcsin2x + arccos2x =^—
32
tenglamani yeching.
Yechish:
arcsinx + arccosx = — bundan
2
arccosx =—- arcsinx
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Yechimlar. Matematika va informatika 2017
12-variant
arcsin2x + (— ~ arcsinx)2 =^—
2 32
arcsinx = у
/(л A 2 5л2
[2 ) 32
2 л2 2 5л2
\Г +----лу + / =---
7 4 32
« 2 .Л 5л
2у - лу +— ---= О
4 32
2У2 -лу + = О kvadrat tenglama
ildizlarini topamiz.
. 3л (л Зл\ Л
sin — = cos-----= cos—
8 (2 8 J 8
. л ( л л} 3л
sin — = cos-= cos--.
8 [28) 8
. , л Зтг
Javob: cos —, cos—.
8 8
29. Qarang: 2-variant 7-savol
(14-bet).
30. {x|x C N, x2 < 23} to'plamning
nechta qism-to‘plamlari mayjud?
4 4 4
1) У-— da arcsinx =—, x = sin —
' 8 8 8
2) y=—da arcsinx =—, x- sin —
y 8 8 8
Yechish:
x2 <23, 1 <x<423 ,x€N
A={1, 2, 3, 4}, A to'plam 4 ta
elementdan iborat.
A to'plamning qism to'plamlarini tuzamiz
1.0c A
3. {1} c A
5. {3} c A
7. {1; 2} c A
9. {1; 4} c A
11. {2; 4} c A
13. {1; 2; 3}c A
15. {1; 3; 4} c A
A to'plamning 16
Javob: 16.
2. A c A
4. {2} c A
6. {4} c A
8. {1; 3} c A
10. {2; 3} c A
12. {3; 4} c A
14. {1; 2; 4} c A
16. {2; 3; 4} c A
qism to'plami bor.
31. Qarang: 5-variant 31-savol (49-bet).
32. Quyidagi keltirilgan misollardan qaysi biri uzluksiz axborot bo'la oladi?
Yechish:
Uzluksiz axborot turi deganda yuqoridan chegaralanmagan axborot
tushuniladi.
telekursatuv - chegarasi bor
vaqt - chegaralanmagan
harorat - chegaralanmagan
dars - chegarasi bor
yozuv - chegarasi bor
ma’ruza - chegarasi bor.
Javob: vaqt, harorat.
33. Qarang: 6-variant 31-savol (60-bet).
34. Paskal dasturlash tilida berilgan ushbu ifodaning qiymatini toping.
trunc(sqrt(sqr(abs(trunc(4,3)-sqrt(100)*round(1,6))))).
108
Yechimlar. Matematika va informatika 2017
13-variant
Yechish:
trunc(sqrt(sqr(abs(trunc(4,3)-sqrt(100)*round(1,6)))))
round(1,6) = 2
SQRTflOO) = 4l00 = 10
trunc(4,3) = 4
abs(4 - 10-2) = abs(4 - 20) = abs(-16) = 16
SQR(16) = 161 2 * = 256
SQRT(256) = 16
trunc(16) = 16.
Javob: 16.
35. MS Excel 2003 dasturida berilgan =И(СТЕПЕНЬ(3;4)>80;
MAKC(15;10;30)<30) formulaning natijasini aniqlang.
Yechish:
Bu formulaning natijasi yolg'on (ложь).
Excel dasturida И funksiya mantiqiy funksiya bo'lib, qavs ichida turgan
mantiqiy ifodalarga bog'liq. И funksiyaning qiymati Ложь yoki Истина bo'ladi.
И funksiyaning sintaksisi quyidagicha:
H(mantiqiy ifoda;mantiqiy ifoda).
Agar har ikkala mantiqiy ifoda rost bo'lsa, и holda bu funksiya qiymati
ИСТИНА bo'ladi. Aks holda ЛОЖЬ bo'ladi. Bizning misolda СТЕПЕНЬ(3;4)>80,
ya’ni 34 > 80 bu rost.
MAKC(15;10;30)<30 ya’ni eng katta son 30 < 30 bu yolg'on.
Shuning uchun berilgan misol natijasi ЛОЖЬ;.
Javob: ЛОЖЬ.
36. Sistemaviy dasturiy ta’minot tarkibi:
Yechish:
Sistemaviy dasturiy ta’minot tarkibiga Operatsion sistema, antiviruslar,
arxivatortar, brauzerlar (tarmoq dasturlari) tashxiz dasturlari kiradi.
Javob: operatsion tizim, tarmoq operatsion tizim, tashxis dasturlari, antivirus
dasturlar, arxivatortar, tarmoq dasturlari.
13-variant
1. A(2; 4); B(3; 6); C(6; 14) nuqtalar
berilgan. |ab + AC| ni hisoblang.
Yechish:
A(2; 4), B(3; 6), C(6; 14),\AB + AC\ = ?
AC =(6-2; 14-4) = (4; 10)
AB = (3-2;6-4) = (1;2)
AB + AC = (5; 12)
| AB + AC| = 4б2+122 = 13.
Javob: 13.
2. Qarang: 3-variant 25-savol
(28-bet).
3. f(x) = 38sin9x cos10x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko'paytmadan yig'indiga otamiz.
1
sin9xcos10x = — (sin(9x + 10x) +
1
+ sin(9x- 10x)) =— (sin19x- sinx)
109
Yechimlar. Matematika va informatika 2017
13-variant
i
2) j(x) = 38- — (sin19x - sinx) =
= 19sin19x- 19sinx
3) F(x) = -cos19x + 19cosx + C =
= 19cosx- cos19x + C.
Javob: 9sin4x - 2sin18x + C.
4. j (x2 + Vx - 7)dx ni hisoblang.
Yechish:
j (x2 + Vx - 7)dx =
1 x3 x^'
= J(x2 +x3-7)dx = ~+ -------
, „ x3 3xl[x 7 „
-7x+C= — +-------7x + C.
3 4
Javob: — + - 7x + C .
3 4
5. f(x)= 2’
x - juft
x+1, x-toq
/(/(/(/(/(17))))) = ?
Yechish:
1) f(17)=x+1 = 17+1 = 18-juft
2)f(18)=X2=1-^ = 9-toq
3) f(9) = x + 1 = 9 + 1 = 10-juft
10
4)f(10)=^ = 5-toq
5) f(5) = x+1 = 5+1 =6.
Javob: 6.
6. Qarang: 5-variant 28-savol
(48-bet).
7. x va 84 sonlarining eng kichik
umumiy karralisi 336 ga, eng katta
umumiy bo'luvchisi esa 12 ga teng,
x ni toping.
Yechish:
x, 84 sonlari
EKUK(x; 84) = 336
EKUB(x; 84) = 12
x = ?
x-84 = EKUK(x; 84) EKUB(x; 84)
x-84 = 336-12
x = 48.
Javob: 48.
—4x
8. у = logi 5—— funksiyaning eng
x +9
katta qiymatini toping.
Yechish:
Hosila yordamida yechamiz.
7 -4x ) _ -4(x2 +9) + 4x-2x _
lx2+9 J x2+9
_ 4x2-9-4
x~2 + 9
4хг - 36 = 0,
x2 = 9, x = ±3
Aniqlanish sohasi -x > 0, x < 0,
bundan x = -3
y °g,,sl4(-3)2+9 J °S,'518
= log,.6| = log,5(1,5)-’=-7
Утах = — 1
Javob: -1.
9. Asosi tepaga qaratib qo'yilgan,
o‘q kesimi teng tomonli uchburchakdan
iborat kontis ichiga suv quyilgan va
unga radiusi г ga teng shar solingan.
Natijada suvning sathi sharga ust
tomondan urindi. Shar suvdan
olingandan keyingi suv sathining
balandligini toping.
Yechish:
ABC teng tomonli uchburchak.
ADB suv sathi. Doira uchburchakka
ichki chizilgan.
110
Yechimlar. Matematika va informatika 2017
13-variant
x2 -5x + 4 + x2 -4 <
x2-4 ~
(8-5x)(2x2 -5x) <
(x2-4)2 ~
x(2x-5)(5x-8)
(x-2)2 (x + 2)2
> 0 son o'qida
AD = R, ЛА=ЛВ=ЛС = 60°,
R = ODtg60° = rj3
H = CD = 3r
Idishdagi suvning hajmi ABC konus
hajmidan shar hajmining ayirmasiga teng.
V=-7tR?H--itr3 =
3 3
_ 1 ^2 4tcR3 5 r3
— тгЗг-Зг -----= —лг
3 3 3
Shar suvdan olinsa, suv NKsathgacha
pasayadi va NKC konusni to'ldiradi.
CF = h, NF = CFtg30° =JL
V3
V=^.-NF2CF =-(—] -h=~
з з (,7з; э
xh3 5тсг3 , 3rr=
---=-----, h - r<J15 .
9 3
Javob: rVl5.
suratning ildizlarini va maxrajning
uzilish nuqtalarini belgilaymiz va har bir
oraliqda kasrning ishorasini aniqlaymiz.
-2 0 1,6 2 2,5
x C [0; 1,6] U (2; 2,5]
Tengsizlik yechimi [0; 1,6] U (2; 2,5].
Bu oraliqda tub son yo‘q.
Javob: 0 ta.
11. 2logsin2xcos2x - 4 +
+ 51од^5х sin2x = 0 tenglamani yeching.
Yechish:
Aniqlanish sohasi:
cos x > 0 Д' Д _ — + 2дп < x < —ь 2лп 2 2
sin 2x > 0 тс тсП < X < — + тгП
cos X Ф 1 sin2x*1 2 X * 2тсп
тс ХФ — + ТСП 4
2logsin2xCOS2x - 4 + 5log^ 5* sin2x = 0
x2 - 5x + 4
x2 -4
< 1 tengsizlikni
nechta tub son qanoatlantirmaydi?
Yechish:
x2 -5x + 4
x2 -4
< 1 tengsizlikning ikki
qismini kvadratga kola rib, qisqa
ko‘paytirish formulasidan foydalanib
yechamiz.
C x2 -5x +4
I x2-4
x2 - 5x + 4-x2 +4
x2-4
4logsin2Xcosx -4+ — logcosxSin2x = 0
5
logsln2xcosx = a
4a-4 +—= 0
a
4a2-4a + 1 = 0
(2a -1)2 = 0,2a-1= 0,
1
2a = 1, a =-
2
, 1
iogsinixcosx =~
cosx - (sin2x)2, cos2x = sin2x
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Yechimlar. Matematika va informatika 2017
13-variant
coszx - sin2x - 0
cosx(cosx - 2sinx) = 0
cosx = 0 yechim emas.
cosx - 2sinx = 0
ctgx = 2, x = arcctg2 + 2m
Javob: arcctg2 + 2лп.
12. Qarang: 2-variant 17-savol
(16-bet).
13. Metall quyma tarkibida 18 kg rux,
36 kg mis, 6 kg qalay bor. Qalay quyma
tarkibiy qismining necha foizini tashkil
qiladi?
Yechish:
rux - 18 kg, mis - 36 kg, qalay - 6 kg.
Quyma = 18 +36+ 6 = 60 kg.
60 kg- 100%
6 kg-x
Javob: 10%.
14. a = -b, c = 2 bo'lsa,
c(a - b)3 + a(b - c)3 + b(c - a)3
c2(b-a) + a2(c-b) + b2(a-c)
ifodaning qiymatini toping.
Yechish:
a =—b
a = 1 desak, b=-1 boladi.
c = 2
c(a - b)3 + a(b - c)3 + b(c - a)3 _
c2(b-a) + a2(c-b) + b2(a-c)
2(1+ 1)3+1(-1-2)3-1(2-1)3
~ 22(-1-1) + 12(2 + 1) + (-1)2(1 - 2) "
_ 2-23+1(-3)3 _ 16-27-1 = -12 2
~ 22 (-2) + 3-1~ -8 + 3-1 ~ -6 ~
Javob: 2.
15. Akvariumning bo'yi 120 sm, eni
70 sm, balandligi 90 sm. Suv sathi
yuqoridan 10 sm pastda bo'lishi uchun
akvariumga necha litr suv quyish
kerak?
Yechish:
H = 90 sm. a = 70 sm, b = 120 sm
CiE = 10, CE = 80 sm, CE = H-i
V = ab-Hi = 80-70-120 sm3 =
= 672-103 sm3 = 672 litr.
Javob: 672.
16. Asosi 4 sm ga, balandligi 6 sm ga
teng bo'lgan teng yonli uchburchakning
yon tomonini diametr qilib yarim
aylana chizilgan. lining asos va yon
tomon bilan kesishish nuqtalari to'g'ri
chiziq bilan tutashtirilgan. Yarim
aylanada hosil bo'lgan ichki chizilgan
to'rtburchakning yuzini toping.
Yechish:
AC = 4, BD = 6
Sadeb - Sabc — Sore
_ _ACBD 6-4 .„
&ЛВС-------— = — = 12
1 1
Sqbc - — Sabc -12 - 6
2 2
Sdec'6 = CE:CB
BC va AC - kesuvchilar.
CE-CB = CD-CA
CD- CA
—-------
CB
o ~CE cCD-CA
Sr„, =6 = 6-----—
ec CB CB2
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Yechimlar. Matematika va informatika 2017
13-variant
22+62 4 0 5
SADeb=12-1,2=10,8.
Javob: 10,8.
17. Agar barcha x, у lar uchun
x3 + 4x2y + axy2 + 3xy - bxcy + 7xy2 +
+ dxy + y2 = x3 + y2 ayniyat bajarilsa,
|a + b + c|(c + d) ni toping, (c > 1)
Yechish:
Ayniyat bo'lganligi uchun o'xshash
darajalar oldidagi koeffitsiyentlardan
foydalanib yechamiz. c-2
x2y oldidagi koeffitsiyent
4 — b=0,b = 4
xy2 oldidagi koeffitsiyent
a + 7 = 0, a = —7
xy oldidagi koeffitsiyent
3 + d = 0, d = -3
a =-7, b = 4, c = 2, d =-3 da
\a + b + c|(c + d) = |—7 + 4 + 2\(2-3) =
= 1(-1)=-1.
Javob: -1.
18. Qarang: 1-variant 9-savol
(5-bet).
19. Rasmda у = f’(x) funksiya grafigi
tasvirlangan. у = f(x) funksiya grafigiga
xo = 0 nuqtasiga o'tkazilgan urinmaning
burchak koeffitsiyentini toping.
Yechish:
xo = 0, к = f'(x0)
У = f'(x) funksiya grafigi berilgan.
f'(0) = 0 bundan k = 0.
Javob: 0.
20. Vx-3 - Vx + 1 + 2 = 0
tenglamaning ildizlari yig'indisini
toping.
Yechish:
Aniqlanish sohasi:
(x-3>0 fx>3
=> J =>x>3
[x + 1>0 [x>-1
yJx-3 = -Jx + 1 -2 tenglikning ikkala
qismini kvadratga kotaramiz.
x - 3 = x + 1 -4yjx + 1 + 4
4 Vx + 1 = 8, -Jx + 1 = 2,
x + 1 = 4, x = 3
Tenglama ildizlariyig'indisi 3 ga teng.
Javob: 3.
21. Radiusi R bo'lgan doiraga bir
burchagi 120° bo'lgan teng yonli
uchburchak tashqi chizilgan. lining
asosini toping.
Yechish:
Z ABC = 120°, AB = BC
ABD uchburchakda
лв_ BD _2R(2 + 43)
sin 30° у/з
AD = BDtg60° =
= y/3R(2+r^ = R(2+y/3)
J3
AC = 2R(J3 + 2).
Javob: 2R(V3 + 2).
22. Piramidaning asosi to'g'ri
to'rtburchak, uning diagonallari orasidag
o'tkir burchagi a ga teng. Piramidaninc
yon qirralari asos tekisligi bilan
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Yechimlar. Matematika va informatika 2017
13-variant
<p burchak hosii qiladi. Shu piramidaga
tashqi chizilgan shaming radiusi R.
Piramida hajmini toping.
Yechish:
EABCD - to'rtburchakli piramida.
E
Z EAC = <p, Z AOiB = a
AC = 2Rsin(180° - 2<p)
AC = 2Rsin<p, AOi = Rsin2<p
AAEOi dan piramida balandligini
topamiz.
EO1 = H = AO1
tg<p = Rsin2<ptg<p
V = ^SasosH
o ACBD .
Sasos =-----=----SU1« =
2
(2Rsin2«>)2 . 2n
= i-----—— sina = 2Fcsin 2<psina
2
V =-| R?sin32<ptg<psina.
2 О О
Javob: — R sin 2<ptg<psina.
23. Tenglama ildizlari yig'indisini
, . . (x2 x'l n
toping: arcsin---= —.
I.v 3 2) 6
Yechish:
1) Aniqlanish sohasi:
-f<—^-<1,-6<2x2-3x<6
3 2
2x2-3x-6<0
[2x2-3x + 6>0
. (x x^ n
2) arcs/n-----= —
V3 2) 6
. (x2 x^l . ir
sinarcsin------= sin —
{3 2) 6
. it 1
sin — = —
6 2
^--- = 1, 2хг-3х = 3
3 2 2
2X2 -3x- 3 = 0, D>0 bolganiigi
3
uchun ildizlariyig'indisiXi + x2 =— = 1,5.
Javob: 1,5.
24. Qarang: 4-variant 26-savol
(38-bet).
25. Qarang: 5-variant 17-savol
(45-bet).
26. Qarang: 5-variant 5-savol
(43-bet).
27. Qarang: 8-variant 13-savot
(73-bet).
28. Qarang: 3-variant 27-savol
(30-bet).
29. Tengsizlikni yeching:
logzCx2 - 6x + 8) - log (x - 1) > 0.
Yechish:
Aniqlanish sohasi:
$x2-6x + 8>0 f(,x-2\x-4)>0
|x-1>0 =>tx>1
fx<2, x>4
|х>1
1 <x <2, x> 4
log2(>? - 6x + 8) - log (x - 1) s 0
logzfx2 ~6x + 8)> 2log2(x- 1)
114
Yechimlar. Matematika va informatika 2017
13-variam
/oc^fx2 - 6x + 8) > /од2(х - 1)2
x — 6x + 8 > x~ 2x+ 1
-4x>-7,x<-
4
(1;1,75].
Javob: (1; 1,75].
30. To‘g‘ri burchakli trapetsiyaning
balandligi h, uning asosiga
perpendikulyar bolmagan tomonini
diametr qilib chizilgan aylana
trapetsiyaning qarama-qarshi tomoniga
urinadi. Katetlari trapetsiyaning asoslari
bo'lgan to'g'ri burchakli
uchburchakning yuzini toping.
Javob: —.
8
31. Qarang: 3-variant 31 -savol (30-bet).
32. A1 = -7, A2 = -1, B1 = 8, B2 = 2 bolsin. Natijasi -7 ga teng bo'ladigan
formulani aniqlang.
Yechish:
A2 = -1;A1 = —7; B1 = 9; B2 = 4.
Natija = 7.
Bu misolda A2, A1, B1, B2 yacheykalarga qiymatlarini joylashtirsak -7 javob
chiqishi lozim.
МИН funksiyasi eng kichik sonni beradi.
-МИН(-7+(-1));8+2.; МИН(-8;16)—8
-ABS(-7*9-(-1)*4)-^> -ABS(-63+4) => =ABS(~59)=59.
=ЦЕЛОЕ((-7+(-1))/(4-9)) ЦЕЛОЕ(-8/-5).
ЦЕЛОЕ funksiyasi sonning butun qismini qaytaradi.
=ЕСЛИ(- 7+4<>(-1)-9; 7*(-1);21 +4)
<> belgi teng emas belgisi
-7 + 40-I-9
—3 <> -8, Bu ifoda rost, shuning uchun javobi 7 (-1) --7.
Javob: =ЕСЛ И(А1 +B2<>A2-B1 ;7*A2;21 +B2).
33. MS ACCESS 2003 dasturida “Kalit”ning vazifasi:
Yechish:
Ma’lumotlar omborida "kalit" tushunchasi mavjud. Kalit yordamida jadvallar
orasidagi bog'lanish o'rnatiladi. Kalit o’rnatilgan jadval ustuni unikalbolishi
lozim, ya’ni undagima’lumotlar qaytarilmasligi lozim. Keyinchalik kalit
o’rnatilgan maydon qiymati o’zgartirilmasligikerak.
Javob: jadvallarni o'zaro bog'laydi.
34. 31322w, 32310 butun sonlarning barchasini yozish mumkin bo'lgan eng
kichik asosli sanoq sistemasida shu sonlar raqamlarining yig'indisini hisoblang.
Yechish:
313221o-1111O1OO1O11O1O2 .
raqamlar yig 'indisi 9
115
Yechimlar. Matematika va informatika 2017
13-variant
3132212___
2 11566112__
11 14 [783012____
10 16 6_ [391512______
13 16 18 2_ 1195712
12 6 18 19 18 I978I2
12 6 3 18 15 8 148912
12 (T) 2 11 14 17 4_ [244I2_
2 10 10 _17 16_ 8 2_jT22!2_
2 10 15 16 18 8_4 12 16112-
© ® 14 ® 18 9 4_ 26 [3OI2_
XD ©84 2752-11512
ф 4 © 10 14Г712
© 10ПЖЗ|2
© ®2®
32310 X2.
X= 1010000112
raqarrtlar yig 'indisl 4
323I2
2 I161|2_
12 16 [80|2~
12 ~©)8 I40I2-
3 ©4 |20|2_
2 ©2 I10I2
ф ©10I5I2
©4|2[2
9 + 4 = 13.
Javob: 13.
35. MS Excel 2003 dasturida berilgan
=СЦЕПИТЬ(“АИо"; nCTP("lnformatika";6;6)) formulaning natijasini aniqlang.
Yechish:
Excel dasturida ПСТР funksiyasi qavs ichida berilgan matnning bir qismini
qirqib olish imkonini beradi.
Lining sintaksisi quyidagicha:
nCTPfmatn; qaysi belgidan boshlab qirqish kerak; nechta belgini qirqib olish
kerak);
Masalan:
nCTP("informatika";6;6) bunda informatika so'zini 6 belgisidan boshlab, oltita
belgini qirqib olish ko‘zda tutiladi.
nCTP("informatika";6;6)=matika.
CLJEnHTb("matn";'!matn") - bu funksiya qavsdagi ikkita matnni bir butun
matnga aylantiradi.
СЦЕПИТЬ("А\Ло"; nCTP("informatika'';6;6))="Avtomatika";
Javob: Avtomatika.
36. Qarang: 5-variant 33-savol (49-bet).
116
Yechimlar. Matematika va informatika 2017
14-variant
14-variant
1. Qarang: 5-variant 3-savol
(42-bet).
2. Qarang: 8-variant 3-savol
(71-bet).
3. tg3x + tg2x > 1 + tgxtengsizlikni
yeching.
Yechish:
tg3x + tg2x > 1 + tgx
tg2x(tgx + 1) - (tgx +1)>0
(tgx + 1)(tg2x- 1) > 0
(tgx + 1)(tgx + 1)(tgx- 1)>0
(tgx+ 1) (tgx - 1)>0
tgx - 1 > 0, tgx > 1
— + 7tn<x< — + 7tn,nez.
4 2
necha marta qatnashishiga bogliq,
chunki 2-5 = 10. 500! da 5 ko‘paytuvchi
2 ko'paytuvchiga nisbatan kamroq
qatnashadi. 5 ko'paytuvchini necha
marta qatnashishini hisoblash yetarli.
Г5001 Г5001
n =
5
500
25
52
500
125
Javob: —t-лп:— + лп , nCZ.
14 2 J
4. Qarang: 5-variant 7-savol
(43-bet).
5. Qarang: 4-variant 12-savol
(34-bet).
6. xls5'5“l9X = 1 tenglamani yeching.
Yechish:
Aniqlanish sohasi: x> 0
al9b = blga
ayniyatdan foydalanamiz.
Y95 _ ggx aynjyaf x> Oda o'rinli.
Javob: (0; oo).
7. 500! soni nechta not bilan tugaydi?
Yechish:
500!= 1-2-3-...-500
Ko‘paytma nechta nol bilan tugallanishi
500! ni tub ko'paytuvchilarga
ajratganda 2 va 5 ko'paytuvchilar
500
~5^~
500
5
= 100 + 20 + 4 = 124
500! soni 124 ta nol bilan tugaydi.
Javob: 124.
8. Qarang: 7-variant 2-savol
(61-bet).
9. Tenglamani yeching:
2 + sinx =---=.
1 + x2
Yechish:
Tenglamani grafik usulida yechamiz.
y = 2 + sinx, у =--------;
1 + x
Funksiya grafiklari kesishmaydi, demak,
tenglama yechimga ega emas.
Javob: 0.
10. Ifoda qiymatining oxirgi raqamini
toping: 2 2O142015 - 3-2013Й14.
Yechish:
20142015 - 4 ning ixtiyoriy toq darajasi
4 bilan tugaydi, juft darajasi 6 bilan
tugaydi. 2014го 5 = ...4
2013ю14 - 3 ning 1-darajasi 3 bilan,
2-darajasi 9 bilan, 3-darajasi 7 bilan,
4-darajasi 1 bilan tugaydi.
2013ю14 = 20134 503+2 =
= (20134)503-20132 = ...1-...9 = ...9
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Yechimlar. Matematika va informatika 2017
14-variant
2-20142015 - 3-20132014 = 2-...4-
-3-...9 = ...8-...7-...1
Demak, ifodaning oxirgi raqami 1.
Javob:1.
11. Qarang: 3-variant 19-savol
(27-bet).
x3
12. f (x) = — - x2 - 35x + 2 funksiya
uchun /'(x) = 0 bo'lsa, x ni toping.
Yechish:
/
1)f’(x) -(x2)'-(35x)'+ (2)'=
= хг-2х-35
2) f'(x)=O, x2-2x-35 = 0,
x = -5, x — 7.
Javob: -5 va 7.
13. Qarang: 3-variant 18-savol
(26-bet).
14. a = -3 bo'lsa,
a+1
j(sin23x + cos23x)dx integralni
a
hisoblang.
Yechish:
1) sin23x + cos23x = 1
Javob:1.
15. у - log3(sin2x + cos2x)
, . . 2016x . .
funksiyaning x =-------nuqtadagi
6
qiymatini hisoblang.
Yechish:
1) sin2x + cos2x = 1
2)y = log31 = 0
3) у [ = 0.
Javob: 0.
16. Qarang: 2-variant 5-savol
(14-bet).
17. Qarang: 7-variant 16-savol
(65-bet).
18. Qaysi jism(lar)ning simmetriya
tekisliklari cheksiz sonda?
1) shar; 2) prizma; 3) konus.
Yechish:
1) Shaming markazidan o'tadigan
tekislik diametral tekislik deyiladi.
Shaming istalgan diametral tekisligi
uning simmetriya tekisligibo'ladi.
Sharda simmetriya tekisligi cheksiz
ko'p bo'ladi.
2) Prizma va konusda simmetriya
tekisligi cheksiz ko'p emas.
Javob:1.
19.
sin x + cos у = 0
1 sistemani
sin2 x + cos2 у - —
2
yeching.
Yechish:
1) sinx + cosy = 0,
sinx = -cosy, -sinx = cosy
2) sin2x + cos2у ,
2 2 "1 2 1
sin x + sin x =— , 2sin x =—
2 2
2 ^~cospx _ 1
2 ~~2’
1 1
1 - cos2x =—, cos2x = —
2 2
2x =±— + 2тгп , x=±- + xn, neZ
3 6
3) x= +— + xn,n = 0dax =+—
6 6
. 2l , ГГ | 2 1
sm ±— + cos у = —
I 6j 2
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Yechimlar. Matematika va informatika 2017
14-variant
о JT 2 *
sin — +cos У = —,
6 2
1 2 1 2 1
— + cos у = — , cos у =—
4 1 2 4
1 + cos2y 1 . _ 1
-------— =—, 1 + cos2y = —
2------4 2
cos2y = -^, 2y =±^~ + 2лк
x = +— + ял у =+— + кк, n, к € Z.
6 3
Javob: ( + — + яп; ± — + як ) n, kCZ.
6 3
20. 324; 255 va 71 sonlarining har
birini qanday natural songa bolganda
qoldiqlari bir xil bo'ladi?
Yechish:
324, 255, 71 sonlar ayirmasini topamiz.
324 - 255 = 69 = 3-23
255-71 = 184 = 8-23
Demak, 324, 255 va 71 sonlarini 23 ga
bo'lsak, qoldiqlari birxil bo'ladi.
Javob: 23.
21. >/36-4х2 +Vl00-4x2 =
= 18x4 + 16 tenglamaning ildizlari
qaysi oraliqqa tegishli?
Yechish:
Aniqlanish sohasi:
|36-4x2>0 Jx2-9<0
\l00-4x2 >0[x2-25z0^
=>-3<x<3
j36-4x2<6, Jl00-4x2 <10
18x4 + 16>16
U holda berilgan tenglama quyidagi
tenglamalar sistemasiga teng kuchli.
\yj36-4x2 + \ll00-4x2 =16
[18x4 +16 = 16
2-tenglamadan 18x4 = 0, x = 0 ildizni
topamiz. Bu ildiz 1-tenglamani
qanoatlantiradi. Tenglamaning ildizi
0 6 [0; 2] oraliqqa tegishli.
Javob: [0; 2].
22. Diagonallarining soni
tomonlarining soniga teng bolgan
qavariq muntazam ko’pburchakning
barcha ichki burchaklari va bitta tashqi
burchagi yig'indisini toping.
Yechish:
D - diagonallar soni, n - tomonlar soni.
D _n(n-3)
2
n = n(n-31 =2,
2
n = 5- muntazam beshburchak.
Ichki burchaklari yig'indisi:
180°(n — 2) = 180°(5-2) = 180°-3 = 540°
Bitta tashqi burchagi
, 360 360"
a =-----=------= 72
n 5
540° + 72° = 612°.
Javob: 612°.
23. у = sinx funksiya grafigi berilgan
bolib, uni parallel ko'chirish yordamida
у = sin(x + a) + b funksiya grafigi hosil
qilingan. Bunday parallel ko'chirishda
koordinata boshi qanday nuqtaga
ko'chadi?
Yechish:
у = sinx funksiya grafigini
у = sin(x + a) + b funksiya grafigiga
parallel ko'chirilsa koordinata boshi
(-a; b) nuqtaga ko'chadi, chunki
У = f(x)’ У ~ f(x + a) + bga parallel
ko'chirilsa koordinata boshi (0; 0)
nuqtadan (-a; b) nuqtaga ko'chadi.
Javob: (-a; b).
24. lx2 - 5ax| = 15a tenglama to'rtta
haqiqiy yechimga ega bo'ladigan
a ning qiymatlarini toping.
Yechish:
lx2 - 5ax| = 15a tenglamada a>0
bolishi kerak.
119
Yechimlar. Matematika va informatika 2017
14-variant
Grafik usulida yechamiz. Tenglama
to'rtta yechimga ega bolishi uchun
O<15a<yo tengsizlik o'rinli bolishi kerak.
О X, 5a
у = |№ - 5ax| funksiyada
Xo = 2,5a, yo = 6,25a2
0 < 15a < 6,25a2, a>0
6,25a2- 15a > 0
a(6,25a -15)>0,a> 2,4
a C (2,4; <x>) oraliqda tenglama 4 ta
yechimga ega bo'ladi.
Javob: (2,4; oo).
25, Qarang: 5-variant 19-savol
(46-bet).
26. Hisoblang: ctg40°ctg20°ctg100°.
Yechish:
ctg40°ctg20°ctg100° =
cos 40” - cos 20° - cos 100°
sin 40° -sin20° sin 100°
1) cos100°cos20°cos40° =
= cos(180° - 80°)cos20°cos40° =
= -cos20°cos40°cos80°
2j -2 sin 20° cos 20" cos 40° cos 80” _
7 2 sin 20° “
-sin40° cos40° cos80’ _
2 sin 20°
-2sin40"cos40”cos80° _
4 sin 20”
-sin 80° cos 80° _
4 sin 20”
-2 sin 80” cos 80” _ - sin 160° _
8 sin 20’ 8 sin 20”
-sin(7 80°-20°) _ -sin 20” = _1_
8 sin 20° 8 sin 20° 8
3) sin40°sin20°sin100° =
= sin20°sin40°sin(180° - 80°) =
= sin20°sin40°sin80° ko'paytmadan
yig'indiga o'tamiz.
sin80°sin20°sin40° =
sin40°(cos(80° - 20°) -
- cos(80° + 20°)) Ц sin40°(cos60° -
1
- cos100°) =— cos60° sin40° -
2
-— sin40°cos100° =— sin40° -
2 2
1 1
sin40° cos((90° + 10°) =-^ sin40° +
1 1
+ — sin40°sin10° = sin40° +
2 4
+ - (cos(40° -10°)- cos(40° + 10°)) =
4
111
=— sin40° +— cos30° cos50° =
4 4 4
= — sin40° + — - — sin40°
4 8 4
„ 1 4з 1 8 1 у/з
8 8 8 43 43 3
27. Asoslarining radiuslari 6 va 7 ga
teng bo'lgan kesik konus va unga
tengdosh silindrning balandliklari bir xil.
Silindr asosining radiusini toping.
Yechish:
r = 6,R=7,H = H1
V = Vi, chunki kesik konus va silindr
tengdosh.
V - kesik konus hajmi.
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Yechimlar. Matematika va informatika 2017
14-variant
1
V =^7tH(R! +Rr + r2)
Vi - silindr hajmi.
Vi = 7tR12H1
^(R2 + Rr + r2} = nHR2
_2 R2 + Rr + r2
rt =-----------
1 3
^=J72+7-6+62 =^=xR-
1 N з N з N з
Javob: ^42^ .
28. Qarang: 4-variant 27-savoi
(38-bet).
29. Qarang: 6-variant 6-savol
(52-bet).
30. x** = 2 tenglamani yeching.
Yechish:
Tenglikning ikkala qismini ikki asosga
ko‘ra logarifmlaymiz.
log2x'' = log22
loga x" = n loga x ga asosan
x'‘ log2x = 1, x'‘ =2 bo'lganligi
uchun 2 1одгХ = 1
1одгх=~, x = 22 = 42 .
Javob: 42 .
31. Qarang: 1-variant 31-savol (11-bet).
32. Paskal. Quyidagi dasturning ekrandagi natijasini aniqlang.
var a, b, s: integer;
Begin a:=2; s:=0; for b:=-10 to 6 do s:=s+a*b;
writein (s); end.
Yechish:
Var a, b, s: integer begin a: = 2 s: = 0
for b: = -10 to 6 do s: = s + a*b; writeln(s); end.
1)b = -10da s: = 0 + 2(-10) = -20.
b=-9 da s: = -20 + 2- (-9) = -20-18 = -38.
b = -8 da s: = -38 + 2- (—8) = -38 - 16 = —54.
b=-7 da s: = -54 + 2- (-7) = —54—14 = -68.
b=-6da s: = -68 + 2-(-6) =-68-12= -80.
b =-5 da s: = -80 + 2- (—5) = -80 — 10 = -90.
b = -4 da s: = -90 + 2- (-4) =—90-8 = -98.
b=-3 da s: = -98 + 2(-3) =-98-6 = -104.
b=-2 da s: = —104 + 2(-2) = -104 -4= -108.
b = -1 da s: = -108 + 2 (-1) = -108 -2 = -110.
b = 0 da s: =-110 + 2-0 =-110 + 0 =-110.
b = 1 da s: = -110 + 2-1 =-110 + 2=-108.
b = 2 da s: = -108 + 2-2 = -108 + 4= -104.
b = 3 da s: = -104 + 23 = -104 + 6 = -98.
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Yechimlar. Matematika va informatika 2017
15-variant
b = 4da s: = -98 н и 2-4 = -98 и - 8 = -90.
b = 5 da s: = —90 н >2-5 = -90 < r 10 = -80.
b = 6 da s; = -80 н >-2-6 = -80< h 12 = -68.
Javob: -68.
33. Qarang: 4-variant 31-savol (39-bet).
34. Brauzerda «x3» yozuvini aks ettirish uchun teglar to'g'ri berilgan javobni
ko'rsating.
Yechish:
x3 ni HTML da yozish uchun <sup> teg ishlatiladi. Ya’nix<sup>3</sup>
yozuvi ekranda x3 ni chiqaradi.
Javob: x<sup>3</sup>.
35. MS Excel 2003 dasturida berilgan =ЦЕЛОЕ(-6,985) + MAKC(15;30;3)
formulaning natijasini aniqlang.
Yechish:
ЦЕЛОЕ(а) funksiyasi qavs ichidagi kasr sonniyaxlit, butun songa o'giradi.
ЦЕЛОЕ(5,2) = 5
ЦЕЛОЕ(—4,9) = -5
Bizning misolda ЦЕЛОЕ(-6,985) = -7.
MAKC(15;30;3) = 30. (eng katta son)
=ЦЕЛОЕ(-6,980)+МАКС(15;30;3) = -7 + 30 = 23.
Javob: 23.
36. Qarang: 6-variant 33-savol (60-bet).
15-variant
1. Qarang: 2-variant 28-savol (20-bet). log x log3 3x
2. Qarang: 5-variant 4-savol (43-bet). = ±lk£.10g252 log3 9x 3
3.1одх31од351одзхЗ = log9X3log925 tenglamaning yechimlari ko'paytmasini toping. 1 11 log, 5 = log3 5 log3 x 1 + log3 x 2 + Iog3 x 1 ' 1
Yechish Aniqlanish x>0 : sohasi: x>0 x*1 log3 x(1 + log3 x) 2 + log3x 1одзх = a 2 + a = a(1 + a) аг + а-а-2 = 0
x?-1 ' 3x^1^' 9x*1 1 x* — 3 1 ХФ- 9 аг = 2, a = ±42 , log3x = ±42 1)log3x=42 ,x = 3^ 2) 1од3х =-42, x = 3^
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Yechimlar. Matematika va informatika 2017
15-variant
Ildizlari ko'paytmasi:
3J2 .3-42=3Ji-j2 = 3o = 1
Javob: 1.
4. у = 1одз(агс1д2х + arcctg2x)
1
funksiyaning x = — nuqtadagi
hosilasining qiymatini toping.
Yechish:
1) arctg2x + arcctg2x
2) у = log3^ - son
3) У = (log3^)'- 0
4у'(£) = о-
Javob: 0.
1 ( 3 I IA
5. 2,(3) +5-: 6--1--8- ni
3 ( 4 3 9j
hisoblang.
Yechish:
1 f 3 1
2,(3) + 5—: 6--1--8- =
3(43 9)
3 16 (27 4 1} 1 16,
9 3(43 9) 33
1
Javob: 8-.
3
6. Qarang: 6-variant 2-savol
(51-bet).
7. Qarang: 6-variant 9-savol
(53-bet).
8. Qarang: 4-variant 21-savol
(37-bet).
9. /(g(2)) + g(/(1)) ning qiymatini
(2, x>1
toping, agar/(x) =( va
(1-x, x<1
[0, x<0
g(x) = ( bo'lsa.
(3х, x>0
Yechish:
f(g(2)) + g(f( 1)) ning qiymatini topamiz.
1)g(2)=3r = 9
f(g(2)) = 2
2)f(1) = 1 -1=0
g(f(1)) = 3P=1
f(g(2)) + g(f(V) = 2 + 1 = 3.
Javob: 3.
10. Diagonallarining soni
tomonlarining soniga teng bo'lgan
ko'pburchakning ichki burchaklari
yig'indisini toping.
Yechish:
D - diagonallar soni, n - tomonlar soni.
D = niD=^n-3)
2
n_n(n-3) n-3
2’2
n-3 = 2, n = 5
Ichki burchaklari yig'indisi:
180°(n-2) = 180°(5-2) =
= 180°-3 = 540°.
Javob: 540°.
11. Uchburchakning uchlari to'g'ri
burchakli dekart koordinatalar
sistemasida quyidagicha berilgan:
A(0; 0), B(1; 5), C(2; 0). Uchburchak
yuzini toping.
Yechish:
1) A(0; 0), B(1; 5), C(2; 0),S=?
2) ДАВС teng yonli.
AC-asos, AC = 2
123
Yechimlar. Matematika va informatika 2017
15-variant
BD - balandlik, BD - 5
o ACBD 2-5 c
2 2
Javob: 5.
12. 7x-1 + x- 7 = 0 tenglamaning
ildizlari yig'indisini toping.
Yechish:
yfx-1 =7 - x
Aniqlanish sohasi:
Tenglikning ikkala qismini kvadratga
kotaramiz.
(4^i)2 = (7-x)2
x- 1 =49- 14X + X2
x2 -15x + 50 = 0
x = 5 va x = 10 tenglama ildizlari.
x = 10 aniqlanish sohasiga tegishli
emas. Tenglama ildizlari yig'indisi x = 5.
Javob: 5.
16. Qarang: 6-variant 20-savol
(56-bet).
17. Qarang: 6-variant 15-savol
(55-bet).
18. Tenglamani yeching:
cos(—-)
1 + tgx + tg2x + tg3x + ... = -=.
^-tg^
Yechish:
1) 1 + tgx + tg2x + tg3x +... progressiya
1 - tg2x > 0, tg2x < 1, |tgx| < 1, demak,
cheksiz kamayuvchi geometrik
progressiya.
bi = 1, b2 = tgx, q = tgx
e_ Ь, _ 1
1-q 1-tgx
( 5лЛ ( ТгА
2) cos--= cos 2n— =
{з I 3
13. Besh yoqli ko‘pyoq(lar)ni aniqlang.
Yechish:
Beshyoqli ko'pburchaklar:
1) * u Yoqlari ABCD, BEFC,
AEFD, ABE, DFC—5ta
3) л---------—u Yoqlari5ta:
ABCD, ABE, AED, DEC, ВЕС.
Javob: 1,3.
14. Qarang: 5-variant 30-savol
(48-bet).
15. Qarang: 5-variant 11-savol
(44-bet).
3) -----= ________
1-tgx 2jl-tg2x
2yjl-tg2x = 1-tgx
(ZyfT^tg^x )2 = (1-tgx)2
4(1 -tg2x)= (1-tgx)2
4(1 - tgx)(1 + tgx) = (1 - tgx)2
(1 - tgx)(4 + 4tgx -1 + tgx) =0
(1-tgx)(5tgx + 3) = 0
1) 1 -tgx = 0, tgx = 1, x =-+xn,neZ
4
3
2) 5tgx + 3 = 0, tgx =--
5
x = -arctgO, 6 + m, n 6 Z
x =— + л-n da tenglama yechimga ega
4
bo'lmaydi.
x = -arctgO,6 + m,nCZ tenglama
yechimi bo'ladi.
Javob: -arctgO,6 +nn, n6 Z.
124
Yechimlar. Matematika va informatika 2017
15-variant
19. Doiraning 6 sm li vatari ajratgan
segmentiga tomoni 2 sm li kvadrat
ichki chizilgan. Doira radiusini toping.
Yechish:
ABCD - kvadrat.
AB = BC = 2, EF- vatar.
EF = 6, R = ?
ОС = OB = OF = OE = R
/ QF va ДВОС teng yonli.
; AEOF da R2 = 32 + OP2
2) ДВОС da f^ = 12 + (2 + OP)2
3) 9 + OF2 =1 + 4 + 40P + OP2
4OP = 4, OP= 1
4) R2 = 32 + 12 = 10, R =y/ld.
Javob: VlO .
20. Qarang: 5-variant 1-savol
(42-bet).
21. Qavariq ko'pburchakning xga
teng bo'lgan bitta burchagidan
tashqari qolgan barcha burchaklari
yig'indisi 2192° ga teng. x burchakning
gradus o'lchovini toping.
Yechish:
Qavariq ko'pburchak ichki burchaklari
yig'indisi: 180°(n-2),
n - tomonlar soni.
Bitta ichki burchagi x =180
n
x + 2192° = 180°(n - 2)
x = 148°.
Javob: 148°.
22. a = 2 bo'lsa,
J(ln(sin23x + cos23x) + 1)dx aniq
a
integralni hisoblang.
Yechish:
1) sin22x + cos22x = 1
2) In1 = 0
a+1 8+1
3) f (0 + 1)dx = J 1dx = x
a a 3
= a + 1- a = 1.
Javob:1.
23. Ifodani soddalashtiring:
1,6+ 5,4________
V^56 -V^64 +^29,16
2,25-1,44 27
1,5-1,2 +10'
Yechish:
a2 - b2 = (a - b)(a + b)
a3 + b3 = (a + b)(a2 -ab + b2)
1) 1,6 + 5,4 =(<б)3 + =
= + ^5)4)(fTfY - ф,6-5,4 +
2) 1,6+ 5,4
^2,56-^8,64+^29,16 ~
_ (</Гб + </5~4)«/2,56 - 1)8,64 + ^29,16
^2,56-^8,64+^29,16
= ^6+^4= ^8-0,2+^27-0,2 =
= 2^2+3^2 =5'f0~2
2,25-1,44 _ (1,5-1,2)(1,5 + 1,2) _
' 1,5-1,2 ~ 1,5-,12
= 1,5+ 1,2 = 2,7
4) 5^2-2,7+ 2,7 = 5^2
Javob: 5^/0,2 .
24. Tengsizlikni yeching:
x2 -3x + 2 < 1
x-1
Yechish:
Tengsizlikning ikkala qismini
kvadratga kolarib, qisqa ko'paytirish
formulasidan foydalanib yechamiz.
x2 -3x + 22
x-1
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Yechimlar. Matematika va informatika 2017
15-variant
(хг-Зх + 2 Л(хг-Зх + 2
x-1 ) x-1 )
f x* 1 2 -4x + 3^l (x2 -2x + 1) < 0
x-1 }x-1 J
(x-1)(x-3)(x-1)2
(x-1)2
x# 7, (x — 1)(x-3)<0
-1 з
1 < x < 3.
Javob: (1; 3].
25. Qarang: 1-variant 27-savol
(9-bet).
26. arctg3 + arctg2 + arctgl ni
hisoblang.
Yechish:
arctg3 + arctg2 + arctgl = x
Tenglikning ikkala qismini
tangenslaymiz.
tg(arctg3 + arctg2 + arctgl) = tgx
tg(arctg3 + (arctg2 + arctgl)) = tgx
tgarctg3 + tg(arctg2 + arctgl) =
1 - tgarctg3 tg(arctg2 + arctgl)
Awal tg(arctg2 + arctgl) =
=-—- = -3 ni hisoblaymiz.
tgarctg3-3 _3-3
1 - tgarctg3 (-3) 1 + 9
tgx = 0, x- n.
Javob: it.
27. Qarang: 6-variant 8-savol .
(53-bet).
28. /(x) = 42cos2xsin5x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko'paytmadan yig'indiga o'tamiz.
sin5xcos2x =~ (sin(5x + 2x) +
1
+ sin(5x - 2x)) =— (sin7x + sin3x)
3) f(x) = 42- (sin7x + sin3x) =
= 21sin7x + 21sin3x
F(x) = -3cos7x~ 7cos3x + C.
Javob: -3cos7x - 7cos3x + C.
29. Qarang: 4-variant 8-savol
(33-bet).
30. Trapetsiyaning 9 ga teng bo'lgan
o'rta chizig' uning yuzini 3:5 kabi
nisbatda bo'ladi. Trapetsiyaning
asoslarini toping.
Yechish:
1) m- o'rta chiziq.
a + b a + b „
m =-----, -----= 9,
2 2
a + b = 18
2) S+S2 = 3:5, Si = 3x, S2 = 5x,S = 8x
„_b + m h
o, —------ — •
1 2 2
c _ m + a h
2 2 2
r. a + b , .
S=------h = mh
2
b + 9 h
' S 9h 8
b + 9 _3
9-4 8
2b+ 18 = 27, 2b = 9,b = 4,5
a + b = 18,
a = 18 -4,5= 13,5
a = 13,5;
b = 4,5.
Javob: 4,5 va 13,5
126
Yechimlar. Matematika va informatika 2017 16-variani
31. Qarang: 6-variant 36-savol (61-bet).
32. 5C9i6, 6ACie butun sonlarning barchasini yozish mumkin bo'lgan eng
kichik asosli sanoq sistemasida shu sonlar raqamlarining yig'indisini hisoblang.
Yechish:
5C916-X2.
Jadval buyicha 5i6-101(2)
Cie— 1100(2)
9ie — 1001(2)
Х(2)=10111001001г
raqamlar yig'indisi 6
6АС16-Хг
616 — 110(2)
Ais — 1010(2)
Cre — 1100(2)
6ACi6- 11010101100,
raqamlar yig'indisi 6
6 + 6=12.
Javob: 12.
33. Qarang: 7-variant 36-savol (70-bet).
34. Informatika faniga qachon asos solingan?
Yechish:
Informatika atamasi XX asrning 60-yillarida ishlatila boshlagan bo'lib, uning
alohida fan sifatida ajralishi 40-50 yillarga to'g'ri keladi. Ya’niXXasrning
ikkinchi yarmida informatika faniga asos solingan.
Javob: XX asrning ikkinchi yarmida.
35. MS Excelda A1=25, B3=144 bo'lsa, “=КОРЕНЬ(ВЗ) + КОРЕНЬ(ВЗ+А1)”
formulaning natijasini aniqlang.
Yechish:
A1 = 25;
B3 = 144.
КОРЕНЬ(ВЗ) => d144 = 12.
K0PEHb(B3+A1) => -J144 + 25 = -Jl69 = 13.
=KOPEHb(B3)+KOPEHb(B3+A1) = 13+12 = 25.
Javob: 25.
36. Qarang: 7-variant 33-savol (69-bet).
16-variant
1. Fermer xo'jaligining to'g'ri 1 kilometrli ariq qazilgan bo'lsa,
to'rtburchak shaklidagi yer maydoni yer maydonining yuzi ko'pi bilan
atrofidan ariq qazishmoqchi. Agar yer qancha gektar bolishi mumkin?
maydonining 3 ta tomoni bo'yicha
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16-variant
Yechish:
a + 2b = 1 km
Smax = ab, 1 km = 103 m
a = 10?-2b = 1000 - 2b
Smax = (1000 - 2b) b = 1000b - 2b2
Eng katta va eng kichik qiymatni topish
uchun hosila yordamida yechamiz.
S' = (1000b - 2b2)' = 1000- 4b
S' = 0, 1000-4b = 0, b = 250
a = 1000 — 2b = 500
Smax = 250-500 = 125000 тг =
= 12,5 gektar.
Javob: 12,5.
2. Qarang: 7-variant 26-savol
(67-bet).
3. Agara(T27; 2л/3; -6) va
b(V20; -4б; 6V2)berilgan bo'lsa,
a b .. .
Л'?2""0ТО
Yechish:
a(427\ 243- -6),
b(420- -4б; 642),
ab
46
a (427 243_ -ff| =
' 4з \ 4з ’ 4з ’ 4з)
= (3;2;-243)
?) ь (420 _4б_ 642}
* 42\42’ 42’ 2 )
= (410; -43; 6)
= (3; 2; -24з )
(4l0;-43;6) =
=з4ю-24з-124з =
=з4ю-144з.
Javob: З/Го - 14л/з .
4. Rasmda у = f(x) funksiya grafigi
tasvirlangan. у = /(x) funksiya grafigiga
xo = 2 nuqtada o'tkazilgan urinmaning
burchak koeffitsiyentini toping.
Yechish:
Xo = 2, f’(xo) = к
f'(2)=-1,k = -1.
Javob: -1.
5. Qarang: 6-variant 25-savol
(58-bet).
6. Qarang: 3-variant 8-savol
(24-bet).
7. у = 7cosx + cos7x funksiyaning
hosilasini toping.
Yechish:
1) (costax + b))' = -asinfax + b)
(cosx)' = -sinx, (cos7x)' = -7sin7x
y' = 7(cosx)’ + (cos7x)’ =
= -7sinx - 7sin7x = -7(sinx + sin7x)
2) sinx + sin7x yig'indidan
ko'paytmaga otamiz.
. „ . x + 7x x + 7x
sinx + sin7x = 2sin----- cos-------=
2 2
= 2sin4xcos3x
3) y' = -7-2sin4x cos3x =
= -14sin4x-cos3x.
Javob: -14sin4xcos3x.
8- /(x)’= 3-2-1 bo'lsa, /’(a) = 4-f(a)
tenglamani yeching.
Yechish:
t) f'(x) = (32*2-1)' = З2"2-11пЗ (24- 1)' =
= 4x-32x2^ In3
2) f'(a) = 4a-32s'2 1 In3
4f(a)=4-32a2-1
3) 4a-32a2-1 In3 = 4-32a2-1
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Yechimlar. Matematika va informatika 2017
16-variant
4a 1пЗ-4-32!‘2' =0
4-32a2~1 (aln3- 1) = 0
32a2-1/0
a ln3 -1 = 0, aln3 = 1
1 Ine .
s =----=----= log,e.
1пЗ ln3
Javob: 1одзе.
9. Qarang: 4-variant 16-savol
(36-bet).
10. +----=------+-----
x-2 x + 7 x-1 x + 1
tenglamani yeching.
Yechish:
Aniqlanish sohasi
x # 2, x # -7, x t1, x #-1
Umumiy maxrajga keltirib yechamiz.
x+7+x-2 _x+1+x-1
(x-2)(x + 7)~ x2—1
(2x + 5VX2 - 1) = 2xf/ + 5x - 14)
2x3 + 5x2-2x-5 = 2x3 + Юх2-28x
5x2-26x +5 = 0
1
x = 5, x= — tenglama ildizlari.
Javob: 0,2 va 5.
11. Tengsizlikni yeching:
tg^ + ^ + 1 > 0.
Yechish:
Keltirish formulasiga asosan
. ( x\ . X
tg\x+-\=tg-
\ V J 'J
tg^+1>0,
7Г X 7Г
-----h ТТЛ < — <---У-7СП
4 3 2
-— + Зпп <x< — + Зяп, nCZ.
4 2
Javob: Г-— + Зкп-— + Зт У n С Z.
4 2 )
12. Qarang: 6-variant 7-savol
(52-bet).
13. Ifoda qiymatining oxirgi raqamini
toping: 20122c'13 + 20132014 - 2014201s.
Yechish:
2O12?013 - 2 ning 1 darajasi 2 bilan,
2 darajasi 4 bilan, 3 darajasi 8 bilan,
4 darajasi 6 bilan tugaydi.
201 z 13 = 20124'503'1 =
= (20124)503 20121 = ...6-2 = ...2
2O132014 = 20134 503+2 =
= (20134)503-20132 = ...1-...9 = ...9
20142015 - 4 ning toq darajasi 4 bilan,
juft darajasi 6 bilan tugaydi.
2O12201" + 20132014 - 2O142015 =
= ...2+ ...9-...4= ...7.
Javob: 7.
14. Agar ctg2a = 1,5 va a C^O;-^
bo'lsa, cos2a - sin2a ni hisoblang.
Yechish:
cos2а - sin2a ni topamiz.
2 , cos2 a - sin2 a
cos a - sm a =------------=
1
cos2 a .
2 . 2 -------1
_ cos a - sm a _ sjn2 a _
cos2 a + sin2 a cos2 a
-7-Г-+1
sm a
Ctg2g-1 1,5-1 0,5 _1
~ ctg2a + 1~ 1,5 + 1~ 2,5~ 5
I u 1
Javob: —.
5
15. {x|x C N, 2 < x2 s 44} to'plamning
nechta qism-to'plamlari mavjud?
Yechish:
2 < x2 < 44, x C N
42<x<444 ,A = {2, 3,4, 5, 6}
A to‘plam 5 ta elementdan iborat. A
to'plamning qism to'plamlarini tuzamiz.
1.0cA 2.AcA
3.{2}cA 4.{3}^A
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16-variant
5. {4}cA
7. {6}cA
9. {2; 4} c A
11. {2; 6} zA
13. {3; 5} c A
15. {4; 5}cA
17. {5; 6} cA
19. {2; 3; 5}cA
21. {2; 4; 5}cA
23. {2; 5; 6}c A
25. {3; 4; 6}cA
27. {4; 5; 6}cA
29. {2; 3; 4; 6}cA
31. {2; 4; 5; 6}cA
A to‘plamning 32 t
Javob: 32.
6. {5}c A
8. {2;3}cA
10. {2; 5} c A
12. {3;4}cA
14. {3; 6}cA
16. {4; 6} :_ A
18. {2; 3; 4}cA
20. {2; 3; 6} c A
22. {2; 4; 6icA
24. {3; 4; 5}<=A
26. {3; 5; 6}cA
28. {2; 3; 4; 5}cA
30. {2; 3; 5; 6} zA
32. {3; 4; 5; 6}cA
qism to'plami bor.
16. Qarang: 10-variant 25-savol
(90-bet).
17. Samandar doskaga ikkita son
yozdi. Uchinchi son sifatida u birinchi
va ikkinchi sonlarning yig'indisini,
to'rtinchi son sifatida ikkinchi va uchinchi
sonlarning yig'indisini va h.k. yozdi,
lekin yettinchi sonni yozmadi. So'ng
dastlabki oltita sonni qo'shdi va bu
yig'indini bilgan holda qo'shiluvchilardan
birini aniq hisoblash mumkinligini
ko'rdi. Bu qaysi qo'shiluvchi edi?
Yechish:
a, b~ ikki son.
c = a + b, d = b + c, e-c + d,
k = d + e
а_нЬ + c + d^ +e+k=c+e+
+ e + d+, e = 3e + c + d = 3e + e = 4e
Bu beshinchiqo'shiluvchi.
Javob: beshinchi.
18. Qarang: 3-variant 14-savol
(25-bet).
19. у > 0 bo'lsin. To'rtburchakning
uchlari to'g'ri burchakli dekart
koordinatalar sistemasida quyidagicha
berilgan: A(0; 0), B(0; у), C(4; y) va
D(6; 0). To'rtburchak diagonallarining
o'rtalari orasidagi masofani toping.
Yechish:
1) у > 0. A(0; 0), B(0; y), C(4; y), D(6; 0),
EK=?
2) EK - diagonalari o'rtasi.
E - AC kesma o'rtasi.
К
£f4±0.0±yVEr2.^
{2'2) { '2)
К - BD kesma o'rtasi.
0 + 6 у + 0
~2~’ 2
= зЛ-
l 2
EK kesma uzunligini topamiz.
I 7 Г7
EK = J(2-3)2+I =Vt + 0=1.
Javob:1.
20. у = 11sin4x + 2sin22x
funksiyaning hosilasini toping.
Yechish:
1) (sin(kx + b))'- kcosfkx + b)
(sin4x)’ = 4cos4x
(sin22x)' = 22cos22x
2) y' = 44cos4x + 44cos22x -
= 44(cos4x + cos22x)
3) cos4x + cos22x yig'indidan
ko'paytmaga otamiz.
cos4x + cos22x -
„ 4x + 22x 4x-22x
= 2cos--------cos---------=
2 2
= 2cos13xcos9x
4) y' - 44-2cos13x cos9x =
= 88cos9xcos13x.
Javob: 88cos9x cos13x.
21. Qarang: 6-variant 16-savoi
(55-bet).
22. Qarang: 6-variant 27-savol
(59-bet).
23. Uchburchakning uchlari to'g'ri
burchakli dekart koordinatalar
sistemasida quyidagicha berilgan:
A(0; 0), B(-3; 0), 0(0; -3). O'tkir
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Yechimlar. Matematika va informatika 2017
16-variant
burchaklar medianalari orasidagi
o'tmas burchak kosinusini toping.
Yechish:
1) A(0; 0), B(—3; 0), C(0; -3), cosa = ?
2) Uchburchak ВАС - to‘g‘ri burchakli.
AB = AC = 3, CB=342
CK = BD- medianalar
CK2 = AC2 + AK2 = 32 +(—
= 9Л=^
4 4
[45 _345
V 4 2
Z BNC burchakni topamiz.
Z BNC = а
BN2 +CN2 -BC2
cos a =----------
2BNCN
BN = CN= — CK = 45
3
(JS)2 +(-J5}2 -(342)2
cosa =--- —p——— =
5 + 5-18 _ 8 _ 4
10 10 ~ 5
4
cosa = — .
5
. 4
Javob: —.
5
24. Qarang: 9-variant 27-savol
(84-bet).
25. Qarang: 8-variant 2-savol
(71-bet).
26. Qarang: 3-variant 6-savol
(23-bet).
27. Hisoblang:
(4 + 4б -yj4l50+ 425 ) 7V6-1 +
+ 1 -45 .
Yechish:
1) 44150 + 425 =
yjy/6-25+5 = 4546+5 = х[б(4б + 1[
2) ^1 + 4б~4б(4б+1) =
= ^1+46(1-45)
з) (1-45)-^+4б-44^1 =
=(1-45)44/6+i)(46-i) =
^(i-45)^4^f^i^(i-45\
46^1 = 45(1-45]
4) 45(1 -45) + 1 -45 = (1 - 45)
(45 + l'] = 1-(45^ ^1-5 = -4.
Javob: -4.
28. Qarang: 4-vsdant 30-savol
(39-bet).
29. 4.7 *’ ifodaning
qiymatini toping.
Yechish:
1) loga1 = 0
2) alO86C = c,OSb“ formuialardan
foydalanib yechamiz.
^log45 __ ^iog^ 3 logg? _
-3i°8"s+7o=O + f = 7.
Javob: 1.
30. Qarang: 13-variant 21-savol
(113-bet).
31. MS Excelda И(-500>5*100; 30-16>3) funksiya natijasini aniqlang.
Yechish:
И(-500>5*100;30-16>3)
-500 > 500 tasdiq yolg‘on.
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Yechimlar. Matematika va informatika 2017
17-variant
30- 16 > 3 bu tasdiq rost.
И mantiqiy funksiyada har ikkala mantiqiy ifodadan loqal bittasi yolg'on
bo'lsa natija ЛОЖЬ.
Javob: ЛОЖЬ.
32. Qanday so’zlar asosida «texnologiya» so’zi tashkil topgan?
Yechish:
Texnologiya lotincha so'zdan olingan bo'lib, <techno> san'at, xunar, soha va
<logos> - fan degan ma’noni anglatadi.
Javob: «techno» va «logos».
33. Hujjatning xossalari (Свойства) oynasida qanday ma’lumotlar beriladi?
Yechish:
Hujjat xossalari deganda, hujjat nomi, turi (qaysi dasturda yaratilgan)
joylashgan joyi, yaratilish sanasi va atributlari hisoblanadi.
Javob: hujjatning nomi, turi, hajmi, joylashgan o’rni, hosil qilingan va
o'zgartirilgan vaqtlari, atributlari.
34. Qarang: 5-variant 36-savol (50-bet).
35. Qarang: 8-variant 33-savol (77-bet).
36. Qarang: 2-variant 31-savol (21-bet).
17-variant
1. Qarang: 12-variant 2-savol (102-bet). 2. Qarang: 7-variant 9-savol (63-bet). Tr । 8 ( . 2 о rr ) 2) / - = In sin 2 — \ ctg2 — = 'Це; ln2 I 8j У 8 = — ln-J=— (—In2) = -8. In2 2 ln2
3. Qarang: 7-variant 8-savol (63-bet). 4. Qarang: 9-variant 25-savol (83-bet). ln2[suii2x) 7Г 5. f(x)=— -bo’lsa, /’(-) In2 8 ni toping. Javob: -8. 6. Ifodani soddalashtiring: V ab 2y[b ay/a + byfb a - b \[q + y[b (y/в + yfb} • (a ~ by Yechish: з — b — (у[з — x/b )( у/э + y/b ) a\[a +b\[b +{\[bj =
Yechish: 1)Г(х)=^ (lnz(sin22x))'= ln2 1 _. , . > 2-2sin2xcos2x = -2ln(sin2x) - = In2 sin 2x = ln(sin2 2x) ctg2x In 2 = [y/a + y/b^a -4ab + bj y/ab 2y[b ay[a+by[b a-b y/a+y/b (Va+Vbj(a-b) 4ab 2y[b ^y[a — \[b^y[a + yfb^ \la+jb
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Yechimlar. Matematika va informatika 2017
17-variant
_(Ja + Jb)(a - Jab + b) _
(Ja + Jb\a-b)
Jab - 2jb(ja - Jb) - (a - Jab + b)
a-b
_ Jab -2Jab + 2b-a + Jab -b
a-b
- b~a - ~(a~_„7
a-b (a-b)
Javob: -1.
7. |OP| = 3, |AC| = |BC| = 8 sm
bo’lsa, aylana radiusini toping.
//oV\
\/
VjSTc
Yechish:
10P| = 3, |AC| = |BC| = 8, R = ?
AC = BC, OH = OP, OH± AC
AH = HC
fj = 32 + 42 = 52, R = 5.
Javob: 5.
8. Qarang: 6-variant 29-savol
(59-bet).
9. x3 - 2ЭХ2 + 4bx - 48 ko'phad
(x - 2) (x - 6) ga qoldiqsiz bo'linsa,
a + b ..
ni toping.
Yechish:
x3 - 2a№ + 4bx - 48
(x -2)(x- 6) ga qoldiqsiz bo'linsa,
x = 2 va x = 6 da ko'phadning qiymati
nolga aylanadi.
x = 2da23- 2a-22 + 4-2b -48 = 0
8- 8a + 8b- 48 = 0 yoki a - b + 5 = 0
x = 6da63 -2a-62 + 4-6b -48 = 0
216 - 72a + 24b -48 = 0
-72a + 24b + 168 = 0 yoki 3a -b-7 = 0
ia-b~--5
l3a b 7 ten3'amalar sistemasidan
a va b ni topamiz.
a = 6,b = 11
a + b 6 + 11 17
2 2 ~2
Javob: 8,5.
10. у = 4sin4x + sin16x
funksiyaning hosilasini toping.
Yechish:
1) (sin(ax + b))'= acos(ax + b)
(sin4x)' = 4cos4x
(sin16x)'= 16cos16x
y' = 4(sin4x)' + (sin16x)' = 16cos4x +
+ 16cos16x = 16(cos4x + cos16x)
2) cos4x + cos16x yig'indidan
ko'paytmaga otamiz.
cos4x + cos16x = 2cos—J- 6x
2
Ду —
cos-------= 2cos10xcos6x
2
3) y' = 16-2cos6x cos10x =
= 32cos6xcos10x.
Javob: 32cos6x cos10x.
11. Qarang: 6-variant 19-savol
(56-bet).
12. Agar 3/(x) = /(x + 1) + f(x - 1),
/(1) = 3, /(2) = 4 bo'lsa, /(5) ning
nechta tub bo'luvchisi bor?
Yechish:
3f(x)=f(x+1)+f(x-1)
f(1) = 3, f(2) =4, f(5) = ?
X = 2 da 3 f(2) = f(3) + f(1),
f(3) = 3f(2)-f(1)
f(3) =34-3=9
x = 3 da 3f(3) = f(4) + f(2),
f(4)=3f(3)-f(2)
f(4)=3-9-4 = 23
x = 4da3f(4)=f(5)+f(3),
f(5)=3f(4)-f(3)
f(5) = 3-23 - 9 = 69 — 9 = 60
60 = 2^-3-5
3 ta tub bo'luvchi bor.
Javob: 3.
13. Qarang: 6-variant 18-savol
(56-bet).
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Yechimlar. Matematika va informatika 2017
17-variant
14. Hisoblang: tg50otg10otg110°.
Yechish:
tg50°tg10°tg110° =
_ sin 50° sin 10” sin 7 7 0°
cos 50° cos 7 0" cos 7 7 0°
Suratiniyechamiz: sin 110° =
= sin(90° + 20°) = cos20°
1) sin10°sin50°sin110° =
= sin10°sin50°cos20° =
= cos20°-cos40°cos80°
2) cos20°cos40°cos80° =
_ 2 sin 20° cos 20° cos 40" cos 80° _
2 sin 20°
_ sin 40° cos 40° cos 80° _
2 sin 20°
_ 2 sin 40° cos 40° cos 80’ _
4 sin 20°
_ sin80°cos80° _ 2 sin 80° cos 80° _
4 sin 20“ 8 sin 20°
_ sin760" _ sin(780° -20°) _ sin20° _ 1
8 sin 20° 8 sin 20° 8 sin 20° 8
3) cos50°cos10°cos110° =
= cos50°cos10°cos(180° - 70°) =
= -cos10°cos50°cos70°
ko'paytmadan yig'indiga otib yechamiz.
4) -cos10°cos50°cos70° =
= -1- cos10°(cos(50° + 70°) +
+ cos(50° - 70°) =~ cos10°(cos120° +
7 7
+ cos20°) =— cos10°(-— + cos20°) =
2 2
= — cos10°~— cos10°cos20° =
4 2
7 7
= — cos10° (cos30° + cos10°) =
4 4
111
= — cos10° cos30° -4 cos10° =
4 4 4
= _1 =
4 2' 8
5) Suratdagi ifoda qiymatini
maxrajdagi ifoda qiymatiga bolamiz.
8\ 8 ) 8\ 43) 43~ 3
Vs
Javob: ----.
3
15. Paraliepipedning asoslari
tomoni 2V2 ga teng kvadratlardan,
barcha yon yoqlari romblardan iborat.
Yuqori asosining uchlaridan biri ostki
asosining barcha uchlaridan baravar
uzoqlikda joylashgan.
Parallelepipedning hajmini toping.
Yechish:
ABCD, A1B1C1D1 - asos
AB = CD = a =242
ABB1A1 - romb
AB = BB1 = B1A1 — AA1 — a = 2^/2
Ai - yuqori asosi uchi
A1B — A1C = A1A = A1D
A^ = H
и L2 f 3 2y[2 _ o
\ 2 J 42 42
Sasos = a2 = (242 )2 = 8
V =SasosH =82 =16.
Javob:'16.
16. x = 7 bo'lsa,
ln(№ + x-6) , z , .
qiymatini toping.
Yechish:
, , ln(№ + x - 6) ,
x =7 da —--------------- - log.y+31(x - 2)
ln(x + 3) ( *
ifodaning qiymatini topamiz.
134
Yechimlar. Matematika va informatika 2017
17-variant
.. Ina .
1) ГТ = Ча
Ino
2) logba - logbc = logb-
c
formulalardan foydalanib yechamiz.
\n(x2 + x-6) , . _
— -------------log,x+,. (x - 2) =
ln(x + 3) * ’
= log(,+3>(^2 + x - 6) - logu+3)(x - 2) =
. x2 + X - 6
" °ёх+3 x-2
log«3 (X+^~2) = logx+3(x + 3) = 1
Ifodaning qiymati 1 ga teng.
Javob: 1.
17. Qarang: 9-variant 8-savol
(80-bet).
18. Agar barcha x, у lar uchun
x3 + 4x2y + axy2 + 3xy - bxcy + 7xy2 +
+ dxy + y2 = x + y2 ayniyat bajariisa,
|a + b + c|(a + b - c) ni toping, (c > 1)
Yechish:
Ayniyatda c = 2. O‘xshash darajalar
oldidagi koeffitsiyentlardan foydalanib
yechamiz.
x2y oldidagi koeffitsiyent
4-b=0, b = 4
xy2 oldidagi koeffitsiyent
a + 7 = 0, a = -7
xy oldidagi koeffitsiyent
3 + d = 0,d = -3
\a + b + c\(a + b - c) ning qiymatini
topamiz.
n =-7, b = 4, c = 2, d =-3
\-7 + 4 + 2|(-7 + 4-2) = \-1\-(-5) = -5.
Javob: -5.
19. Qarang: 1-variant 16-savoi
(6-bet).
20. 3-5'91 - 3lg25 + 5lg9 ifodaning
qiymatini toping.
Yechish:
1) Ig1 = 0
2) a'sb = b''"’ formulalardan foydalanib
yechamiz.
3is25 = 3^52 = 32,g5 — glgS —
з-^- d" + d99 = 3-1 = 3.
Javob: 3.
21. t ning nechta qiymatida
1-(t + 1)x + (t2-3t-4)x2 = 0
tenglama yagona yechimga ega
bo'ladi?
Yechish:
(t2-3t-4)x2-(t + 1)x + 1 = 0
tenglama yagona yechimga ega
bolishi uchun D = 0 bolishi kerak.
D = b2-4ac = (-(t+1))2-
— 4-1 (t2 - 3t-4) = 0
? + 2t+ 1 -4? + 12t+ 16 = 0
-3? + 14t+17 = 0
3f - 14t — 17 = 0 kvadrat tenglamada
D=142- 4-3 (—17) = 196 + 204 = 400
D> 0 bo'lgan ligi uchun 2 ta yechimga
ega bo'ladi.
Ikkinchidan, ax2 + bx + c = 0
tenglamada a = 0 bo'lsa, tenglama
bitta yechimga ega bo'ladi.
Demak, f-3t-4 = 0
t = -1 vat = 4
Tenglamaning ildizlari t = -1,
t ning 3 ta qiymatida.
Javob: к < 0, к > 1.
22. Tenglama ildizlari nechta?
731og2(-x) - log2 Jx2 = 0
Yechish:
Aniqlanish sohasi:
-x >0, x<0 va log2(-x) >0,x<-1
4x2 =| x |, x < -1 bo'lganligi sababli
|x| = -X.
.y31og2(-x) =log2(-x) kvadratga
kotaramiz.
135
Yechimlar. Matematika va informatika 2017
17-variant
3log2(-x) = log22(-x)
log2(-x)(log2(-x) -3) = 0
1) log2(-x) =0,-x = 2°, -x = 1, x=-1
2) log2(-x) = 3, -x = 23,-x = 8, x = -8
x = -1, x = -8 aniqlanish sohasiga
tegishli. Tenglama 2 ta yechimga ega.
Javob: 2 ta.
23. Muntazam uchburchakli
piramida asosining tomonidan unga
ayqash yon qirraga perpendikulyar
bo'lgan tekislik o'tkazilgan. Kesuvchi
tekislik yon qirrani uchidan
hisoblaganda 3:2 nisbatda kesadi.
Asos tomoni 6-J2 ga teng bo'lsa,
piramida yon sirtining yuzini toping.
Yechish:
SABC - muntazam uchburchakli
piramida.
AB = бл/2 , SA va BC - ayqash
qirralar. Z ADF = 90°.
AD.DS = 2:3
18J2SF
» =---------
yon q
AADF to'g'ri burchakli.
SO = H, DE = h, AO = R, OF = r, AE = x
h=DE=jAEEF -
- иУб _2V66
~V 5 ’ 5 " 5
Н=4б6 , SF=6T2
SyOn = 9^2-6^2 = 54-2 = 108.
Javob: 108.
24. Diagonallarining soni
tomonlarining sonidan 2 marta kam
bo'lgan qavariq muntazam
ko'pburchakning bitta ichki burchagini
toping.
Yechish:
D - diagonallar soni, n - tomonlar soni.
D=n n(n-3)_n
2’ 2 2’
n- 3 = 1, n — 4 — muntazam
to'rtburchak. Muntazam to'rtburchak-
bu kvadrat. Bitta ichki burchagi 90°.
Javob: 90°.
25. Qarang: 11-variant 2-savol
(94-bet).
26. Sonning 8 foizi 40 foizining
necha foizini tashkil qiladi?
Yechish:
x-son
Sonning 8% i 40% ining
40%x- 100%,
8%x - Xi
8x-100 ono/
Xi =------= 20% .
40x
Javob:20%.
y[3 J3
r = -/6 , AF = 3x/6
AAOS va AED o'xshash, bundan
H=R = 5
h~ x~ 2
2R 4-J6
X~ 5 " 5
5 5
27. 9s'"2* - Э""2' = 8 tenglamaning
[0; 2л] kesmadagi ildizlari yig'indisini
toping.
Yechish:
sin2x + cos2x = 1,
cos2x = 1 - sin2x
.2 2 2
gsm X _ фСО8 X _ Q gsin X _ Q
a-- = 8, a2-8a-9 = 0
a
a =-1, a = 9
136
Yechimlar. Matematika va informatika 2017
17-variant
gSin* 2 * *x _ у tenglama haqiqiy ildizga
ega emas.
9s 6”2 x = 9, sin2x = 1,
x = — + яп,п eZ
2
n = 0da x =—
2
,, Зтг
n = 1 dax = —
2
Tenglama ildizlariyig'indisi:
. ЗТГ
— +-----2тг.
2 2
Javob: 2л = 360°.
28. Qarang: 8-variant 17-savol
(88-bet).
29. Qarang: 5-variant 13-savol
(45-bet).
30.
2tv
Х + У = 7Г
sistemani yeching.
sin x _
sin у
Yechish:
a + b c+d , , , , ., ,
----=------dan foydalanib yechamiz.
a-b c-d
sinx + sin у _ 2 + 1
sin x - sin у 2-1
______2_____2__ = 3
„ x+y . x-y
2 cos-— sm —
2 2
. x+y , x-y „
tg—^-ctg—^- = 3
2я .я . x-y „
x + y= —, tg-ctg—^- = 3
. x-y x-y
ctg — = V3, - = — + яп
2 2 6
Я
х-у = — + 2яп
3
f 2
J 3
x - у = — + 2xn
I 3
2х = тс + 2яп
71 71
X =— + 7ГП , V =------7tn,n + Z .
2 6
Javob: х=— + яп,
2
л
у = — тгп,n e_Z.
6
31. MS Excelda ИЛИ(-36>=1; 81<30) funksiya natijasini aniqlang:
Yechish:
-36 > 1 bu tasdiq yolgon.
80 < 30 bu tasdiq ham yolg'on.
ИЛИ funksiyasida har ikkala mantiqiy ifoda yolg'on bo'lsa, natija ЛОЖЬ
bo'ladi.
Javob: ЛОЖЬ.
32. Windows operatsion tizimi (sistemasi)da fayl nomi to'g'ri berilgan javobni
toping.
Yechish:
Fayl nomlashda quyidagi belgilar ishlatish mumkin emas.
<> - otkir katta, kichik belgilari
[] - to'rtburchak qavslar
* - yulduzcha
l\ - qiya to'g'ri chiziqlar.
Javob: lnformatika.doc.
137
Yechimlar. Matematika va informatika 2017 18-variant
33. Tenglik o'rinli bo'lishi uchun sonlarning asosi qanday bo'lishi kerak?
24005(X) + 2003(X) = 26010M.
Yechish: 8 lik sanoq sistemasida go‘shish jadvali.
4 1 2 3 5 6 7
0- -0 1 2 3 £ 6 7
1 1 2 3 4 t 7 10
2- Q -3- —4— 1 10 11
34 U- —4— -5- -6- -7- —4 ? 11 12
4 4 5 6 7 10 11 12 13
5 5 6 7 10 11 12 13 14
6 6 7 10 11 12 13 14 15
7 7 10 11 12 13 14 15 16
260iC
Javob: sakkiziik.
34. Qarang: 4-variant 36-savol (41-bet).
35. Qarang: 11-variant 31-savol (101-bet).
36. Qarang: 2-variant 36-savol (22-bet).
18-variant
1. Qarang: 4-variant 9-savol 1 2
(33-bet). 4 |T;1|_1-|X + 1|_2
2. Qarang: 1-variant 26-savol (9-bet). 3. Rasmda berilganlarga ko'ra R ni toping. Bunda BC = 4. / \ \ / ! o\ \ \/X__— в x" 4 7 , Yechish: Aylanaga ichki chizilgan burchak markaziy burchakning yarmiga teng.' Z BOC = 2 Z ВАС = 60° Uchburchak BOC teng yonli, chunki OB = OC = R. Z BOC = 60° bolganligi sababli Z OBC = Z BCO = 60° Bundan OB = BC = ОС = 4 R = 4. Javob: 4. tengsizlikning eng katta manfiy butun yechimini toping. Yechish: |x + 1\ = a belgilash kiritamiz. -J —>0 a-1 a-2 a — 2 — 2a + 2 > (a-1)(a-2) Z? >o (a-1)(a-2) < 0 tengsizlikni oraliqlar (a-1Xa-2) usuli bilan yechamiz. a <0,l1< a <2 1) |x + 1\ < 0, x + 1 = 0, x = -1 2) 1 <\x+1\< 2
138
Yechimlar. Matematika va informatika 2017
18-variant
|x + 1\ > 1 va |x + 1\ < 2
|x +1 > 1
1 va -2 < x + 1 < 2
]x + 1<-1
fx>0
< va -3 < x < 1
x<-2
-3-2 0 1
Tengsizlikning yechimi
(-3,-2) U (0; 1) U{-1).
Tengsizlikning eng katta manfiy butun
yechim -1.
Javob: -1.
5. Tengsizliklar sistemasini yeching:
|6(2x-2,5) + 1,5(2-4x)>6
[x2-8x + 12<0
Yechish:
Har bir tengsizlikni yechib olamiz.
1) 6(2x — 2,5) + 1,5(2 — 4x) > 6
12x-15 + 3-6x>6
6x>6+12
x > 3
2) хг-8х+12<0
(x-2)(x — 6)<0
2<x<6
3) Ikkala tengsizlikni bir vaqta
qanoatlantiruvchi oraliqni topamiz.
2 3 6
Demak, x 6 (3; 6) tengsizlik yechimi
boladi.
Javob: (3; 6).
6. Qarang: 7-variant 21 -savol
(65-bet).
7. Qarang: 7-variant 12-savol
(64-bet).
8. To'g'ri burchakli uchburchakning
katetlariga tushirilgan medianalari
V52 va v73 ga teng bo'lsa,
gipotenuzani toping.
Yechish:
m3 = 452 , mb = 473 , c = ?
m =-л/2Ь2 +2c2 -a2
a 2
m, = --j2a2 + 2c2 -b2
2
4m2 = 2b2 + 2c2 - a2
+4m2b =2a2 +2c2 -b2
4(52 + 73) = 4c2 + a2 +b2 = 5c2
5сг = 4-125, c2 = 100, c = 10.
Javob: 10.
9. Qarang: 4-variant 24-savol
(38-bet).
10. Bir ayol bog'ga olma tergani kirdi.
Bog'dan u 4 ta eshik orqali chiqishi
kerak edi. Har bir eshik oldida qorovul
turgan bo'lib, ayol birinchi qorovulga
tergan olmalarining yarmini berdi.
Ikkinchi qorovulga esa qolgan
olmalaming yarmini berdi. Uchinchi va
to'rtinchi qorovullarni ham xuddi shunday
siyladi. Oxirida o'zida 10 ta olma qoldi.
Ayol bog'dan necha dona olma uzgan?
Yechish:
x ta olma
10ta olma qolgan.
x
1-eshik oldidagi qorovulga —
X X
2-qorovulga = —
— -— = — qolgan olma.
2 4 4
X X
3-qorovulga = —
XXX
-----= — qolgan olma.
4 8 8
X X
4-qorovulga----= —
8 2 16
139
Yechimlar. Matematika va informatika 2017
18-variant I
XXX x
— + — + — + — + 10 = x
2 4 8 16
x(8 + 4 + 2 + 1) . n
—-------- + 10 = x
16
15x .n .cn
x------= 10, x = 160.
16
Javob: 160.
11. a = -3 bo'lsa,
a+1
J (ln(sin2 2x + cos2 2x) + 1)dx aniq
a
integralni hisoblang.
Yechish:
1) sin22x + cosz2x = 1
2) In1 = 0
a+7 a+< a +1
3) f (0 + 1)dx = f 1dx = x
a
= a+1~a = 1.
Javob: 1.
13. у > 0 bo'lsin. To'rtburchakning
uchlari to'g'ri burchakli dekart
koordinatalar sistemasida quyidagicha
berilgan: A(0; 0), B(0; y), 0(6; y) va
D(8; 0). To'rtburchak diagonallarining
o'rtalari orasidagi masofani toping.
Yechish:
1) у > 0, A(0; 0), B(0; y), C(6; y),S=?
2) AC, BD - diagonallar. E - AC
kesma o'rtasi. К - BD kesma o'rtasi.
E(o+6_q+y} = Ef jq
V 2 2 J I 2)
<0+8 O+yA V
< 2 2 ) I 2)
EK kesma uzunligini topamiz.
I 7 77 ____
EK = . (3-4)2+ =Jl + 0=1.
V [2 2)
12. {x|x G N, -2 < x S 5} to'plamning
nechta qism-to'plamlari mavjud?
Yechish:
-2 < x < 5, x C N
A = {1; 2; 3; 4; 5}
A to‘plam 5 ta elementdan iborat.
A to'plamning qism to'plamlarini tuzamiz.
1. QcA
3. {1}rzA
5. {3}^A
7. {5}cA
9. {1; 3}cA
11. {1; 5}cA
13. {2; 4}<± A
15. {3;4}cA
17. {4; 5}cA
19. {1; 2;4}cA
21. {1;3;4}c.A
23. {1; 4; 5}cA
25. {2; 3; 5}cA
27. {3; 4; 5}c.A
29. {1; 2; 3; 5} c A
31. {1; 3; 4; 5}cA
A to'plamning 32 ta
2. A = A
4. {2}cA
6. (4}cA
8. {1;2}cA
10. {1; 4} c A
12. {2; 3}c A
14. {2; 5} a A
16. {3; 5} c A
18. {1; 2; 3}cA
20. {1;2;5}cA'
22. {1; 3; 5}cA
24. {2; 3; 4} c A
26. {2; 4; 5}cA
28. {1; 2; 3; 4} c A
30. {1; 2; 4; 5}cA
32. {2; 3; 4; 5}cA
qism to'plami bor.
Javob: 32.
Javob: 1.
14. Qarang: 7-variant 27-savol
(67-bet).
15. Tenglama nechta yechimga ega?
(2sin2x - 1) л/9-х2 = 0
Yechish:
Aniqlanish sohasi:
9-^>0,^-9<0
(x - 3)(x + 3)<0,-3<x<3
1
1) 2sin2x- 1=0, sin2x = —
n = 0 da x =—
12
. , 5я
n = 1 da x =—
12
n =-1 da x = -
7тг
12
140
Yechimlar. Matematika va informatika 2017
18-variant
„ , 11л
n = -2 da x =----
12
2) ^9-x2 =0, 9 — x2 = 0
x2 = 9, x = +3
Tenglama 6 ta yechimga ega.
Javob: 6 ta.
16. Qarang: 8-variant 3-savol
(71-bet).
17. Qarang: 10-variant 6-savol
(86-bet).
18. |AB| = 6, |BC| = 4 bo'lsa, |CD| ni
toping.
Yechish:
AB - urinma, BD - kesuvchi.
AB2 = BCBD, 62 = 4(4 + CD)
CD = 5.
Javob:5.
19. Qarang: 12-variant 6-savol
(103-bet).
20. Qarang: 7-variant 28-savol
(68-bet).
21. Agar m = 64 bo'lsa,
log7( ^ + 27 . tfn-3^ + 9 }
л/т-2л/т-15 Vm-25
ni hisoblang.
Yechish:
Qisqa ko'paytirish formulasidan
foydalanib yechamiz.
- ( - V
1)-Jm + 27 = m2 +27 = m6 +33 =
1 1.2 1
= (m6+3)(m6 -3 m6 +32) =
= (i/m + 3)(^m-3^/m + 9)
2) tfm - 25 = (>/m - 5)(y/m + 5)
3)y/m-2y/rn-15 = (\/m + 3)(y/m-5)
.('Im + 3)(y/m - З'/т + 9)
(y/m -5)(\fm +5) 6l—
3r- 'Sr-^ = Wm + 5)
5) m = 64 da log7( ^64 + 5) =
= log7(2 + 5)= log77 = 1.
Javob:1.
22. Kichik asosi 2 ga teng bo'lgan
teng yonli trapetsiyaga radiusi 2 ga
teng bo'lgan aylana ichki chizilgan.
Trapetsiyaning yuzini toping.
Yechish:
b = 2, r= 2, S=?
и M C
BC = b*OE = r,BK= 2OE = 4
BM = BN =- = 1
2
AN=AE=—
2
Uchburchak AKB to‘g‘ri burchakli.
AB2 = BK2 + AK2
AB = BN + NA = 1 +-
2
AK=—-1
2
z \ 2
2
2
2
fa a .If a a
12 2 Д2 2 ,
2 a = 16, a = 8
S=^-h = ^-4 = 20.
2 2
Javob: 20.
= 16
141
Yechimlar. Matematika va informatika 2017
18-variant
23. VT- 3 - Vx + 1 + 2 = 0
tenglamaning ildizlari ko'paytmasini
toping.
Yechish:
Aniqlanish sohasi: x > 3.
4x~- 3 = 4x + 1 - 2
(4^3 )2 = (4x4i - 2)2
x-3=x+1-4 44+1 + 4
4 4x + 1 = 8, 4x+1 = 2
x+1 = 4, x = 3
x = 3 tenglama ildizi. Ildizlari
ko'paytmasi 3 ga teng.
Javob: 5.
24. 32x x2 + 5x ~ 6 <;
< x2 + 5x-32x- 2-32x+1 tengsizlikning
eng katta manfiy butun yechimining
moduli bilan eng kichik musbat
yechimi ko'paytmasini toping.
Yechish:
32x / + 5x - 6 2 x2 + 5x-32" - 2'32x+)
З^-х2-5x-32x + 2-3?x+1 - x2 + 5x - 6 <.0
32x(x2 - 5x + 6) - fx2 - 5x + 6) < 0
(x2 - 5x + 6)(32x - 1) < 0 tengsizlik
jx2-5x + 6>0
tengsizhklar
[3!x-1<0
sistemasiga teng kuchli.
[x2-5x + 6<0 |2<x<
^}32Х>7- Дх>0
2<X<:3
2)
[x2-5x + 6>0 (x>3,x<2
[з2х <1 |x < 0
x<0
Tengsizliklar yechimlari [2; 3] U (-«>; 0]
Eng katta butun manfiy yechim moduli
H| = 1.
Eng kichik musbat yechimi 2.
Yechimlar ko'paytmasi 1-2 = 2.
Javob: 2.
25. Qarang: 4-variant 18-savol
(36-bet).
26. (x2 + 10x+10)-
(x2 + x + 10) = Юх2 tenglama haqiqiy
ildizlari yig'indisini toping.
Yechish:
x2 + 10 = a belgilash kiritamiz.
(x2 + 10+ 10x)(x2 + 10+ x) = 10x2
(a + 10x)(a + x) = Юх2
a2 + 11xa + Юх2 = Юх2
a2 + 11xa = 0, a(a + 11x) = 0
a = 0,
a=-11x
4 + 10 = 0 tenglama haqiqiy ildizga
ega emas.
x + 10 = -11x
x2 + 11x + 10 = 0 haqiqiy ildizlari
yig'indisi x-t + x? = -11.
Javob: -11.
27. Qarang: 3-variant 15-savol
(26-bet).
28. /(x) = 70sinx sin6x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko'paytmadan yig'indiga ofamiz.
1
sinx-sin6x =— (cos(x - 6x) -
-I
- cos(x + 6x)) =— (cos5x - cos7x)
2) f(x) = (cos5x- cos7x) =
= 35cos5x- 35cos7x
Boshlang'ich funksiyasini topamiz.
F(x) = 7sin5x- 5sin7x + C.
Javob: 7sin5x - 5sin7x + C.
29. Qarang: 8-variant 7-savol
(72-bet).
30. Beshta ai, a2, аз, ад, as tub
sonlar ayirmasi 6 ga teng bo'lgan
arifmetik progressiyani tashkil qiladi.
ai + as ni toping.
142
Yechimlar. Matematika va informatika 2017
18-variant
Yechish:
ai, a2, аз, a4, as - tub sonlar.
d = 6
2, 3, 5, 7, 11, 13, 17, 19, 23...-tub
sonlar.
ai = 5, d = 6 bolsa a2 = 11, a3 = 17,
a4 = 23, a® = 29.
a4 + a5 = 23 + 29 = 52.
Javob: 52.
31. MS Excel 2003 dasturida berilgan =ЗНАК(-1256)+МИН(15;16;17)
formulaning natijasini aniqlang.
Yechish:
Excel dasturida ЗНАК funksiyasi qavs ichidagi sonning ishorasiga bogliq
bo'ladi, ya’ni 3HAK(son);
- agar son musbat bo'lsa 3HAK(son) = 1
- agar son manfiy son bo'lsa 3HAK(son) = -1
- agar son 0 ga teng bo'lsa 3HAK(0) = 0
qiymatlar bolishi mumkin.
МИН(а;Ь;с) funksiyasi qavs ichida berilgan sonlarning eng kichigini beradi.
3HAK(-1256) = -1.
МИН(15;16;17) = 15.
-1 + 15 = 14.
Javob: 14.
32. Paskal dasturlash tilida berilgan ushbu ifodaning qiymatini toping.
trunc(sqrt(abs(trunc(9,5)+sqrt(100)*round(3,5)))).
Yechish:
trunc(sqrt(abs(trunc(9,5) +sqrt( 100) *round(3,5))))
round(3,5) = 4
SQRT(100) =4100 = 10
trunc(9,5) = 9
abs(9 + 10-4) = abs(49) = 49
SQRT(49) = 449 = 7.
trunc(7) = 7.
Javob: 7.
33. A1 = -7, B1 = 9, B2 = 4 bo'lsin. Quyidagi formula natijasi 7 ga teng
bolishi uchun A2 katakka kiritilishi kerak bo'lgan qiymatni aniqlang.
=ЕСЛИ(И(А1 +B2>=A2*B1 ;A1 *B1 <0);
Yechish:
A1 = -7; B1 =9, B2 = 4, A2=?
Natija = 7.
i =ЕСЛИ(И(А 1+B2>=A2*B1;A 1 *B1<0);A 1-B2+B1-A2;A1*B1+B2-A2).
Bu misolda mantiqiy ifoda A1 + B2> = A2*B1; va A1*B1 < 0.
-7 + 4>A2-9 -7 9 <0.
-3>A2 9. -63 <0
rost
Bu ikkala mantiqiy ifoda rost bo'lsa, -7-4 + 9-A2 = 7 bo'ladi.
‘ A2=-9.
143
Yechimlar. Matematika va informatika 2017
19-variant
Aks holda yolg'on bo'lsa,
-7-9 + 4 —A2= 7
-63 + 4 - 7 = A2
A2 = -63 - 3= -66 bunday javob yo'q.
A2 = —9.
Javob: -9.
34. Qarang: 7-variant 31-savol (68-bet).
35. Qarang: 11-variant 33-savol (102-bet).
36. Qarang: 7-variant 36-savol (70-bet).
19-variant
1. To'g'ri burchakli ABC
uchburchakda Z A = 20°. BC = 12 sm li
katetni diametr qilib aylana chizilgan.
Aylananing uchburchak ichida bo'lgan
qismining uzunligini toping.
Yechish:
Z A = 20°, BC - 12, R = 6
Yechish:
c
Z C = 90°, AB = 52°
CD - mediana, CE - bissektrisa
Z DCE ni topamiz.
CD - mediana, CD = DB, Z DCB = 52°
CE - bissektrisa, Z ACD = Z ECB = 45°
ZDCE = .'DCB-YECB =
= 52° - 45° = 7°.
Javob: 7°.
CD yoy uzunligini topish kerak.
Uchburchak BOD da Z В = 70°,
2D = 70°, chunki OB=OD= R.
Л О = 40°
CD yoy uzunligi yarim aylana uzunligidan
DB yoy uzunligini ayirganimizga teng.
. 2itR TtRa c tl-6-40
2 180° 180°
. 4л 14л
= 6л-----=-----.
2. Qarang: 7-variant 16-savol
(65-bet).
3. To'g'ri burchakli uchburchakning
bir burchagi 52° ga teng bo'lsa, to'g'ri
burchak uchidan tushirilgan
bissektrisa va mediana orasidagi
burchakni toping.
4. Hisoblang: 1-
1
1 зУз
2V3
Yechish:
n 3 6-3 3 Уз
J 243 = 243 ~ 2^3 ~ 2
. з4з . „ -
2) 1-=^ = 1-6 =-5
4з
2
1
3) 1—— = 1 + 0,2 = 1,2.
-5
Javob: 1,2.
5. Qarang: 13-variant 21-savol
(113-bet).
6. Qarang: 5-variant 1-savol
(42-bet).
144
Yechimlar. Matematika va informatika 2017
19-variant
7. Sx2 + 4x3 - 3x - 7 = 0 tenglama
ildizlari ko'paytmasini toping.
Yechish:
4x3 + 8x? - 3x - 7 = 0
ax3 + bx2 + ex + d = 0 tenglama
ildizlari ko'paytmasi.
, . .3 d d .
Xrx2x3 = (-1) — = — teng.
a a
Shunga asosan a = 4, d = -7
4x3 + 8x2- 3x-7 = 0
d -7 7 .3 .
хгхгх3 = — =—-=-=1-= 1,75.
a 4 4 4
Javob: 1,75.
8. Qarang: 5-variant 7-savol
(43-bet).
9. Qarang: 12-variant 27-savol
(107-bet).
10. Rasmda у = f'(x) funksiya
grafigi tasvirlangan у = /(x) funksiya
grafigi Xi = 2 va x2 = 3 abssissali
nuqtalarida o'tkazilgan urinmalar
orasidagi o'tkir burchakni toping.
Yechish:
fW = tga
Xo = 2, f'(2) = -1 = tga = kj
Xo = 3, f'(3) = 0 = tgfi = k2
Urinmalar orasidagi burchak:
tg<p =
1 + k2-k1
0 + 1
1 + 0-I-1)
= 1
Tt
V = ~4-
Javob: —.
4
11. Tenglamani yeching:
6.5‘<® * + 2 • 5‘°82 = 12 x2 + 2 41082 21
Yechish:
Aniqlanish sohasi: x> 0
Э'О82 *2 y2
jflog2 x-1 _ q2 log2 x-2 _ r __ л_
22 4
x2
6 5‘°82 * + 2 • 51082 *-1 = 12x2 + 2 • —
4
25x2
5iog2^ (6.5 + 2) = ^y-
! x 125x2
5 82 =-------- kasr maxraji
64
64 bolganligi uchun x = 8 tenglama
ildizi boladi.
Javob: 8.
12. Qarang: 10-variant 24-savol
(90-bet).
13. cos2x < cos6x tengsizlikni
yeching.
Yechish:
cos6x - cos2x > 0 ayirmadan
ko'paytmaga 0‘tamiz.
cos6x - cos2x -
„ . 6x + 2x . 6x-2x
= -2sin------sin---------=
2 2
= -2sin4xsin2x
-2sin4x sin2x > 0, -sin4x sin2x s 0
Tengsizlik quyidagi tengsizliklar
sistemasiga teng kuchli.
„ [sin4x>0 fsin4x<0
1) \ 2) I
sin 2x<0 sin 2x>0
fsin 4x>0
[sin2x < 0
(2тгп < 4x <л + 2лп
[-л: + 2тт <2x< 2лп
7ГП Tt ТГП
—<x<—+---
2 4 2
---F 7ГП < X < 7ГП
2
ЯП _ It 7СП
—<x<—+----
va — + xn< x <яп
2 4 2
145
Yechimlar. Matematika va informatika 2017
19-variant
sin4x<0
sin 2x > 0
—л + 2лп < 4x < 2лп
2лп <2х<л + 2лп
л лП лП
— + — < x<---
4 2 2
2
л ЛП ЛП
—+—<x<--------
4 2 2
лп < x <~ + nn yechimlarni
umumlashtirib tengsizlikning yechimini
topamiz.
1) ning yechimi
1
2 4
я
5я
71 TV i • I 7L ЗтГ
— + тгл;— + яп yokl\—+niT.,— + тгп
2 4
2) ning yechimi
2 4
x 3x
2 T
о
я
Зя
1 1
3) —1 2 cos2a £ 1, у --cos2a
7 2 4
11 11
утах=---с0&2а = ---(-T)-
= 1 + 1 =1=0,75
2 4 4
_1 1 „ _1 1 .1 1 _
y™„----cos2«-- --1-2 --
=—=0,25
4
Eng katta qiymati 0,75.
Javob: 0,75.
15. 2lgx1 2 - (lg(-x))2 = 4 tenglamani
yeching.
Yechish:
Aniqlanish sohasi: -x> 0, x <0
2lg(-x)2-lg2(-x) = 4
4lg(-x) - lg(-x) = 4, lg(-x) = a
4a - a2 = 4
a2 -4a + 4 = 0
(a-2)2 = 0, a = 2
lg(-x) = 2,-x = 10?, x = -100.
Javob: -100.
Л
2
Л
— + лП;—t-лп
\_4 2 j
Demak, tengsizlikning yechimi:
f— + яп;— + лп ,nCZ.
J4 4
Javob:
7Г 3/Г
—+ 7ГП1-----+ 7ГП , П C Z.
4 4
14. Ifodaning eng katta qiymatini
1
toping: —cos2a + sin2a .
4
Yechish:
1) Darajani pasaytiramiz:
. 2 7-cos2a 1 1 „
in a =--------=------cos 2a
2 2 2
1 11 11
2)—cos2a +------cos2« =---—cos2a
4 2 2 2 4
fg2tgx+cosy __ 3
16. t sistemani
=2
yeching.
Yechish:
f^2^x+cosy — Гg4tgx+2cosy _
1 gcosy _ = 2 32cosy _ gttgx _ 2
Birinchi tenglamadan
4tgx + 2cosy = 1
2cosy = 1 - 4tgx topib ikkinchi
tengiamaga qo'yamiz.
__ g4tgx _ 2
34tsx = a, l-a = 2,
a
a2 + 2a- 3 = 0
a = -3, a = 1
3f"sx = -3 tenglama yechimga
34tgx = = 4tgx = 0
146
Yechimlar. Matematika va informatika 2017
19-variant
tgx = 0, x = Tin, n 6 Z
2cosy = 1 - tgx,
tgx = 0 bo'lganligi uchun
1
cosy =—
у = +~ + 2лк, к e Z
X = я:П
у = ±— + 2xk'
Г 3
Javob: x = Tin, у = ±— + 2лк, n, к C Z.
3
17. Jx2~6x < 2x + 8 tengsizlikning
nomusbat butun yechimlari nechta?
Yechish:
Tengsizlik quyidagi
x2-6x>0
2x + 8>0 =>
ylx2 - 6x <2x + 8
x(x-6)>0
x>-4
(Vx2-6x)2 <(2x + 8)2
x<0,x>6
x > -4
(x2 -6x)<(2x + 8)2
tengsizliklar sistemasiga teng kuchli.
Sistemadagi uchinchi tengsizlikning
ikki qismini kvadratga kotaramiz.
(Vx2-6x)2 < (2 x + 8)2
x2 - 6x < 4x2 + 32x + 64
Зх2 + 38x + 64 > 0
x<-~,x>-3
3
32 -3
3
Sistemadagi birinchi va ikkinchi
tengsizliklar uchun
(-4; 0] U [6; oo ) oraliqdagi x ning
qiymati yechim bo'ladi.
Tengsizlikning umumiy yechimi (-3; 0].
3
Tengsizlikning nomusbat butun
yechimi 2 ta.
Javob: 2 ta.
18. Qarang: 2-variant 11 -savoi
(15-bet).
19. у = Iog2(arctg3x + arcctg3x)
1
funksiyaning x - — nuqtadagi
hosilasining qiymatini toping.
Yechish:
1) arctg3x + arcctg3x ;
2)y = log2—; y' = (log2-)' = 0;
2 2
И?] = а
Javob: 0.
20. Radiusi 10 ga teng О markazli
aylana berilgan. Agar AD = 4, BD = 9
bo'lsa, OD kesma uzunligi topilsin.
Yechish:
R = 10, AD = 4, BD = 9, OD = ?
Kesishuvchi vatarlarning xossasiga
ko'ra, AD DB = ED DC, ED = x,
DC = 20-x, DO = 10-x
4-9=x(20-x),
x2 - 20x + 36 = 0, x = 2, x = 18
OD = 10-x= 10-2=8.
Javob: 8.
147
Yechimlar. Matematika va informatika 2017
19-variant
21. Qarang: 9-variant 2-savol
(78-bet).
22. Qarang: 8-variant 30-savol
(76-bet).
23. 2 -зГ—"l =0
Vx-V Ix-V
tenglamaning ildizlari ko'paytmasini
toping.
Yechish:
z \ 2
------ = a beigiiash kiritamiz.
{.x-1 J
2 + a* 2 - За = 0, a2 - 3a + 2 = 0
tenglama ildizlari a = 1 va a = 2.
=7, -4— = +1
\x-1) x-1
1) —— = 7, x = x- 1, 0 = -1
x-1
tenglama ildizga ega emas.
2) — = -1,x=-x + 1,2x=1,x=-
x — 1 2
lx-7J
-*- = ±72
x-1
1) --- = ^2 ,x=42x-42
x-1
[2
x(>/2-1) = 42 , x=^—
42-1
2) ~4t- = -42 ,x=-42 x+42
x-1
1-1-42
x(42 + 1)-42 , x = —j=—
42+1
Tenglama ildizlari ko'paytmasi
1_
2'42-1 4г+1~
2((424-1) 2-1
Javob: 1.
24. Qarang: 7-variant 30-savol
(68-bet).
25. Qarang: 13-variant 8-savol
(110-bet).
26. Geometrik progressiya
hadlarining soni juft son. lining
hamma hadlari yig'indisi toq o'rindagi
hadlari yig'indisidan 3 marta katta.
Progressiya maxrajini toping.
Yechish:
bi + b2 + Ьз + b4 = 3(bi + Ьз)
bi(1 + q + q2 + q3) = 3^(1 + q2)
1 + q + q2 + q3 = 3 + 3q2
q3 -2q2 + q-2 = 0
q2(q - 2) +(q - 2) = 0
(q - 2)(q2 + 1) = 0
q -2 = 0,
q = 2.
Javob: 2.
27. Qarang: 8-variant 1 -savol
(70-bet).
28. 7(x) = x (x + 1) bo'lsa,
Д1) +7(2)+ /(3) + ...+ДЗЗ)
yig'indining qiymatini toping.
Yechish:
f(1)+ f(2) +f(3) + ...+f(33) =
= T2 + 2-3 + 3-4 + ... +33-34
yig'indini hisoblaymiz.
1-2+ 2-3 + 3-4+ ... +n(n+ 1)
yiglndini topish formulas! quyidagicha:
1-2 + 2-3 + 34+ ...+n(n+ 1) =
_ n(n + 7)(n + 2)
Shunga asoslanib yechamiz.
1-2 + 2-3 + ... + 33-34 =
_ 33 (33 + 7)(33 +2) _
3
= 33:34 35 = 11-34-35= 13090
3
Demak, f(1) + f(2) +f(3) + ...+ f(33) =
= 1-2 + 2-3 + 3-4 + ... + 33-34 = 13090.
Javob: 13090.
148
Yechimlar. Matematika va informatika 2017
19-variant
29. Aniq integralni hisoblang:
f sin4xdx.
Yechish:
1) sinx + cosx = 42 s/h x + —
Yechish:
1
1) Jsin4xdx =—cos4x + C
72 1 12
2) fsin4x =—cos4x =
* 4
4 7
7 . ж 7 . ( it
= —cos4----p—cos4- —
4 72 4 I 4
л 1 7
=----cos— + —cos л: =---1-
3 4 4
1 , 11
f-(-7) =---
4 8 4
2
3 '
. U 2
Javob: —
3
30.
4sin2x-1
л/З- (sinx + cosx)
tengsizlikni yeching.
-7 <sin\ x+— | < 7
I 4)
-42 ±42 sin\ x + — 15 V2
I 4 J
a = 4з - (sinx + cosx) =
= 43 - 42 sin \ x+ —\
I 4 J
43-42<а<4з+42
Ixtiyoriy x da musbat, shuning uchun
4sin2x - 1 > 0.
. 2 . _ _ , 7~cos2x . „
2) 4sin x - 1 >0, 4-------7 > 0
2(1 - cos2x) - 7 > 0,
cos2x <—, — + 2яп <2x< — + 2яп
2 3 3
я 5я
— + яп <х< — + xn,neZ.
6 6
Javob: | — + яп;~ + яп |,neZ.
16 6 J
31. MS Excel 2003 dasturida berilgan =ИЛИ(СТЕПЕНЬ(3;4)=80;
MAKC(15;10;30)<30) formulaning natijasini aniqlang.
Yechish:
=ИЛИ(СТЕПЕНЬ(3:4)=80;МАКС(15;10;30)<30).
Bu funksiya sintaksisi
ИЛИ(А;В). Ava В- mantiqiy ifoda.
Agar A va В ifodalar rost bolsa, ИЛИ(А;В) qiymati ИСТИНА;
Agar A va В ifodalardan bin rost biri yolg'on bolsa, unda ham ИЛИ(А;В)
qiymati ИСТИНА;
Agar A va В ifodalaming har ikkalasi yolg'on bo'lsa, ИЛИ(А;В) qiymatiЛОЖЬ.
СТЕПЕНЬ(3;4)=80 yolg'on.
MAKC(15;10;30)<30 yolg'on.
Bu misol natijasi ЛОЖЬ.
Javob: ЛОЖЬ.
32. |(34 + 91/2)| ifodaning Microsoft Excel 2003 dasturidagi to'g'ri formula
ko'rinishini toping.
Yechish:
Son yoki biror ifodaning modulini Excel dasurida ifodalash uchun ABS
funksiyasidan foydalaniladi.
149
Yechimlar. Matematika va informatika 2017 20-variant
Masalan: |a - 2| ifoda =ABS(a-2) kabi yoziladi.
Sonni darajaga kofarish uchun Степень funksiyasidan foydalaniladi.
Masalan: 52 Excel dasturida yozish =Степень(5;2)
Sonni ildizdan chiqarish uchun esa КОРЕНЬ funksiyasi ishlatiladi.
Masalan 436 yozish uchun -КОРЕНЬ(Зб)
\34 + 91/2\ Excel dasturida ko'rinishi:
34 -+ =Степень(3;4)
91/2 = Jg =K0PEHb(9)
Bundan foydalanib,
=АВ5((Степень(3;4)+КОРЕНЬ(9)))); kelib chiqadi.
Javob: =АВ8((Степень(3;4)+Корень(9))).
33. (162 -43)4 + |5 + (-3)3| ifodaning Microsoft Excel 2003 dasturidagi
formula ko'rinishini toping.
Yechish:
Microsoft Excel 2003 dasturida an darajaga ko'tarish uchun степень nomli
funksiyadan foydalaniladi. Buning uchun Вставка —> Мастер функций buyruq
bajariladi.
CmeneHb(son;daraja);
Masalan: 23 excel 2003 dasturida yozish quyidagicha amalga oshiriladi.
~степень(2;3);
Sonning absolyut qiymati, ya’ni moduli ABC funksiyasidan foydalaniladi.
Masalan: |2 - 4\ yozish uchun
=ABS(2-4);
(162- 4j + 15 + (-3/| formulani yozish quyidagicha amalga oshiriladi.
-Степень(( 16*16-Степень(4;3));4)+АВ5(5+(-3)*Степень(-3;2));.
Javob: =Степень(16*16-Степень(4;3);4) + АВ8(5+(-3)*Степень(-3;2)).
34. Qarang: 6-variant 36-savol (61-bet).
35. O'zbekistonda qachondan boshlab Internet provayderlar xizmat ko'rsata
boshladi?
Yechish:
O‘zbekiston Respublikasida 1997 yildan beri internet xizmatini ko'rsatuvchi
provayderlar o‘z ishini boshlagan. Hozirgi kunda yildan yilga provayderlar soni
oshib bormoqda.
Javob:1997.
36. Qarang: 2-variant 36-savol.(22-bet).
20-variant
1. Bir arifmetik progressiyaning
ikkinchi hadi 14, uchinchi hadi 16.
Shunday geometrik progressiya tuzingki,
uning maxraji arifmetik progressiyaning
ayirmasiga teng bo'lib, ikkala
progressiyaning oldingi uchta
hadlarining yig'indilari teng bolsin.
Yechish:
аг = 14, аз = 16
d = аз - аг = 16- 14 = 2
ai = a2 - d = 14 - 2- 12
ai + a2 + a3 = 12+ 14 + 16 = 42
Geometrik progressiyada
d = q = 2
150
Yechimlar. Matematika va informatika 2017
20-variant
bj + b2 + Ьз — 42
bi(1 + q + q2) = 42
bi=6,q = 2,b2 = 12, b3 = 24
Demak, 6; 12; 24.
Javob: 6; 12; 24.
2. 9х - 5х - 4х = 2v2(F tenglama
nechta yechimga ega?
Yechish:
9х — 5х — 4х = 2^207
20х =4X-5X, у/20* =4? -52
Bundan 9х = 5х + 2-5~2 -42 + 4х
9х = (5? + 42 /yOki(92)г = (52+42 )2
дг = 5! +47
х
2
= 1 bo'lganda, tenglama yechimga
х
еда boladi, demak, ~= 1, x -2.
x -2 tenglama yechimi. Tenglama
1 ta yechimga ega.
Javob: 1 ta.
3. Iogx256 = 4(2 -V2 +
V2
2-V2
—-— + ...) tenglamani yeching.
Yechish:
Aniqlanish sohasi: x> 0, x + 1
» (2-V2)1 + -t= + - + ...
I 42 2
1 1
1 + -/= + — +... yig mdini hisoblaymiz.
bi = 1,b2=^=, b3=t
42 2
s_ b, _ 1 42
1-q 7____!_ V2-1
42
logx256 = 4 (2 - 42 ) ~—
\2 -1
/2
logx256 = 442 (42 - l)—¥-
42-1
logx256 = 8, 256 = xa, 2a = xa,x = 2.
Javob: 2.
4. Qarang: 9-variant 18-savol
(82-bet).
5. Qarang: 10-variant 12-savol
(87-bet).
2
6. Rasmda у = a +-----funksiya
bx + c
grafigi tasvirlangan. Quyidagilardan
qaysi biri noto‘g‘ri?
Yechish:
2
у = a +-----
bx + c
Funksiya grafigi faqat II, IV choraklarda
joylashgan, shuning uchun a =0,
c-0,b<0.
Javoblardan A javob noto'g'ri, chunki
masala shartida a = 0, b <0.
Javob: A.
7. |sinx|t9X + |cosx|t9X = 1.
Yechish:
Bunda 3 ta holni ko‘rish mumkin.
1) tgx = 2
2) tgx > 2
151
Yechimlar. Matematika va informatika 2017
20-variant
3) tgx < 2
1) tgx = 2,x = arctg2 +лп, n CZ
|s/nx|2 + |cosx|2 = 1
sin2x + cos2x = 1
2) tgx >2 da |s//?x|3 + |cosx|3 = 1
a) sinx = ±1 da cosx = 0, lekin tgx da
cosx + 0.
b) cosx = ±1 da, sinx - 0, bundan tgx = 0,
lekin (f + 1° Z 1, 0° - aniqmaslik.
2) va 3) hollarda tenglama yechimga
ega emas. Demak, tenglama yechimi.
x = arctg2 + лп, n e Z.
Javob: arctg2 + тгп, n Q Z.
8. |3‘9“ - 31,9"x| > 2 tengsizlikni
yeching.
Yechish:
3®" _ 31-t»x > 2 yokj yg» _ 31-tgxx < _?
у __ ^l-tgrrx > 2
3®“ _ > 2 3®" = a
^tgirx ’
a-~>2, 3—.—
a a
(a~3)(a + 1)^0
a
-1 s a < 0, a a 3
-1 s 3tg" < 0 tengsizlik yechimga ega
vnas.
,f9“ > 3, tgnx > 1,
'- + лк < лх < — + лк
г . 2
+k^x<— + k,k&Z
2
^tgxx _ 31-tgxx < _2 3tg« _ a
-- + 2 < 0, < о
a a
+ 3Xa-1)'Q
a
-3
3®”* < -3. X = 0
0 < З,3'я < 1, tgnx 0,
2
1
— + n < x < n.n eZ
2
Г7 , 1 ,} ( 1
— + k;— + k u —
L4 2 J I 2
Г1 1 W 1
Javob: — + k;~ + k и — + n;n ,
L4 2 J I 2 J
n, кez.
n, кez.
9.4—-ff27 - 22—1:1--2-|-3-
21 (X 16 J 4 8) 3
ni hisoblang.
Yechish:
4 ((
4— 27-22
21 Ц
-27-
8
_,2_88
3 21
-3^
3
21'
n2
^=M(4lLi_27-
3 21\ 16 5 8
= ^.(H 1-2-кз- =
21{.1б '5 8J 3
( 2--2^-\-3^- = ——-3~ =
[4 8 J 3 21 8
3
2 2
= 3--3- = 0.
3 3
Javob: 0.
+ a3 ni
10.-ь=+а;ь/.2а;ь:
a 3b 2 -b 3a-2
soddalashtiring. (a b)
Yechish:
Eng kichik darajani qavsdan
tashqariga chiqarib yechamiz.
1) a 3b 2 -a 2b 3 =a~2b~2 - (a3 -b3)
2) a 3b 2 - b 3a~2 = a~2b 2 (a3 - b3 )
152
Yechimlar. Matematika va informatika 2017
20-variant
3)
a2b~2(a3 -b3)
\ J (a3-b3)(a3+b3)
a3 -b3
= a3 - b3 +a3 +b3 = 2a3.
Javob: 2a3 .
11. Qarang: 15-variant 19-savol
(125-bet).
12. /(x) = x -Vx +2 funksiyaning
[-2; 2] kesmadagi eng katta qiymatini
toping.
Yechish:
1) f (-2) = -2 - Vr2+ 2 = -2
1
24x72
1
247+2 = 1,
Vx + 2 =
2
1 7
x + 2 =71 x =-—€ [2;-2]
4 4
4) 4~zl = “7'/4+2 =
< 4) 4 Y 4
= -L_l^-.?_ = -2,25
4 2 4
5)f(-2)=-2,f(2)=0,
= -2,25
fmax ~ 0,
fmin = -2,25.
Javob: 0.
13. Qarang: 13-variant 22-savoi
(113-bet).
14. a = 3 bo'lsa,
a+1
J (sin2 3x + cos2 3x)dx integralni
a
hisoblang.
Yechish:
1) sin23x + cos23x = 1
Javob: 1.
15. a ning qanday butun qiymatida
у = -x2 + 2x + a funksiya faqat 1 ta
nomanfiy butun qiymatga ega?
Yechish:
у = -x2 + 2x + a funksiya 1 ta nomanfiy
butun qiymatga ega bolishi uchun
—x2 + 2x + a = 0 da D = 0 bolishi kerak.
D = 22-4(-1)-a = 0
4 + 4a =0, a =-1.
Javob: -1.
16. З2"12*1 >3i-2ooS2x tengsizlikni
yeching.
Yechish:
a = 3 > 1 demak,
2cos2x-6> cosx
2cos2x-7 7-2cos2x
2cos2x-6> cosx
2cos2x-7 2cos2x-7
2 cos2 x -6 cosx
---------+----------> о
2cos x-1 2cos x-1
2cos2 x + cosx-6 _
-------------> (j
2cos2 x-1
( 3^l
2(cosX + 2) cosx —
-----------L-----'LL.
(42 cosX-7^V2cosx + tj
3
(cosx + 2)(cosx~— ) <0,
chunki -1 s cosx < 1
Bundan
(42 cosx - 1)( 42 cosx + 1) <0
153
Yechimlar. Matematika va informatika 2017
20-variant
Javob: — + лп;— + яп ,n eZ .
<4 4 )
17. 2e '°г“ - xlnx = e4 tenglamani
yeching.
Yechish:
2eь,2х - x"“ = e4
2(elnx)lnx - xlnx = e4, elnx = x
2x,nx - x,nx = e4
x'“ - e4 tenglikning ikki qismini
logarifmlaymiz.
Inknx = Ine4
Inx Inx - 4lne
ln2x = 4, Inx - ±2
1) Inx = 2, x = e2
2) Inx = -2, x = e~2.
Javob: e2, e“2.
18. Har qanday x C (xi, x2) uchun
У = f(x) funksiya hosilasi manfiy bo'lsin.
(x1f x2) oraliqqa tegishli ixtiyoriy a va b
(a > b) uchun qanday tengsizlik o'rinli?
Yechish:
x C (x-i; хг). uchun f(x) <0. a 6 (xi; x2),
b C (xi; x2).
(a >b) uchun f (a) < f(b) tengsizlik o'rinli.
Chunk! hosila manfiy bo'lsa, funksiya
kamayuvchi bo'ladi. a > b uchun esa
f(a) <f(b) tengsizlik to'g'ri bo'ladi
funksiya kamayuvchi bo'lganligi uchun.
Javob: /(a) < f(b).
19.1 + log?sinx + log \ sinx +
+ log f sinx +
2
=—tenglamani yeching.
3
Yechish:
Aniqlanish sohasi:
sinx > 0, 2nn <x <n + 2xn
Geometrik progressiya bi - 1,
b2 = log2sinx, q = log2sinx,
q < 1 - cheksiz kamayuvchi
s . b, _ 1
1 - q 1 - log2 sin x
1 2
1 - log2 sin x 3
3 = 2- 2log2sinx
log2sinx = -1
sinx = 2?
1 -
sinx = —, x = (-1)” q + лп, n € Z.
Javob: (-1 )n — + лп, n C Z.
6
20. ABC uchburchakning
AC tomonida D nuqta olindi. Agar
Z ABC = Z BDC bo'lib, ЗАВ = 4BD va
BC = 6 sm bo'lsa, AC kesma
uzunligini (sm) toping.
Yechish:
Z ABC = Z BDC
ЗАВ = 4BD, BC = 6, AC = ?
В
Sinuslar teoremasiga ko'ra ABC
l.,. c. ,
uchburchakda-----=-----.
sin у sin p
onz- U I o,75x BC
BDC uchburchakda-------=------.
sin у sin fl
x AC BC 6-4
------=----, ли =------=----= о .
0,75x BC 0,75 3
Javob: 8.
21. lx2 - 5ax| = 15a tenglama
a ning qanday qiymatlarida kamida
2 ta ildizga ega?
154
Yechimlar. Matematika va informatika 2017
20-variant
Yechish:
|x2 - 5ax| modul doimo musbat
bo'lganligi uchun 15a 2 0, a > 0.
Tenglamani grafik usulida yechamiz.
Tenglama kamida 2 ta yechimga ega
bo'lishi uchun a > 0 bo'lishikerak.
a - Oda tenglama 1 ta yechimga ega
bo'ladi.
grafiklari kamida ikkita nuqtada
kesishishiuchun a >0bolishikerak.
Javob: (0; oo).
22. О markazli aylanada AB vatar
diametrini N nuqtada 60° li burchak
ostida kesib o'tadi. Agar AN = 10 sm,
BN = 4 sm bolsa, ON kesma
uzunligini toping.
Yechish:
AB - vatar, AN = 10, NB = 4,
/. ANO = 60°, ON = ?
О nuqtadan AB vatarga
perpendikulyar tushiramiz.
AB = 14, AP = PB = 7,
NP =7-4 = 3
Z PNO = 60°, Z NPO = 90
Z NOP = 30° bundan
NP
sin.30°=——,
ON
NP
ON = -!?—= 3-2 = 6.
sin 30°
Javob: 6.
23. у = 5cos4x + cos20x
funksiyaning hosilasini toping.
Yechish:
1) (cos(ax + b))' = -asinfax + b)
(cos4x)' = -4sin4x
(cos20x)' = -20sin20x
y' = 5(cos4x)' + (cos20x)' = -20sin4x -
- 20sin20x = -20(sin4x + sin20x)
2) sin20x + sin4x yig'indidan
ko'paytmaga o'tamiz.
. „„ , . . „ . 20x + 4x
sin20x + sin4x = 2sm--------
2
-cos—-^~—— = 2sin12xcos8x
2
3) y' = -20-2sin12xcos8x =
= -40sin12xcos8x.
Javob: -40sin12x-cos8x.
24. Doira ichida berilgan nuqtadan
o‘tkazilgan vatar bolaklarining biri
ikkinchisidan 2 marta katta, diametr
bo'laklari esa 4 va 12,5 ga teng. Vatar
bo'laklarini toping.
Yechish:
D
Kesishuvchi vatarlar xossasidan
foydalanib yechamiz.
AP = 12,5; PB = 4; DP = 2PC
AP-PB = DP-PC
12,5-4 = 2PC2, PC2 = 25, PC = 5
DP = 2PC = 10
10 va 5.
Javob: 10 va 5.
25. Qarang: 11-variant 19-savol
(98-bet).
1
26. Hisoblang: cos(2arcctg —).
4
Yechish:
1
cos(2arcctg—) =>
155
Yechimlar. Matematika va informatika 2017
20-variant
„ 1-tga ctga-1
=> cos2a =------ ---------
1 + tg a ctg a +1
•/
1 ctg2arcctg—-1
cos(2arcctg— ) =----------—
4 ctg2arcctg -+1
Javob:
1-16
1 + 16
15
17 '
15
17
27. Uchburchakning uchlari to'g'ri
burchakli dekart koordinatalar
sistemasida quyidagicha berilgan:
1
A(0; 0), B(~; -4), C(-1; 0).
Uchburchak yuzini toping.
Yechish:
1) A(0; 0), C(-1; 0), S = ?
2) Uchburchak ABC - teng yonli.
AC- asos, AC - 1, BD- balandlik,
BD = 4
Javob: 2.
28. Qarang: 10-variant 27-savol
(91-bet).
29. arcsinzx < arccos2x tengsizlikni
yeching.
Yechish:
1) (arcsinx)2 - (arccosx)2 < 0
0 < (arcsinx)2
(arcsinx - arccosx)-
(arcsinx + arccosx) < 0
2) arcsinx + arccosx =— > 0
2
arcsinx - arccosx < 0
arcsinx < arccosx
3) Tengsizlikning ikki qismini
sinuslaymiz.
x
x>
x2 <1-x2
\x>0
1 1
0^x<-~, [0;-^)
42 42
fx<0 fx<0
2) < => < =>
(1-x2>0 [-1<x<1
-1 <x <0,0.
1
Javob: [0; —).
42
30. (arcsinx)3 + (arccosx)3 = л3.
Yechish:
(a + b)3 - 3ab(a + b)=a3 + b3
a + b = arcsinx + arccosx = —
2
/ V3
I It | л 7Г - 3
_ -J — arcsin x • arccos x = тг
V2j -2
7 2
arcsinx arccosx ----x
12
arcsinx = у
arcsinx + arccosx - —
arccosx =— - у
2
(x ) 7 ,
/-----у =------л-
U ) 12
156
Yechimlar. Matematika va informatika 2017
20-variant
n 7 2
— у----
2 7 12
- 0 kvadrat tenglamani
yechamiz.
it fir2 28ir2
у -2~'VT + ^2~
У 1,2 2
2 У 12
2
Bunda ildizning absolyut qiymatlari
— dan kattadir.
2
Demak, tenglama yechimga ega
emas.
Javob: 0.
31. Noto'g'ri mulohazani aniqlang.
Yechish:
Birinchi avlod mashinalari XX asrning 50-yillarida ishlab chiqarilib, uning
komponentlari eletron lampalardan iborat bolgan.
Javob: birinchi EHM avlodlari tranzistorlar asosida ishlangan.
32. 100101111(2) soni 16 lik sanoq sistemasida nechaga teng boladi?
Yechish:
100101111(2)- X(ib). Ikkilik sanoq sistemasidan o‘n oftilik sanoq sistemasiga
о . sh uchun ikkilik sanoq sistemasida sonni eng kichik razryaddan boshlab
tetradlarga ajratib chiqamiz. Eng katta razryadga yetmagan raqamlarni nollar
bilan to'ldiramiz.
0001 0010 11112 tetradlarga ajratdik.
1-jadvaldan foydalanib
00012 — lie
00102 — 2ib
11112-F16
0001001011112->12Fit,
Javob: 12Fi6.
33. “O‘zbekiston - Vatanim manim” iborasining hajmini toping (qo'shtirnoq
belgisi hisobga olinmasin).
Yechish:
Kompyuterda axborot Ova 1 raqamlardan iborat ikkilik sanoq sistemasida
kodlanadi.
Har bir belgiga 8 bit (1 bayt) mos keladi.
Masalan, “NON” so‘zi 3 bayt yoki 38 = 24 bitdan iborat.
1 bayt - 8 bit
1 Kbayt = 210 bayt = 1024 bayt
1 Mbayt = 210 Kbayt = 1024 Kbayt
1 Gbayt = 2W Mbayt = 1024 Mbayt
Bundan “O‘ zbekiston -Vatanim manim ”so‘zida 25 ta belqi bor.
1 2 3 4 5 S 7 8 9101112131415161718 19 20 21 22 23 24 25
Bu so'z 25-8 bit = 200 bitdan iborat.
Javob: 200 bit.
34. MS Excel 2003 dasturida formula bajarilishi natijasida #3нач xatoligi
sodir bo'lsa, bu xatolik turini aniqlang.
Yechish:
Agar Excel dasturida formula bilan ishlanganda son o'rniga matn yoki harf
ishlatilgan bolsa, #3HA4 xatoligi yuz beradi.
157
Yechimlar. Matematika va informatika 2017 21-variant
Masalan:
A1 yacheykaga a ni kiritsak,
A2 yacheykaga formula kiritsak,
=К0РЕНЬ(А1) и holda #3HA4 xatoligiyuz beradi. Chunki A1 yacheyka
qiymatida son o‘miga harfli ifoda noaniq ifoda mavjud.
Javob: argument sifatida sonning o'rnida matn turibdi.
35. Mikroprosessorlarning yaratilish davri:
Yechish:
Birinchi mikroprosessor Intel firmasining Intel - 4004 mikroprosessori bo'lib,
1970-yilda ixtiro qilingan. U 4 bitli so'zlar ustida sekundiga 8000 amal bajara
olish qobiliyatiga ega.
Intel- 4004 4 Kbayt hajmlimikroprosessorlar dasturlovchikalkulyator uchun
mo'ljallangandir.
Javob: 1970.
36. Qarang: 8-variant 36-savol (78-bet).
21-variant
1. Qarang: 3-variant 16-savol
(26-bet).
2. Qarang: 9-variant 29-savol
(84-bet).
3. Shunday to'rtta musbat son
topingki, ulaming oldingi uchtasining
yig'indisi 12 ga teng bo'lib arifmetik
progressiyani tashkil qilsin, keying!
uchtasining yig'indisi 19 ga teng bo'lib
geometrik progressiyani tashkil qilsin.
Yechish:
a, >0, a2> 0,
a3> 0, a4> 0
ai + a2 + a3 -12- arifmetik
progressiya.
a 2 + a3 + a4 = 19- geometrik
progressiya.
Arifmetik progressiyada ai + a2 + a3-
= 3a2 = 12, a2 = 4, a3 + a4- 15.
Geometrik progressiyada a2- 4
bo'lgani uchun a2 q + a2 q2 = 15
2 15 2 15 n
q2 + q=—,q+9~~ = 0
4 4
Berilgan sonlar musbat, shuning uchun
3 3
q =—, bundan a3 = a2-q = 4 —= 6.
2 2
a4=a2q2 = 4-~ = 9
4
ai + a2 + a3 = 12
a2 = 4, a3 = 6, a-i = 2
Demak, bu sonlar 2; 4; 6; 9.
Javob: 2; 4; 6; 9.
2-h/x+x
i— 71 ''i 2+2Л
4. V3-3Wx- - =81
<3;
tenglamaning ildizlari yig'indisini
toping.
Yechish:
an-ak = an*
Bir xil asosga keltirib yechamiz.
1 x 2+Jx+x
^2 . $1+4x . 3 2+2-Jx — 33 4
1 X 2+у/к+х
32 1+Jx 2(1+Vx) __ 34
1 x 2 + 4x + x
2 1 + Vx 2(1+ Vx)
1 +Vx + 2x-2-yfx ~x = 8(1 +y[x )
158
Yechimlar. Matematika va informatika 2017 21-vahani
x- 1 = 8 + 8Vx x - 8Vx -9-0 y/x = a belgilash kiritamiz. a2 -8a-9-0 tenglama ildizlari. a=-1, a = 9 y/x = -1 tenglama haqiqiy ildizga ega emas. 4x-9,x = 81 Tenglama ildizlari yig'indisi 81 ga teng. Javob: 81. 5. x2 - 4x + y2 + 6y = 12 tenglama bilan berilgan aylana uzunligini toping. 3) 1 - cos2a = cos2a + sin2 a - - cos2a + sin2a - 2sin2a 4) 1 + cos2a = cos2a + sin2a + + cos2a - sin2a - 2cos2a 5) sin2a = 2sinacosa 1 + sin2a-cos2a _ 1 + sin 2a + cos 2a _ sin 2a + 2 sin2 a _ sin 2a+ 2 cos2 a _ 2sinacosa + 2sin2a _ 2sinacosa + 2cos2 a _ 2sina(cosa + sina) sin a
Yechish: x2 -4х + У2 + 6y =12 (x - a)2 + (y- b)2 = R2 aylana tenglamasi x2-4x+y2 + 6y=12 (x - 2)2 - 4 + (у + 3)2 — 9 = 12 (x-2)2 + (y + 3)2 = 25 R2 = 25, R = 5 Aylana uzunligi C = 2itR = 10я. Javob: 10л. 2cosa(sina + cosa) cosa Javob: tga. 11. ABC uchburchakning AB, BC, CA tomonlarida mos ravishda shunday M, N, P nuqtalar olinganki, AM:AB = BN:BC = CP:CA = 1:3 munosabat o'rinli. MNP uchburchak yuzasi 2 ga teng bolsa, ABC uchburchak yuzasini toping. Yechish:
6. Qarang: 2-variant 28-savol (20-bet). 7. Qarang: 11-variant 3-savol (95-bet). 8. Qarang: 13-variant 16-savol (112-bet). AM:AB = BN.BC = CP:CA = 1:3 Smnp = 2, Sabc = ? c р/\ / _sXn AM 8 AC-AB
9. Qarang: 4-variant 22-savol (37-bet). 10. Soddalashtiring: S№r = . sin a , AC = 3y, AB = 3> Abb ‘ „ АРАМ o ... Samp = sm a, AP = 2y, AM = X rWlr ' J '
f Зтг 1 + sin2a + sin 2a V 2 ) 1 + sin2a-sin| ^ + 2a | I 2 ) Yechish: Sabc ^C AB _3y-3x _9 Samp AP AM 2y X 2 Xuddi shunday . J Q о с о ^BNM Smnp - Sabc — Samp — Scpn — Sbnm
1) sin | — - 2а | = -cos2a I 2 ) 2) sin f + 2</j = -cos2a 2 _ 1 Smnp = Sabc _ 3- — Sabc Sabc 9 3 Sabc - 3-Smnp - 3-2 = 6. Javob: 6.
159
Yechimlar. Matematika va informatika 2017
21-variant
12. Soddalashtiring:
^22^1272+-^1-—.
42+1
Yechish:
1) J22+ 1242 = ^22 + 2-2-342 =
= ^18 + 2-342-2 + 4 =
= ^з42 + 2^ =342 + 2
242-1 J242-1)(42-1)
42+1 (42+l](42-l)
4-342+1 c , X
=----------= 5-3v2
2-1
3) 342 + 2 + 5-342 = 7.
Javob: 7.
13. x va у sonlar ayirmasining
uchlanganini yozing va shu ifodaning
x = -0,37, у = -0,42 bo'lgandagi son
qiymatini toping.
Yechish:
x= -0,37, у = -0,42
3(x-y) = 3-(-0,37 - (-0,42)) =
= 3(-0,37 + 0,42) = 3-0,05 = 0,15.
Javob: 0,15.
14. Qarang: 18-variant 18-savol
(141-bet).
15. Qarang: 4-variant 25-savol
(38-bet).
16. Qarang: 1-variant 28-savol
(10-bet).
17. Qarang: 12-variant 19-savol
(105-bet).
18. Qarang: 18-variant 8-savol
(139-bet).
19. Qarang: 11-variant 2-savol
(94-bet).
20. Qarang: 7-variant 2-savol
(61-bet).
21. To'g'ri burchakli uchburchakning
bir kateti 5 ga teng. lining medianalari
kesishish nuqtasidan ikkinchi
katetigacha bo'lgan masofani toping.
Yechish:
Z C = 90°, CB = 5,
OD - ?
CE = EB=~ = 2,5
2
ДАСЕ va AADO o‘xshash. DO\\CE,
AE - mediana.
AO = 2x, OE = x, AE = 3x
СЕ AE
DO AO
„„ СЕ-AO 2,5-2x 5 .2
AE 3x 3 3
. l. -2
Javob: 1—.
3
22. Beshta а,, а?, аз, a4, as tub
sonlar ayirmasi 6 ga teng bo'lgan
arifmetik progressiyani tashkil qiladi.
a4 + as ni toping.
Yechish:
a-i, a2, аз, a4, a5 - tub sonlar.
d = a2 — a-j = Эз — a2 = a4 — аз — as — a4
Tub sonlar ayirmasi 6 teng bolishi
kerak. 5, 11, 17, 23, 29 ketma-ketlik
arifmetik progressiya tashkil qiladi.
a4 = 23, a5 = 29
a4 + a$ = 23 + 29 = 52.
Javob: 52.
23. a ning qanday qiymatlarida
x2 - 4|x| - a + 3 = 0 tenglama ildizlari
yig'indisi nolga teng?
Yechish:
Tenglama yechimga ega bolishi
uchun D>0 bolishi kerak.
160
Yechimlar. Matematika va informatika 2017 21-variant
D = 42 - 4(-a + 3)>0 16-4(-a + 3)>0 4 - (-a + 3) > 0, a + 1 > 0, a>-1 1 + cos(9x- 13x)) =— (cos22x + cos4x)
a >-1 da tenglama yechimga ega boladi. Ikkinchidan, a>-1 da tenglama ildizlari yig'indisi nolga teng bo'ladi. Masalan, 1) a =-1 da x2 - 4\x\ +4 = 0 flx| - 2)2 = 0, |x| = 2, x = ±2,-2+ 2 = 0 2) a = 0 da x2- 4|x| +3 = 0 |x| = 1, |x| = 3, x = +1, x = ±3 ildizlari yig'indisi 1 + (-1) + 3 + (-3) = 0. Javob: a s -1. 2) f(x) = 88- (cos22x + cos4x) = = 44cos22x + 44cos4x Boshlang'ich funksiyasi: F(x) = 2sin22x + 11sin4x + C. Javob: 2sin22x + 11sin4x + C. 29. (costcx + ) lg(3x - 2X2) = 0 ildizlari yig'indisini toping.
24. a va b sonlar qanday bo'lganda 2y + ax = b to'g'ri chiziq abssissa o'qi musbat yo'nalishi bilan 135° li burchak hosil qilib, (0; -2) nuqtadan o'tadi? Yechish: Yechish: Aniqlanish sohasi: Зх-2хг>0, 2x2-3x<0 x(2x-3)<0
2y + ax = b to'g'ri chiziqda a b . a у=—x + —,к=— 2 2 2 1) k = tga=tg135° = -1 -—=-1,a = 2 2 2)(0;-2) 2 b b у = —x + — = -x + — 2 2 2 -2 = 0+-,b=-4 2 Demak, a = 2, b = -4. Javob: 2; -4. 25. Qarang: 3-variant 20-savol (27-bet). 26. Qarang: 10-variant 21-savol (89-bet). 27. Qarang: 6-variant 1-savol (51-bet). 28. /(x) = 88cos9xcos13x uchun boshlang'ich funksiyani toping. Yechish: 0<x<1,5 Tenglamani yechamiz. 1 1 1) cosrrx +— = 0, cosnx = — 2 2 2% „ тех = + г 2лп 3 2 x = ±— + 2n,neZ 3 2 n=0dax=^ C(0; 1,5) 2 4 n = 1 da x =-— + 2 = —€(0; 1.5) 3 3 ' 2) lg(3x - 2X2) = 0 Зх-гх2^, 2x2-3x+ 1 = 0 x = 1, x =^ aniqlanish sohasiga tegishli yechimlar. 2 4 1 Ildizlari yig ‘indisi: —+— + 1 + — = 3,5. 3 3 2 Javob: 3,5.
1) Ko'paytmadan yig'indiga o'tamiz. 1 cos9xcos13x =— (cos(9x + 13x) + 30. л/il + 9 soni x2 + mx + n ko'phadning ildizi bo'lsa, m va n butun sonlar yig'indisini toping.
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Yechimlar. Matematika va informatika 2017
21-variant
Yechish:
Xi = V77 + 9 bo'lsa x2 = 9-JTl
-m = Xi + Хг
-m = 9+4Tl + 9-JTi = 18
m = -18
П = Xi x2
n = (9+ Jl1)(9-4l1) =
= 81—11 = 70
m + n =-18 + 70 = 52.
Javob: 52.
31. Nuqtalar o'rniga joylashtirish mumkin bo'lgan javobni aniqlang.
Axborotga old 1 bit orqali ... ifodalanadi.
Yechish:
Axborotni kodlashda ikkilik sanoq sistemasidan foydalaniladi. Axborotda bir
belgini ifodalash uchun 8 bit ishlatiladi. Bitlar ikkilik sanoq sistemasidagi 0 yoki
1 raqamlari orqjli ifodalanadi.
Javob: 0 yoki 1.
32. 4 megabayt necha baytga teng?
Yechish:
1 Mbayt = 210 Kbayt
1 Kbayt = 210 bayt.
4 Mbayt = 4-210-210 bayt = ?-210-210 bayt = = 222 bayt.
Javob: 222.
33. 25 + 23 + 22 + 2“1 berilgan yoyilmaning ikkilik sanoq sistemasidagi
ko'rinishini aniqlang.
Yechish:
2s + 2Э + 22 + 2~’ = 32 + 8 + 4 +- = 44 + 0,5 = 44,5;
44,5io—>X2.
44 + 0,5-+X2.
O'nlik sanoq sistemasidagi sonni ikkilik sanoq sistemasiga otkazish uchun
shu soni 2 ga ketma-ket bo'lib boriladi, toki bo'linma 2 dan kichik bo'lganga
qadar, keyin sonlar bo'linmadan boshlab barcha qoldiqlar yozib olinadi.
44|2_
4^Г2212_
4 2_ITi[2
4 2 101512
@ 2 ^41212
© ®2ф
44io~+ 101100(2).
O'nlik kasr sonni ikkilik sanoq sistemasiga otkazish uchun shu sonni 2 ga
ko'paytirib boramiz toki butun qismi 1 bo'guncha.
0,510-+X2.
5
*2
0
*2
0
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Yechimlar. Matematika va informatika 2017 22-variant
Demak, 0,5™—> 1г-
44,5ю^ 101100,1(2).
Javob: 101100,1(2).
34. Ma’lumotlar bazasining xotira fayli turini aniqlang:
Yechish:
Ma’lumotlar ba’zasining fayllari .dbf (Data Base file), dbt- ma’lumotlar ombori
dasturida memo - matn maydonlari uchun ishlatiluvchi kengaytmali fayllar.
Fayl kengaytmasi faylning turini, qaysi dasturda yaratilganligini bildiradi.
Javob: dbt, dbf.
35. Axborot tizimining ta’minoti:
Yechish:
Axborot tizimining ta’minotideganda texnik, matematik, axborot ta’minoti,
huquqiy ta’minot, tashkiliy-dasturiy lingvistik ta’minotlarga ajratilishiqabulqilingan.
Axborot ta’minoti axborot tizimlarida ma’lumot omboriniyaratish
hujjatlashtirish bir xil tartibga keltirilgan tizimlarini ichiga olgan axborotni
kodlashtirish, joylashtirish bo’yicha uslub va vositalar yig'indisidir.
Javob: texnik, matematik, axborot ta’minoti, huquqiy ta’minot, tashkiliy-
dasturiy.
36. Qarang: 3-variant 36-savol (31-bet).
22-variant
1 (4X+4 + Д~х) 41°82 х-1о82(5х3+6х2) < f
tengsizlikning eng katta va eng kichik
butun yechimlari nisbatini toping.
Yechish:
+ 4-*) 41о82х-1о82(5><3+6хг) < у
ko'rsatkichli tengsizlikning asosi
4X+4 + > 1, shuning uchun
4log2X - log2(5x3 + бх2) < 0
log2x4 < log2(5x3 + бх j
хГ< 5x3 + бх2
^()3-5x-6)<0
x2(x+ 1)(x-6) <0
x C (~1;°0) U (0; 6)
Aniqlanish sohasi:
(x>0 fx>0
|5x3 + 6x2 > 0|x2(5x + 6) > 0
=$x>0
Demak, tengsizlikning yechimi (0; 6)
oraliq. (0; 6) oraliqdagi eng katta
yechim 5, eng kichik yechim 1,
ularning nisbati 5:1 =5 ga teng.
Javob: 5.
2. Qarang: 7-variant 24-savoi
(66-bet).
3. Qarang: 17-va<4ant 7-savol
(133-bet).
4. Tengsizlikni yesteig:
5 3, 7
-----1------>-------- *
x-1 2x + 7 3x 3
Yechish:
Umumiy maxrajga keltiramiz.
5 3 7
x-1 2x + 7 3(x-1)
15(2x + 7) + 9(x -1) - 7(2x + 7) > Q
3(x-1)(2x + 7)
30x + 105 + 9x-9-14x-49 >^)
3(x - 1)(2x + 7) "
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Yechimlar. Matematika va informatika 2017
22-variant
25x + 47
(x-1)(2x + 7)
> 0 son o‘qida
suratning ildizini va maxrajning uzilish
nuqtalarini belgilaymiz va har bir oraliqda
kasrning ishorasini aniqlaymiz.
7 _47 1
2 25
47
25
u>(1;x).
Yechish:
f 1 )
1) A(0; 0), , C(-1; 0), S = ?
2) Uchburchak ABC - teng yonli.
AB - BC. AC - asos, AC = 1,
BD - balandlik, BD = 6
Л’
Javob:
47
25
u(1; co) .
5. Qarang: 17-variant 30-savol
(137-bet).
6. Qarang: 1-variant 30-savol
(10-bet).
ACBD1-6
О - --- — - — О .
2 2
x 3
7. f(— + 1) = -- x + 7, /(1) ni toping.
2 4
Yechish:
x a —-i
— — Q — I , X “ 2.3 — £
2
f(a)^ -2(a-1) + 7 (a - 1) + 7 =
= 1,5a- 1,5+ 7 = 1,5a + 5,5
f(a) = 1,5a + 5,5
f(1) = 1,5-1 + 5,5= 7.
Javob: 7.
8. Qarang: 7-variant 26-savol
(66-bet).
9. Qarang: 13-variant 30-savol'
(115-bet).
10. Uchburchakning uchlari to'g'ri
burchakli dekart koordinatalar
sistemasida quyidagicha berilgan:
i
A(0; 0), B(-j;-6), C(-1; 0).
Uchburchak yuzini toping.
Javob: 3.
1 2
11. log33- = -tenglamadan x ni
x 3
toping.
Yechish:
Aniqlanish sohasi: x > 0
1 , 12
1.3,x = l
x 3
Javob: -.
3
'og л
3 32 X
1 ~-
L = 323
X
12. Qarang: 3-variant 25-savol
(28-bet).
13. у = 13cos3x - 3cos13x
funksiyaning hosilasini toping.
Yechish:
1) (cos(kx + b))'--ksin(kx + b)
(cos3x)' = -3sin3x
(cos13x)' = -13sin 13x
2) y'= (13cos3x - 3cos13x)' =
= -3-13sin3x + 3-13sin13x = -39sin3x +
+ 39sin13x - 39(sin13x - sin3x)
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Yechimlar. Matematika va informatika 2017
22-variant
3) Ayirmadan ko‘paytmaga o'tamiz.
sin13x - sin3x = 2sin ——
2
13x + 3x „ . c „
cos--------= 2sin5xcos8x
2
4) y' = 39-2sin5xcos8x =
= 78sin5xcos8x.
Javob: 75sin5xcos8x.
14. Agar f(x) = x - 3 va
/(g(x)) = x1 2 3 4 - 6x + 6 funksiyalar
berilgan bo'lsa, g(x) funksiyaning
ko'rinishini aniqlang.
Yechish:
f(x) =x - 3 va f(g(x)) -x2- 6x + 6
f(g(x)) = g(x) - 3
g(x) -3 = x-6x + 6
g(x) -)3 — 6x + 9 = (x-3)2
g(x) = (x-3)2.
Javob: (x - 3)2.
15. Qaysi jism(lar)ning simmetriya
o'qlari chekli sonda?
1) shar; 2) prizma; 3) konus; 4) kub.
Yechish:
17. R radiusli sferaga muntazam
to'rtburchakli piramida ichki chizilgan.
Uchidagi yassi burchak 45° ga teng
bo'lsa, piramida yon sirtining yuzini
toping.
Yechish:
SABCD - muntazam to'rtburchakli
piramida.
SOj = H, SO = AO = R,
Z ASB = a, Syon = ?
A E a/2 В
S = 4aha =2ah
yon 2 a
ДА SB dan apofem an i topamiz.
S /-
tg-
2
ha=-^~
2tg~
*2
a
. а
2sm—
2
1) Shaming simmetriya o'qlari cheksiz
ko'p. Shaming diametri uning
simmetriya o'qi bo'ladi.
2) Prizma, ya’ni parallelepipedda
diagonallari kesishgan nuqta uning
simmetriya markazi bo'ladi, bundan
simmetriya o'qlari chekli ekanligi kelib
chiqadi.
3) Konusda simmetriya o'qi chekli.
4) Kubda to'qqizta simmetriya tekisligi
bo'lganligi uchun simmetriya o'qlari
chekli bo'ladi.
Javob: 2, 3, 4.
16. Qarang: 10-variant 6-savol
(86-bet).
AASOt dan H = J/2 -— = -aVc-a
’ 2 2sin-
2
AO = R, OO-i = H - R,
AO^^l
2
R2 =(H-R)2 +y
D a2+2H2
l \ —----
4H
2 „ a2 cosa
a + 2----------
. . 2 а
4sin -
2
4-------
2sin“
2
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Yechimlar. Matematika va informatika 2017
22-variant
a
. °- I----
4sm-->/cos«
. a I------
a - 4Rsin - \' cos а
Javob: 4R2.
18. Oldingi uchta toq hadining
yig'indisi 42 ga, oldingi uchta juft
hadining yig'indisi 84 ga teng bo'lgan
geometrik progressiyaning beshinchi
hadini toping.
Yechish:
(Ь} + b3 + b5 = 42
[b2 + b4 + b6 = 84
(bf+b.q2 + b,q4 =42
[fyq + b,q3 + b,q5 = 84
(bf(1 + q2 + q4) = 42
[b.qd + q2 +q4) = 84
bt(1 + q2 +q4) 42
b1q(1 + q2 + q4) 84’
b_ 42 _ 42 42 _2
1~1 + q2 + q4 ~1 + 4 + 16~ 21 ~
b5 = bi-q*=2-24 = 25 = 32.
Javob: 32.
19. (x2 - 9x + 15) = 7 tenglama
yechimlari ayirmasining modulini toping.
Yechish:
x2 - 9x + 15 = 7
x2 - 9x + 8 = 0 tenglama ildizlari
xi = 1, x2 - 8.
|xy-x2| =|1-8| = 7.
Javob: 7.
20. у = x2014 + 2014* * * * * * * * х xi * funksiya
hosilasining x = 0 nuqtadagi qiymatini
toping.
Yechish:
1) darajali funksiya hosilasi (>?)' = n^~1
2) ko'rsatkichli funksiya hosilasi
(ax)' = a* Ina
3) y' = 2014X2013 + 2014”ln2014
y’(0) = 20140 + 2014?ln2014 = ln2014.
Javob: ln2014.
21. Qarang: 3-variant 18-savol
(26-bet).
22. Qarang: 15-variant 18-savol
(124-bet).
23. a = -b, c = 1 bo'lsa,
c(a - bf + a(b - c)3 + b(c - a)3
c2(b - a) + a2(c - b) + b2(a - c)
ifodaning qiymatini toping.
Yechish:
Qulay usul bilan yechish uchun a = 2,
b = -2 deb olamiz.
a = 2, b = -2, c = 1 bo‘lsa, ifodaning
qiymatini topamiz.
1(2+ 2)3 + 2(-2-I)3-2(1-2)3
12(-2 - 2) + 22(1 + 2) + (-2)2(2 - T) ~
64-54 + 2 12 4
=-----------= — = 7 .
-4+12 + 4 12
Javob:1.
24. Qarang: 3-variant 1-savol
(22-bet).
25. Qarang: 8-variant 4-savol
(71-bet).
26. Qarang: 2-variant 2-savol
(13-bet).
27. Qarang: 4-variant 6-savol
(33-bet).
28. Qarang: 11-variant 7-savol
(96-bet).
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Yechimlar. Matematika va informatika 2017
22-variant
29. >?8x-4 + V2x-1 = 9
tenglamaning ildizlari quyidagi
oraliqlardan qaysi biriga tegishli?
Yechish:
Aniqlanish sohasi:
(8x-4>0 1
[2x-1>0^X~2
y]4(2x-1)+у/2х-1 =9
2y/2x-1 +y/2x-1 =9
3s/2x-1 = 9, 42x-1 = 3
2x - 1 - 9, 2x = 10, x = 5
Tenglama ildizlari x = 5,
5 6 [5; 9].
Javob: [5; 9].
30. Qarang: 8-variant 7-savol
(72-bet).
31. Nuqtalar o'rniga joylashtirish mumkin bo'lgan javobni aniqlang. Axborotni
kodlashda faqat... ishtirok etgan kodlash usuli ikkilikda kodlash usuli deyiladi.
Yechish:
Axborotni kodlash uchun ikkilik sanoq sistemasidagi 0 yoki 1 raqamlaridan
foydalaniladi.
Javob: 0 va 1.
32. a4 x3 + (1 - y2)2 ifodaning Paskal dasturlash tilida to'g'ri berilgan yozuvni
toping.
Yechish:
a2 => sqr(a) kabi Paskalda yoziladi.
(a-b)2 => sqr(a-b).
a4 x3 + (1 - y2)2 - (a2)2x3 + (1 - y2)2;
Paskal tilida bu formula quyidagicha sqr(sqr(a))*x*x*x+sqr(1-sqr(y));.
Javob: sqr(sqr(a))*x*x*x + sqr(1 - sqr(y)).
33. 457(8) sonining 16 lik sanoq sistemasidagi qiymatini aniqlang.
Yechish:
457(8)-Xie- 8lik sanoq sistemasidan o‘n oltilik sanoq sistemasiga olish uchun
awal 8 lik sanoq sistemasidan 2 lik sanoq sistemasiga olamiz. Keyin 16 lik
sanoq sistemasida yozamiz.
457(S)-X2. _______________
2 Hk sanoq sistemasi 8 lik sanoq sistemasi
000 0
001 1
010 2
011 3
100 4
101 5
110 6
111 7
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Yechimlar. Matematika va informatika 2017
23-variant
7e^ 111(2).
5a —* 101(2).
4g —> 100(2).
457в—100101111(2).
Endi 100101111(2) sonlarni tetradlarga ("4 ta raqamdan iborat guruhlarga"
ajratamiz) (1-jadvaldan).
000100101111 -> 12F1B.
•--.-----.->—---'(2)
216 HQ
Javob: 12Fie.
34. Qarang: 12-variant 36-savol (109-bet).
35. Fayl atributi nima?
Yechish:
Fayl atributi deganda fayl haqidagi barcha ma’lumotlar aks etadigan fayl
xossalari tushuniladi, ya’ni fayl nomi, kengaytmasi (qaysi dasturda
yaratilganligi), tashkil etilgan sanasi va vaqtidir.
Javob: fayl nomi, kengaytmasi, tashkil etilgan sanasi va vaqti.
36. Qarang: 4-variant 33-savol (41-bet).
23-variant
1. Arifmetik progressiyada
S26 125 .
—— =------bo Isa, a8 ni toping.
S17 17
Yechish:
s ^±£25..25
2b
s +al7 17
17 2
+ ^25 OR
~2 _125
a> + a<7 -f 7 17
2
(ai+a25)-25 _125
(a,+a17)-17 17
ai +a25
a<+a„
ai + a2s = 5(ai + ai7)
ai + a25 = 5(ai + ai7) -ат va d orqali
ifodalaymiz.
ат + ат + 24d = 5(ат + ат + 16d)
2ат + 24d = 5(2ат + 16d)
ai + 12d = 5(ат + 8d)
4ai + 28d = 0
ai + 7d = 0,
aa = 0.
Javob: 0.
2. Qarang: 2-variant 8-savol
(15-bet).
3. Qarang: 4-variant 8-savol
(33-bet).
4. Qarang: 1-variant 2-savol
(3-bet).
5. Qarang: 2-variant 13-savol
(16-bet). ’
6. Hisoblang: cos40° cos20° cos100°.
Yechish:
1) cos100Q = cos(180° - 80°) = -cos80°
2) cos20° cos40°-cos80° =
_ 2 sin 20° cos 20° cos 40" cos 80" _
" 2sin20° ”
_ sin 40 cos 40° cos 80° _
" 2sin20°
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Yechimlar. Matematika va informatika 2017
23-variant
_ 2 sin 40 cos 40“ cos 80° _
4 sin 20°
_ sin80° cos80“ _ 2sin80°cos80" _
4 sin 20° 8 sin 20°
_ sin 7 60° _ s'm(180° - 20°) _ sin 20° _ 1
~ 8sin20° ~ 8sin20" ” 8sin20“ ~8
> ... 1
Javob: —.
8
7. Qarang: 12-variant 2-savol
(102-bet).
8. Qarang: 2-variant 29-savol
(20-bet).
z л \ Vx 2
I I I X
9. I — I > tengsizlikning butun
yechimlari nechta?
Yechish:
x > 0
2~Зу/х+3^/х > ^,2
2°>x2, x2< 1,-1 <x< 1
Aniqlanish sohasiga ko'ra x>0,
demak, Oix* 1, x C[0; 1]
Tengsizlik 2 ta butun yechimga ega.
x = 0va x- 1.
Javob:2.
10. 2012-2011 -2009-2014 ni
hisoblang.
Yechish:
2009 = a belgilash kiritamiz.
(a + 3)fa + 2) - a(a + 5) =
= a2 + 5a+ 6-a -5a = 6.
Javob: 6.
Yechish:
x2 + 7 - a belgilash kiritamiz.
(x2 +7x + 7)(x2 + x+7) = 7x2
fa + 7x)(a + x) = 7X2
a2 + 8xa + 7X2 = 7X2
a2 + 8xa = 0,
a(a + 8x) = 0
a = 0,
a =-8x
x2 + 7 = 0 tenglama haqiqiy ildizga ega
emas.
>? + 7=-8x
x2 + 8x + 7 = 0 tenglama haqiqiy
ildizlari yig‘indisi
Xi +x2 = —8.
Javob: -8.
13. Qarang: 3-variant 6-savol
(23-bet).
14. Hisoblang: log3 (д/з • )
+1083(5/3^27^)
Yechish:
1) logab + logac = logabc
2) an-ak - a™ formulalardan
foydalanib yechamiz.
log3^-9^ -27^ +
+log, p2 З6 - 324 / = log3 p'1 312 324 j
• 32 36 -324 = logj Зз+к+ з+г+г+’2
1-17 41
= log33 24 = 1—log33 = —.
3 24 Вз 24
41
Javob: —.
24
11. Qarang: 19-variant 16-savol
(146-bet).
12. (x2 + 7x + 7)(x2 + x + 7) = 7X2
tenglama haqiqiy ildizlari yig'indisini
toping.
л fx2+ (y+ a)2-4 = 0
15. Agar 2
[x2 + у = -2
tenglamalar sistemasi yechimga ega
bo'lmasa, a ning eng katta manfiy
butun qiymatini toping.
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Yechimlar. Matematika va informatika 2017
23-variant
Yechish:
\x2 + (y + a)2-4 = 0
< tenglamalar
(_x2 + у = -2
sistemasini grafik usulida yechamiz.
1) x? + (y + a)2 = 4- aylana
tenglamasi. Markazi (0; a) nuqtada,
radiusi 2 ga teng.
2) у --2-х2 - parabola tenglamasi.
у
a = 0 da tenglama 1 ta yechimga ega.
a =-1 da tenglama 2 ta yechimga ega.
a = 1 da tenglama yechimga ega emas.
aS-7, a z 1 da tenglamalar sistemasi
yechimga ega emas. Chunk! bu
qiymatlarda funksiya grafiklari
kesishmaydi. Sistema yechimga ega
bo'lmaydigan a ning eng katta manfiy
butun qiymati-7.
Javob: -7.
16. Qarang: 4-variant 28-savol
(39-bet).
17. Qarang: 2-variant 7-savol
(14-bet).
18. Samandarning o‘g‘il bola
sinfdoshlari soni qiz bola sinfdoshlari
sonidan 7,taga ko‘p. Sinfda o‘g‘il
bolalar soni qiz bolalar sonidan 2 marta
ko‘p. Diyora - Samandarning sinfdoshi.
Diyoraning sinfdosh dugonalari nechta?
Yechish:
x - o'g'ii bolalar soni, y-qiz bolalar soni.
x = y + 7- Samandarning sinfdoshlari
soni
1 + x -2y- Samandar bilan birgalikda
(x = y + 7 (2y-1 =y + 7
[x = 2y-1 [x = 2y-1
(y = 8-qizlarsoni
[x = 15-o'д'il bolalar
Diyoraning sinfdosh dugonalari 7 ta.
Javob: 7 ta.
19. Qarang: 3-variant 27-savol
(29-bet).
[3 + ax-7a>0
20. , tengsizliklar
(3 - ax + 3a > 0
sistemasi a ning qanday qiymatlarida
yechimga ega bolmaydi?
Yechish:
(3 + ax-7a>0 (ax>7a-3
[з2 -ax + 3a>0 [ax<9 + 3a
=> 7 a - 3 < ax i 3a + 9
7a - 3 < ax < 3a + 9 qo'sh tengsizlik
yechimga ega bo'lmasligi uchun
3a + 9 <7a - 3 bo'lishi kerak.
3a -7a < -9- 3
-4a <-12 yoki a > 3.
Javob: (3; <»).
21. у = kx + n funksiyaning grafigi
faqat I va III chorakda yotishi uchun
к va n qanday bolishi kerak?
Yechish:
у = kx + n to'g'ri chiziq faqat I va III
chorakdan otishi uchun n = Ova k> 0
bo'lishi kerak. у = kx.
Javob: n = 0, к > 0.
22. 3-6+ 12-24 + 48-
- ... + 3072 ni hisoblang.
Yechish:
3-6 + 12-24 + 48- ...+3072
geometrik progressiya yig'indisini
topamiz.
bl = 3,b2 = -6, q = —2, bn = 3072
170
Yechimlar. Matematika va informatika 2017
23-variant
3072 = br(f~1
3072:3 = cf-1,
1024 = cT1
1024 = (-2)"-1,
(-2)10 = (~2)n~1
10 = n - 1, n = 11
, w->> З'И’Ч.
’’ q-1 -2-1
= 211 + 7.
Javob: 211 + 1.
23. Diagonallarining soni
tomonlarining soniga teng bo'lgan
qavariq muntazam ko'pburchakning
ichki burchaklaridan birini toping.
Yechish:
D - diagonallar soni, n - tomonlar
soni. D = n.
n^n(n-3)
2 ’
n-3 = 2,
n = 5- muntazam beshburchak.
Bitta tashqi burchagi:
, 360 360
а =----=-------= 72
n 5
Bitta ichki burchagi:
а = 780° - a' = 780° - 72° = 108°.
Javob: 108°.
24. Qarang: 10-variant 16-savol
(88-bet).
25. Qarang: 20-variant 16-savol
(153-bet).
26. Asoslarining radiuslari 2 va 5 ga
teng bo'lgan kesik konus va unga
tengdosh silindrning balandlikiari bir
xil. Silindr asosining radiusini toping.
Yechish:
r = 2, R = 5- kesik konus radiuslari.
H - kesik konus balandligi. Kesik
konus va silindr tengdosh, demak,
hajmlari teng. V - kesik konus hajmi,
Vi - silindr hajmi.
H = Hi
V = ^tcH[R2 + Rr + r2},
Vi = nR12H1
V= Vi
~(R2 + Rr + r2) = R2
, ^R2+Rr + r2
1 3
l22+2-5 + 52 _
N з
Javob: Vl3 .
27. Tenglamani yeching:
2log/2(x-2) = log/-(x + 6)~
-log^(x-5).
2
Yechish:
Aniqlanish sohasi:
x-2>0 [x>2
< x + 6 > 0=> < x > -6 =>x>5
x-5>0 x>5
21og^(x - 2) = log ^(x + 6) +
+ log^(x-5)
log^(x-2)2= log^(x + 6)(x-5)
(x-2)2 = (x + 6)(x-5)
x2-4x + 4 = x2+x-30
-5x =-34
x-3-±-6,8.
5
Javob: 6,8.
28. Qarang: 14-variant 3-savol
(117-bet).
29. Qarang: 12-variant 15-savol
(104-bet).
30. Qarang: 11-variant 17-savol
(98-bet).
171
Yechimlar. Matematika va informatika 2017 23-variant
31. O'nlik sanoq sistemasidagi 37 sonini ikkilik va sakkizlik sanoq
sistemalaridagi qiymatlari to'g'ri berilgan javobni aniqlang.
Yechish:
37w X2.
37 [2 _
~2_|18I2
17 181912
“16 (Q)8[4|2
(D (Т>4|2'2
©2©
©
X2 = 100101.
37(10; —* Xa.
3718
32@
ID
X8 = 45;.
Javob: 100101; 45.
32. Agar a = 16, b = 7 va c = 36 bo'lsa, trunc(sqr(a mod b) + sqrt(c))
ifodaning natijasini toping. (Paskal dasturlash tilida).
Yechish:
trunc funksiyasi sonning butun qismini qaytaradi.
Masalan: trunc(3,2)=3
trunc(2,6)=2
a = 16, b = 7, c = 36. trunc(sqr(a mod b)+sqrt(c))
mod - funksiyasi a ni b ga bo'lib qoldiqni qaytaradi.
16 mod 7 = 2.
Chunki 16 = 7-2 + 2;
sqrt(c) => 4c ni beradi. sqrt(36))= 436 = 6.
trunc(sqr(16mod7)+sqrt(c))
trunc(sqr(2)+sqrt(36))=trunc(sqr(2)+6);
sqr(2) + 6 = 22+ 6 = 4 + 6= 10.
trunc (10) = 10.
Javob: 10.
33. Qarang: 8-variant 36-savol (78-bet).
34. Qarang: 2-variant 33-savol (21-bet).
35. MS Excel 2003 dasturida to'g'ri yozilgan formulani ko'rsating:
Yechish:
=A1+4*B5
Excel dasturida formula yozish uchun quyidagilar inobatga olinadi.
1) Formula albatta tenglik (=) belgisidan boshlanadi.
2) Formula yozganda arifmetikaning to'rt amallar quyidagicha yoziladi.
* ko'paytma
+ yig'indi
172
f Yechimlar. Matematika va informatika 2017 24-variant
- ayirma
/ bo'lin ma
3) Yacheyka nomida avval ustun nomi keyin satrraqami ko'rsatiladi.
Javob: =A1+4*B5.
36. Agar a = 6, b = 3vac=16 bo'lsa, trunc(sqr(a div b) + sqrt(c)) ifodaning
natijasini toping. (Paskal dasturlash tilida).
Yechish:
a div b funksiyasia ni b ga bo'lib butun qismini qaytaradi, ya’nibo'linmani
qaytaradi.
6 div 3
bu yerda bo'linma 2.
sqr(a) => a2ni bildiradi.
sqrt(a) => у/a ni bildiradi.
trunc(a) => a sonining butun qismi.
trunc(sqr(a div b)+sqrt(c))=8.
sqrt(16)=4
6 div 3 = 2
sqr(2) = 2? = 4
trunc(4+4)=8.
Javob: 8.
24-variant
1. Qarang: 4-variant 12-savol (34-bet). q1° + 1 =L 4 4 4
2. Qarang: 7-variant 27-savol (67-bet). ч ^(q40-1) -1) q-7 q-1
3. Geometrik progressiyada — = —bo'lsa, —"ni toping. S10 4 S20 Yechish: - bAq20-i)(q20+1) д-1 _аго + 1 q-1 b^-T) 3) q20 + 1 = (q10)2 + 1 = ffV . 1 .17
— - — bo'lsa, — ni topamiz. Sw 4 S20 Geometrik progressiyada: } 20 q-1 ’ 10~ q-1 _ b,(q2°-1). b,(q№ -1) 20' w q-1 ' q-1 _ b^q10-1\q1°+1) q-1 _ ,0 , q-1 Ь^0-!) \4) 16 16 Javob: —. 16 4. Nomanfiy x, у sonlar uchun a = Зх + — у vab = 2^xy bo'lsin. Qaysi tengsizlik har doim o'rinli? Yechish: x, у - noma’lum sonlar.
173
Yechimlar. Matematika va informatika 2017
24-variant
a = 3x +^y va b= 2y[xy
Doimo o'rinli bo'ladigan tengsizlikni
topamiz.
a = 3x+^y > 2^3x~y = 2y[xy = b
Bundan a> b ekanligi kelib chiqadi.
Javob: a s b.
5. cos2x > cos6x tengsizlikni yeching.
Yechish:
cos2x - cos6x>0
1) ayirmadan ko'paytmaga otamiz.
„ „ „ . 2x-6x
cos2x - cos6x = -2sin------
2
sin^Lt^L - 2sin2x sin4x
2
2) 2sin2xsin4x > 0 tengsizlikni yechamiz.
sin2xsin4x > 0 tengsizlik quyidagi
tengsizliklar sistemasiga teng kuchli.
[sin2x>0 _. fsin 2x<0
1) < 2/1
[sin4x>0 [sin4x<0
[sin2x>0
1) =v
[sin4x > 0
(2лп< 2x <л + 2яп
j 2лп <4x <л + 2лп
ЯП < X <-t-ЯП
2
ЯП Я ЯП
—<х<—+----
2 4 2
Birinchi sistemaning yechimi:
ЯП Я" ЯП
—<х< —+--
2 4 2
sin2x < О
sin4x < О
—я + 2яп < 2х < 2яп
-я + 2яп <4х< 2яп
~— + ЛП<Х< ЯП
2
Я ЯП ЯП
---1-—<х<-
.42 2
Ikkinchi sistemaning yechimi:
л лп лп
—+—<х<—
4 2 2
Yechimlarni umumlashtirib
tengsizlikning yechimini topamiz.
П ЯП ЛП
— + — <x< — ,n6Z
4 2 2
X 0 я
*4 4
Tengsizlikning yechimi:
ЛП Л ЛП 'i
2 ’4
, , (ЛП Л лп\ „
Javob: —;— +— , n C Z.
I 2 4 2 J
6. Qarang: 6-variant 6-savol
(52-bet).
7. Qarang: 10-variant 30-savol
(92-bet).
8. Qarang: 14-variant 7-savol
(117-bet).
9. Qarang: 4-variant 21 -savol
(37-bet).
10. Diagonallarining soni tomonlari
sonidan 3,5 barobar ko‘p bo'lgan
qavariq muntazam ko'pburchakning
tashqi burchaklaridan biri topilsin.
Yechish:
D - diagonallar soni, n - tomonlar
soni.
D = 3,5n.
n(n^^7n 3 = 7
2 2
n = 10- muntazam o'nburchak.
a' - tashqi burchagi.
, 360° 360°
а =-----=-------= 36 .
n 10
Javob: 36°.
174
Yechimlar. Matematika va informatika 2017
24-variant
11. Uchburchakning uchlari to'g'ri
burchakli dekart koordinatalar
sistemasida quyidagicha berilgan:
A(0; 0), B(1; -4), C(2; 0). Uchburchak
yuzini toping.
Yechish:
1) A(0; 0), B(1;-4), C(2; 0), S=?
2) Uchburchak ABC - teng yonli.
AB = BC. AC - asos, AC = 2,
BD - balandlik, BD = 4
„ ACBD 2-4 .
o —-------—-----— 4 .
2 2
Javob: 4.
12. 4log2x + 8 = (2log2x)2
tenglamadan x ning qiymat(lar)ini
toping.
Yechish:
Aniqlanish sohasi: x> 0
4log2x + 8 = 4log22x,
log2x + 2 = log2x
log2x = a, a2-a-2 = 0
a — —1, a — 2
7
1) a --1 da log2x =-1,x=—
2) a = 2 da log2x = 2, x = 4.
Javob: 4 va —.
13. Tengsizlikni yeching:
z к (JT6 7-2x3+1)1/2 z y-x
<2 J
Yechish:
a , 0 < a < 1, ab < ac, b> c
2
(x6 - 2x3 + 1)2 > 1 - x
(x3 - 1) 2 > 1 -X
O 2-
(x3 * *-* 1) 2>1 -X
|x3- 1| > 1 -X
1) 1 - x > 0, x < 1 da
ix3-1>1-x [x3 +x-2>0
( , => ( , =>
[x-1<x-1 (x -x<0
hx-1)tx2 +x + 2)>0
(x(x2-1)<0
a) (x - 1)(t3 + x + 2) > 0, x> 1
b) x(x- 1)(x + 1) < 0, x <-1, 0 <x < 1
x <-1 va 0 < x < 1
2) 1 -x< 0bo'lsa, ya’nix > 1 da
|x3 - 1| > 1 - x tengsizlik doimo o'rinli.
x> 1
Tengsizlik yechimlari
(-oo;-1)U(0; 1)U(1; <xj
oraliqlardan iborat.
Javob: (-oo;-1)U(0; 1)U(1; «>).
14. Qarang: 20-variant 8-savol
(152-bet).
15. Qarang: 4-variant 30-savol
(39-bet).
16. Qarang: 8-variant 29-savoi
(76-bet).
17. Qarang: 11-variant 30-savol
(101-bet).
18. Qarang: 8-variant 30-savol
(76-bet).
19. Qarang: 12-variant 27-savol
(107-bet).
20. Qarang: 6-variant 25-savol
(58-bet).
21. ai, a2.a8 ketma-ketlikda
ixtiyoriy uchta ketma-ket hadining
yig'indisi 40 ga teng. Agar ketma-
ketlikning uchinchi hadi 6 ga teng
bo'lsa, birinchi va sakkizinchi
hadlarining yig'indisi nechaga teng?
Yechish:
Ixtiyoriy uchta ketma-ket hadning
yig'indisi 40. Demak, ai + a2 + a2 =
175
Yechimlar. Matematika va informatika 2017
24-variant
= Эг + Эз + 34 = Эз + 34 + as = ... =
= a6 + a? + as = 40
аз = 6 bundan ai + аг + a3 = 40,
ai + az — 34
az + аз + Э4 = 40, az + a4 = 34
аз + a4 + as = 40, Э4 + a5 = 34
ai + a8 yig‘indisi Э4 + a5 yig'indiga
teng, chunki ak + an-ac + am, agar
к + n = c + m teng bo'lsa. Shunga
asosan ai + a8 = a4 + as = 34.
Javob: 34.
22. у > 0 bo'lsin. To'rtburchakning
uchlari to'g'ri burchakli dekart
koordinatalar sistemasida quyidagicha
berilgan: A(0; 0), B(0; y), C(2; y) va
D(4; 0). To'rtburchak diagonallarining
o'rtalari orasidagi masofani toping.
Yechish:
1) у > 0, A(0; 0), B(0; y), 0(2; y),
D(4; 0), EN = ?
2) ABCD - trapetsiya.
AC, BD - diagonallari. E nuqta - AC
kesma o'rtasi. E 1;~
l 2
N nuqta - BD kesma o'rtasi.
Javob: -1.
\y2 + xy = 12
23. f tenglamalar
[x2 + xy = 4
sistemasini yeching.
Yechish:
\y2 +xy = 12 (y(y + x) = 12
[x2+xy = 4 [x(x + y) = 4
УА*+У1ЛгУ=3,у = зх
x(x + y) 4 x
x2 + xy = 4, у = 3x
x2 + 3x2 = 4, 4X2 = 4, x2 = 1,
x = ±7, у = ±3
(1;3) va(-1;-3).
Javob: (1; 3) va (-1; -3).
24. Teng yonli trapetsiyaning
diagonali o'rta chizig'ini 1,5 va 7,5 ga
teng kesmalarga ajratadi.
Trapetsiyaning yuzasi 72 ga teng
bo'lsa, uning yon tomonini toping.
Yechish:
EP = 1,5, PK = 7,5, AE = EB, S= 72,
Uchburchak ABC da EP o'rta chiziq.
ВС = 2EP = 3. Uchburchak ACD da
PK o'rta chiziq. AD = 2PK =15
s=AD+BC cn
2
72 = Y-^- CN, CN = 8
2
ND = ^-BC = 15-3=6
2 2
Uchburchak CND to'g'ri burchakli.
CD2 = CN2 + ND2 = 82 + 62= 102,
CD = 10
CD=AB= 10.
Javob: 10.
25. Qarang: 7-variant 21-savol
(65-bet).
26. Qarang: 19-variant 30-savol
(149-bet).
27. Qarang: 7-variant 12-savol
(64-bet).
28. Aylananing A nuqtasidan
o'tkazilgan AB va AC vatarlarning
uzunliklari mos ravishda 8 va 9 teng.
176
Yechimlar. Matematika va informatika 2017 24-variant
Agar ularning ikkinchi uchlari tutashtirilsa, yuzi 18 ga teng uchburchak hosil bo'ladi. AB va AC vatarlar orasidagi burchakni toping. x2-x- 1 = о _1 + 4l + 4 1±V5 %1,2 „ _ 2 2
Yechish: AB, AC - vatarlar. AB = 8, AC = 9, SABC = 18 ZA = ? 2 1 + 45 ll + 45 q = , q = . 2 V 2 2 1-45 q = yechimga ega emas. 2
, . b, bt q 1 + 45 tgA = q = J——- =
„ ABAC . . Sabc- 2 -smA . . 2-Sabc 2-18 1 sm A = = = — ABAC 8-9 2 Л A = 30°. Javob: 30°. b, b, V 2 = l2 + 2^ -^2 + 245 \ 4 2 , u s/2 + 245 Javob: . 2 30. P(-2; 2) nuqtadan o'tuvchi va a (6; 4) vektorga perpendikulyar bo'lgan
29. To'g'ri burchakli ABC uchburchakda Z В = 90°. AB, BC va AC tomonlari uzunliklari geometrik progressiya tashkil qiladi. tgA ning qiymatini toping. to'g'ri chiziq tenglamasini toping. Yechish: у = kx + b - to‘g‘ri chiziq tenglamasi. 1)P(-2;2),2 = -2k + b 2 2
Yechish: tgA=— ZB = 90°, AB AB = bi, BC = b2, AC = b3 tgA = ? A 2) a (6; 4), y = |x, k1=4 kJ a vay= kx + b perpendikulyar k ki =-1, k = —- = k, 2
\ 3J2 = -2-f--l + b, b =-1 L 2)
.La To‘g‘ri burchakli uchburchakda b32 = bi2 + b22 bi2-q4 = b2 + b2<f q4-q2- 1=0, <f = x 3 у = — x - 1 2 2y + 3x + 2 = 0 yoki 3x + 2y + 2 = 0. Javob: 3x + 2y + 2 = 0.
31. Papka nomiga qo'yish mumkin bo'lmagan belgilarni ko'rsating.
Yechish:
Papkani nomlashda quyidagicha belgilarni ishlatish mumkin emas.
<> - ochilgan yopilgan katta, kichik belgilar
l\ - qiya chiziqlar
? - so'roq belgisi
* - yulduzcha
177
Yechimlar. Matematika va informatika 2017
25-variant
"- qo'shtirnoq
: - ikki nuqta
{} - figurali qavs belgisi
% - foiz belgisi.
Javob::" / \ * ? < >.
32. Algoritm natijasini aniqlang: x:=4; x:=x+11; x:=x*x;.
Yechish:
operatori o'zlashtirish opeatoridir.
x:=4 (bu qatorda x o'zgaruvchiga 4 soniyuklandi)
x:=x+11; (x o'zgaruvchiga qaytadan x+ 11 ya’ni 4 + 11 yuklandi).
x:=x*x; (x o'zgaruvchisida endi 4 emas yangi 4+11 = 15 soni hosil bo'ldi va
x:=x*x=15*15=225).
Javob: 225.
33. Agar A-‘Modulyator - raqamli signallarni analog signallarga o'tkazuvchi
qurilma”, B=“0 va 1 raqamlar faqat ikkilik sanoq sistemasida qo'llaniladi",
C=“Not(1610 = 11112)”, D=“Fayl nomida * belgisini ishlatish mumkin” sodda
mulohazalar berilgan bo'lsa, ularning qiymati asosida -,(Д л S) v (С л D)
mantiqiy ifoda qiymatini aniqlang.
Yechish:
Mantiqiy elementlar
v - dizyunksiya ("yoki" amali)
л - konyuksiya ("va” amali)
= - ifoda inkori. (agar A rost bo'lsa ->A = yolg'on bo'ladi va aksi).
A va В tasdiqlar rost
D tasdiq yolg'on
C - aniq bo'lmagan tasdiq.
Shunga -<(ArdB)v(OD) aniqlab bo'lmaydi.
Javob: sodda mulohazalardan birining qiymatini aniqlab bo'lmaydi.
34. Qarang: 5-variant 36-savol (50-bet).
35. Qarang: 10-variant 31-savol (92-bet).
36. Qarang: 1-variant 36-savol (12-bet).
25-variant
1. у = 3cos4x - cos12x
funksiyaning hosilasini toping.
4x — 12x
sin4x - sin12x = 2sin------
2
Yechish:
1) (cosfax + b))' = -asin(ax + b)
( sos4x)‘ = -4sin4x
'cos12x)' = -12sin 12x
4x+12x , я x о
cos-------= 2sin(-4x)cos8x
3) У' --12 (-2sin4x cos8x) =
= 3(cos4x)' - (cos12x)' = -12sin4x +
12sin12x = -12(sin4x - sin12x)
2) sin4x - sin12x ayirmadan
ko'paytmaga o'tamiz.
= 24sin4xcos8x.
Javob: 24sin4x cos8x.
2. Qarang: 1 -variant 22-savol
(8-bet).
178
Yechimlar. Matematika va informatika 2017
25-variant
3. Qarang: 14-variant 30-savol
(121-bet).
4. x3 - Зах2 + bx - 15 ko'phad
(x - 1 ) (x - 3) ga qoldiqsiz bo'linsa,
a va b ni toping.
Yechish:
x3 - Зах2 + bx - 15 ko‘phad
(x - 1)(x - 3) ga qoldiqsiz bo'linsa,
x = 1 va x = 3 da kophadning qiymati
nolga teng bo'ladi.
x = 1 da 1 - 3a + b - 15 = 0
-3a + b= 14
x = 3da 27-27a + 3b -15 = 0
-27a + 3b =-12
9a- b =4
(3a-b = -14
19 b 4 teiglamalar sistemasidan
a vab ning qiymatini topamiz.
3a-b=-14
9a-b = 4 ,a = 3,b = 23.
-6a = -18
Javob: 3 va 23.
5. Qarang: 3-variant 17-savol
(26-bet).
6. Qarang: 8-variant 13-savol
(73-bet).
7. Qarang: 20-variant 2-savol
(151-bet).
8. Qarang: 3-variant 19-savol
(27-bet).
9. Geometrik progressiyani tashkil
qiluvchi uchta sonning yig'indisi 93. Bu
soniarni arifmetik progressiyaning
, birinchi, ikkinchi va yettinchi hadlari deb
qarash mumkin. Bu soniarni toping.
Yechish:
bi + bi + Ьз = 93
bi + biq + biq2 = 93
Masala shartiga ko'ra ai = bi,
a2=b2 = biq, a7-b3 = brq2
1) d = a2 - a1 = bi(q - 1)
2)6d = a7-a1=bi(q2- 1)
b
6(q-1)=q2-1
q -6q + 5 = 0, q = 5 yoki q = 1
bi + biq + biq2 = 93, q = 5 da
bi(1 + 5 + 25) = 93,
bi =3,b2= 15, b3 = 75
q = 1 da bi(1 + 1 + 1) = 93
bi = 31, b2 = 31, b3 = 31.
Javob: 3; 15; 75va31; 31; 31.
10. 2234 va 2235 sonlarining
umumiy natural bo'luvchilari nechta?
2234
1117
1
Yechish:
2234 va 2235 sonlarini tub
ko'paytuvchilarga ajratamiz.
2235 3
745 5
149 149
1
EKUB(2234; 2235) = 1
Bu sonlarning umumiy natural
bo'luvchisi 1 ta. Bu son 1.
Javob: 1.
11. /(x) = 96cos5x sin11x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko'paytmadan yig'indiga otamiz.
1
sin11xcos5x = — (sin(11x + 5x) +
1
+ sin(11x- 5x)) =— (sin16x + sin6x)
2) f(x) =- -96(sin16x + sin6x) =
= 48sin16x + 48sin6x
Boshlang'ich funksiyasini topamiz.
F(x) = -3cos16x - 8cos6x + C.
Javob: -3cos16x - 8cos6x + C.
12. Qarang: 5-variant 8-savol
(43-bet).
179
Yechimlar. Matematika va informatika 2017
25-variant
13. Teng yonli trapetsiyaning
diagonali o'rta chizig'ini 1,5 va 7,5 ga
teng kesmalarga ajratadi. Agar
trapetsiyaning yon tomoni 10 ga teng
bo'lsa, balandligini toping.
Yechish:
ABCD - teng yonli trapetsiya.
AD - a, BC = b, EK-o'rta chiziq,
AC - diagonal, CP - balandlik.
EN = 1,5; NK = 7,5; EK =9
EN ABC uchburchakning o'rta chizig'i.
BC = 2EN = 3
NK ACD uchburchakning o'rta chizig'i.
AD = 2NK = 15
ABCD teng yonli trapetsiya.
„„ AD-BC 15-3 .
2 2
CPD to'g'ri burchakli uchburchak.
CP2 = CD2 -PD2 = 1O2-62 = ^
CP = 8.
Javob:8.
14. Rasmda у = /'(x) funksiya
grafigi tasvirlangan. у = /(x)
funksiyaning grafigiga xo = -2
nuqtasida o'tkazilgan urinmaning
burchak koeffitsiyentini toping.
Yechish:
Rasmda у = f'(x) funksiya grafigi
tasvirlangan.
f'(x0) = k, xo--2, f'(-2) = 1 = k.
Javob:1.
15. у = 4sin22x + 4л/з sinxcosx +
+ 1,5cos4x + 1,5 — 2л/з funksiyaning
eng kichik qiymatini toping.
Yechish:
1) 4 y[3 sinxcosx = 2 43 sin2x
2) cos4x = 1 - 2sin22x
3) y = 4sin22x + 2^3 sin2x + 1,5 -
- 3sin22x + 1,5-243 =
= sin22x + 24з sin2x + 3 - 2^3 =
= (sin2x +43 )2 - 243
4) -1 < sin2x s 1 bo'lganligi sababli
sin2x = -1 da
у — (—1 +4з)2-24з = 1 -243 +
+ 3-243 =4-4^3
sin2x= 1 day = (1 +4з)2-24з =
= 1 +243 +3-243=4
Eng katta qiymati у = 4. Eng kichik
qiymatiу - 4 - 443 .
Javob: 4 - 4л/3 .
16. Qarang: 6-variant 20-savol
(56-bet).
17. 4sin2x - 5sinx < 0 tengsizlikning
[0; 2л] kesmadagi yechimlari to'plamini
toping.
Yechish:
4sin2x - 5sinx < 0
sinx(4sinx - 5) <0
-1 < sinx s 1 bo'lganligi sababli
4sinx - 5 ifodaning qiymati manfiy son
bo'ladi.
4sinx - 5<0
Demak, sinx(4sinx - 5) <0, sinx > 0
sinx > 0 tengsizlikning yechimi
2im < x <jr + 2rtn
[0; 2л] oraliqdagi yechim (0; л) bo'ladi.
Javob: (0; л).
18. Qarang: 1 -variant 12-savol
(6-bet).
19. Quyidagi ko'pyoqlardan qaysi
birida 5 ta yoq, 8 ta qirra va 5 ta uchi
bor?
180
Yechimlar. Matematika va informatika 2017
25-variant
Yechish:
5 ta yoq, 8 ta qirra va 5 ta uchi bor
ko‘pyoq bu 3 chizmadagi ko'pyoq.
5 ta yoq - ABCD, ABE, ВЕС, DEC,
AED
8 ta qirra AB, BC, AD, DC, EA, EB,
EC, ED.
5 ta uchi А, В, C, D, E.
6. (4}cA
8. {1;3}cA
10. {2 ;3}cA
12. {3;4}cA
14. {1; 2; 4} c A
16. {2; 3; 4}cA
5. {3}cA
7. {1;2)cA
9. {1; 4}cA
11. {2; 4} c A
13. {1; 2; 3}cA
15. {1; 3; 4}czA
A to'plamning 16ta qism to'plami bor.
Javob: 16.
Javob: 3.
20. у = 3sinx - sin3x funksiyaning
hosilasini toping.
Yechish:
1) (sin(ax + b))' = acos(ax + b)
(sinx)' - cosx
(sin3x)' = 3cos3x
y' = 3(sinx)' - (sin3x)' =
- 3cosx - 3cos3x = 3(cosx - cos3x)
2) cosx - cos3x ayirmadan
ko'paytmaga otamiz.
cosx - cos3x = -2sin x + ^x
2
x
cos—-— = -2sin2xsin(-x) =
= 2sin2xsinx
3) y' = 3-2sin2x sinx = 6sinx-sin2x.
Javob: 6sinxsin2x.
21. Qarang: 2-variant 30-savol
(20-bet).
22. Qarang: 1 -variant 5-savol
(4-bet).
23. {x|x e N, x2 < 17} to'plamning
nechta qism-to'plamlari mavjud?
Yechish:
X2<17,xCN, 1<x<Jl7
A = {1, 2, 3, 4}. A to'plam 4 ta
elementdan iborat. A to'plamning qism
to'plamlarini tuzamiz.
1. 0cA 2. AcA
3. {1}cA 4. (2}cA
24. Qarang: 3-variant 7-savol
(24-bet).
25. Qarang: 18-variant 15-savol
(140-bet).
26. Qarang: 10-variant 28-savol
(91-bet).
27. Qarang: 2-variant 19-savol
(17-bet).
28. Qarang: 2-variant 25-savol
(19-bet).
29. Tenglamani yeching:
5sin4x + 4cos4x = 4.
Yechish:
z 2 x2 ff + cos2xY (1 + cos2x)2
(cos2x) =l---------I
4
4
, . 2 .2 ff-cos2xY (1-cos2x)2
(sin2x)2=l-2---I --------i-
cos2x = a
5.^L+4S1^L=4
4 4
5-10a + 5a2 + 4 +8a + 4a2 = 16
9a2 -2a-7 = 0,
. 7
a = 1, a = —
9
1) cos2x =1, 2x = 2пп, х=яп, nCZ
2) cos2x = ,
9
o 1- tg2x
cos2x =--
1 + tg x
1-tg2x_ 7
1 + tg2x 9
181
Yechimlar. Matematika va informatika 2017
26-variant
9 - 9tg2x = —7 - 7t<fx
2tq2x = 16,
tg2x = 8,
tgx-=+2^2 ,
x = ±arctg2 y/2 + лп, n C Z.
Javob: x = Ttn, x = ±arctg 2л/2 + лп.
30. Qarang: 8-variant 17-savol
(73-bet).
31. Paskal dasturiash tilida berilgan ushbu ifodaning qiymatini toping.
trunc(sqrt(abs(trunc(4,3)-sqrt(100)*round(1,6))))
Yechish:
round(1,6) = 2
sqrt(100) => y/Тод = Ю
trunc(4,3) = 4
abs(4 - 10-2) = abs(-16) = 16 (son moduli)
sqrt( 16)=*. 4l6 = 4.
trunc(4) = 4
Javob: 4.
32. Ali aytdi: “Aralashtirilgan alifbo usuli tekis kodlash usulidir”. Vali aytdi:
‘Aralashtirilgan alifbo usuli notekis kodlash usulidir”. Ularning fikrlari haqida
nima deya olasiz?
Yechish:
Aralashtirilgan alifbo usuli tekis kodlash usuli. Bundan Ali aytgan firk to‘g‘ri.
Javob: faqat Ali to'g'ri fikr aytgan.
33. Ali aytdi: “1 petabayt 1024 gigabaytga teng”, Vali aytdi: “1 bit 8 baytga
teng”. Ularning fikrlari haqida nima deya olasiz?
Yechish:
Ali aytgan fikr noto'g'ri, chunki 1024 gigabayt 1 terrabaytdir.
Vali aytgan fikr ham noto ‘g'ri.
1 bit 8 bayt emas. 1 bayt - 8 bit.
Har ikkala fikr noto'g'ri.
Javob: ikkalasi ham noto'g'ri fikr aytgan.
34. Qarang: 5-variant 31-savol (49-bet).
35. Qarang: 20-variant 34-savol (157-bet).
36. Qarang: 1-variant 31-savol (11-bet).
26-variant
1. Qarang: 4-variant 5-savol
(33-bet).
2. Qarang: 6-variant 7-savol
(52-bet).
3. Qarang: 16-variant 8-savol
(128-bet).
4. Qarang: 21-variant 29-savol
(161-bet).
5. Qarang: 2-variant 4-savol
(14-bet).
6. у = sinx funksiya grafigi berilgan
bo'lib, uni parallel ko'chirish yordamida
182
Yechimlar. Matematika va informatika 2017
26-variant
у = sin(x - a) + b funksiya grafigi hosil
qilingan. Bunday parallel ko‘chirishda
koordinata boshi qanday nuqtaga
ko'chadi?
Yechish:
у = sinx funksiya grafigi у = sin(x -a) + b
funksiya grafigiga parallel ko‘chirish
orqali hosil qilingan.
у - f(x) funksiya grafigini (a; b)
vektorga parallel kochirilsa,
у = f(x-a) + b funksiya hosilbo‘ladi.
Bunda koordinata boshi (a; b) nuqtaga
ko'chadi.
Javob: (a; b).
7. Qarang: 9-variant 8-savol
(80-bet).
8. Qarang: 5-variant 13-savol
(45-bet).
9. 2-д/х-л/х + 8 = 0, tenglamaning
haqiqiy ildizlari yig'indisini toping.
Yechish:
yjx-Jx + 8 = 2 kvadratga kotaramiz.
x-Jx + 8 - 4, *Jx + 8 = x - 4
x-4^0, x> 4
x + 8 = x2 - 8x + 16
x2 - 9x + 8 = 0, x = 1, x = 8
x>4 bo'lganligi uchun x = 8
tenglamaning yechimi bo'ladi. Haqiqiy
ildizlari yig'indisi 8 ga teng.
Javob: 8.
10. У = ^/logo 5(x-3)-1 funksiyaning
aniqlanish sohasini toping.
Yechish:
Aniqlanish sohasini topamiz.
flog25(x-3)-7>0_
[x-3>0
/(logo5(x- 3) - 1)(logo5(x- 3) +1) > 0
[x>3
Quyidagi tengsizliklar sistemasini
yechamiz.
riog0,5(x-3)-f>0
( yoki
[^og0fi(x-3) + 1>0
[logo,5(x-3)-1<0
llog0,5(x-3) + 1<0
flog,(x-3)>7 f
fj 2 =>Г’
|log,(x-3)>-7 1
x<3—
2
1
x > 3 bo'lganligi sababli
3<x<3-
2
flog0i5(x-3)<1
|log0,5(x-3)<-1^
x-3>2
x>5
x> 3 bo'lganligi sababli [5; да ).
Demak, tengsizlikning yechimi
(3;3^]U[5; да/
1
Javob: (3; 3^] U [5; да).
11. Qarang: 20-variant 7-savol
(151-bet).
12. /(x) = 22cos5xcos6x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko'paytmadan yig'indiga otamiz.
1
cos5x cos6x =— (sin(5x - 6x) +
1
+ cos(5x + 6x)) =— (cosx + cos11x)
1
2) f(x) =— -22(cosx + cos11x) =
= 11cosx+ 11cos11x
183
Yechimlar. Matematika va informatika 2017
26-variant
Boshlang'ich funksiyasl:
F(x) = 11 sinx + sin11x + C.
Javob: 11sinx + sin11x + C.
13. Qarang: 4-variant 2-savol
(32-bet).
14. ABCD parallelogrammda CD
tomonni D uchidan boshlab
hisoblaganda 1 ;2 nisbatda bo’luvchi
AN to'g’ri chiziq o'tkazilgan. Agar ABC
uchburchakning yuzi 3 ga teng bo’lsa,
ANC uchburchakning yuzini toping.
Yechish:
ABCD - parallelogramm.
DN:NC = 1:2, Sabc = 3,
Sanc = ?
В
A
Sadc - Sabc = 3,
DN — x, DC = 3x
o ADDN .
SADN =-?---SID“
o ADDC .
SADC =-----Sln“
ADDN .
SAm.r~2 sm a DN x
*ODC DC 3x
2
-> _ §арс = 1
>AND 3 3
wc = Sadc — Sand = 3—1=2.
Javob: 2.
1_
3
15. Uchta teng kesmalardan tashkil
>gan shakl eng ko’pi bilan nechta
imetriya o'qiga ega bo’lishi mumkin?
Yechish:
3 teng kesmadan muntazam
iburchak yasaymiz. Muntazam
•burchak eng ko'pl bilan uchta
metriya o'qiga ega.
Javob: 3.
16. Qarang: 5-variant 11-savol
(44-bet).
17. Qarang: 2-variant 16-savol
(16-bet).
18. x3 - (л/б + 1 Jx2 + 6 ko’phadni
ko’paytuvchilarga ajrating.
Yechish:
Qavslarni ochib soddalashtiramiz.
x3-4б x2 -x2 + 6 =^(x-46 )-
- (x2 -6) = х2(х-4б ) -
-(х-4б )(х+4б ) =(х-4б )
(>?-(х+4б )) = fx-V6 )-(х?-х-4б ).
Javob: (х - Тб )(х2 - х - Тб ).
19. Qarang: 7-variant 28-savol
(68-bet).
20. Tub sonlar qatorini toping,
a) 3, 5, 7, 9; b)1,2, 3, 5, 7;
c) 2, 3, 5, 7, 21; d) 2, 3, 5, 7, 19.
Yechish:
1 dan va o'zidan boshqa natural
bo'luvchlga ega bo'lmagan 1 dan katta
natural son tub son deyiladi. 2, 3, 5, 7,
19-tub senlar.
Javob:d.
21. Soddalashtiring:
I T~1 Я-ft
\a-2a2b2 +b+—--------r ; (a S b > 0;
a2 + b2
a2 + b2 0).
Yechish:
Qisqa ko'paytirish formulasidan
foydalanamiz.
184
Yechimlar. Matematika va informatika 2017
26-variant
1) a- b - (a2 -b2)(a2 +b2)
2) a-2a2b2 + b = (a2-b2)2
^a-2a2b2 +b +-?—^- =
a2 +b2
п~ | a2 — b2 || a2 + b2 |
= a^-bH Л--------------------L
Vk J a2+b2
L 1 L 1
= a2 -b2 +a2 -b2
a>b>0 bo'lgan ligi uchun
a2 - b2 + a2 - b2 = 2a2 - 2b2 =
= 2-Ja-2jb .
Javob: 2>/a-2>/b .
22. Qarang: 20-variant 3-savol
(151-bet).
23. x log20,125 = -2 tenglamadan
x ni toping.
Yechish:
xlog20, 125 = -2
tog20,125 = log2 125 = log2 - =
1000 8
= !одг2Гэ = -3
2
x(-3)=-2,x=j.
I U 2
Javob: —.
3
24. Qarang: 20-variant 19-savol
(154-bet).
25. Qarang: 20-variant 30-savol
(156-bet).
26. ABCD parallelogrammning AD
tomonidan P nuqta shunday olinganki,
AP:AD - 1:5. Ac va BP to'g'ri chiziqlar
_ . . . . AC . .
Q nuqtada kesishsa,---ni toping.
AQ
Yechish:
дс
AP:AD = 1:5, — = ?
AQ
Uchburchak ACD va AQP da
A burchak umumiy, AD tomon umumiy,
. . . AC AQ ..
shunmg uchun— = — yoki
ы AD AP
AC AD 5 c
-----— — — — о .
AQ AP 1
Javob: 5.
27. (m - 2)y - (m2 - 9)x - 3 = 0,
m € R tenglama bilan berilgan to'g'ri
chiziqlar m ning qanday qiymatlarida
Ox o'qiga parallel bo'ladi?
Yechish:
(m - 2)y - (m2 - 9)x - 3 - 0
(m2 -9)x + 3 . , , . ...
у = i----l-----f0 g n chiziq Ox
m-2
o'qiga parallel bo'lishi uchun m2 -9 = 0
bo'lishi kerak.
m2 - 9 = 0, m2 = 9, m = ±3.
Javob: +3.
28. a + 2, a + b, ab sonlar arifmetik
va geometrik progressiyani tashkil
. 3 .. . ,
etadi. — ning qiymatini toping.
b
Yechish:
a + 2, a + b, ab sonlar arifmetik va
geometrik progressiya.
Arifmetik progressiyada
a + 2 + ab = 2(a + b)
Geometrik progressiyada
(a + 2)ab = (a + b)2
(a + 2 + ab-2a-2b = 0
[(a + 2)ab = (a + b)2
(ab-a-2b + 2 = 0
}(a + 2)ab = (a + b)2
ab-2b-a + 2 = 0
b(a -2)-(a-2) = 0
185
Yechimlar. Matematika va informatika 2017
27-variant
(a — 2)(b — 1) = 0, a = 2, b = 1
— 2
b~ 1~
Javob: 2.
29. Qarang: 5-variant 4-savol
(43-bet).
30. Qarang: 9-variant 29-savol
(84-bet).
31. Paskal dasturlash tilida berilgan ushbu ifodaning qiymatini toping.
trunc(abs(sqr(5)-sqrt(25)*round(5,6))).
Yechish:
trunc(abs(sqrt(5)-sqrt(25) *round(5,6))).
SQR(5) = о = 25.
SQRT(25) =y[5 =5.
round(5,6) = 6 (eng yaqin butun songacha yaxlitlash)
abs(25 - 5-6)=abs(-5)=5 (sonning moduli)
trunc(5)=5.
Javob: 5.
32. Axborotning to'liqlik xususiyati bajarilmaydigan javobni aniqlang.
Yechish:
28 fevral oyning oxirgi kuni axboroti to'liq emas. Chunki har 4 yilda, kabisa
yilida fevral oyi 29 kundan iborat.
Javob: 28 fevral oyining oxirgi kunidir.
33. Axborot uzatish tezligi berilgan javobni aniqlang.
Yechish:
Bod - signal yuborish tezligining birligi bo'lib,
1024 bod - 819,2 bit/sek
1024 bit/sek = 1,28-103 bod.
Axborot uzatish tezligining birligi 1 bod.
Javob: 1 bod.
34. Qarang: 6-variant 31-savol (60-bet).
35. Qarang: 10-variant 33-savol (94-bet).
36. Qarang: 3-variant 31-savol (30-bet).
27-variant
1. Qarang: 4-variant 9-savol
(34-bet).
2. Qarang: 7-variant 6-savol
(62-bet).
3. Tomoni 6 sm bo'lgan kvadrat
berilgan. Kvadrat tomoni o'rtalari
biriashtirib kvadrat hosil qilingan. Yana
xuddi shunday kvadratlar hosil qilindi.
9 kvadrat perimetrini toping.
Yechish:
ai = 6, ai - birinchi kvadrat tomoni.
аг - ikkinchi kvadrat tomoni. аз - uchinchi
kvadrat tomoni. ag - to'qqizinchi kvadrat
tomoni. P = 4ag ni topishimiz kerak.
Bl C2 Cl
186
Yechimlar. Matematika va informatika 2017
27-variant
2
Э1 = 6, З2 =~i—•= З5/2 , Q3 = 3
V2
a2 3V2 V2
g = —=----= —
af 6 2
z у—ч8
a9=af-qs=6-(-^) =
„ f 1 Y 6 _ 3
\42) 16 8
— . .33
P = 4ag = 4—=—.
8 2
4. Ifodani soddalashtiring:
sin10a-2sin25a + 1.
Yechish:
1) Darajani pasaytiramiz:
. 2c 1-coslOa
sin 5a =---------
2
„ 1-coslOa .
2) sm10a - 2---------+ 1 =
= sinlOa - 1 + coslOa + 1 =
= sinlOa + coslOa =
= cos — — 10a + coslOa
<2 )
3) Yig'indidan ko'paytmaga otamiz.
cos --10а + coslOa =
k.2 J
— ~10a + 10a
= 2cos-----------
2
--10a-10a
cos—----------=
2
= 2cos — cosf— -10a | =
4 <4 )
= 2-^cos(45°- 10a) =
= y/2 cos(45° - 10a) =
= ^2 cos(10a-45°).
Javob: V2 cos(10a-45°).
5. sin4x + cos4x = a tenglama a ning
qanday qiymatlarida yechimga ega?
Yechish:
1) sin4x + cos4x = (sin2x + cos2x)2 -
- 2sin2xcos2x = 1 - — sin22x
2
2) 1 -- sin22x = a
2
— sin22x = 1 - a
2
sin22x = 2-2а, 0< sin22x < 1
0<2-2a<> 1,-2<-2a<-1,-<a<1.
2
1
Javob: — < a < 1.
6. Qarang: 17-variant 16-savol
(134-bet).
7. Qarang: 8-variant 19-savol
(74-bet).
8. Qarang: 10-variant 26-savol
(90-bet).
9. Qarang: 4-variant 24-savol
(38-bet).
10. Natural boluvchilari soni eng ko‘p
boladigan uch xonali natural sonni
toping.
Yechish:
840 soni natural boluvchisi eng ko'p
bo'lgan son.
1.1
187
Yechimlar. Matematika va informatika 2017
27-variant
840 = 23-3-5-7
n = (3 + 1)(1 + 1)-(1 + 1)-(1 + 1) =
= 4-2-2-2 = 32
840 ning 32 ta natural bo'luvchisi bor.
908 = z-227
n = (2 + 1)(1 + 1)=6
480 = 25-3-5
n = (5 + 1)(1 + 1)(1 +1) = 21
804 = 2* 1 2 3-3-67
n = (2+ 1)(1 + 1)(1 + 1) = 12.
Javob: 840.
11. Rasmda у - /(x) funksiya
grafigi berilgan. у = /(x) funksiya
ekstremum nuqtalari koordinatalari
yig'indisini toping, (x 6 [-3; 1])
V y=fte)
Yechish:
у - f(x) funksiya
\ '3 / 1 \ 1 Л
dan “+”ga o'zgarsa - minimum
nuqta.
“+” dan ga o'zgarsa - maksimum
nuqta.
x =-3 min
x =-1 max
x = 1 min
Ekstremum nuqtalari yig'indisi:
-3 + (-1) + 1=-3.
Javob: -3.
12. {x|x € N, 6 < x2 < 39} to'plamning
nechta qism-to'plamlari mavjud?
Yechish:
6 < x2 < 39, x e N
4б <x<439 , A={3; 4; 5; 6}
A to'plam 4 ta elementdan iborat.
A to'plamning qism to'plamlarini tuzamiz.
1.0cA 2.AcA
3.{3}<~A 4.{4}cA
5. {5}c:A 6. {6}cA
7.{3;4}cA 8.{3;5}cA
9. {3; 6}cA
11. {4; 6}cA
13. {3; 4; 5}cA
15. {3; 5; 6} cA
A to'plamning 16
Javob: 16.
10. {4; 5} c A
12. {5; 6} c A
14. {3; 4; 6}-A
16. {4; 5; 6}cA
qism to'plami bor.
x tg — 1
13. 2sin —=—2----tenglamaning
2 ctg^-1
(180°; 540°) oraliqqa tegishli ildizlari
ayirmasining modulini toping.
Yechish:
tg-^-1 tg--1
1}—2-----= ---=
' x . 1 .
ctg--l —-f
tg2
x . x
— 1 -to —
2 ) 2 . x
----< = -tg —
. . x ^2
1~t9^
2) 2sin — = -tg— , 2sin —
'222
. x
sm —
—2- = o
x
cos—
2
. x
sin —
2
2 +
= 0
1
x
cos—
2 J
. X n X 1
sin — = 0, cos— = —
2.22
3) sin -X- = 0,^ = xn, x = 2nn yechim
X X
emas, chunki ctg— da sin — # 0
2 2
bo'lishi kerak.
x 1 x 2л
4)cos—=—, — =±—+2тсп,
2 2 2 3
4л .
x-+— + 4лп
3
188
Yechimlar. Matematika va informatika 2017
27-variant
n=O,x=^e (180°; 540°)
8тг
n = 1 da x=—e (180°; 540°)
3
Ildizlari ayirmasining moduli:
4?r 8?r _ _ 240"
3 3 ~ 3 ~
Javob: 240°.
14. Qarang: 7-variant 25-savol
(66-bet).
15. Qarang: 5-variant 1-savol
(42-bet).
16. Qarang: 1 -variant 9-savol
( bet).
17. Qarang: 9-variant 20-savol
(82-bet).
18. Qarang: 6-variant 18-savol
(56-bet).
19. Qarang: 4-variant 16-savol
(36-bet).
20. Qarang: 5-variant 28-savol
(48-bet).
21. у = x2 - 6x + 13 parabolaning
uchi koordinatalar boshidan qanday
masofada joylashgan?
Yechish:
Parabola uchi koordinatalari:
Yechish:
1) sin22x + cos22x = 1
2) In1 = 0
a+1 a+1
3) J (0 + 1)dx = j 1dx = x
a a
a + 1
a
= a + 1-a = 1.
Javob:1.
23. (a'/5~2,?: al78+2)2 j ifodaning
a =42 bo'lgandagi qiymatini toping.
Yechish:
Daraja xossasidan foydalanib yechamiz.
1. an:ak = an-k
2. (an)k - an k
1) а(^~2? a^+2^2 - q(J5-2)2-(45+2)2 _
_ „5-445+4-5-445-4 „-845
— a — d
Г Г
(a-875)-0^ =a 1 2>=a20
3) a =42 da a20 = (42 )20 =
—20 4
= 25 = 24= 16.
Javob: 16.
24. /(x) = 60sin2x sin8x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko paytmadan yigi'ndiga otamiz.
1
sin2xsin8x = — (cos (2x- 8x) -
уо = 3г-6-3+ 13 = 4
A(3; 4) parabola uchi. 0(0; 0)
koordinata boshi.
AO = ^(3-0)2 +(4-0)2 =
= 44+16 = 425 = 5.
Javob: 5.
22. a = 2 bo'lsa,
a+1
J (ln(sin2 2x + cos2 2x) + 1)dx aniq
a
integral™ hisoblang.
1
- cos(2x + 8x)) =— (cos6x - coslOx)
1
2) f(x) =— -60(cos6x - coslOx) =
= 30cos6x - 30cos10x
f(x) = ЗОсозбх - 30cos10x ning
boshlang'ich funksiyasi:
F(x) = 5sin6x - 3sin10x + C.
Javob: 5sin6x - 3sin10x + C.
25. Qarang: 5-variant 3-savol
(42-bet).
189
Yechimlar. Matematika va informatika 2017
27-variant
26. Qarang: 6-variant 17-savol
(55-bet).
27. Qarang: 5-variant 2-savol
(42-bet).
28. To'g'ri burchakli uchburchakning
gipotenuzasi 25 ga, o'tkir burchagining
sinusi 0,6 ga teng bo'lsa, gipotenuzaga
tushirilgan balandlikni toping.
Yechish:
c = 25, sina = 0,6
hc~?
sina = — , а = csina = 25-0,6 = 15
c
b2 = с2 - а2 = 252 — 152 = 40-10 = 400
b = 20
h =^- = ^^- = 12.
c c 25
Javob: 12.
29. Qarang: 1-variant 25-savol
(9-bet).
30. 1 - 2x + >/l6-6x + 3x2 = 0
tenglamaning ildizlari ko'paytmasini
toping.
Yechish: * 1
Vl6-6x + 3x2 = 2x- 1
2x-1>0,
16-6x + Зх2 =4x2-4x+ 1
x2 + 2x-15 = 0,
x=-5,
x = 3
1
x>- bo'lganligi uchun x =-5
tenglama yechimi bo'la olmaydi.
x = 3 tenglama ildizi. Tenglama ildizlari
ko'paytmasi 3 ga teng.
Javob:3.
31. Paskal dasturlash tilida yozilgan ifodaning qiymatini toping:
sqr(trunc(4,9)).
Yechish:
trunc(4,9) = 4 (sonninq butun qismi)
SQR(4) = 42 = 16.
Javob: 16.
32. Bir sanoq sistemasida berilgan sonning boshqa sanoq sistemasidagi
ko'rinishini aniqlang: 111000101011101(2).
Yechish:
Sonni to'rtta raqamdan iborat guruhlarga ajratamiz (tetradalarga). Ajratishni
eng kichik razryaddan boshlaymiz.
1-jadvaldan quyidagilarni topamiz.
0111000101011101 = 715D16 .
7« lie 0,6 2
Javob: 715Di6.
33. Bir mamlakat hududidagi foydalanuvchilarni birlashtiradigan tarmoq
qanday ataladi?
Yechish:
Tarmoqlar lokal, global, mintaqaviy turlarga ajraladi
- lokal tarmoq - bir xonada yoki bir muassasada kompyuterlarni
bog'lanishini ta’minlovchi tarmoq.
190
Yechimlar. Matematika va informatika 2017 28-variant
- mintaqaviy - bir mamlakat hududida foydalanuvchilarni birlashtiradi.
- global- bir nechta mamalakatlarnibir-biribilan bog’lovchi tarmoqdir.
Javob: mintaqaviy.
34. Qarang: 1-variant 33-savol (12-bet).
35. MS ACCESS 2003 dasturida Maydon turlarini aniqlang:
Yechish:
Ma’lumotlar omborida har bir ustun, ya’ni maydon o’zining turiga ega, hamda
shu ustunda faqat shu turdagi ma’lumotlar saqlanadi. Har bir maydonga
quyidagi turni o’rnatish mumkin: 01E, MeMO, Matnli, sonli, vaqt va sana,
mantiqiy, pul birlikli (денежный), schyotchik.
Javob: OIE, Memo, Matnli, sonli, vaqt va sanani ifodalovchi, mantiqiy, pul
birliklarida ifodalangan, schyotchik.
36. Qarang: 4-variant 36-savol (41 -bet).
28-variant
1. cos2x > cos6x tengsizlikni yeching.
Yechish:
cos2xs cos6x
1) cos2x - cos6x > 0 ayirmadan
ko’paytmaga o’tamiz.
„ _ „ . 2x-6x
cos2x - cos6x - -2sm-------
Я ЯП ЯП
—+—<х<—
4 2 2
ЯП Я ЯП Я ЯП ЯП
— <х< —+— va — + — <х< —
2 4 2 4 2 2
yechimlarni umumlashtiramiz.
. 2x + 6x „ . . „ . . .
sin—-— - -2sin(-2x) sm4x =
= 2sin2xsin4x
2) 2sin2x-sin4x > 0 tengsizlikni
yechamiz.
2sin2x sin4x > 0 tengsizlik quyidagi
tengsizliklar sistemasiga teng kuchli.
,, fsin2x>0 , fsin2x<0
1) ( 2) \
sin4x>0 sin4x<0
------H ЯП'. H ЯП
4-----4
Javob:
я я
---+ яП;— + яП
4 4
sin 2x > 0
sin 4x > 0
=> sin4x s 0
2яп <4х^л + 2яп
яп _ я яп
—<x<—+—
2)
sin2x < 0
sin4x < 0
=> sin4x < 0
2. Qarang: 1-variant 24-savol
(9-bet).
3. Qarang: 20-variant 1-savol
(150-bet).
4. J x 7x2 + 1 dx aniqmas integralni
toping.
Yechish:
j x Jx2 +1dx = y/x2 + 1d(x2 +1) =
1 (X +1)2
— - •---------"Г —
2 4 2
—я + 2яп - 4x < 2яп
191
Yechimlar. Matematika va informatika 2017
28-variant
= ---J(x2 +1f+C =
2
Л{х2 +1)44^1+C .
Javob: ^-(x2 + 1)Vx2 +1 + C .
5. Qarang: 22-variant 18-savol
(166-bet).
6. Qarang: 6-variant 29-savol
(59-bet).
7. Qarang: 25-variant 9-savol
(179-bet).
8. Qarang: 6-variant 2-savol
(51-bet).
9. Qarang: 11-variant 25-savol
(100-bet).
.. д 79 148 49 . ..
11. Agar— +----+ — = n bo Isa,
41 51 61
3 5 12 . ,
— + — + —ni n orqah ifodalang.
41 51 61
Yechish:
79 148 49
41 + 51 + 61
bo'lsa,
3 5 12
— + — + — ni orqali ifodalaymiz.
41 51 61
79 148 49 82-3
----1----1-----------h
41 51 61 41
153-5 61-12 „ 3 „
51 61 41
5 „ 12
---+ 1---= n
51 61
6-~
41
^__12
51 61
10.
|x + 2|-|x|
л/4-х2
>0 tengsizlikning
yechimi bo'lmaydigan eng katta
manfiy butun yechimini toping.
Yechish:
Kasrning maxraji doimo musbat.
Shuning uchun ||x + 2| - )x|| > 0.
44-х2 ifodaning aniqlanish sohasini
topamiz.
4 - x2 >0,
4-4<0
(x - 2)(x + 2) < 0
-2 <x< 2
|[x + 2| - |x|| > 0 har qanday sonning
moduli manfiy son emas. Bundan
|x + 2| — |x| - 0 yoki
|x + 2| = |x|.
x + 2 = +x
1)x + 2 = x, 2 = 0 0
2) x + 2 = -x, -2x = 2
x=-1 <= (-2; 2).
Tengsizlikning yechimi (-2; 2).
Tengsizlikning yechimi bo'lmaydigan
eng katta butun yechimi -3.
Javob: -3.
3 5 12 c
— + — + — = 6-n.
41 51 61
Javob: 6 - n.
12. Qarang: 4-variant 26-savol
(38-bet).
13. Tengsizlikni yeching:
log , (x-4) + log2(2x2 - 11x + 12) <
< log2(x + 3).
Yechish:
Aniqlanish sohasi:
x-4>0
• 2x2-11x + 12>0=>
x + 3>0
x >
(x-4)(2x-3)>0^
x>-3
3
x<—,x >4 => x> 4
2
x>-3
192
Yechimlar. Matematika va informatika 2017
28-variant
x e (4; co)
Bir xil asosga keltirib yechamiz.
-Iog2(x -4) + log2(2>c - 11x + 12) <
< log2(x + 3J
log2(24 - 11x + 12) < log2(x + 3) +
+ log2(x - 4)
log2(2x2 - 11x + 12) < log2(x + 3)(x- 4)
2? - 11x + 12<, (x + 3)(x - 4)
(2x -3)(x-4)-(x + 3)(x-4)<0
(x-4) (2x-3-x-3)SO
(x - 4)(x -6)<.O
4 < x s 6.
Javob: (4; 6].
14. logx(1/25) - -2/3 tenglamadan
x ni toping.
Yechish:
, 1 2
log,— = —, x> 0
x 25 3
x = 52 = 53 = 125.
Javob: 125.
15. у = 4x3 - 18X2 + 24x - 10 egri
chizig'i qaysi nuqtalarida o'tkazilgan
urinmalar Ox o'qiga parallel bo'ladi?
Yechish:
1)y' = (4x3 - 18X2 + 24x-10)'=
= 12x2-36x + 24
2) у' = 0, 12X2 -36x + 24 = 0
4 - 3x + 2 = 0, x = 1, x = 2
3)x=1 da y = 4-18 +24-10 = 0,
(1; 0)
x = 2 da у = 4'8 —18-4 + 24-2 -10 = -2,
(2;-2)
(1; 0) va (2; -2) nuqtalarda o'tkazilgan
urinmalar Ox o'qiga parallel bo'ladi.
Javob: (2; -2) va (1; 0).
16. Qarang: 6-variant 27-savol
(59-bet).
17. 4x2 -Vx* - 3 = 0 tenglama
ildizlari ko'paytmasini toping.
Yechish:
4X2-44 -3 = 0
4X2 - |x| - 3 = 0, x2 = |x|2
|x| = a belgilash kiritamiz.
4a2 - a - 3 = 0, a = 1, a =-—
4
]x| = 1 dan x = +1
3
|x| = — tenglama haqiqiy ildizga ega
4
emas.
x = ±1, ildizlari ko'paytmasi
1(-1)=-1.
Javob: -1.
18. Qarang: 6-variant 8-savol
(53-bet).
19. Sharga asosining tomoni
5>/2 ga, balandligi 5 ga teng bo'lgan
muntazam to'rtburchakli piramida ichki
chizilgan. Shar radiusini toping.
Yechish:
s
AB = a =542
SO = H= 5
R - shar radiusi.
AO = R, AAOjO to'q'ri burchakli.
OOi=H-R
R2=(H-R)2 +
AC = y/AB2+BC2 =a42=10
R2=(5-R)2+^
F? = 25- 10R+14 + 25
10R = 50, R = 5.
Javob: 5.
193
Yechimlar. Matematika va informatika 2017
28-variant
20. f (x) = 3x - 5 va g(x) = 3x + 5
bo'lganda, f(g(x)) - g(f(x)) ni toping.
Yechish:
1) f(g(*)) = 3g(x) — 5 = з-(3х + 5) -
— 5 = 9x + 10
2) g(f(x)) = 3f(x) + 5 = 3(3x - 5) +
+ 5 = 9x- 10
f(g(x))-g(f(x)) = 9X+io-
- (9x- 10) = 9x + 10-9x+ 10 = 20.
Javob: 20.
21. Qarang: 5-variant 19-savol
(46-bet).
22. Qarang: 19-variant 26-savol
(148-bet).
23. a = 2 bo'lsa,
a+1
j (sin22x + cos22x)dx integralni
a
hisoblang.
Yechish:
1) sinz2x + cos22x = 1
a + 1
a
a+1
2) \ldx = x
=a+1-a=1.
Javob:1.
24. Qarang: 3-variant 29-savol
(30-bet).
25. Qarang: 6-variant 14-savol
(54-bet).
26. Qarang: 1-variant 8-savol
(5-bet).
27. Qarang: 12-variant 6-savol
(103-bet).
28. Qarang: 21-variant 3-savol
(158-bet).
29. Diagonallarining soni
tomonlarining soniga teng bo'lgan
qavariq muntazam ko'pburchakning
ichki burchaklaridan biri va har bir
uchidan bittadan olingan tashqi
burchaklari yig'indisi topilsin.
Yechish:
D - diagonallar soni, n - tomonlar
soni.
D = n.
n(n-3) „ „ c
-i------ = n, n-3 = 2, n = 5
2
a' = = 72- - bitta tashqi
burchagi.
а = 180° - а'= 180° - 72° =
= 108° - bitta ichki burchagi.
Tashqi burchaklari yig'indisi 360°.
360° + 108° = 468°.
Javob: 468°.
30. Qarang: 8-variant 2-savol
(71-bet).
31. Paskal tilida 50 ta elementdan iborat butun turdagi chiziqli massiv to'g'ri
tavsiflangan javobni ko'rsating.
Yechish:
Paskal tilida butun tur integer kabi belgilanadi.
“array” kalit so'z bilan massiv belgilanadi.
Var- o'zgaruvchilarni e’lon qilish bo'limi.
Var C:array[5...54] of integer;
Bir o'lchovli massiv. Bu massiv elementlari soni 50 ta va hammasi butun
turdagi.
Massiv elementlari soni
(54-5)+ 1 = 50 ta.
Javob: Var C: array[5..54] of integer;.
194
I Yechimlar. Matematika va informatika 2017 29-variant
32. Faqat antivirus dasturlari to'g'ri berilgan qatorni toping.
Yechish:
Antivirus dasturlari NOD32, AVP, Kasperskiy, Dr. Web bo'lib, qolgan
dasturlar WinRar, WinZip - arxivatorlar. Opera, Chrome, Firefox - brauzer
dasturlar.
Javob: 14.
33. Dastur interfeysida maxsus buyruqlar va amallar yig'indisi biror-bir
xususiyatlari asosida birlashishiga ... deyiladi.
Yechish:
Dasturlar interfeysida barcha buyruqlar va barcha instrumentlar yig'indisi
menyular qatorida joylashadi.
Javob: menyu.
34. Qarang: 2-variant 31-savol (11-bet).
35. Qarang: 13-variant 33-savol (115-bet).
36. MS ACCESS 2003 dasturida “запись” nimani bildiradi?
Yechish:
MS Access ma’lumotlar ombori dasturida ma’lumotlar ba’zasi tuzilganda и
bir nechta o'zaro bog'langan yoki bog'likka ega bo'lmagan jadvallardan iborat
bo'ladi. Har bir jadval, ma’lumki, satr va ustundan iborat.
Ma’lumotlar omborida satr - “yozuv”har bir ustun "maydon" deb ataladi.
Javob: jadvaldagi satr.
29-variant
1. Qarang: 1-variant 18-savol
(7-bet).
2. Qarang: 10-variant 25-savol
(90-bet).
3. Qarang: 6-variant 9-savol
(53-bet).
4. Qarang: 2-variant 5-savol
(14-bet).
5. Qarang: 3-variant 8-savol
(24-bet).
6. Agar barcha x, у lar uchun
x3 + 4x"y + axy2 + 3xy - bxcy + 7xy2 +
+ dxy + y2 = x3 + yz ayniyat bajarilsa,
|a + b + c|(a + b + d) ni toping, (c > 1)
Yechish:
c- 2. Ayniyat bo'lganligi uchun
o'xshash darajalar oldidagi
koeffitsiyentlardan foydalanib topamiz.
x2y oldidagi koeffitsiyent
4-b = 0,b = 4
xy2 oldidagi koeffitsiyent
a + 7 = 0, a = -7
xy oldidagi koeffitsiyent
3+d = 0,d = -3
a = -7, b = 4, c = 2, d = -3 da
|a + b + c|(a + b + d) =
= \-7+ 4 + 2\(—7 + 4-3) = HH-6) = -6.
Javob: -6.
7. Qarang: 6-variant 19-savol
(56-bet).
8. Qarang: 5-variant 29-savol
(48-bet).
9. Kollejda 4 ta guruhda 126 ta
o'quvchi bor. Birinchi guruhda jami
3
o'quvchilarning—qismi, ikkinchi
195
Yechimlar. Matematika va informatika 2017
29-variant
5
guruhda—qismi, uchinchi guruhda
21
11
— qismi, qolganlari to'rtinchi guruhda
o'qiydi. Har bir guruhda qanchadan
o'quvchi bor?
Yechish:
4 ta guruhda 126 ta o'quvchi.
1-guruhda 126—= 27
14
к
2-guruhda 126 — = 30
11
3-guruhda 126 — = 33
4-guruhda 126 -27-30 = 36
27, 30, 33, 36.
Javob: 27, 30, 33, 36.
.n x2+6x + 21 , . .
10. у =-------- funksiyaning
11 + 6X+X
eng kichik va eng katta butun
qiymatlari nisbatini toping.
Yechish:
_ x2 +6x + 21 _
У~ x2 +6x + 11 ~
. Ю . 10
— 1 "I-5------— 1 4----------
x +6x+ 11 (X + 3)2 + 2
10
ymax=1+y=1 + 5 = 6
Funksiyaning qiymatlar sohasi (1; 6].
Ут/п ~ 2, Утах — 6
У 2 1
у 6~3’
J max
I u 1
Javob: —.
3
11. Qarang: 24-variant 3-savol
(173-bet).
12. Vx + 7x-Vl-x = 1 tenglamani
yeching.
Yechish:
•Jx-Jl^x = 1-Jx
1 -4x>o, Jx 1, x< 1
x -Vl-x = 1 - 2Jx + x
<Jl-x = 24x - 1
2jx- 1>0, 2jx>1, xb-
4
~<x< 1
4
1 - х = 4х-4у/х + 1
5x-4jx = 0,
4x(54x~4) = 0
4x = 0, x = 0
5 yjx - 4, yj x — , x-
5 25
7
— < x < 1 bo'lganligi sababli
4
x= — tenglama ildizibo'ladi.
. u 16
Javob: —.
25
13. Qarang: 5-variant 5-savol
(43-bet).
14. ABC uchburchakning AB, BC,
CA tomonlarida mos ravishda
shunday M, N, P nuqtalar olinganki,
AM:AB = BN:BC = CP:CA = 1:3
munosabat o'rinli. MNP uchburchak
yuzasi 2 ga teng bo'lsa, ABC
uchburchak yuzasini toping.
Yechish:
AM:AB = BN:BC = CP:CA = 1:3,
Smnp = 2, Sabc = ?
C N B
Sabc = Samp + Sbmn + Scnp + Smnp
o AM AP .
1) SAMP=---------sma
196
Yechimlar. Matematika va informatika 2017
29-variant
AM = x, AP = 2y
_ AC-AB .
5ЛВС =--------sina
AB = 3x, AC = 3y
S№P _ AM-AP _ x-2y _ 2^
Здвс " AC-AB ~ 3x-3yl
2
О __c
AMP ~ g °4SC
2
f oMN g ABC ’
2
Q - Q
°CNP “ g °ABC
2
§ABC ~ 3 • — SABC + SMNp
Sabc = 3Smnp = 2-3 = 6
Sabc = 6.
Javob: 6.
15. Di(/) to’plamda /(x) va D2(g)
to’plamda g(x) funksiyalari berilgan
bo’lsa, bunda D,(/) fl D2(g) # 0.
Quyidagi qaysi sohada /(x) va g(x)
funksiyalar bo hnmasi F(x) =^--^
g(x)
aniqlangan bo’ladi?
Yechish:
Di(f) to‘plamda f(x) funksiya berilgan,
D2(f) to'plamda g(x) funksiya berilgan.
Di(f) n D2(g) # 0
Funksiyalar bo'linmasi
F(X)
g(x)
1) g(x) # 0 bo'lishi kerak.
2) Drff) n D2(g) # 0 bo'lganligi sababli
Dj(f) U D2(g) bo'lishi kerak.
Demak, g(x) 0 ga aylanadigan x larni
o'z ichiga olmagan.
D-i(f) U D2(g) sohada
f(X)
F(x) aniqlangan bo'ladi.
g(x)
Javob: g(x) 0 ga aylanadigan
x larni o’z ichiga olmagan
Di(g)(JD2(f) sohada.
16. Qarang: 8-variant 1-savol
(70-bet).
17. Qarang: 5-variant 17-savoi
(45-bet).
18. Agar/(x) = 4*xbo’lsa, /’(x) = 0
tenglamani yeching.
Yechish:
1) Ko'paytmadan hosila olamiz:
f(x) = (4*-x)' = 4?ln4x + 4х = 4f(x-ln4 + 1)
2) f'(x) = 0, 4x(xln4 + 1) = 0
4х # 0, x-ln4 + 1= 0, xln4 = -1
1 Ine ,
ln4 ln4
Javob: -Iog4e.
19. Qarang: 4-variant 27-savol
(38-bet).
( ( 1A ( 1
- (-3). -5- -(-1): — +
\ \ 'J/ \ ' b J J
+ (-5,12):(—1,28)-(-8,3 + 0,35-
- 2,05 + 7) ni hisoblang.
Yechish:
( ( 1\ ( 1 \\
“k3)- -5irH):r^ +
\ \ J \ ib J)
+ (-5,12):(-1,28)-(-8,3 + 0,35-
34 22 16 . .„'I
- £.,05 + 7) =----3--------1-16 +
11 17 I 3 )
5 12
+±l±.^3') = -4-0 + 4-(-3)=-16.
1,28
Javob: -16.
21.
1 4
- + -=7,
a b
4 1
- + - = 3
a b
bo’lsa, a soni b
sonidan necha marta katta?
197
Yechimlar. Matematika va informatika 2017 29-variant
Yechish: 26. x = -y, z = -2 bo'lsa,
Birinchi tenglamani 4 ga ko‘paytirib, ikkinchi tenglamadan ayiramiz. \— + — = 28 a b 4 1 ._75_3 h+b~3 ’ 25 5 — = 25 b 1 4 4 20 1 a b 3~ 3 ~ 3 5 ? 5 a =3, a:b = 3:- = 3 - = 5 5 3 5 marta katta. Javob: 5 marta. x3 +y3 + z3 -3xyz .. , . —7 ifodaning X + у + z - xy - xz - yz qiymatini toping. Yechish: Qulay usul bilan yechish uchun x= 1, у = -1 deb olamiz. x = 1, у = -1, z=-2 da ifodaning qiymatini topamiz. 13+ (-1)3+ (-2)3-3-1-{-1)-(-2) 12 + (-1)2 + (-2)2 -1-(-1)- -1 (-2) - (-1) (-2) ~ _ 1-1-8-6 _-14 _ _2 ~ 1 + 1 + 4 + 1 + 2-2~ 7 Javob: -2.
22. Qarang: 23-variant 1-savol (168-bet). 27. f(x) = 42sin9x sin12x uchun boshlang'ich funksiyani toping.
23. Agar - - — = 0 bo'lsa a 2 24 a2 ... .. a2 + (_д)2 ni hisoblang. Yechish: 1) Ko'paytmadan yig'indiga olamiz. 1 sin9xsin12x =— (cos(9x- 12) - 1 - cos(9x + 12)) =— (cos3x - cos21 x)
Yechish: — - — = 0 tenglamadan a ning a 2 qiymatini topamiz. — = — , a2 - 16, a - ±4 a 2 24 a2 16 16 „ a2 (-4) 16 16 Javob:2. 1 2) f(x) = 42 — (cos3x- cos21x) - = 21cos3x - 21cos21x Boshlang'ich funksiyasi: F(x) = 7sin3x - sin 21 x + C. Javob: 7sin3x - sin21x + C. 28. Qarang: 26-variant 28-savol (185-bet).
24. Qarang: 23-variant 22-savol (170-bet). 29. Qarang: 1-variant 21-savol (8-bet).
25. Qarang: 24-variant 29-savol (177-bet). 30. Qarang: 9-variant 25-savol (83-bet).
31. Paskal tilida 63 ta elementdan iborat ikki o'lchovli belgili massiv to'g'ri
tavsiflangan javobni ko'rsating.
Yechish:
Belgili massivlar "char" belgi turi bilan belgilanadi.
198
Yechimlar. Matematika va informatika 2017 29-variant
Ikki o'lchovli massiv sintaksisi quyidagicha:
Var B: array[ 1..10;1..10] of integer.
Bu yerda 10ta ustun va 10ta satr elementlaridan iborat butun turdagi ikki
o'lchovli massiv.
Var M:array[3..11,2.. 8] of cher;.
Javob: Var M: array[3..11,2..8] of char;.
32. Hisoblang va javobini 16 lik sanoq sistemada ifodalang:
1001011(2)*100(2) + 254(8).
Yechish:
1001011(2)-100(2) + 254(8)
1001011
X 100
0000000
0000000
1001011
100101100(2)
254(-Sj -> xp) (1-chi jadval)
2 5 4
4* 4* 4*
010 101 100
254(8)^ 10101100(2).
100101100(2) + 10101100(г) = 111011000(2).
100101100
10101100
111011000(2)
1+0 = 1
1 + 1 = 10
8765432 1 0
111011000, -»xW).
1-28 + 1-27 + 1-26 + 0-25 + 1-24 + 1-23 + 0-22 + 0-2* 1 * iii + 02° =
= 256 + 128 + 64 + 0 + 16 + 8 + 0 + 0 + 0 = 472(10).
1001011(2)100(2) + 254(8) = 111011000(2) ^ X1e.
Ikkilik sanoq sistemasidagi bu sonni oxiridan boshlab tetradlarga ajratib,
chiqamiz.
(1-chi jadvaldan)
0001 1101 1000->1D8lte .
iii
1 D 8
Javob: 1D8.
33. Hisoblang va javobini 10 lik sanoq sistemada ifodalang:
1001011 (2)*100(2) + 254(8).
Yechish:
1001011(2)' 100(2) + 254(8)
199
Yechimlar. Matematika va informatika 2017
30-variam
1001011
100
0000000
0000000
1001011
100101100(2)
254(8) -»X(2) (1-chi jadval)
2 5 4
J. 'L Ф
010 101 100
254(8) 10101100(2).
100101100(2) + 10101100(2) = 111011000(2).
100101100
' 10101100
iiioiiooo{2)
1+0=1
1 + 1 = 10
87 6 5 43 2 1 0
1110110002^X{10).
1-28 + 1-27 + 1-2S + 0-25 + 1-24 + 1-2? + 0-22 + 0-21 + 0-2° =
= 256+128 + 64 + 0+16 + 8 + 0 + 0 + 0 = 472(io>.
Javob: 472(ю).
34. Qarang: 7-variant 31-savol (68-bet).
35. Qarang: 21-variant 34-savol (163-bet).
36. Qarang: 4-variant 31-savol (39-bet).
30-variant
1. Qarang: 16-variant 27-savol (131-bet). 2. Qarang: 10-variant 24-savol (90-bet). 3. Qarahg: 13-variant 5-savol (110-bet). 4. Rasmda у = a +—— funksiya x + c grafigi tasvirlangan. Quyidagilardan qaysi biri doim o’rinli? b J\ b<0 ' I o4> I _ I I / I/ Yechish: b у = a + x + c Bunda a = 0, b < 0, c < 0. Doimo o'rinliboladigan tenglik: be-a5 > 0. Javob: be - a5 > 0. 5. (x2 — 1) -•- (2X2 — 3) + ... + + (Юх2 - 19) = T20 tenglama yechimlari yig’indisini toping. Yechish: a-, = x2 - 1, a? = 2X2 - 3 d = a2-a1 = 2x2-3-xz+ 1 =x2-2 S = x?-1 + 10.xlr19.10 = 120 1° 2
200
Yechimlar. Matematika va informatika 2017
30-variant
11x2-20 = 24, 11)? = 44,
x2 = 4, x = ±2
Tenglama ildizlariyig'indisi 2 + (-2) = 0.
Javob: 0.
6. Qarang: 4-variant 28-savol
(39-bet).
7. Qarang: 3-variant 2-savol
(22-bet).
8. Qarang: 17-variant 5-savol
(132-bet).
9. Qarang: 6-variant 16-savol
(55-bet).
10. у = 2cos18x - cos36x
funksiyaning hosilasini toping.
Yechish:
1) (cos(ax + b))’ = -asin(ax + b)
(cos18x)’ = -18sin18x
(cos36x)' = -36sin36x
y’ = 2(cos18x)' - (cos36x)' =
= -36sin18x + 36sin36x =
= -36(sin18x - sin36x)
2) sin18x - sin36x ayirmadan
ko'paytmaga o'tamiz.
. . 18x-36x
sm18x - sm36x = 2sin----------
2
18x + 36x „ . , n .
cos--------- = 2sin(-9x)-cos27x
3) y' = -36(-2sin9xcos27x) =
= 72sin9xcos27x.
Javob: 72sin9xcos27x.
I, 15
11. J4 +-—-= 16 tenglamaning
haqiqiy ildizlari yig'indisini toping.
Yechish:
— = a-1
a
yj4x-2+15-4i2 =16
J16-4X-2 =16
x-2 x-2
4-4 2 = 42,4 2 =4
v — 2
?—±- = 1,x-2 = 2,x = 4
2
x = 4 tenglama ildizi. Ildizlari yig'indisi
4 ga teng.
Javob: 4.
12. Quyidagi ko'pyoqlardan qaysi
birida 4 ta yoq bor?
Yechish:
4 ta yoqi bor ko'pyoq bu uchburchakli
piramida ABC, ADC, ADB, BDC.
D
Javob: 2.
13. Kasrning maxrajini
irratsionallikdan qutqaring.——Д——
3 + V3 + V9
Yechish:
(Va - "Jby^a2 + tfab + ^/b2) =
= (Va)3 - (^/b)3 = a - b formuladan
foydalanib yechamiz.
6 6
3+V3 + V9 "V3(V?+1 + V3)_
6(V3-t)
~ V3(V3-i)(V?+V3 +f)-
e(V3-i) e(V3-i)
~V3((V3)3-i)”W-7r
e(V3-i) 3(tf3-i)
2-^/3 V3
201
Yechimlar. Matematika va informatika 2017
30-variant
= 4?(43-1) = 3-№ = 3-49.
Javob: 3-79 .
14. Qarang: 15-variant 9-savol
(123-bet).
15. Qarang: 8-variant 4-savol
(71-bet).
16. Qarang: 2-variant 1-savol
(13-bet).
17. Qarang: 22-variant 7-savol
(164-bet).
18. Qarang: 9-variant 27-savol
(84-bet).
19. Qarang: 6-variant 15-savol
(55-bet).
20. Qarang: 10-variant 28-savol
(91-bet).
21. Agar a (-4; 8; -12) va
b (-6; -3; 9) berilgan bo'lsa,
1 1 -
—a + — b ni hisoblang.
Yechish:
1 -
1) la=(-2;4;-6)
। la | =7(-2)2+42+(-6)2 =
= 44 + 16 + 36 = 456 = 24l4
1 -
2) ~b = (-2;-1;3)
•J
| lb I =V(-2)2 + (-7)2 + 32 =
•J
= 44 + 1 + 9 = 4l4
3) ||a| + |lb | =
= 24l4 + 4k4 = з4Й .
Javob: 3Vl4 .
22. Kvadratga ikkita doira ichki
chizilgan. Radiusi 2,25 ga teng bo'lgan
birinchi doira kvadratning ikkita qo'shni
tomonlariga urinadi, radiusi 3,75 ga
teng bo'lgan ikkinchi doira kvadratning
qolgan ikkita tomoni va birinchi doiraga
urinadi. Kvadratning yuzini toping.
Yechish:
ABCD kvadrat
r = 2,25; R = 3,75, SAbcd=?
A F D
AC - diagonal, AB = a,
AC=a42,
OF = OE = AE = FA- r,
AO=r42
OiN = NC=CM = OjM = R
OiC = R42
OOi = r + R
AC=a42 =r42 + r+ R+R42
a42 = r + R + 42 (R + r)
(r + R\42+1) _
42 “
(2,25 + 3,75)(V2+1) _
42
=342(42+1)
S = a2 = (342(42+ 1)4 =
= 18(3 + 242).
Javob: 18(3 + 242).
23. (x2 + 28x + 28)-
(x2 + x + 28) = 28X2 tenglama haqiqiy
ildizlari yig'indisini toping.
Yechish:
x2 + 28 = a belgilash kiritamiz.
(>4 + 28 + 28x)(>4 + 28 + x) = 28x4
202
Yechimlar. Matematika va informatika 2017
30-variant
(a + 28x)(a + x) = 28/
a2 + 29xa + 28/ = 28/
a2 + 29xa = 0, a(a + 29) =0
a =0, a = -29x
/ + 28 = 0 tenglama haqiqiy ildizga
ega emas.
x?+28 = -29x
x2 + 29x + 28-0 tenglama haqiqiy
ildizlari yig'indisi
Xi +x2 = -29.
Javob: -29.
24. /(x) = 30cos7x cos8x uchun
boshlang'ich funksiyani toping.
Yechish:
1) Ko'paytmadan yig'indiga otamiz.
cos7xcos8x =— (cos(7x - 8x) +
1
+ cos(7x + 8x)) =— (cosx + cos15x)
1
2) f(x) = 30-~ (cosx + cos15x) =
= 15cosx + 15cos15x
Boshlang'ich funksiyasini topamiz.
F(x) = 15sinx + sin15x + C.
Javob: 15sinx + sin15x + C.
25. Qarang: 5-variant 30-savol
(48-bet).
26. Qarang: 19-variant 28-savol
(148-bet).
27. Qarang: 2-variant 3-savol
(13-bet).
28. Tenglamani yeching:
|3x + 9| - |2x + 6| = |x - 5|
Yechish:
Soddalashtirib yechamiz.
|3x + 9| -|2x + 6| = |x—5|
3|x + 3| - 2|x + 3| = |x - 5|
|x + 3| = |x- 5|
|a| = |b| da a = ±b bo'ladi.
|x + 3| = |x - 5|
x + 3 = +(x-5)
1)x + 3 = x-5
3--50
2)x + 3 = 5-x
2x = 2
x-1.
Javob: 1.
29. Qarang: 3-variant 14-savol
(25-bet).
30. Agar 1 + 2f(x - 1) = 2/(x) va
/(0) = 0 bo'lsa, f(2014) ni toping.
Yechish:
1 + 2f(x — 1) = 2f(x)
f(0) = 0, f(2014) = 2
x = 1 da 1 + 2f(1 - 1) = 2f(1)
1 + 2f(0) = 2f(1)
1
2f(1) = 1 + 20=1, f(1)=-L
x = 2da1 + 2f(1) = 2f(2)
1
2f(2) = 1 + 2-= 2, f(2) = 1
f(2014) ni topamiz.
ai=f(1)=-^,a2 = f(2) = 1
агон = a< + 2013 d =- + 2013-- =
2 2
1 1
= 7- (1 + 2013) =± -2014 = 1007
a2oi4 = f(2014) = 1007.
Javob: 1007.
31. Paskal tilida berilgan ifodaning qiymatini toping:
trunc(abs(-15)/round(4,5)).
Yechish:
round(4,5) = 5. abs(-15) = 15.
trunc(15/5) = 3.
Javob: 3.
203
Yechimlar. Matematika va informatika 2017 30-variant
32. Hisoblang va javobini 16 lik sanoq sistemada ifodalang:
20(16*10(8) + 254(8).
Yechish:
20(1в)-10(8) + 254(8; —* Xi6-
20(1в) —> 00100000(2) 1-chijadval bo’yicha
10(8) —* 001000(2)
254(8) 010101100(2)
1) 100000
1000
000000
000000
000000
100000
100000000(2|
2) 100000000
10101100
110101100(2,
000110101100.,-X1S
1 A C
Xie = 1AC(16).
Javob: 1AC.
33. Hisoblang va javobini 16 lik sanoq sistemada ifodalang:
101011(2*10(2)+ 237(8).
Yechish:
101011
x 10
1010110
Ю1011(2)-10(2) = 1010110(2).
237(8) = 010011111(2). (1-chijadvaldan).
1010110
010011111
011110101(2)
011110101 = F5m.
F 5
Javob: F5.
34. Qarang: 3-variant 33-savol (31-bet).
35. MS ACCeSS dasturida Memo maydoni qanday ma’lumotlarni saqlaydi?
Yechish:
MS Access dasturining memo tor maydonida matnli ma’lumotlarni saqlash
uchun mo'ljallangan. Bu maydonga 255 dan yuqori belgidan iborat matnlarni
saqlash mumkin. Chunk! Accessning “текстовый” tur maydonida faqat
255 gacha bolgan matnlarniyozish mumkin.
Memo maydonining afzalliklari, unda kata hajmdagi matnlar saqlanadi,
xususan, 256 da 65535 ta simvolya’ni belgidan iborat matnlarni saqlash mumkin.
Javob: 256 tadan 65535 tagacha belgidan iborat bo'lgan matnli
ma’lumotlarni faylda saqlaydi.
204
Yechimlar. Matematika va informatika 201 / 30-variant
36. Intranet nima?
Yechish:
Intranet - bu internet texnologiyasi hisoblanib, dastur ta’minoti va protokollar
asosida tashkil etilgan tarmoqdir. Intranet ma’lumotlar ba’zasi va elektron
jadvallar bilan jamoa bo’lib ishlash imkonini beruvchi korxona yoki tashkilot
miqyosidagi yangi axborot muhitinl tashkil etuvchi kompyuter tarmog’idir.
Intranet- bu tarmoq internet global tarmog‘in ing cheklangan turidir. Chunki
Intranet tarmog'i foydalanuvchisi undagima’lumotlardan foydalanishi uchun
ularning qaysi serverda, qaysi katalogda, qanday nom bilan
saqlanayotganligini, ularga kirish usul va shartlarini bilishi zarur bo’ladi.
Internetda esa bunday noqulayliklar oldi olingan.
Javob: katta global tarmoqning cheklangan turi.
205
Ilova
1-jadval
O'nlik sanoq sistemasi Ikkilik sanoq sistemasi Sakkizlik sanoq sistemasi O‘n oltilik sanoq sistemasi
0 0 0 0
1 01 1 1
2 10 2 2
3 11 3 3
4 100 4 4
5 101 5 5
6 110 6 6
7 111 7 7
8 1000 10 8
9 1001 11 9
10 1010 12 A
11 1011 13 В
12 1100 14 C
13 1101 15 D
14 1110 16 E
15 1111 17 F
16 10000 20 10
17 10001 21 11
18 10010 22 12
19 10011 23 13
20 10100 24 14
21 10101 25 15
206
MUNDARIJA
1-variant.............................................................3
2-variant............................................................13
3-variant............................................................22
4-variant............................................................32
5-variant............................................................42
6-variant............................................................51
7-variant.......................................................... 61
8-variant............................................................70
9-variant............................................................78
10-variant...........................................................86
11-variant...........................................................94
12-variant..........................................................102
variant.........................................................109
14-variant..........................................................117
15-variant..........................................................122
16-variant..........................................................127
17-variant..........................................................132
18-variant..........................................................138
19-variant..........................................................144
20-variant..........................................................150
21-variant..........................................................158
22-variant..........................................................163
23-variant..........................................................168
24-variant..........................................................173
25-variant..........................................................178
26-variant..........................................................182
27-variant..........................................................186
28-variant..........................................................191
29-variant..........................................................195
30-variant..........................................................200
llova...............................................................206