46. Agar log
47. y=
funksiya grafigini 3 birlik o‘nga surgach , 2 birlik pastga sursak , so‘ngra ordinatalarini 2 marta uzaytirsak
20
A) −4
B)−8
C)−2
D)−5
48.Ifodani soddalashtiring.
a+b
√a2
3
− √b2
3
+
√ab2
3
− √a2b
3
√a2
3
−2 √ab
3
+ √b2
3
√a
6
− √b
6
─√b
6
A)1 B)
√b
6
C)
√a
6
D)
√a
6
─2
√b
6
49.y=
x
2
+10 parabola va shu parabolaga (0 ;1) nuqtadan o‘tkazilgan urinmalar bilan chegaralangan shakl yuzini toping
?
A) 29 B)17 C)18 D)22
50. Hisoblang. 1∙3+3∙9+5∙27+7∙81+…+43∙3
22
A) 3
22
∙21+6 B) 3
43
∙27+3 C) 3
23
∙22+3 D) 3
23
∙21+3
51.
α + β + γ = π cos α cos β csc γ=−
1
2
bo‘lsa sin
2
α+sin
2
β+sin
2
γ=?
A)1 B)√
2 + √2 C)2 D)−√2
52. ABCD parallelogramm uchta uchining koordinatalari ma’lum: A(0;1), B(1;3), C(9;3). ABCD parallelogramm
yuzini toping.
A) 14 B) 16 C) 25 D) 24
53. Agar tg4α=−
1
3
bo‘lsa, ctgα−tgα−2tg2α ning qiymatini toping.
A) −6 B) −2 C) −8 D) −12
54. a, b manfiy butun sonlar uchun a=b+5 va a+b─c=13 bo‘lsa, c ning eng katta qiymatini toping.
A) ─17 B) ─18 C) ─19 D) ─20
55. Yuk tashish mashinasi 240 km yo‘lni bosib o‘tishi kerak edi. Mashina yo‘lning o‘rtasida 30 daqiqa to‘xtab qolgach
tezligini 20km/soat ga oshirib, belgilangan joyga o‘z vaqtida yetib keldi. Mashina yo‘lning ikkinchi yarmini bosib
o‘tishiga ketgan vaqtni (soat) toping.
A) 1,5 B) 2 C) 1,2 D) 1,4
56. Koordinatalari A(─2;0), B(4;0) va C(2;3) nuqtalarda bo‘lgan uchburchakning Ox o‘qi atrofida aylantirilishidan
hosil bo‘lgan jismning hajmini toping.
A) 18π B) 15π C) 16π D) 12π
57.
y = −2√x va y = −2x
3
egri chiziqlar bilan chegaralangan soha yuzini toping.
A)
5
3
B)
5
6
C)
5
4
D)
5
12
58. ABCD parallelogramm berilgan. M nuqta BD dioganalda yotadi, bunda MD : BM = 2 : 1. Agar ADCM
to‘rtburchak yuzi 8 ga teng bo‘lsa, ABCD parallelogramm yuzini toping.
A) 24 B) 16 C) 12 D) 8
59.
Tenglamani yeching. x + 1 +
x+
x+
…..
x+1
x+1
x+1
=21
A)19;0 B)─19;0 C)19 D)0
60. P ( x ) ko‘phadni x – 2 ga bo‘lgandagi qoldiq 5 ga,
x
2
− 2x + 3 ga bo‘lgandagi qoldiq 3x – 10 ga teng bo‘lsa, bu
ko‘phadni
(x − 2)(x
2
− 2x + 3) bo‘lgandagi qoldiqni toping.
A)2x
2
− 3x + 3 B) 3x
2
− 3x − 1 C) 2x
2
− 5x + 7 D) 3x
2
− 5x + 3
61. 3 , 12 , 30 , 57 , 93 , …………………….. ketma ketlikni nechanchi hadi 34455 ga teng bo‘ladi.
A)88
B) 77
C) 66
D) 55
62.
3x =
π
4
bo‘lsa,
cos11x+cos9x
sin5x∙sin8x
=?
A) –2 B) –1 C) 0 D) 1
63. Agar
α + β + γ = π bo‘lsa ctg α ctg β + ctg βctgγ + ctg α ctg γ ni hisoblang.
A)0.5
B)0
C)
−0.5
D)1
21
64. Teng yonli trapetsiyaning dioganali o‘tkir burchak bissektrisasidir. Trapetsiyaning asoslari uzunliklari 2:3 kabi
nisbatda, perimetri esa 12 ga teng. Trapetsiyaning o‘rta chizig‘ini toping.
A)
7
2
B)
10
3
C)
7
3
D) 3
65.
𝑥
3
+ 𝑥
2
+ 𝑥 = −
1
3
tenglamaning nechta haqiqiy ildizi bor.
A) 3 B) 1 C)
∅ D) 2
66. 5 xil kitobdan necha usul bilan 8 kitobdan iborat to‘plam yozish mumkin.
A) 336
B) 56
C) 495
D) 385
67. muntazam oltiburchakli prizma yon yog‘ining dioganali d ga teng va u asos bilan
α burchak
tashkil etadi. Prizma hajmin toping.
A)
3√3
2
d
3
sin
2
acosa B)
3√3
4
d
3
sin2acosa
C)
3√3d
3
sin2acosa D)
3√3
2
d
3
sinacos2a
68.Tengzizlikni yeching.
1+log
2
x
1−log
4
x
≤ 2
A) (
−∞; √2]∪ (4; ∞)
B)[
√2; 4)
C) (
0; √2]∪ (4; ∞)
D)
(4; ∞)
69.Har birining yuzi 6 ga teng ikkita muntazam uchburchak bo‘lib, ulardan ikkinchisi birinchisini markazi
atrofida 30
0
ga burishdan hosil bo‘lgan shu uchburchaklar kesishmasining yuzini toping.
A)
6(√3 − 1)
B)
3(√2 − 1) C) 9(2√3 − 2) D) 2(√3 + 1)
70. O‘suvchi arifmetik progressiya dastlabki 3ta hadi yig‘indisi 15 ga teng. Agar shu progressiyaning birinchi ikkita
hadidan bir ayrilsa, uchinchi hadiga esa bir qo‘shilsa, u holda hosil bo‘lgan uchta son geometrik progressiya tashkil
qiladi. Arifmetik progressiya dastlabki 10 ta hadining yig‘indisini toping.
A) 109 B)120 C)96
D)136
71. Muntazzam tetraedrning balandligi 2 ga teng bo‘lsa, uning to‘la sirtini toping.
A)
3√3 B) 8√3 C) 6√3 D) 12√3
72.
{
|x + 4| ≤ 9
|2x + 5| ≥ 15
tengsizliklar sistemasi nechta butun yechimga ega?
A) 5 B) 4 C) 6 D) 7
73. Tekislikni kesib o‘tuvchi kesmaning uchlari tekislikdan 4 va 6 masofada tursa, berilgan kesma o‘rtasidan
tekislikkacha bo‘lgan masofani toping.
A) 4 B) 1 C) 3 D) 2
74.
x
lg
2
x−4lgx+1
> 1000 tengsizlikning eng kichik natural yechimini toping.
A) 10001 B) 100 C) 10000 D) 1001
75. tg(0,5arccos0,6−3arcctg(−2)) ni Hisoblang
A)
−
24
7
B)
−
28
7
C)
−
20
7
D)
−
28
7
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