ahmadjonov azamat unco academy axborotnoma 2003

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ahmadjonov azamat unco academy axborotnoma 2003 1 a.03 abituriyentlar uchun variantlashtirilgan testlar to’plami o’qituvchi: ahmadjonov azamat foydalanuvchi: _______________________ ahmadjonov azamat unco academy axborotnoma 2003 2 ahmadjonov azamat unco academy axborotnoma 2003 3 1-variant 1. 2 2 3 2 5 2 6 2 3                  ni soddalashtiring. a)   5 2 2 3 6    b) 2 3 c) 5 6  d)  2 2 3  2. 3 52 3 4 4 5 7x x x x    tenglamani yeching. a) ∅ b) 1,75 c) -1,75 d) 1,25 3. 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 tenglamaning koeffitsiyentlari 𝑏 = 𝑎 + 𝑐 tenglikni qanoatlantiradi. agar 𝑥1 𝑣𝑎 𝑥2 berilgan kvadrat tenglamaning ildizlari bo’lsa, 1 2 2 1 x x x x  ni hisoblang. a) 2 2a …

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ng. a) 0; 2   b) 0; 1 c)    0; 1 1; 2 d) 0; ; 2 2 2             17.  2 3cos 8 7y x   funksiyaning eng kichik musbat davrini toping. a) 2𝜋 b) 𝜋 2 c) 𝜋 3 d) 𝜋 4 18. 2 sin 1 4 x x   tengsizlikni yeching. a) ∅ b) 2 ; 2 , 2 2 n n n z              c) ;    d)  ;   ahmadjonov azamat unco academy axborotnoma 2003 4 19. uchburchakning tomonlarining uzunliklari sin30 ; sin40 sin60va   ga teng. shu uchburchakning turini aniqlang. a) o’tkir burchakli b) o’tmas burchakli c) to’g’ri burchakli d) aniqlab bo’lmaydi 20. agar 5 11log 2 log 3x …

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c) 0,392 d) 0,416 3. 6 61 sin 22,5 cos 22,5    ni hisoblang. a) 3 1 2  b) 6 5 4  c) 10 3 2 8  d) 16 7 2 16  4. 2 1 cos sin 0 2 3 3 x x tg x                       tengsizlikni yeching. a) ; , 3 2 n n n z              b) ; , 3 3 n n n z              c) ; , 3 6 n n n z              d) ; , 6 3 n n n z …

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d) 3 3 8 16. teng tomonli silindr shaklidagi g’o’ladan eng katta hajmdagi shar yo’nib olindi. g’o’laning qancha foizi chiqindiga ketgan? a) 25 b) 30 c) 33 1 3 d) 37 2 5 17. qavariq n burchakning diagonallari soni 25 tadan kam emas 30 tadan ko’p emas. n nechaga teng bo’lishi mumkin? a) 7 b) 8 c) 9 d) 10 18. 𝑡𝑔5550 ni hisoblang. a) 3 6 b) 3 1 c) 2 3 d) 2 3 19. muntazam o’nikkiburchakli piramidaning apofemasi 2√2 ga teng, barcha yoqlari asos tekisligiga 450 li burchak ostida og’ishgan. uning hajmini toping. a) 56 − 30√2 b) 64 − 32√3 c) 68 − 48√2 d) 64 − 30√3 20. 4sin 2 . ?y x y   a) 22sin 2 sin4x x b) 24sin 4 sin2x x c) 24sin2 sin 4x x d) 24sin 2 sin4x x 21. cos 2 x x     …

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a) 12 b) 14 c) 15 d) 16 3-variant 1. a ning qanday haqiqiy qiymatlarida 4 2 2x a x a   tenglama uchta turli haqiqiy ildizga ega bo’ladi? a) (0: 4) b) 2 c) 0 va 1 d) [0; 1] 2. agar 2 2 2 2 1m n p q    va mp+nq=0 bo’lsa, mn+pq ning qiymatini toping. a) 1 b) 0 c) 2 d) 4 3.  na ketma-ketlikning dastlabki n ta hadining yig’indisi 211 4ns n  formula bo’yicha hisoblanadi. 5 6a a ning qiymatini toping. a) 60 b) 80 c) -80 d) -60 4. lg , lg 3a b va sonlar ko’rsatilgan tartibda arifmetik progressiyani tashkil etadi. agar 4 2a b bo’lsa, a b ning qiymatini toping. a) 1000 b) 300 c) 101 d) 110 5.  nb geometrik progressiyada 4 2 2 324 6b b va b b    bo’lsa, …

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ahmadjonov azamat unco academy axborotnoma 2003 1 a.03 abituriyentlar uchun variantlashtirilgan testlar to’plami o’qituvchi: ahmadjonov azamat foydalanuvchi: _______________________ ahmadjonov azamat unco academy axborotnoma 2003 2 ahmadjonov azamat unco academy axborotnoma 2003 3 1-variant 1. 2 2 3 2 5 2 6 2 3                  ni soddalashtiring. a)   5 2 2 3 6    b) 2 3 c) 5 6  d)  2 2 3  2. 3 52 3 4 4 5 7x x x x    tenglamani yeching. a) ∅ b) 1,75 c) -1,75 d) 1,25 3. 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 tenglamaning koeffitsiyentlari …

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