2019 Matematika savollari Uchta tengdosh prizmaning balandliklari
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80 o 4 12 12 x Chizmaga ko’ra 𝑥 ni toping. A) 4 B) 6 C) 8 D) 12 292.
o 40 o 4 x 10 2 Chizmaga ko’ra 𝑥 ni toping. A) 20 B) 22 C) 24 D) 18 15
293. (3𝑥 − 1) 2 (3𝑥 + 1)
2 ko’phadlarni ko’paytiring. A) 81𝑥
2 − 18𝑥 + 1 B) 81𝑥 4 + 18𝑥
2 + 1 C) 81𝑥 4
2 + 1 D) 81𝑥 4 − 18𝑥 + 1 294. 𝑓(𝑥) 6-darajali funksiya bo’lsa, (𝑥 − 4)
2 ∙ 𝑓(𝑥) + 4𝑥 ni 𝑥 0 = 4 nuqtadagi hosilasini qiymatini toping. A) 0 B) 4 C) 2 D) 1 295. 𝑓(𝑥) 5-darajali funksiya bo’lsa, (𝑥 − 5)
2 ∙ 𝑓(𝑥) ni 𝑥 0 = 5 nuqtadagi hosilasini toping. A) 5 B) 4 C) 2 D) 0 296. 𝑓(𝑥) 9-darajali funksiya bo’lsa, (𝑥 − 1)
2 ∙ 𝑓(𝑥) + 1 ni 𝑥 0 = 1 nuqtadagi hosilasini toping. A) 0 B) 1 C) 2 D) 9 297. 2
2 𝑥2−6𝑥
= 1 bo’lsa, 𝑥 ning qiymati 14 dan qanchaga kam? A) 8 B) −6 C) 6 D) 20 298.
𝑙𝑜𝑔 3 12+𝑙𝑜𝑔 4 12 𝑙𝑜𝑔 3 12∙𝑙𝑜𝑔
4 12 + 1 2 305. Uchburchakning burchaklari 1:2:3 nisbatda. Agar uchburchakning eng kichik tomoni 6 ga teng bo’lsa, uning eng katta tomonini toping. A) 6√3 B) 12 C) 9 D) 6√2 306. Uchburchakning ikki tomoni uzunligi 10 va 14 ga teng. Agar ularga tushirilgan medianalar o’zaro perpendikulyar bo’lsa, uchburchakning uchinchi tomoni uzunligini toping. A) 2√7,8 B) 4√7,8 C) 2√15,6 D) 4√15,6 307. Uchburchakning ikki tomoni uzunligi 10 va 8 ga teng. Agar ularga tushirilgan medianalar o’zaro perpendikulyar bo’lsa, uchburchakning uchinchi tomoni uzunligini toping. A) 2√8,2 B) 4√8,2 C) 4√4,1 D) 2√4,1 308. 4
2𝑥−2 𝑥 = √2 𝑥+2 tenglamaning haqiqiy ildizlari nechta? A) 1 B) 0 C) 2 D) 3 309. Agar 44𝑛 (𝑛𝜖𝑁) ko’paytma biror sonning kvadrati bo’lsa, 𝑛 + 13 ning eng kichik qiymatini toping. A) 13 B) 24 C) 17 D) 57 310. Agar 95𝑛 (𝑛𝜖𝑁) ko’paytma biror sonning kvadrati bo’lsa, 𝑛 + 18 ning eng kichik qiymatini toping. A) 18 B) 37 C) 23 D) 113 311. O’q kesimi muntazam uchburchak bo’lgan konusning to’la sirti yuzi 30 ga teng bo’lsa, uning asosi yuzini toping. A) 20 B) 10 C) 10 𝜋
20 𝜋
312. O’q kesimi muntazam uchburchak bo’lgan konusning to’la sirti yuzi 48 ga teng bo’lsa, uning asosi yuzini toping. A) 16 B) 32 C) 16 𝜋
32 𝜋
313. 𝑓(𝑥) = (4 𝑥− 1 2 𝑙𝑜𝑔
2 12 ni hisoblang. A) 3 + 𝑙𝑜𝑔 2 3 B) 4 + 𝑙𝑜𝑔 2 3 C) 𝑙𝑜𝑔 4 48
2 3 299. 𝑓(𝑥) = 𝑘𝑥 + 3 funksiya uchun 𝑓(1) = 1 o’rinli bo’lsa, 𝑓(−1) ni toping. A) 3 B) 4 C) 5 D) 1 300. 𝐴(−7; 11) nuqtaga koordinatalar boshiga nisbatan simmetrik nuqtani toping. A) (−11; 7) B) (7; 11) C) (−7; −11) D) (7; −11) 301. (3𝑥 2 − 27)(𝑥 + 5) = (𝑥 − 3)(6𝑥 + 30) tenglamaning barcha ildizlari yig’indisini toping.A) 2 B) −6 C) −2 D) −3 302. Agar 𝑠𝑖𝑛𝛽 − 𝑐𝑜𝑠𝛽 = −1,35 bo’lsa, 𝛽 qaysi chorakda yotadi? A) 𝐼 chorak B) 𝐼𝐼 chorak C) 𝐼𝐼𝐼 chorak D) 𝐼𝑉 chorak 303. Agar 𝑠𝑖𝑛𝛽 + 𝑐𝑜𝑠𝛽 = −1,35 bo’lsa, 𝛽 qaysi chorakda yotadi? A) 𝐼 chorak B) 𝐼𝐼 chorak C) 𝐼𝐼𝐼 chorak D) 𝐼𝑉 chorak 304. Agar −𝑠𝑖𝑛𝛽 + 𝑐𝑜𝑠𝛽 = −1,35 bo’lsa, 𝛽 qaysi chorakda yotadi? A) 𝐼 chorak B) 𝐼𝐼 chorak C) 𝐼𝐼𝐼 chorak D) 𝐼𝑉 chorak − 2 𝑥+1
) ∙ 1 𝑙𝑛2 − 3𝑥 + 2 funksiyaga o’tkazilgan urinma 𝑦 = 5𝑥 to’g’ri chizig’iga parallel bo’lsa, urinma tenglamasini toping. A) 𝑦 − 5𝑥 + 14 = 0 B) 𝑦 − 5𝑥 − 14 = 0 C) 𝑦 − 5𝑥 + 12 = 0 D) 𝑦 − 5𝑥 − 12 = 0 314. Agar 7 𝑎 + 7 −𝑎 = 7 bo’lsa, 7 2𝑎 − 6 ∙ 7
𝑎 + 7
−𝑎
ning qiymatini toping. A) 7 B) −7 C) 6 D) −6 16
315. Agar 4 𝑎 + 4
−𝑎 = 4 bo’lsa, 4 2𝑎 − 3 ∙ 4
𝑎 + 4
−𝑎
ning qiymatini toping. A) 3 B) −3 C) 4 D) −4 316. Agar 𝑛 𝑎 + 𝑛
−𝑎 = 𝑛 bo’lsa, 𝑛 2𝑎
𝑎 + 𝑛
−𝑎 ning qiymatini toping. A) 𝑛 B) −𝑛 C) 𝑛 − 1 D) 1 − 𝑛 317. 𝐴 to’plamning elementlari soni 48 ning natural bo’luvchilari soniga teng bo’lsa, 𝐴 to’plamning elementlari sonini toping. A) 4 B) 8 C) 10 D) 5 318. 𝐴 to’plamning elementlari soni 60 ning natural bo’luvchilari soniga teng bo’lsa, 𝐴 to’plamning elementlari sonini toping. A) 6 B) 2 C) 12 D) 10 319. Ketma-ketlikning ixtiyoriy ketma-ket kelgan 2 ta hadi yig’indisi 10 ga teng. Agar 𝑎 3 = 7
bo’lsa, ketma-ketlikning dastlabki 9 ta hadi yig’indisini toping. A) 36 B) 35 C) 42 D) 47 320. Uchlari va yoqlari yig’indisi 38 ga teng bo’lgan piramidaning asosining dioganallari sonini toping. A) 135 B) 104 B) 170 D) 152 321. Uchlari va yoqlari yig’indisi 40 ga teng bo’lgan piramidaning asosining dioganallari sonini toping. A) 152 B) 170 C) 189 D) 209 322. Ixtiyoriy uchtasi bir nuqtada yotmaydigan 11 ta nuqta orqali nechta turli kesma hosil qilish mumkin? A) 55 B) 44 C) 66 D) 77 323. Agar 𝑓(𝑥) = { 𝑥 2
𝑥, 𝑥 > 1 bo’lsa, ∫ (3𝑥 − 𝑓(𝑥))𝑑𝑥 2 −1 ni hisoblang. A) 1
1 3 B) 1 2 3 C) 2 1 3 D) 2 2 3
324. Agar 𝑡𝑔𝛼 = 1 2 bo’lsa, 𝑠𝑖𝑛𝛼+𝑠𝑖𝑛3𝛼+𝑠𝑖𝑛5𝛼+𝑠𝑖𝑛7𝛼 𝑐𝑜𝑠𝛼−𝑐𝑜𝑠3𝛼+𝑐𝑜𝑠5𝛼−𝑐𝑜𝑠7𝛼
ifodaning qiymatini toping. A) 1 2 B) 2 C) − 1 2 D) −2 325. Agar 𝑡𝑔𝛼 = 5 6
𝑠𝑖𝑛𝛼+𝑠𝑖𝑛3𝛼+𝑠𝑖𝑛5𝛼+𝑠𝑖𝑛7𝛼 𝑐𝑜𝑠𝛼−𝑐𝑜𝑠3𝛼+𝑐𝑜𝑠5𝛼−𝑐𝑜𝑠7𝛼
ifodaning qiymatini toping. A) 1,2 B) −1,2 C) 5 6 D) − 5 6 326. Agar 𝑡𝑔𝛼 = 4 3
𝑠𝑖𝑛𝛼+𝑠𝑖𝑛3𝛼+𝑠𝑖𝑛5𝛼+𝑠𝑖𝑛7𝛼 𝑐𝑜𝑠𝛼−𝑐𝑜𝑠3𝛼+𝑐𝑜𝑠5𝛼−𝑐𝑜𝑠7𝛼
ifodaning qiymatini toping. A) 4 3 B) − 4 3 C) 3 4 D) − 3 4 327. To’g’ri silindrning asosi radiusi 4 sm. Yon sirtidan 𝐴 va 𝐵 nuqtalar olingan. 𝐴 va 𝐵 nuqtalardan asos tekisligigacha bo’lgan masofalar mos ravishda 4 sm va 1 sm. Agar 𝐴𝐵 kesma uzunligi 6 sm bo’lsa, silindr o’qidan 𝐴𝐵 kesmagacha bo’lgan masofani toping. A)
√33 2 B) √37 2 C) 2√2 D) √35 2
328. To’g’ri silindrning asosi radiusi 4 sm. Yon sirtidan 𝐴 va 𝐵 nuqtalar olingan. 𝐴 va 𝐵 nuqtalardan asos tekisligigacha bo’lgan masofalar mos ravishda 2 sm va 3 sm. Agar 𝐴𝐵 kesma uzunligi 5 sm bo’lsa, silindr o’qidan 𝐴𝐵 kesmagacha bo’lgan masofani toping A) √10 B) √5 C) 2√3 D) 3 329. 𝐴(6; 0) nuqtadan 𝑦 = √25 − (𝑥 − 5) 2 funksiya grafigigacha bo’lgan eng qisqa masofani toping. A) 5 B) 4 C) 6 D) 2√5 330. 𝐴(8; 0) nuqtadan 𝑦 = √25 − (𝑥 − 5) 2 funksiya grafigigacha bo’lgan eng qisqa masofani toping. A) 2 B) √5 C) 8 D) 3 17 Download 438.47 Kb. Do'stlaringiz bilan baham: 
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