Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №4 Matematika attestatsiya Matematika attestatsiya №4 InfoMaster Yanvar 21, 2022 44 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 1 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №4 1 / 40 tengsizlikni yeching. A) 1 B) 3 C) 4 D) 2 2 / 40 tengsizlikni yeching. A) x∈(-3;1) B) x∈(-∞;5)∪(-3;1) C) x∈(-5;1) D) x∈(-∞;5) 3 / 40 Kamayish tartibida joylashtiring. a)5√6 b)12 c)4√10 A) c>a>b B) b>a>c C) a>c>b D) a>b>c 4 / 40 tenglamalar sistemasi yechimga ega bo‘ladigan a ning barcha qiymatlari to‘plamini toping. A) 2 B) 1 C) 4 D) 3 5 / 40 Agar ning qiymatini toping. A) 75/23 B) 23/75 C) 23/25 D) 3 6 / 40 tenglamani yeching. A) 3 B) log(3/5)2 C) 5 D) log(5/3)2 7 / 40 y=f(x) funksiya uchun tenglik o'rinli bo'lsa, f(π/4)=? A) 4 B) 1/3 C) 1/5 D) 1/4 8 / 40 20132015 ni 10 ga bo'lgandagi qoldiqni toping. A) 3 B) 7 C) 9 D) 1 9 / 40 f(x)=26sin6x·sin7x funksiya uchun boshlang'ich funksiyasini toping. A) -13cosx- cos13 x +C B) 13sinx-sin13x + C C) 13cosx-cos13x + C D) 13sinx+sin13x + C 10 / 40 P(x), Q(x) va R(x) ko'phatdalar berilgan. Bunda P(x) ko'phadning odoz hadi Q(x) ko'phadning ozod hadidan ikki marta katta va P(0)≠0. P(x)=Q(x)·R(x+1) bo'lsa, R(x) ko'phadning koeffitsiyentlarining yig'indisini toping. A) 4 B) 3 C) 1 D) 2 11 / 40 Tengsizlikni qanoatlantiradigan eng kichik ikkita butunsonning yig`indisini toping? A) 6 B) 5 C) 4 D) 3 12 / 40 AM||KL , ∠BCD=70°,∠CDE=40°, ∠DEF =45°,∠EFG=80°, ∠FGK=70°∠MBC=x ga teng bo`lsa x ni toping ? A) 10° B) 40° C) 30° D) 20° 13 / 40 Hisoblang: A) 3 B) 1 C) 2 D) 4 14 / 40 Hisoblang 200−199+198−197+⋯+4−3 A) 99 B) 101 C) 98 D) 100 15 / 40 Quyidakilardan qaysi biri juft: 200956+200855 31210+0! 55!+877 222+333+4444 A) 4 B) 3 C) 2 D) 1 16 / 40 a,b,c -1 dan katta musbat sonlar uchun a³=b² va a4=c5 bo'lsa, logbc ni toping. A) 5/6 B) 1/3 C) 8/15 D) 1/2 17 / 40 ABC o‘tkir burchakli uchburchakda sinA=4/5 va sinB=15/13 bo'lsa, sinC ni qiymatini toping. A) 40/63 B) 48/65 C) 56/65 D) 36/65 18 / 40 Tenglamaning natural ildizining butun bo‘luvchilari nechta? A) 2 B) 5 C) 4 D) 6 19 / 40 To‘g‘ri burchakli uchburchakning bir burchagi 52° ga teng bo‘lsa, to‘g‘ri burchak uchidan tushirilgan balandlik va mediana orasidagi burchakni toping. A) 7° B) 24° C) 14° D) 17° 20 / 40 108 sonining natural bo‘luvchilari ko‘paytmasi quyidagilardan qaysi biriga teng? 212·318 22·33 28·310 26·39 A) 3 B) 4 C) 1 D) 2 21 / 40 ABCD trapetsiyaning AC diagonali CD yon tomonga perpendikulyar. Agar ∠D=69° va AB = BC bo‘lsa, B burchakni toping. A) 142° B) 138° C) 135° D) 132° 22 / 40 Radiusi 25/4 bo‘lgan sferaga balandligi 8 ga teng bo‘lgan konus ichki chizilgan. Konusning hajmini toping. A) 144π B) 72π C) 96π D) 192π 23 / 40 Agar bo‘lsa, m/n ni toping. A) 2 B) (1+√5)/2 C) 3 D) 4 24 / 40 Tenglamani yeching: A) 6/5 B) 9/5 C) 5*4 D) 5/9 25 / 40 Ayniyatdan foydalanib x + y + z ni toping: A) 9 B) 12 C) 3 D) 6 26 / 40 Quyidagi sonini 9 ga bo‘lganda qoladigan qoldiqni toping. A) 0 B) 5 C) 8 D) 1 27 / 40 Uchburchakning 3 va 4 ga teng bo‘lgan tomonlariga o‘tkazilgan medianalar o‘zaro perpendikulyar bo‘lsa, bu uchburchakning uchinchi tomonini toping. A) √6 B) √5 C) 2,4 D) 2,5 28 / 40 Dastlabki n ta hadining yig‘indisi formula bilan aniqlanadigan arifmetik progressiyaning 6-hadini toping. A) 9 B) 10 C) 8 D) 6 29 / 40 Tenglamalar sistemasidan foydalanib x + y + z ni toping: A) 6 B) 10 C) 8 D) berilganlar yetarli emas 30 / 40 1, 1, 2, 3, 5, 8, 13, … ketma-ketlikning umumiy hadi an bo‘lsa, quyidagilardan qaysi biri to‘g‘ri? A) 4 B) 1 C) 2 D) 3 31 / 40 Raqamlari yig‘indisi 10 bo‘lgan nechta turli 3 xonali son bor? A) 0,216 B) 0,432 C) 0,144 D) 0,288 32 / 40 Bo‘yi 130 cm, eni 90 cm, balandligi 60 cm bo‘lgan idishdagi suvning 234 litri olindi. Idishda qolgan suvning (sathi) balandligini toping. A) 25 B) 40 C) 20 D) 35 33 / 40 Aylana tashqarisidagi nuqtadan aylanaga kesuvchi o‘tkazilgan. Berilgan nuqtadan aylanani kesgan nuqtalarigacha bo‘lgan masofalar mos ravishda 9 va 45 ga teng bo‘lsa, shu nuqtadan aylanaga o‘tkazilgan urinmaning urinish nuqtasigacha bo‘lgan masofa uzunligini toping. A) 9√5 B) 6√5 C) 8√3 D) 12√3 34 / 40 Burchagi 120° ga teng bo‘lgan doiraviy sektorga ichki doira chizilgan. Berilgan doiraning radiusi R ga teng bo‘lsa, yangi doiraning radiusi topilsin. A) 1,5R B) √3R(2-√3) C) R(√3-√2) D) 3R 35 / 40 x>0 bo‘lsa, 128x+1/x² yig‘indining eng kichik qiymatini toping. A) 16 B) 32 C) 48 D) 64 36 / 40 Agar x =17 bo‘lsa, quyidagi ifodaning qiymatini toping. A) -√17 B) √17 C) 4 D) -4 37 / 40 7/3 kasrning o‘nli kasr yoyilmasida verguldan keyingi 100- raqamni toping. A) 1 B) 5 C) 4 D) 2 38 / 40 Funksiyaning aniqlanish sohasini toping. A) (-∞;0)∪(0;∞) B) (-∞;3]∪[2;∞) C) x∈R D) [-3;2] 39 / 40 tenglama intervalda nechta ildizga ega? A) 2 B) 4 C) 1 D) 3 40 / 40 Integralni hisoblang: A) 4 B) 3 C) 2 D) 1 O'rtacha ball 0% 0% Testni qayta ishga tushiring Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
Istaklar ro'yxatiga qo'shildiIstaklar ro'yxatidan olib tashlandi 13 Matematika fanidan attestatsiya savollari №16
