Namangan Viloyat Xalq ta’limi boshqarmasi Viloyat metodika markazi
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⎪ 7 10 13 40 8 5
⇒ 1 1 7 10 1 1 13 40 1 1 8 5 y x z y z x + =
+ = + =
⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ quyidagicha belgilashlar kiritamiz: 1
= , 1 y b = ,
1 z c = . b a c b c a + =
+ = + =
⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ 7 10 13 40 8 5 bu sistemani yechsak, a=79/80. Bundan x=80/79. To’g’ri javob: A 78(2003-9.48). f(x)=x-1-ctg 2
toping. A)
x 2 2 -ctgx+C B) x 2 2 +ctgx+C C) x 2 2 -tgx+C D) x 2 2 +tgx+C E) x 2 +ctgx+C ## Funksiya formulasini quyidagicha qayta yozamiz: f(x)=x-1-ctg 2
1 2 sin x . Bu funksiyaning boshlang’ich funksiyasi: F(x)= x 2 /2+ctgx+C. To’g’ri javob: B. 79(2003-10.13). x x − +
− = 2 1 2 tenglamani yeching. A) ∅ B) 2 C) 1,2 D) 0,4 E) 0,9 ## Bu tenglamani yechishdan oldin uning aniqlanish sohasini topish maqsadga muvofiq: x x − ≥
− ≥ ⎧ ⎨ ⎩ 2 0 1 0
⇒ x x ≥ ≤ ⎧ ⎨ ⎩ 2 1
⇒ sistema yechimga ega emas. Demak, tenglamaning aniqlanish sohasi, shuningdek, yechimlar to’plami ham bo’sh to’plam - ∅. To’g’ri javob: A.
80(2003-11.9). S n arifmetik progressiyaning dastlabki n ta hadining yig’indisi bo’lsa, S 5 -3S 4 +3S
3 -S 2 ning qiymatini toping. A) 0 B) -2a 1 C) 2a 1 D) 3a 1 E) -3a 1
## S 5 -3S
4 +3S
3 -S 2 =S 5 -S 4 -2(S
4 -S 3 )+S 3 -S 2 =a 5 -2a 4 +a 3 = =a 1 +4d-2(a 1 +3d)+a 1 +2d=0. To’g’ri javob: A. 81(2003-11.15). y=sin(sinx) funksiyaning eng katta qiymatini aniqlang. A) sin1 B) 1 C) 1/2 D) aricsin1 E) π/2
## t=sinx deb olsak, -1 ≤ t ≤ 1 bo’ladi. Bu oraliqda y=sin(sinx)=sint funksiya o’suvchi. Shuning uchun y max
= sin1. To’g’ri javob: A. 82(2003-12.58). 5 5
log log
a a a − ifodani soddalashtiring. A) a B) a 2 C) 5a D) 1 E) 0 ## 5 5 5 log
log a a a − = a a a a a log
log log
5 1 5 5 ⋅ − = a a log 5
- a a log 5
=0 To’g’ri javob: E. 83(2003-12.77).
2 2 2 2 2 7 24 24 1 7 24 24 π π π π − ⎛ ⎝⎜ ⎞ ⎠⎟ − ⋅ ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ : ni hisoblang. A) 1/9 B) 9 C) 1/3 D) 1 E) 3 ##
tg tg tg tg 2 2 2 2 2 7 24 24 1 7 24 24 π π π π − ⎛ ⎝⎜ ⎞ ⎠⎟ − ⋅ ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ : = = = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 2 2 2 2 2 2 2 2 24 24 7 1 24 24 7 1 24 24 7 24 24 7 24 24 7 1 24 24 7 π π π π π π π π π π π π
tg tg tg tg tg tg tg tg tg tg tg
= tg tg tg tg tg tg tg tg 7 24 24 1 7 24 24 7 24 24 1 7 24 24 2 π π π π π π π π + − ⋅ − + ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ = tg tg π π 3 4 2 ⎛ ⎝⎜ ⎞ ⎠⎟ =3.
To’g’ri javob: E.
84(2003-7.61). Uchburchakning ikki tomoni 7 va 11 ga teng, uchinchi tomoniga o’tkazilgan medianasi 6 ga teng. Uchburchakning uchinchi tomonini toping. A) 12 B) 8 C) 14 D) 10 E) 13 ## Berilgan: ABC uchburchakda AB=7, B D AC=11, AO=6 (AO-mediana). BC=? O Yechilishi: ABDC parallelogrammga to’ldiramiz. AD
2 +BC
2 =2(AB
2 +AC
2 ); A C (2 ⋅6)
2 +BC
2 =2(7
2 +11
2 )
Bundan BC=14. To’g’ri javob: C. 85(2003-9.36). Agar 17 ,
= = b a r r va 35 3 = − b a r r bo’lsa, b a r r + ning qiymatini toping. A) 19 B) 20 C) 8 3 D) 9
2 E) 4
6
## (a+b) 2 +(a-b) 2 =2(a 2 +b 2 ), (a+b) 2 =2(a 2 +b 2 )-(a-b) 2 = =2(49+289)-315=676-315=361. a+b=19 To’g’ri javob: A.
86(2003-9.52). To’g’ri burchakli uchburchakka ichki chizilgan aylananing urinish nuqtasi gipotenuzani 2:3 nisbatda bo’ladi. To’g’ri burchak uchidan aylana markazigacha bo’lgan masofa 2 2
ga teng.Berilgan uchburchakning yuzini toping.. A) 12 B) 16 C) 18 D) 20 E) 24 ## AF=AD=2x, BE=BD=3x. B (3x+2) 2
2 =(5x)
2 ; 3x
2 -5x-2=0
x 1 =2; x 2= -1/3. D AC=2 ⋅2+2=6; BC=3⋅2+2=8. E S ABC=
24. C F A 87(2003-10.48). To’g’ri burchakli uchburchakning katetlaridan biri ikkinchisidan ikki marta katta.Shu uchburchakning gipotenuzasiga tushirilgan balandligi 12 ga teng. Uchburchakning yuzini toping. A) 180 B) 84 C) 120 D) 90 E) 108 ## AC=x; BC=2x deb olsak, C AB = x x x 2 2 2 5 + = ( )
A D B S= 1 2 AC ⋅BC=
1 2 AB ⋅CD. AC⋅BC=AB⋅CD x ⋅2x=x 5
x=6 5 . S ∆ = 1 2 AC ⋅BC= 1 2 6 5 ⋅12 5 =180. To’g’ri javob: A 88(2003-10.55). To’g’ri burchakli uchburchakning uzunligi 14 va 18 ga teng katetlariga tushirilgan medianalari uni uchta uchburchakka va to’rtburchakka ajratadi. To’rtburchakning yuzini toping. A) 56 B) 64 C) 48 D) 72 E) 42 ## Berilgan uchburchakning yuzi S bo’lsin. S=(18 ⋅14):2=126. S 1
Shakldan S S S S S S 1 2 1 2 2 2 2 2 + = + = ⎧ ⎨ ⎩ / / S 2
S 1 =S 2 ; 3S 1 =S/2 S 3
S 1 =S/6.
S shakl
= 42 3 6 2 6 2 2 1 = = = − = − S S S S S S
To’g’ri javob: E. 89(2003-10.61). To’rtburchakli piramidaning barcha yon qirralari asos tekisligi bilan 60 ° li burchak hosil qiladi. Uning asosi teng yonli trapetsiyadan iborat. Trapetsiyaning diagonallari uning o’tkir burchaklarining bissektrisalaridir. Piramidaning balandligi 4 3 ga teng. Trapetsiyaning katta asosini toping. A) 4
3 B) 8 C) 8 3 D) 12 E) 3 6
## Masala shartidan ∠ACD=90° ekani kelib chiqadi. Piramidaning yon qirralari asos tekisligiga bir xil og’ishganligi uchun, piramida S balandligining asosi ∆ACD ning gipotenuzasi AD ning o’rtasida yotadi.
∆ASE dan AE=SE⋅ctg60°= B C = 4 3
1 3 4 ⋅ = ; AD=2AE=8. To’g’ri javob: B A E D
90(2003-10.63). Ikki vektor yig’indisining uzunligi 20 ga, shu vektorlar ayirmasining uzunligi 12 ga teng. Shu vektorlarning skalyar ko’paytmasini toping.
A) 16 B) 48 C) 24 D) 64 E) 32 ##
20 = + b a r r ; 12 = − b a v r larning har birini kvadratga ko’taramiz:
400 2 2 2 = + + b b a a r r r r (1) 144
2 2 2 = + − b b a a r r r r (2) (1) dan (2) ni hadlab ayiramiz: 256
4 =
a r r ; 64 = b a r r To’g’ri javob: D. 91(2003-12.31). Parallelogrammning diagonali 8 2 ga teng. Shu parallelogrammga ichki va tashqi aylanalar chizish mumkin bo’lsa, parallelogrammning yuzini toping. A) berilganlar yetarli emas B) 32 C) 64 D) 128 E) 256 ## Tashqi chizilgan to’rtburchakning qarama-qarshi tomonlari yig’indisi teng. Agar u parallelogramm bo’lsa, bu parallelogramm romb bo’ladi. Ichki chizilgan to’rtburchak qarama-qarshi burchaklari yig’indisi 180 ° ga teng bo’ladi. Agar bu to’rtburchak romb bo’lsa, u holda bu romb kvadrat bo’ladi. Diagonali 8 2 ga teng bo’lgan kvadratning tomoni 8 ga, yuzi esa 64 ga teng bo’ladi. To’g’ri javob: C. 92(2003-12.41). Parallelepipedning bir uchidan chiquvchi uchta qirrasining o’rtalari orqali o’tkazilgan tekislik undan hajmi 6 ga teng piramida kesib ajratadi. Parallelepipedning hajmini toping. A) 120 B) 144 C) 180 D) 288 E) 276 ## Hosil bo’lgan piramidani uchburchakli prizmaga to’ldiramiz. Bu prizma hajmi piramida hajmi- ning 3 baravariga teng. Asosi piramida asosiga teng va Balandligi parallelepipedning Balandligiga teng uchburchakli prizma hajmi piramida hajmining 6 baravariga tengligi shakldan ko’rinib turibdi. Parallelepiped diagonali orqali o’tgan tekislik uni ikkita uchburchakli prizmaga ajratadi. Ularning har birining hajmi avvalgi prizma hajmining 2 2 =4 baravariga teng. Shunday qilib, berilgan parallelepipedning hajmi 2 ⋅4⋅6⋅6=288 ga teng. To’g’ri javob: D. 93(2003-5.43). cos cos
cos x x x = − 2 1 tenglama [ π; 2π] kesmada nechta ildizga ega? A) 1 B) 2 C) 3 D) 4 E) ∅ ## cosx ning qiymati [ π; 1,5π) oraliqda manfiy bo’lgani uchun, bu oraliqda berilgan tenglama -1=cos2x-1 ko’rinishda bo’ladi. Bundan cos2x=0; x= π/4+πn/2 (Bulardan 5π/4 berilgan oraliqqa tegishli)
cosx ninq qiymati (1,5π; 2π] oraliqda musbat bo’lgani uchun, bu oraliqda berilgan tenglama 1=cos2x-1 ko’rinishda bo’ladi. Bundan cos2x=2. bu tenglama yechimga ega emas. To’g’ri javob: A. 94(2003-9.44). A(1;4) nuqtadan y=-2-2/x funksiya grafigiga ikkita urinma o’tkazilgan. Urinish nuqtalari abssissalarining yg’indisini toping. A) –1 B) 1 C) 1/3 D) 2/3 E) –2/3 ## A(1;4) nuqtadan o’tadigan to’g’ri chiziq tenglamasi y=kx+b ko`rinishda bo’lsin. Bunga A nuqtaning koordinatalarini qo’yib, topamiz: 4=k ⋅1+b;
b=4-k. bu to’g’ri chiziq urinma bo’lgani uchun k=y ′ bo’lishi kerak. y ′=2/x
2 .
b va y ning ifodalarini to’g’ri chiziq tenglamasiga qo’yamiz: y x x x x x = ⋅ + − = − + +
2 4 2 2 2 4 2 2 2 . Bu chiziq bilan berilgan funksiya grafigining umumiy nuqtalarini topamiz: y x y x x = − −
= − + +
⎧ ⎨ ⎪⎪ ⎩ ⎪ ⎪ 2 2 2 2 4 2 bundan − + + = − − 2 2 4 2 2 2 x x x . Bu tenglamani yechib, x 1 =-1 va x 2 =1/3 ga ega bo’lamiz. x 1 +x
=-2/3.
To’g’ri javob: E. 95(2003-11.20) y=3- ⎪x-3⎪ funksiya grafigi va OX o’qi bilan chegaralangan shaklning yuzini toping. A) 9 B) 8 C) 12 D) 6 E) 10 ## Modulning ta’rifidan foydalanib, bu funksiyani quyidagicha yozish mumkin: y= ⎩ ⎨
≥ + − ≤ 3 ; 6 3 ; x x x x
shunga ko’ra, grafikni x ≤3 va x≥3 oraliqlarda chizamiz.
3 S= 1 2 6 3 9 ⋅ ⋅ = 3 6 To’g’ri javob: A.
96(2003-1.40). To’g’ri burchakli ACB uchburchakning katetlari 8 ga va 10 ga teng. Shu uchburchakning C –to’g’ri burchagi uchidan CE-mediana va CD-bissektrisa o’tkazilgan. CDE uchburchakning yuzini toping.
A) 2
2 9 B) 2 2 7 C) 2 3 8 D) 3 2 5 E) 3 2 3 C ## S ACE
=S ∆ /2=(8 ⋅10/2)/2=20 AB= 8 10 2
+ =2 41 A D E B h c
Bissektrisa xossasidan AD:DB=AC:BC AD:BD=8:10=4:5 AD=4x; BD=5x; 4x+5x=2 41 ; x=2 41 /9; AD=8 41 /9; S ACD =(8 41 /9)(40/ 41 )/2=160/9. S DCE =S ADE
-S ACD
=20-160/9=20/9= 2 2 9 . To’g’ri javob: A. 97(2003-7.83). Kubning ostki asosidagi tomonlarining o’rtalarini ketma- ket tutashtirildi. Hosil bo’lgan to’rtburchakning uchlari kub ustki asosining markazi bilan tutashtirildi. Agar kubning qirrasi a ga teng bo’lsa, hosil bo’lgan piramida- ning to’la sirtini toping. A) 2a 2 /3 B) 3a 2 C) 1,5a 2 D) 2a 2 E)
2 3 3 2 a
## Hosil bo’lgan piramidaning asosi kvadrat bo’lib, uning diagonali kub qirrasiga teng. Shuning uchun piramida asosining yuzi S asos =a 2 /2.
Asos tomoni b=a/ 2 a = a a a 2 2 8 3 2 2 + = / ; S yon =4 ⋅ 1 2 2 3 2 2 3 2 2 ⋅ ⋅ = a a a S
t =
a a 2 2 2 2 3 2 2 + = . To’g’ri javob: D. 98(2003-9.7). Ikkita ishchi birgalikda ishlab, ma’lum ishni 12 kunda tamomlaydi. Agar ishchilarning bittasi shu ishning yarmini bajargandan keyin, ikkinchi ishchi qolgan yarmini bajarsa, shu ishni 25 kunda tamomlashlari mumkin. Ishchilardan biri boshqasiga qaraganda necha marta tez ishlaydi? A) 1,2 B) 1,5 C) 1,6 D) 1,8 E) 2 ## 1-ishchi yolg’iz o’zi butun ishni x kunda, 2-si esa y kunda tamomlaydi. 1 1 1 12 0 5 0 5 25
y x y + =
+ = ⎧ ⎨ ⎪ ⎩⎪ , , bu sistemani yechsak, x=20; y=30 30/20=1,5 marta tezroq ishlaydi. To’g’ri javob: B. 99(2003-11.4). x x x x x x x − + − + − + + = 1 2 3 1 4 ... tenglamaning ildizi 10 dan nechta kam? A) 1 B) 2 C) 3 D) 4 E) 5 Kasrlarning surati birinchi hadi x-1 ga, ayirmasi -1 ga teng bo’lgan arifmetik progressiya tashkil etadi. S= (
( ) ( ) x x x x − +
⋅ − =
− 1 1 2 1 1 2 bo’lgani uchun, berilgan tenglamani x x x ( ) − = 1 2 4 ko’rinishda yozish mumkin. Bu tenglamani yechib, x=9 ga ega bo’lamiz. U 10 dan 1 ta kam. To’g’ri javob: A. 100(2003-9.17). 2 7
7 1 7 7 1 7 7 1 2 ⋅ − ≥ − − + x x x x x x tengsizlikni yeching. A) (0; ∞) B) (-∞;0) C) (-∞;0] D) (-1; 1) E) (1; ∞) ## 2 7 7
7 7 1 7 7 1 0 2 ⋅ − − − + + ≥ x x x x x x
2 7
7 7 1 7 7 1 7 1 0 2 ⋅ − + +
− − ≥ x x x x x x ( ) ( )
1 7 1 0 2 x − ≥ 7 1 2 x − >0
1 7 2 >
x>0 To’g’ri javob: A.
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