Namangan Viloyat Xalq ta’limi boshqarmasi Viloyat metodika markazi
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## sinx=t deb belgilasak, cost<0 bo’ladi, bundan π π
3 2
t yoki
π π 2 3 2 < < sin x . Bu tengsizlikning esa yechimi yo’q. To’g’ri javob: E. 53(2002-2.35).
≤ +
∫ 4 0 tengsizlikni qanoatlantiruvchi a ning qiymatlari oralig’i uzunligini toping. A) 6 B) 5 C) 4 D) 8 E) 7 ##
xdx x a a a = = ∫ 2 0 2 0 2 2 dan
a a 2 2 4 ≤ + ; a 2 -2a-8 ≤0; (a+2)(a-4)≤0
[-2;4] oraliqning uzunligi 6 birlik. -2 4 To’g’ri javob: A. 54(2002-2.51). 2sin6x(cos 4 3x-sin 4 3x)=sinkx tenglik hamma vaqt o’rinli bo’lsa, k ni toping. A) 12 B) 24 C) 6 D) 18 E) 4 ## 2sin6x(cos 4 3x-sin 4 3x)q 2sin6x(cos 2 3x-sin 2 3x)(cos 2 3x + sin 2 3x)= =2sin6x(cos 2 3x-sin 2 3x)=2sin6xcos6x=sin12x. sin12x=sinkx dan x=12. to’g’ri javob: A. 55(2002-3.9) Agar a(x-1) 2 +b(x-1)+c=2x 2 -3x +5 bo’lsa, a +
+
nechaga teng bo’ladi? A) 7 B) 8 C) 6 D) 4 E) 5 ## a(x-1) 2 +b(x-1)+c=ax 2 -2ax+a+bx-b=ax 2 -(2a-b)x+a-b+c. Bundan a a b a b c = − = − + = ⎧ ⎨ ⎪ ⎩ ⎪ 2 2 3 5 Demak , b=1; c=4. a+b+c=7. To’g’ri javob: A. 56(2002-3.78). cos 2
2 -2 tenglama a ning qanday qiymatlarida yechimga ega? A) [
− 2 2 ; ] B) [0; 2 ] C) [0;2] D) (-2;2) E) [1;0) ## max(cos2x+6sinx)=6 va min(cos2x+6sinx)=-6 bo’lgani uchun. 4 2 6 4 2 6 2 2 a a − ≥ −
− ≤ ⎧ ⎨ ⎩
⇒ 4 4 4 8 2 2 a a ≥ −
≤ ⎧ ⎨ ⎩
⇒ a a 2 2 1 2 ≥ − ≤ ⎧ ⎨ ⎩
⇒ a 2 ≤2; a ≤ 2 ; − ≤ ≤
2 2
. To’g’ri javob: A. 57(2002-3.80). (8x-1)(x+2)ctg πx=0 tenglama [-2;2] kesmada nechta ildizga ega? A) 5 B) 4 C) 6 D) 7 E) 3 ##
a) 8x-1=0; x 1 =1/8. bu ildiz [-2;2] kesmaga tegishli. b) x+2=0; x=-2. bu chet ildiz, chunki ctg2 π aniqlanmagan. c)
ctg πx=0; πx=π/2+πn; x=1/2+n; bu ko’rinishdagi sonlardan -3/2; -1/2; 1/2; 3/2 lar [-2;2] kesmaga tegishli. Shunday qilib, tenglamaning [-2;2] kesmadagi ildizlari 5 ta. To’g’ri javob: A. 58(2002-4.17). Geometrik progressiyaning mahraji 1/2 ga teng. b 1 (b 2 ) -1 b 3 (b 4 ) -1 ⋅
⋅
A) 64 B) 32 C) 16 D) 128 E) 256 ## b
⋅
⋅
1 2 3 4 13 14 7 1 ⋅ ⋅ ⋅ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =
... 2 7 =128. To’g’ri javob: D. 59(2002-4.19). Arifmetik progressiya hadlari uchun
=a 2 +a 4 +...+a 20 +15 tenglik o’rinli bo’lsa, a 11 ni toping. A) 11 B) 13 C) 15 D) 17 E) 19 ## a 1 +(a 3 -a 2 )+(a 5 -a 4 )+...+(a 21 -a 20 )=15; a 1 +10d=15; a 11 =15.
To’g’ri javob: C. 60(2002-4.41). f x x x ( )
/ = − + − 5 1 25 funksiyaning aniqlanish sohasiga tegishli barcha butun sonlarning o’rta arifmetigini toping. A) -2 B) -1 C) 0 D) 1 E) 2 ##
5 1 25 0 0
x − ≥ − ≥ ⎧ ⎨ ⎪ ⎩⎪
⇒ 5 5 0 2
x ≥ ≤ ⎧ ⎨ ⎩ − ⇒
x ≥ −
≤ ⎧ ⎨ ⎩ 2 0 ⇒ -2≤x≤0.
− + − +
= − 2 1 0 3 1 ( ) . To’g’ri javob: B. 61(2002-4.44). 1 2 + ≤ +
x arccos(
) tengsizlikning eng katta butun yechimini toping. A) -2 B) -1 C) 0 D) 1 E) 2 ##
1 0 1 2 1 + ≥ − ≤ + ≤ ⎧ ⎨ ⎩ x x
⇒ x x ≥ −
− ≤ ≤ − ⎧ ⎨ ⎩ 1 3 1
⇒ x= -1. To’g’ri javob: B. 62(2002-5.14).
+ −
+ + + + + = 3 14 3 14 4 bo’lsa, x/(x+1) ning qiymatini toping. A) 2/3 B) -2/3 C) 3 D) 3/2 E) -3/2 ##
x t + = 14 deb belgilasak, x=t 2 -14 bo’ladi. Bundan t t t t 2 2 11 11 4 − − +
− + = . Bu tenglamani yechsak: t 1,2 = ±4. a) x + = − 14 4 tenglama yechimga ega emas. b) x + = 14 4 ; x+14=16; x=2. To’g’ri javob: A. 63(2002-6.14). Nechta natural (x;y) sonlar jufti x 2 -y 2 =53 tenglikni qanoatlantiradi? A)
∅ B) 1 C) 2 D) 3 E) 4 ## x 2 -y 2 =1 ⋅53; (x-y)(x+y)=1⋅53 x y x y − =
+ = ⎧ ⎨ ⎩ 1 53 ⇒ x=27; y=26. (27;26) Bitta juftlik. To’g’ri javob: B. 64(2002-6.20). log ( , )
...) 128
1 3 1 9 1 27 0 25 6 + + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ni hisoblang. A) 2/7 B) 3/8 C) 1/14 D) 2/5 E) 1/12 ## 1
1 9 1 27 + +
+... - birinchi hadi 1 3 va mahraji 1 3 ga teng bo’lgan cheksiz kamayuvchi geometrik progressiyaning yig’indisi. 1 3
9 1 27 + + +...=
1 2 . Buni berilgan ifodaga qo’yamiz: log
( , ) log ( ...)
128 1 3 1 9 1 27 0 25
16 + +
+ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = log ( / ) log (
) log
( / ) log
/ 128
1 2 2 2 1 2 1 4 1 7 4 16 4 = =
= 1 7 2 1 7 1 2 2 1 2 log
/ = ⋅ =
1 14 . To’g’ri javob: C. 65(2002-2.46). Doiradan markaziy burchagi 90 ° bo’lgan sektor qirqib olingach, uning qolgan qismi o’ralib konus shakliga keltirildi. Bu konus diametrining yasovchisiga nisbatini toping. A) 3/2 B) 2 C) 5/4 D) 2/3 E) 4/5 sektor yoyining uzunligiga, konusning yasovchisi esa sektorning radiusiga teng. C konus
=1,5 πR sektor ;
d konus
=1,5 R sektor
; L konus
= R sektor
D konus
:l konus
=1,5R:R=1,5=3/2. 66(2002-3.57). To’g’ri burchakli uchburchakning α va β o’tkir burchaklari uchun cos α+sin(α-β)=1 tenglik o’rinli bo’lsa, β ning qiymatini toping. A) 30
° B) 45° C) 60° D) 75° E) aniqlab bo’lmaydi. ##
α+β=90° dan α=90°-β bo’ladi. cos
α+sin(α-β)=cos(90°-β)+sin(90°-2β)=sinβ+cos2β=sinβ+1-2sin 2 β. sin β+1-2sin
2 β=1; 2sin 2 β-sinβ=0; sinβ(2sinβ-1)=0; sin β=0 bo’la olmaydi, shuning uchun 2sinβ-1=0; β=30°. To’g’ri javob: A. 67(2002-4.49). Uchburchakning burchaklari 1:2:3 kabi nisbatda. Uchburchak katta tomonining kichik tomoniga nisbatini toping. A) 1 B) 2 C) 3 D) 4 E) 5 ## ∠A:∠B:∠C=1:2:3 dan ∠A=30°; ∠B=60°; ∠C=90°. c=a/sinA=a/sin30 °=a/0,5=2a. c:a=2a:a=2. To’g’ri javob: B 68(2002-4.50). Uchburchakning tomonlar 10, 13 va 17 ga teng. Bu uchburchakka tashqi chizilgan aylananing markazi qayerda bo’lishini aniqlang. A) uchburchak ichida B) uchburchakning kichik tomonida C) uchburchak tashqarisida D) aniqlab bo’lmaydi E) uchburchakning katta tomonida ## 10 2
2 <17 2 bo’lgani uchun bu uchburchak o’tmas burchakli. O’tmas burchakli uchburchakka tashqi chizilgan aylananing markazi uchburchak tashqarisida joylashadi. To’g’ri javob: C. 69(2002-4.52). Piramidaning asosi gipotenuzasining uzunligi 2 bo’lgan to’g’ri burchakli uchburchakdan iborat. Piramidaning qirralari asos tekisligi bilan α burchak tashkil qiladi. Agar uning balandligi 5 ga teng bo’lsa,
tg α ning qiymatini toping. A) 1 B) 2 C) 3 D) 4 E) 5 ## Piramidaning barcha yon qirralari asos tekisligiga bir xil og’ishganda piramida balandligining asosi, piramida asosiga tashqi chizilgan aylana
markazi bilan ustma-ust tushadi. Piramida asosi to’g’ri burchakli uchburchak bo’lgani uchun balandlikning asosi gipotenuzaning o’rtasida yo’tadi. Shunday qilib, tg α=5/1=5. To’g’ri javob: E. 70(2003-1.8). 2 3 4 2 − + > −
x x tengsizlikning eng kichik butun yechimini toping. A) 0 B) -1 C) -2 D) -3 E) -5 ## Arifmetik kvadrat ildizning qiymati doimo nomanfiy son bo’lgani uchun, bu tengsizlik 2 3
0 − + ≥ x x tengsizlikka teng kuchli. Bundan (2-3x)(x+4) ≥0
-4 2/3 (-4; 2/3] Eng kichik butun yechimi: -3. To’g’ri javob: D. 71(2003-1.52).
+ ∫ 1 1 2 ni hisoblang. A) 2+ln(1/2) B) 1+ln(2/3) C) 3-ln(2/3) D) 1-ln(2/3) E) 2-ln(2/3) ##
+ ∫ 1 1 2 = ( ) x x dx + −
+ ∫ 1 1 1 1 2 = ( ) 1 1 1 1 2 − + ∫ x dx = (x-ln(x+1))⏐ 1 2
=(2-ln(2 +1))-(1-ln(1+1))=2-ln3-1+ln2=1+ln(2/3). To’g’ri javob: B. 72(2003-6.5). Agar 29 31 38 41 47 51 + + = a bo’lsa, 2 31
41 4 51 + + quyidagi- larning qaysi biriga teng? A) 3-
a B) 4-a C) 5-a D) 3-a/2 E) 4-a/2 ##
2 31 3 41 4 51 + + = 31 29 31 41 38 41 51 47
51 − + − + − = 1 29 31 1 38 41 1 47 51 − + − + − = 3-( 29 31 38 41 47 51 + + )=3- a. To’g’ri javob: A. 73(2003-6.15). x x x + + + = ... 3 3 3 4 tenglamani yeching. A) 56 B) 48 C) 60 D) 54 E) 64 ## Har ikki qismini kubga ko’taramiz:
x x x x + + + + = ... 3 3 3 64 .
x x x + + + = ... 3 3 3 4 ni e’tiborga olsak, x+4=64 bo’ladi. Bundan
x=60. To’g’ri javob: C 74(2003-6.26). Agar sin37 °=a bo’lsa, sin16° ni a orqali ifodalang. A)
B)
a-1 C) 2a 2 -1 D) 1-2a 2 E) aniqlab bo’lmaydi. ## sin37 °=sin(45°-8°)=sin45°cos8°-cos45°sin8°= 2 2 cos8 °- 2 2 sin8
°= = 2 2 (cos8
°-sin8°). Demak: 2 2 (cos8 °-sin8°)=a; cos8°-sin8°= 2
(cos8 °-sin8°) 2 =( 2 a) 2 ; cos 2 8 °-2cos8°sin8°+sin 2 8 °=2a 2 .
Bundan sin16 °=1-2a 2 . To’g’ri javob: D. 75(2003-7.30). Agar f(x)= 7 2 x ax b x + + funksiya grafigi (2;0) nuqtada absissalar o’qiga urinib o’tsa, a+b nimaga teng? A) 0 B) 20 C) -21 D) 28 E) -56 ## Shartga ko’ra f(2)=0 bo’lishi kerak, shuning uchun: 7 2 2
0 2 ⋅ + ⋅ + =
b . Bundan b= -2a-28. Ikkinchi tomondan urinma OX o`qiga parallel (aniqrog’i OX o`qining o’zi) bo’lgani uchun f ′
bo’lishi kerak. Shuning uchun ( )
7 0 2 2 x a x x ax b x + − − − = 14 7 0 2 2
ax x ax b + − − − =
a =-28
b =-28-2
⋅(-28)=28 a +b=-28+28=0. To’g’ri javob: A.
76(2003-8.34). Agar x y xy x y xy + −
= + + = ⎧ ⎨ ⎪ ⎩⎪ 7 133 2 2 bo’lsa, xy ning qiymatini toping.
A) 36 B) 42 C) 25 D) 81 E) 16
⎪⎩ ⎪ ⎨ ⎧ + = + + = + xy y x xy y x 133
) ( 7 2 =>
xy + = + 133
) 7 ( =>
xy xy xy + = + + 133 14 49
6 = xy
To`g`ri javob: A
77(2003-8.40). xy x y yz y z xz x z + = + = + = ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ 10 7 40 13 5 8 tenglamalar sistemasidan x ni toping. A) 80/79 B) 5/7 C) 7/13 D) 79/80 E) 7/5 ## Tenglamalar sistemasini quyidagicha qayta yozamiz:
y xy y z yz x z xz + = + = + = ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ Download 254.39 Kb. Do'stlaringiz bilan baham: 
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