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18-BILET
1. Integralni hisoblang:
∫ (𝒙
𝟐
+ 𝟐𝒙 + 𝟑)
𝟏
−𝟐
𝒅𝒙
∫(𝑥
2
+ 2𝑥 + 3)
1
−2
𝑑𝑥 = (
𝑥
3
3
+ 2 ∙
𝑥
2
2
+ 3𝑥) |
1
−2
= (
𝑥
3
3
+ 𝑥
2
+ 3𝑥) |
1
−2
=
= (
1
3
3
+ 1
2
+ 3 ∙ 1) − (
(−2)
3
3
+ (−2)
2
+ 3 ∙ (−2)) =
1
3
+ 1 + 3 +
8
3
− 4 + 6 =
=
9
3
+ 6 = 3 + 6 = 9 . 𝐽𝑎𝑣𝑜𝑏: 9 .
2. Tenglamani yeching:
𝟒
√𝟐𝒙−𝟒
= 𝟔𝟒 ∙ 𝟐
√𝟐𝒙−𝟒
4
√2𝑥−4
= 64 ∙ 2
√2𝑥−4
; 2
2√2𝑥−4
= 2
6
∙ 2
√2𝑥−4
;
2
2√2𝑥−4
2
√2𝑥−4
= 2
6
;
2
2√2𝑥−4−√2𝑥−4
= 2
6
; 2
√2𝑥−4
= 2
6
; √2𝑥 − 4 = 6 ;
(√2𝑥 − 4)
2
= 6
2
; 2𝑥 − 4 = 36 ; 2𝑥 = 36 + 4 ; 2𝑥 = 40 ;
𝑥 =
40
2
; 𝑥 = 20 . 𝐽𝑎𝑣𝑜𝑏: 𝑥 = 20 .
3.
𝒇(𝒙) = 𝒆
𝟐𝒙
− 𝟐 ∙ 𝒆
𝒙
funksiya statsionar nuqtalarini toping.
𝑓′(𝑥) = (𝑒
2𝑥
− 2 ∙ 𝑒
𝑥
)
′
= 2𝑒
2𝑥
− 2𝑒
𝑥
; 2𝑒
2𝑥
− 2𝑒
𝑥
= 0 ; 𝑒
𝑥
= 𝑡 ;
2𝑡
2
− 2𝑡 = 0 ; 𝑡
2
− 𝑡 = 0 ; 𝑡 ∙ (𝑡 − 1) = 0 → {
𝑡 = 0
𝑡 − 1 = 0
→ {
𝑡 = 0
𝑡 = 1
→
→ {
𝑒
𝑥
= 0
𝑒
𝑥
= 1
→ {
𝑒
𝑥
= 0
𝑒
𝑥
= 𝑒
0
→ {
∅
𝑥 = 0
𝐽𝑎𝑣𝑜𝑏: 𝑥 = 0 .
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4. Uchburchakning uchlari toʻgʻri burchakli dekart koordinatalar sistemasida
quyidagicha berilgan:
𝑨(𝟎; 𝟎), 𝑩(−𝟏; 𝟓), 𝑪(−𝟐; 𝟎)
. Uchburchak yuzini toping.
2) 𝑥
1
= −1 ; 𝑦
1
= 5 ; 𝑥
2
= −2 ; 𝑦
2
= 0 .
𝐵𝐶 = √(𝑥
2
− 𝑥
1
)
2
+ (𝑦
2
− 𝑦
1
)
2
= √((−2) − (−1))
2
+ (0 − 5)
2
=
= √(−2 + 1)
2
+ (−5)
2
= √1 + 25 = √26 . 𝐴𝐵 = 𝐵𝐶 = √26 ; ∆𝐴𝐵𝐶 𝑡𝑒𝑛𝑔 𝑦𝑜𝑛𝑙𝑖
1) 𝑥
1
= 0 ; 𝑦
1
= 0 ; 𝑥
2
= −2 ; 𝑦
2
= 0 .
𝐴𝐶 = √(𝑥
2
− 𝑥
1
)
2
+ (𝑦
2
− 𝑦
1
)
2
= √(−2 − 0)
2
+ (0 − 0)
2
= √4 + 0 = √4 = 2 .
𝐴𝐷 =
𝐴𝐶
2
=
2
2
= 1 ; 𝐵𝐷 = √𝐴𝐵
2
− 𝐴𝐷
2
= √(√26)
2
− 1
2
= √25 = 5 ;
𝑆
𝐴𝐵𝐶
=
1
2
∙ 𝐴𝐶 ∙ 𝐵𝐷 =
1
2
∙ 2 ∙ 5 =
10
2
= 5 . 𝐽𝑎𝑣𝑜𝑏: 𝑆
𝐴𝐵𝐶
= 5 (𝑘𝑣𝑎𝑑𝑟𝑎𝑡 𝑏𝑖𝑟𝑙𝑖𝑘) .
5. Qirrasi
√𝟐 + 𝟏
boʻlgan kubdan uchlari kub uchlarida boʻlgan sakkizta
muntazam piramidalar shunday kesib olindiki, qolgan jismning sirti sakkizta
muntazam sakkizburchak va sakkizta muntazam uchburchaklardan iborat.
Hosil boʻlgan jism hajmini toping.
𝐵𝑒𝑟𝑖𝑙𝑔𝑎𝑛: 𝐴(0; 0), 𝐵(−1; 5), 𝐶(−2; 0).
𝑇𝑜𝑝𝑖𝑠ℎ 𝑘𝑒𝑟𝑎𝑘: 𝑆
𝐴𝐵𝐶
=?
1) 𝑥
1
= 0 ; 𝑦
1
= 0 ; 𝑥
2
= −1 ; 𝑦
2
= 5 .
𝐴𝐵 = √(𝑥
2
− 𝑥
1
)
2
+ (𝑦
2
− 𝑦
1
)
2
=
= √(−1 − 0)
2
+ (5 − 0)
2
= √1 + 25 = √26 .
𝐾𝑢𝑏𝑛𝑖𝑛𝑔 8 𝑡𝑎 𝑢𝑐ℎ𝑖𝑑𝑎𝑔𝑖 𝑞𝑖𝑟𝑟𝑎𝑠𝑖
𝑎
2
𝑔𝑎 𝑡𝑒𝑛𝑔
𝑝𝑖𝑟𝑎𝑚𝑖𝑑𝑎𝑙𝑎𝑟𝑛𝑖𝑛𝑔 ℎ𝑎𝑗𝑚𝑖 𝑦𝑖𝑔‘𝑖𝑛𝑑𝑖𝑠𝑖𝑛𝑖 𝑘𝑢𝑏 ℎ𝑎𝑗𝑚𝑖𝑑𝑎𝑛
𝑎𝑦𝑖𝑟𝑎𝑚𝑖𝑧. 𝑎 = √2 + 1 ; 𝑉
𝑘𝑢𝑏
= 𝑎
3
;
𝑉
1 𝑝𝑖𝑟𝑎𝑚𝑖𝑑𝑎
=
1
3
∙ (
𝑎
2
)
3
=
𝑎
3
24
;
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8 ∙ 𝑉
1 𝑝𝑖𝑟𝑎𝑚𝑖𝑑𝑎
= 8 ∙
𝑎
3
24
=
𝑎
3
3
; 𝑉
𝑗𝑖𝑠𝑚
= 𝑉
𝑘𝑢𝑏
− 8 ∙ 𝑉
1 𝑝𝑖𝑟𝑎𝑚𝑖𝑑𝑎
= 𝑎
3
−
𝑎
3
3
=
=
3𝑎
3
− 𝑎
3
3
=
2𝑎
3
3
=
2
3
∙ (√2 + 1)
3
=
2
3
∙ (2√2 + 3 ∙ (√2)
2
∙ 1 + 3√2 ∙ 1
2
+ 1
3
) =
=
2
3
∙ (2√2 + 3 ∙ 2 + 3√2 + 1) =
2
3
∙ (5√2 + 7) =
10√2 + 14
3
.
𝐽𝑎𝑣𝑜𝑏: 𝑉
𝑗𝑖𝑠𝑚
=
10√2 + 14
3
.
19-BILET
1. Moddiy nuqta
𝒔(𝒕) = 𝒕
𝟐
+ 𝟑𝒕
qonun boʻyicha harakatlanyapti. Vaqtning
qaysi paytida nuqtaning harakat tezligi
𝟏𝟓
𝒎
𝒔
ga teng boʻladi?
𝑣(𝑡) = 15
𝑚
𝑠
; 𝑠′(𝑡) = (𝑡
2
+ 3𝑡)
′
= 2𝑡 + 3 ; 2𝑡 + 3 = 𝑣(𝑡); 2𝑡 + 3 = 15 ;
2𝑡 = 15 − 3 ; 2𝑡 = 12 ; 𝑡 =
12
2
; 𝑡 = 6 𝑠 . 𝐽𝑎𝑣𝑜𝑏: 𝑡 = 6 𝑠 .
2. Tenglamani yeching:
𝟑 ∙ 𝒕𝒈
𝟐
𝒙 − 𝟒 ∙ 𝒕𝒈𝒙 + 𝟏 = 𝟎 .
𝑡𝑔𝑥 = 𝑦 ; 3𝑦
2
− 4𝑦 + 1 = 0 ; 3𝑦
2
− 3𝑦 − 𝑦 + 1 = 0 ;
3𝑦(𝑦 − 1) − (𝑦 − 1) = 0 ; (3𝑦 − 1)(𝑦 − 1) = 0 ; {
3𝑦 − 1 = 0
𝑦 − 1 = 0
→
→ {
3𝑦 = 1
𝑦 = 1
→ {
𝑦 =
1
3
𝑦 = 1
→ {
𝑡𝑔 𝑥 =
1
3
𝑡𝑔 𝑥 = 1
→ {
𝑥 = 𝑎𝑟𝑐𝑡𝑔
1
3
+ 𝜋𝑘, 𝑘𝜖𝑍
𝑥 =
𝜋
4
+ 𝜋𝑘, 𝑘𝜖𝑍
𝐽𝑎𝑣𝑜𝑏: 𝑥
1
= 𝑎𝑟𝑐𝑡𝑔
1
3
+ 𝜋𝑘, 𝑘𝜖𝑍 ; 𝑥
2
=
𝜋
4
+ 𝜋𝑘, 𝑘𝜖𝑍 .
3. Hisoblang:
𝐥𝐨𝐠
𝟕
𝟏𝟒−
𝟏
𝟑
∙ 𝐥𝐨𝐠
𝟕
𝟓𝟔
𝐥𝐨𝐠
𝟔
𝟑𝟎−
𝟏
𝟐
∙ 𝐥𝐨𝐠
𝟔
𝟏𝟓𝟎
log
7
14 −
1
3
∙ log
7
56
log
6
30 −
1
2
∙ log
6
150
=
log
7
(2 ∙ 7) −
1
3
∙ log
7
(8 ∙ 7)
log
6
(5 ∙ 6) −
1
2
∙ log
6
(25 ∙ 6)
=
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=
log
7
2 + log
7
7 −
1
3
∙ (log
7
8 + log
7
7)
log
6
5 + log
6
6 −
1
2
∙ (log
6
25 + log
6
6)
=
log
7
2 + 1 −
1
3
∙ (log
7
2
3
+ 1)
log
6
5 + 1 −
1
2
∙ (log
6
5
2
+ 1)
=
=
log
7
2 + 1 −
1
3
∙ log
7
2
3
−
1
3
log
6
5 + 1 −
1
2
∙ log
6
5
2
−
1
2
=
2
3
+ log
7
2 −
1
3
∙ 3 ∙ log
7
2
1
2
+ log
6
5 −
1
2
∙ 2 ∙ log
6
5
=
=
2
3
+ log
7
2 − log
7
2
1
2
+ log
6
5 − log
6
5
=
2
3
1
2
=
2
3
∙
2
1
=
4
3
. 𝐽𝑎𝑣𝑜𝑏:
4
3
.
4. Aylanaga yon tomoni
𝟏𝟎
ga, asosi
𝟏𝟎 √𝟏𝟏
𝟑
ga teng boʻlgan teng yonli uchburchak
ichki chizilgan. Aylana radiusini toping.
𝑎
sin 𝛼
= 2𝑅 ; 𝑅 =
𝑎
2 ∙ sin 𝛼
=
10
2 ∙
5
6
=
10
5
3
=
30
5
= 6 . 𝐽𝑎𝑣𝑜𝑏: 𝑅 = 6 (𝑏𝑖𝑟𝑙𝑖𝑘).
5.
𝑨(−𝟔; 𝟖; 𝟒)
va
𝑩(𝟒; −𝟕; 𝟏)
nuqtalar berilgan.
𝑨𝑩
kesmaning oʻrtasidan
𝑶𝒙
o‘qigacha boʻlgan masofani toping.
𝐵𝑒𝑟𝑖𝑙𝑔𝑎𝑛: 𝐴(−6 ; 8 ; 4) , 𝐵(4 ; −7; 1) . 𝑇𝑜𝑝𝑖𝑠ℎ 𝑘𝑒𝑟𝑎𝑘: 𝑑 =?
𝐶(𝑥; 𝑦; 𝑧); 𝑥
1
= −6 ; 𝑦
1
= 8 ; 𝑧
1
= 4 ; 𝑥
2
= 4 ; 𝑦
2
= −7 ; 𝑧
2
= 1 ;
𝑥 =
𝑥
1
+ 𝑥
2
2
=
−6 + 4
2
= −
2
2
= −1 ; 𝑦 =
𝑦
1
+ 𝑦
2
2
=
8 + (−7)
2
=
8 − 7
2
=
1
2
;
𝑧 =
𝑧
1
+ 𝑧
2
2
=
4 + 1
2
=
5
2
; 𝐶 (−1;
1
2
;
5
2
) 𝑛𝑢𝑞𝑡𝑎𝑑𝑎𝑛 𝑂𝑥 𝑜‘𝑞𝑖𝑔𝑎𝑐ℎ𝑎 𝑚𝑎𝑠𝑜𝑓𝑎 ∶
𝐵𝑒𝑟𝑖𝑙𝑔𝑎𝑛: 𝑎 = 10 ; 𝑏 =
10√11
3
. 𝑅 =?
cos 𝛼 =
𝑏
2
𝑎
=
𝑏
2𝑎
=
10√11
3
2 ∙ 10
=
10√11
3 ∙ 20
=
√11
6
;
sin 𝛼 = √1 − cos
2
𝑥 = √1 − (
√11
6
)
2
=
= √1 −
11
36
= √
36 − 11
36
= √
25
36
=
5
6
;
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𝑑 = √𝑦
2
+ 𝑧
2
= √(
1
2
)
2
+ (
5
2
)
2
= √
1
4
+
25
4
= √
1 + 25
4
= √
26
4
=
√26
2
.
𝐽𝑎𝑣𝑜𝑏: 𝐶 𝑛𝑢𝑞𝑡𝑎𝑑𝑎𝑛 𝑂𝑥 𝑜‘𝑞𝑖𝑔𝑎𝑐ℎ𝑎 𝑚𝑎𝑠𝑜𝑓𝑎 ∶ 𝑑 =
√26
2
.
20-BILET
1. Funksiyaga teskari funksiyani toping:
𝒚 = −
𝟏
𝟑
𝒙 −
𝟐
𝟑
𝑦 = −
1
3
𝑥 −
2
3
; 𝑥 = −
1
3
𝑦 −
2
3
;
1
3
𝑦 = −𝑥 −
2
3
;
𝑦
3
=
−3𝑥 − 2
3
;
𝑦 = −3𝑥 − 2 . 𝐽𝑎𝑣𝑜𝑏: 𝑦 = −3𝑥 − 2 .
2.
𝒇(𝒙) = (𝒙 + 𝟐) ∙ √𝒙
𝟑
funksiya hosilasini toping.
𝑓′(𝑥) = ((𝑥 + 2) ∙ √𝑥
3
)
′
= (𝑥 + 2)
′
∙ √𝑥
3
+ (𝑥 − 2) ∙ (√𝑥
3
)
′
=
= (1 + 0) ∙ √𝑥
3
+ (𝑥 − 2) ∙ (𝑥
1
3
)
′
= √𝑥
3
+ (𝑥 − 2) ∙
1
3
∙ 𝑥
1
3
−1
=
= √𝑥
3
+ (𝑥 − 2) ∙
1
3
∙ 𝑥
−
2
3
= √𝑥
3
+ (𝑥 − 2) ∙
1
3
∙
1
𝑥
2
3
= √𝑥
3
+ (𝑥 − 2) ∙
1
3√𝑥
2
3
=
= √𝑥
3
+
𝑥 − 2
3√𝑥
2
3
=
3√𝑥 ∙ 𝑥
2
3
+ 𝑥 − 2
3√𝑥
2
3
=
3√𝑥
3
3
+ 𝑥 − 2
3√𝑥
2
3
=
3𝑥 + 𝑥 − 2
3√𝑥
2
3
=
4𝑥 − 2
3√𝑥
2
3
.
𝐽𝑎𝑣𝑜𝑏:
4𝑥 − 2
3√𝑥
2
3
.
3. Tengsizlikni yeching:
(𝒙
𝟐
− 𝟒) ∙ 𝐥𝐨𝐠
𝟎,𝟓
𝒙 > 𝟎
(𝑥
2
− 4) ∙ log
0,5
𝑥 > 0 ; −(𝑥 − 2) ∙ (𝑥 + 2) ∙ log
2
𝑥 > 0 ;
(𝑥 − 2) ∙ (𝑥 + 2) ∙ log
2
𝑥 < 0 ; 𝑥 + 2 𝑚𝑢𝑠𝑏𝑎𝑡
(𝑥 − 2) ∙ log
2
𝑥 < 0 → {
𝑥 − 2 < 0
log
2
𝑥 > 0
→ {
𝑥 < 2
𝑥 > 1
𝐽𝑎𝑣𝑜𝑏: (1 ; 2).
+
////
−
/////
+
1 2
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4. Oʻnsakkizburchakning yuzi
𝟒
ga, unga ichki chizilgan doiraning yuzi
ga teng.
Oʻnsakkizburchakning perimetrini toping.
𝐵𝑒𝑟𝑖𝑙𝑔𝑎𝑛: 𝑆
18
= 4 ; 𝑆
𝑑𝑜𝑖𝑟𝑎
= 𝜋 . 𝑇𝑜𝑝𝑖𝑠ℎ 𝑘𝑒𝑟𝑎𝑘: 𝑃
18
=?
𝑆
𝑑𝑜𝑖𝑟𝑎
= 𝜋 ; 𝑆
𝑑𝑜𝑖𝑟𝑎
= 𝜋𝑟
2
; 𝜋 = 𝜋𝑟
2
; 𝑟
2
=
𝜋
𝜋
; 𝑟
2
= 1 ; 𝑟 = 1 ;
𝑆
18
=
1
2
∙ 𝑃
18
∙ 𝑟 ; 4 =
1
2
∙ 𝑃
18
∙ 1 ; 𝑃
18
= 4 ∙ 2 = 8 . 𝐽𝑎𝑣𝑜𝑏: 𝑃
18
= 8 .
5. Konus balandligi va yasovchining uzunliklari
𝟒 ∶ 𝟓
kabi nisbatda, hajmi esa
𝟗𝟔
ga teng. Toʻla sirtini toping.
1
3
𝜋𝑅
2
ℎ = 𝑉 ;
1
3
∙ 𝜋 ∙ (3𝑥)
2
∙ 4𝑥 = 96𝜋 ;
1
3
∙ 9𝑥
2
∙ 4𝑥 = 96 ;
12𝑥
3
= 96 ; 𝑥
3
=
96
12
; 𝑥
3
= 8 ; 𝑥
3
= 2
3
; 𝑥 = 2 ;
ℎ = 4𝑥 = 4 ∙ 2 = 8 ; 𝑙 = 5𝑥 = 5 ∙ 2 = 10 ; 𝑅 = 3𝑥 = 3 ∙ 2 = 6 ;
𝑆
𝑦𝑜𝑛
= 𝜋 ∙ 𝑅 ∙ 𝑙 = 𝜋 ∙ 6 ∙ 10 = 60𝜋 (𝑘𝑣𝑎𝑑𝑟𝑎𝑡 𝑏𝑖𝑟𝑙𝑖𝑘) ;
𝑆
𝑎𝑠𝑜𝑠
= 𝜋 ∙ 𝑅
2
= 𝜋 ∙ 6
2
= 36𝜋 (𝑘𝑣𝑎𝑑𝑟𝑎𝑡 𝑏𝑖𝑟𝑙𝑖𝑘) ;
𝑆
𝑡𝑜‘𝑙𝑎
= 𝑆
𝑎𝑠𝑜𝑠
+ 𝑆
𝑦𝑜𝑛
= 36𝜋 + 60𝜋 = 96𝜋 (𝑘𝑣𝑎𝑑𝑟𝑎𝑡 𝑏𝑖𝑟𝑙𝑖𝑘) .
𝐽𝑎𝑣𝑜𝑏: 𝑆
𝑡𝑜‘𝑙𝑎
= 96𝜋 (𝑘𝑣𝑎𝑑𝑟𝑎𝑡 𝑏𝑖𝑟𝑙𝑖𝑘) .
𝐵𝑒𝑟𝑖𝑙𝑔𝑎𝑛: ℎ = 𝑆𝑂 = 4𝑥 ; 𝑙 = 𝑆𝐴 = 5𝑥 ; 𝑉 = 96𝜋 .
𝑇𝑜𝑝𝑖𝑠ℎ 𝑘𝑒𝑟𝑎𝑘: 𝑆
𝑡𝑜‘𝑙𝑎
=?
𝑅 = 𝐴𝑂 = √𝑙
2
− ℎ
2
= √(5𝑥)
2
− (4𝑥)
2
=
= √25𝑥
2
− 16𝑥
2
= √9𝑥
2
= 3𝑥 ; 𝑅 = 3𝑥 ;
Do'stlaringiz bilan baham: |