72
73
33–36
MASALALAR YECHISH
112.
Moddiy nuqta harakatining qonuni
s=s
(
t
) ga ko‘ra
uning eng katta
yoki eng kichik tezligini toping:
1)
s=
13
t
; 2)
s=
17
t –
5;
3)
s=t
2
+
5
t+
18; 4)
s=t
3
+
2
t
2
+
5
t+
8;
5)
s=
2
t
3
+
5
t
2
+
6
t+
3;
6)
s=
13
t
3
+
2
t
2
;
7)
s=t
3
+t
2
+
3.
113.
Berilgan funksiya grafigiga: 1)
x
0
=
–1;
2)
x
0
=
2,2; 3)
x
0
=0 abssissali
nuqtada o‘tkazilgan urinmani toping:
1)
f
(
x
)
=
12
x
2
+
5
x+
1;
2)
f
(
x
)
=
13
x+
4; 3)
f
(
x
)
=
60;
4)
f
(
x
)
=x
3
+
4
x
.
114.
Berilgan funksiya uchun
y=
–7
x+
2 to‘g‘ri chiziqqa parallel bo‘lgan
urinma tenglamasini yozing:
1)
f
(
x
)
=
5
x
3
–
2
x
2
+
16; 2)
f
(
x
)
=–
4
x
2
+
5
x+
3; 3)
f
(
x
)
=–
8
x+
5.
115.
Berilgan
f
(
x
) va
g
(
x
) funksiyalar grafiklarining
urinmalari parallel
bo‘ladigan nuqtalarini toping:
1)
f
(
x
) = 2
x
2
–3
x
+4,
g
(
x
)=12
x
–8;
2)
f
(
x
) = 18
x
+19,
g
(
x
)=–15
x
+18;
3)
f
(
x
) = 2
x+
13,
g
(
x
)=4
x
–19;
4)
f
(
x
) = 2
x
3
,
g
(
x
)=4
x
2
;
5)
f
(
x
) = 2
x
3
+3
x
2
,
g
(
x
)=15
x
–17;
6)
f
(
x
) = 2
x
4
,
g
(
x
)=4
x
3
.
116.
1)
x
y
1
=
funksiya grafigining
2
1
−
=
x
nuqtadan o‘tuvchi
urinmasi
tenglamasini tuzing.
2)
2
x
y
=
parabolaning
1
=
x
va
3
=
x
abssissalarga mos nuqtalari
tutash tirilgan. Parabolaning ushbu 2 nuqtani tutashtiruvchi kesmaga
parallel bo‘lgan urinmasi qaysi nuqtadan o‘tadi?
3)
Moddiy nuqta
2
( )
sin
3
9
2
t
s t
π
= ⋅
+
qonuniyat bilan harakatlanmoqda
(s – santimetrda,
t
– sekundda). Moddiy nuqtaning 1-sekunddagi tezla-
nishini toping.
117.
Funksiyaning ko‘rsatilgan nuqtadagi hosilasini hisob lang:
1)
f
(
x
)=
x
2
–15,
2
1
0
−
=
x
;
2)
x
x
f
cos
3
)
(
=
,
0
x
= −π
;
74
75
122.
1)
4
x
y
=
; 2)
2
1
+
−
=
x
x
y
; 3)
20
y x
x
= −
; 4)
2
ln
y x
x
=
;
5)
x
x
y
sin
3
=
; 6)
y=e
x
sin
x
; 7)
2
4
1
x
x
y
+
=
; 8)
2(10 1)sin
y
x
x
=
−
.
123.
Berilgan
funksiyalar uchun
'(
), '( )
2
4
f
f
π
π
−
sonlarni hisoblang:
1)
x
e
x
f
x
cos
)
(
=
; 2)
1
3
)
(
+
=
x
x
f
;
3)
3
2
)
(
2
+
+
=
x
x
x
f
;
4)
2
sin
)
(
x
x
x
f
+
=
; 5)
x
x
x
f
cos
sin
)
(
+
=
; 6)
x
x
f
sin
)
(
=
;
7)
4
( ) cos
f x
x x
=
+
; 8)
x
x
x
f
3
cos
3
sin
)
(
+
=
.
124.
Moddiy nuqta
3
2
( )
6
15
6
t
x t
t
= − +
+
qonuniyat bilan harakatlanmoqda.
1) tezlanish nol bo‘lgan
t
0
vaqtni; 2) shu
t
0
vaqtdagi tezlikni toping.
125*.
2
( )
13
2
f x
x
x
=
−
+
funksiya
Ox
o‘qi bilan
qanday burchak ostida kesi-
shadi?
126.
f
′(0) sonni toping: 1)
( )
4
4
3
6
+
−
=
x
x
x
f
; 2)
( ) (
)
6
10
f x
x
=
+
.
127.
y
′
(
x
) ni toping: 1)
( )
x
x
y
2
sin
=
; 2)
( )
x
x
y
2
cos
=
; 3)
( )
2
tg
y x
x
=
.
128.
Funksiyaning o‘sish va kamayish oraliqlarini toping:
1)
( )
x
x
f
7
3
+
=
; 2)
( )
3
17
f x
x
x
=
+
;
3)
( )
1
18
4
f x
x
=
+
;
4)
( )
21
x
f x
x
+
=
; 5)
2
( )
5 14
f x
x
x
=
+
−
; 6)
( )
(
)
8
2
+
=
x
x
x
f
;
7)
( )
6
4
2
+
−
−
=
x
x
x
f
; 8)
( )
2
1
x
x
f
−
=
;
9)
( )
3
2
12
17
23
f x
x
x
x
=
−
−
−
; 10)
( )
4
3
3
18
6
f x
x
x
=
+
−
;
11)
3
2
( )
5
19
22
f x
x
x
x
=
−
+
+
; 12)
( )
2
4
7
x
x
x
f
+
=
.
129.
Funksiyaning statsionar nuqtalarini toping:
1)
( )
9
7
3
2
+
−
=
x
x
x
f
; 2)
( )
3
1
19
7
f x
x
x
=
−
;
3)
( )
x
x
f
5
=
x
3
;
4)
( )
2
8
x
x
f
=
;
5)
( )
7 14
f x
x
=
−
;
6)
( )
7
+
=
x
x
f
27–
x
3
;
7)
( )
3
2
12
13
16
f x
x
x
=
+
−
;
8)
f
(
x
)=
x
3
–6
x
2
+9.