Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If y = 2xn+1 + (3/xn) ,then x2 (d2y/dx2) is
Q. If
y
=
2
x
n
+
1
+
x
n
3
,then
x
2
d
x
2
d
2
y
is
6121
159
KCET
KCET 2020
Continuity and Differentiability
Report Error
A
6n(n+1)y
15%
B
n(n+1)y
48%
C
x
d
x
d
y
+
y
27%
D
y
9%
Solution:
y
=
(
2
x
)
(
n
+
1
)
+
(
3
x
)
(
−
n
)
⇒
d
y
/
d
x
=
2
(
n
+
1
)
x
n
−
(
3
n
x
)
(
−
n
−
1
)
⇒
(
d
2
y
)
/
(
d
x
2
)
=
2
n
(
n
+
1
)
x
(
n
−
1
)
+
3
n
(
n
+
1
)
x
(
−
n
−
2
)
⇒
x
2
(
d
2
y
)
/
(
d
x
2
)
=
n
(
n
+
1
)
[
(
2
x
)
(
n
+
1
)
+
3/
x
n
]
⇒
x
2
(
d
2
y
)
/
(
d
x
2
)
=
n
(
n
+
1
)
y