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JEE Main Math Class 12 Continuity and Differentiability Part 1 Questions
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© examsnet.com
Question : 95
Total: 100
If
f
(
x
)
=
x
n
,
then the value of
f
(
1
)
−
f
′
(
1
)
1
!
+
f
"
(
1
)
2
!
−
f
"
′
(
1
)
3
!
+
.
.
.
.
.
.
(
−
1
)
n
f
n
(
1
)
n
!
is
[2003]
1
2
n
2
n
−
1
0.
Validate
Solution:
Given that
f
(
x
)
=
x
n
⇒
f
(
1
)
=
1
f
′
(
x
)
=
n
x
n
−
1
⇒
f
′
(
1
)
=
n
f
"
(
x
)
=
n
(
n
−
1
)
x
n
−
2
⇒
f
"
(
1
)
=
n
(
n
−
1
)
.................................
f
n
(
x
)
=
n
!
⇒
f
n
(
1
)
=
n
!
f
(
1
)
−
f
′
(
1
)
1
!
+
f
"
(
1
)
2
!
−
f
′
′
′
(
1
)
3
!
+
.
.
.
.
.
.
.
.
+
(
−
1
)
n
f
n
(
1
)
n
!
=
1
−
n
1
!
+
n
(
n
−
1
)
2
!
−
n
(
n
−
1
)
(
n
−
2
)
3
!
+
.
.
.
.
.
+
(
−
1
)
n
n
!
n
!
=
n
C
0
−
n
C
1
+
n
C
2
−
n
C
3
+
.
.
.
.
.
.
.
+
(
−
1
)
n
n
C
n
=
(
1
−
1
)
n
=
0
© examsnet.com
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