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JEE Main Math Class 12 Continuity and Differentiability Part 1 Questions
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Question : 5
Total: 100
The value of
c
in the Lagrange's mean value theorem forthe function
f
(
x
)
=
x
3
−
4
x
2
+
8
x
+
11
,
when
x
∈
[
0
,
1
]
is:
[Jan. 7, 2020 (II)]
4
−
√
5
3
4
−
√
7
3
2
3
√
7
−
2
3
Validate
Solution:
Since,
f
(
x
)
is a polynomial function.
∴
It is continuous and differentiable in [0,1]
Here,
f
(
0
)
=
11
,
f
(
1
)
=
1
−
4
+
8
+
11
=
16
f
′
(
x
)
=
3
x
2
−
8
x
+
8
∴
f
′
(
c
)
=
f
(
1
)
−
f
(
0
)
1
−
0
=
16
−
11
1
=
3
c
2
−
8
c
+
8
⇒
3
c
2
−
8
c
+
3
=
0
⇒
c
=
8
±
2
√
7
6
=
4
±
√
7
3
∴
c
=
4
−
√
7
3
∈
(
0
,
1
)
© examsnet.com
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