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KEAM 2012 Math Solved Papers
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© examsnet.com
Question : 2
Total: 120
Let
f
(
x
)
=
x
3
−
x
+
p
(
0
≤
x
≤
2
)
where
p
is a constant. The value
c
of mean value theorem is
√
3
2
√
6
3
√
3
3
√
2
3
2
√
3
3
Validate
Solution:
Given,
f
(
x
)
=
x
3
−
x
+
p
,
0
≤
x
≤
2
Here,
a
=
0
,
b
=
2
f
′
(
c
)
=
f
(
b
)
−
f
(
a
)
b
−
a
......(1)
⇒
f
′
(
c
)
=
f
(
2
)
−
f
(
0
)
2
−
0
∴
3
c
2
−
1
=
(
2
)
3
−
2
+
p
−
(
0
2
−
0
+
p
)
2
−
0
⇒
3
c
2
−
1
=
6
2
=
3
⇒
3
c
2
=
4
⇒
c
=
±
2
√
3
=
±
2
√
3
3
∴
c
=
2
√
3
3
(
∵
−
2
√
3
3
∉
[
0
,
2
]
)
© examsnet.com
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