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Manipal Entrance Test 2015 Math Paper
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© examsnet.com
Question : 4
Total: 70
The value of
c
prescribed by Lagrange's mean value theorem, when
f
(
x
)
=
√
x
2
−
4
,
a
=
2
and
b
=
3
, is
2.5
√
5
√
3
√
3
+
1
Validate
Solution:
Clearly,
f
(
x
)
=
√
x
2
−
4
is continuous on [2, 3] and differentiable on (2, 3).
So, by Lagrange's mean value theorem, there exists
c
∈
(
2
,
3
)
such that
f
′
(
c
)
=
f
(
3
)
−
f
(
2
)
3
−
2
⇒
c
√
c
2
−
4
=
√
5
−
0
[
∵
f
(
x
)
=
√
x
2
−
4
⇒
f
′
(
x
)
=
x
√
x
2
−
4
]
⇒
c
2
=
5
(
c
2
−
4
)
⇒
4
c
2
=
20
⇒
c
=
√
5
© examsnet.com
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