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Question

If f(x) is differentiable in [a,b] such that f(a)=2, f(b)=6, then there exists at least one c, a<c<b, such that (b3a3)f(c)=

A
c2
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B
2c2
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C
3c2
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D
12c2
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Solution

The correct option is D 12c2
Let h(x)=f(x)f(a)+A(x3a3)
where A is obtained from the relation h(b)=0
h(b)=4+A(b3a3)=0
A=4b3a3
Also,
h(a)=0
h(a)=h(b)=0
Clearly, h(x) is continuous on [a,b] and differentiable in (a,b).
Hence, Rolle's theorem holds for h(x) in [a,b], so there exists c(a,b) such that h(c)=0
f(c)+A(3c2)=0
f(c)=3c2(4b3a3)
(b3a3)f(c)=12c2
Hence, option D.

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