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Question

If f:RR is a differentiable function and f(2)=6, then limx2f(x)62t dt(x2) is :

A
0
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B
2f(2)
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C
12f(2)
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D
24f(2)
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Solution

The correct option is C 12f(2)
limx2f(x)62t dt(x2)
=limx21(x2)[t2]f(x)6
=limx2[f(x)]262(x2) (00 form)

Applying L'Hospital Rule, we get
limx22f(x)f(x)1
=2f(2)f(2)
=12f(2)

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