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Question

Let f:RR be a differentiable function and f(1)=4. Then the value of limx1f(x)42tx1dt is

A
8f(1)
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B
4f(1)
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C
2f(1)
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D
f(1)
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Solution

The correct option is A 8f(1)
limx1f(x)42tx1dt=limx1f(x)42tdtx1
Applying L'Hospital's rule
=limx12f(x)f(x)1=f(1)f(1)=8f(1)

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