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Question

If f(x)=xksin(1x),x00,x=0 is differentiable at x=0, then (where k is an integer)

A
k>0
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B
k>1
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C
k1
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D
k0
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Solution

The correct option is B k>1
At x=0
R.H.D. =limh0f(0+h)f(0)h
= limh0hksin1h0h
= limh0hk1sin1h

For this limit to exist, exponent of h must be positive.
i.e., k1>0
k>1

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