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Question

The function f(x)={xsin(lnx2), x00,x=0 is:

A
non differentiable at x=0
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B
differentiable at x=0
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C
discontinuous at x=0
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D
none of the above
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Solution

The correct option is A non differentiable at x=0
Given : f(x)={xsin(lnx2), x00,x=0

For continuity,
R.H.L=limx0+xsin(lnx2)=0 (sinθ[1,1])L.H.L.=limx0xsin(lnx2)=0f(0)=f(0+)=f(0)=0
Hence, f(x) is continuous at x=0

For differentiability,
f(0+)=limh0+f(h)f(0)hf(0+)=limh0+hsin(lnh2)0hf(0+)=limh0+sin(lnh2)f(0+)=Not defined
Similarly, L.H.D. is also not defined
Hence, f(x) is not differentiable at x=0

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