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Question


Let f(x)={xnsin1x,x00,x=0 , then f(x) is continuous but not differentiable at x=0 if
  1. n(0,1]
  2. n=0
  3. n[1,)
  4. n(,0)

A
n(0,1]
B
n=0
C
n[1,)
D
n(,0)
Solution
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Since, f(x) is continuous at x=0, therefore, limx0f(x)=f(0)=0

limx0xnsin(1x)=0n>0

f(x) is differential at x=0 if limx0f(x)f(0)x0 exists finitely

limx0xnsin(1x)0x exists finitely

xn1sin(1x) exists finitely.

n1>0n>1

If n1, then limx0xn1sin(1x) does not exists and hence, f(x) is not differentiable at x=0

Hence, f(x) is continuous but not differentiable at x=0 for 0<n1

i.e., n(0,1]

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