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Question

Show that the function fx=xm sin1x , x00, x=0

(i) differentiable at x = 0, if m > 1
(ii) continuous but not differentiable at x = 0, if 0 < m < 1
(iii) neither continuous nor differentiable, if m ≤ 0

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Solution

Given:
f(x) = xm sin1x0 x≠0 , x=0

(i) Let m=2, then the function becomes f(x) = x2 sin1x0 , x≠0, x=0

Differentiability at x=0:
limx0 f(x) - f(0)x-0 = limx0 f(x)x = limx0 x sin1x =0.
[ ∵ limx0 x sin1x = 0 , as x sin1x - 0 = x sin1x = x sin1x x (∵sinθ1 for all θ) and hence x sin1x<0 when x-0<εx-0<ε ]
So, f'(0) = 0, which means f is differentiable at x=0.
Hence the given function is differentiable at x=0.

(ii) Let m=12, 0<m<1. Then the function becomes
f(x) = x120sin1x , x≠0 , x=0

Continuity at x=0:
(LHL at x=0) = limx0- f(x) = limh0 f(0-h) = limh0 (-h)12 sin10-h = limh0 h12 sin1h = limh0 h32 = 0.
(RHL at x=0) = limx0+ f(x) = limh0 f(0+h) = limh0 h12 sin1h = limh0 h32 = 0.
and f(0) = 0
LHL at x=0 = RHL at x=0 = limx0 f(x),
Hence continuous.
Now Differentiabilty at x=0 when 0<m<1.
(LHD at x=0) = limx0- f(x) - f(0)x-0 = limh0 f(0-h) - f(0)0-h-0 =limh0 (-h)12 sin1-h-h


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