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Question

Let g(x) = x.f(x), where f(x) = f(x)={x sin1x,x00,x=0. at x = 0

A
g is differentiable but g' is not continuous
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B

g is differentiable while f is not

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C

Both f and g are differentiable

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D
g is differentiable and g' is continuous
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Solution

The correct options are
A g is differentiable but g' is not continuous
B

g is differentiable while f is not


f(x)={x sin1x,x00,x=0.g(x)=f(x)={x2 sin1x,x00,x=0.
L f'(0) = limh0f(0h)f(0)h
=limh0f(0h)sin(1h)(0)h=limh0sin(1h)
= a quantity which lies between - 1 and 1
R f'(0) = limh0f(0+h)f(0)h
=limh0(0+h)sin1h0h=limh0sin1h
= a quantity which lies between - 1 and 1
Hence L f'(0) R f'(0)
f(x) is not differentiable at x = 0
Now L g'(0) = limh0f(0h)f((0)0h
Lg(0)=limh0(0h)2sin(1h)0h=limh0h sin(1h)
L g'(0) = 0×(1sin1h1)Lg(0)=0 and
R g'(0) = =limh0f(0+h)f(0)h=limh0(0+h)2sin1h0h=limh0h sin(1h)=0×(1sin(1h)1)=0
L g'(0) = R g'(0) then g(x) is differentiable at x = 0
Now g(x)=x2sin1xg(x)=2x sin1x+x2 cos1x×1x2
g(x)=2x sin1xcos1x g(x)=2f(x)cos1x
So, g'(x) is not differentiable at x = 0.

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