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Q.

Let g(x)=xf(x), where f(x)=xsin1x,x00,x=0.At x=0

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a

g is differentiable but g' is not continuous

b

g is differentiable while f is not

c

both f and g are differentiable

d

g is differentiable and g' is continuos

answer is A, B.

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Detailed Solution

limx0f(x)=limx0xsin1x=0×( any value between 1 and 1)=0

f0+=limh0hsin1h0h=limh0sin1h

which is not a fixed value.

Hence, f(x) is non–differentiable at x = 0.

 Clearly, g(x)=xf(x)=x2sin1x,x00,x=0

is continuous at x = 0.

Also g(x) is differentaible at x = 0.( as f(0) = 0) with g(0)=0

 If x0,g(x)=x2cos1x1x2+2xsin1x

=cos1x+2xsin1x which exists for x0

 If x=0 then g(0)=limx0x2sin(1/x)0x0=limx0xsin1x=0

g(x)=cos1x+2xsin1x,x00,x=0

 At x=0,cos1x is not continuous, 

therefore, g(x) is not continuous at x = 0.

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Let g(x)=xf(x), where f(x)=xsin⁡1x,x≠00,x=0.At x=0