Question

Let g(x) = x.f(x), where f(x) = f(x)={x sin1x,x≠00,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 option is C g is differentiable but g' is not continuous f(x)={x sin1x,x≠00,x=0.g(x)=f(x)={x2 sin1x,x≠00,x=0. L f'(0) = limh→0f(0−h)−f(0)−h =limh→0f(0−h)sin(−1h)−(0)−h=limh→0−sin(1h) = a quantity which lies between - 1 and 1 R f'(0) = limh→0f(0+h)−f(0)h =limh→0(0+h)sin1h−0h=limh→0sin1h = 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) = limh→0f(0−h)−f((0)0−h Lg′(0)=limh→0(0−h)2sin(−1h)−0−h=limh→0h sin(1h) L g'(0) = 0×(−1≤sin1h≤1)⇒Lg′(0)=0 and R g'(0) = =limh→0f(0+h)−f(0)h=limh→0(0+h)2sin1h−0h=limh→0h sin(1h)=0×(−1≤sin(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 sin1x−cos1x⇒ g′(x)=2f(x)−cos1x So, g'(x) is not differentiable at x = 0.

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