Question

Let f(x) = x|x|, g(x) = sin x and h(x) = (gof) (x). Then

• A. h(x) is not differentiable at x = 0.
• B. h(x) is differentiable at x = 0, but h'(x) is not continuous at x = 0
• C. h'(x) is continuous at x = 0 but it is not differentiable at x = 0
• D. h'(x) is differentiable at x = 0

Answer 1

Answer: (C)

Let f (x) = x|x| = x|x|, g(x) = sin x and h (x) = gof (x) = g[f (x)]
$\therefore h(x)=\{\begin{array}{ll}\sin x^2,& x\ge 0\\ -\sin x^2,& x<0\end{array}$
Now, $h^{\prime }(x)=\{\begin{array}{ll}2x\cos x^2,& x\ge 0\\ -2x\cos x^2,& x<0\end{array}$
Since, L.H.L and R.H.L at x = 0 of h' (x) is equal to 0 therefore h' (x) is continuous at x = 0
Now, suppose h' (x) is differentiable
$\therefore h^{\prime \prime }(x)=\{\begin{array}{ll}2(\cos x^2+2x^2(-\sin x^2),& x\ge 0\\ 2(-\cos x^2+2x^2\sin x^2),& x<0\end{array}$
Since, L.H.L and R.H.L at x = 0 of h" (x) are different therefore h" (x) is not continuous.
$\Rightarrow$ h"(x) is not differentiable
$\Rightarrow$ our assumption is wrong
Hence h'(x) is not differentiable at x = 0.