Lt $⨍ (x)=x |x|,g (x) =sin x$ and $h (x) = (go ⨍) (x) .$ Then

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$h (x)$ is not differentiable at $x = 0$

$h (x)$ is differentiable at $x = 0,$ but $h^{'} (x)$ is not continuous at $h = 0$

$h^{'} (x)$ is continuous at $a t x = 0$ but it is not differentiable at $x = 0$

$h^{'} (x)$ is differentiable at $x = 0$

answer is C.

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Detailed Solution

$f (x) = x |x| g (x) = \sin x, gof (x) = g (f (x)) = \sin f (x) = \sin x |x| gof (x) = \{\begin{array}{l} - {sinx}^{2}; x < 0 \\ {sinx}^{2}; x \geq 0 \end{array}, (gof) (x) = \{\begin{cases} - 2 x {cosx}^{2}; x < 0 \\ 2 x {cosx}^{2}; x \geq 0 \end{cases}$ ${gof}^{1} (0^{-}) = 0, {(gof)}^{1} (0^{+}) = 0 ∴ gof is differentiable at x = 0 derivative of gof (x) is continuous at x = 0 {(gof)}^{n} (x) = \{\begin{cases} 4 x^{2} {sinx}^{2} - 2 {cosx}^{2}; x < 0 \\ - 4 x^{2} {sinx}^{2} + 2 {cosx}^{2}' x > 0 \end{cases} {(gof)}^{''} (0^{-}) = - 2, {(gof)}^{11} (0^{+}) = 2 ∴ (gof) (x) is not twice differentiable at x = 0$