Let f(x) = x|x|,g(x) = sin x and h(x) = (g ci | vedclass

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(C) Given $f(x) = x|x|$ and $g(x) = \sin x$.$h(x) = g(f(x)) = \sin(x|x|)$.Since $x|x| = x^2$ for $x \ge 0$ and $-x^2$ for $x < 0$,we have $h(x) = \beg

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$h'(x)$ is continuous at $x = 0$ but it is not differentiable at $x = 0$.

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(C) Given $f(x) = x|x|$ and $g(x) = \sin x$.$h(x) = g(f(x)) = \sin(x|x|)$.Since $x|x| = x^2$ for $x \ge 0$ and $-x^2$ for $x Now,$h'(x) = \begin{cases} 2x \cos(x^2) & x \ge 0 \ -2x \cos(x^2) & x At $x = 0$,$LHL = \lim_{x 	o 0^-} (-2x \cos(x^2)) = 0$ and $RHL = \lim_{x 	o 0^+} (2x \cos(x^2)) = 0$. Since $h'(0) = 0$,$h'(x)$ is continuous at $x = 0$.Now,check differentiability of $h'(x)$ at $x = 0$ by finding $h''(x)$:$h''(x) = \begin{cases} 2 \cos(x^2) - 4x^2 \sin(x^2) & x > 0 \ -2 \cos(x^2) + 4x^2 \sin(x^2) & x $LHD = \lim_{x 	o 0^-} (-2 \cos(x^2) + 4x^2 \sin(x^2)) = -2$.$RHD = \lim_{x 	o 0^+} (2 \cos(x^2) - 4x^2 \sin(x^2)) = 2$.Since $LHD 
eq RHD$,$h'(x)$ is not differentiable at $x = 0$.

Column $I$	Column $II$
$A$. $x\|x\|$	$I$. Strictly increasing and continuous in $(-1,1)$
$B$. $\sqrt{\|x\|}$	$II$. Continuous but not differentiable in $(-1,1)$
$C$. $x+[x]$	$III$. Differentiable in $(-1,1)$
$D$. $\|x-1\|+\|x+1\|+\|x\|$	$IV$. Differentiable in $(-1,0) \cup (0,1)$
	$V$. Strictly increasing and not differentiable in $(-1,1)$

Let $f(x) = x|x|$,$g(x) = \sin x$ and $h(x) = (g \circ f)(x)$. Then

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