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 $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+\|x-1\|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

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

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