Let f(x) = x|x| and g(x) = sin x.Statement-1: | vedclass

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(D) We have $f(x) = x|x|$ and $g(x) = \sin x$.Therefore,$(gof)(x) = \sin(x|x|) = \begin{cases} -\sin(x^2), & x < 0 \ \sin(x^2), & x \ge 0 \end{cases}

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Statement-$1$ is true,Statement-$2$ is false.

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(D) We have $f(x) = x|x|$ and $g(x) = \sin x$.Therefore,$(gof)(x) = \sin(x|x|) = \begin{cases} -\sin(x^2), & x $LHD$ of $gof$ at $x=0$: $\lim_{x 	o 0^-} \frac{-\sin(x^2) - 0}{x - 0} = \lim_{x 	o 0^-} -x \left(\frac{\sin(x^2)}{x^2}ight) = 0$.$RHD$ of $gof$ at $x=0$: $\lim_{x 	o 0^+} \frac{\sin(x^2) - 0}{x - 0} = \lim_{x 	o 0^+} x \left(\frac{\sin(x^2)}{x^2}ight) = 0$.Since $LHD = RHD = 0$,$gof$ is differentiable at $x=0$ and $(gof)'(0) = 0$.The derivative is $(gof)'(x) = \begin{cases} -2x\cos(x^2), & x Since $\lim_{x 	o 0^-} (gof)'(x) = 0$ and $\lim_{x 	o 0^+} (gof)'(x) = 0$,the derivative is continuous at $x=0$. Thus,Statement-$1$ is true.Now,check for second differentiability at $x=0$:$LHD$ of $(gof)'$ at $x=0$: $\lim_{x 	o 0^-} \frac{-2x\cos(x^2) - 0}{x - 0} = -2\cos(0) = -2$.$RHD$ of $(gof)'$ at $x=0$: $\lim_{x 	o 0^+} \frac{2x\cos(x^2) - 0}{x - 0} = 2\cos(0) = 2$.Since $LHD 
eq RHD$,$gof$ is not twice differentiable at $x=0$. Thus,Statement-$2$ is false.

Let $f(x) = x|x|$ and $g(x) = \sin x$.
Statement-$1$: $gof$ is differentiable at $x=0$ and its derivative is continuous at that point.
Statement-$2$: $gof$ is twice differentiable at $x=0$.

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Let $f(x) = x|x|$ and $g(x) = \sin x$.Statement-$1$: $gof$ is differentiable at $x=0$ and its derivative is continuous at that point.Statement-$2$: $gof$ is twice differentiable at $x=0$.