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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 $x=0$.
Options:
Statement-1 : gof is differentiable at $x=0$ and its derivative is continuous at that point.
Statement-2 : gof is twice differentiable at $x=0$.
Solution:
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Verified Answer
The correct answer is:
Statement-1 is true, Statement-2 is false
Statement-1 is true, Statement-2 is false
$$
\begin{aligned}
& f(x)=x|x| \text { and } g(x)=\sin x \\
& g \circ f(x)=\sin (x|x|)=\left\{\begin{array}{cc}
-\sin x^2 & , x < 0 \\
\sin x^2 & , x \geq 0
\end{array}\right. \\
& \therefore \text { (gof })^{\prime}(x)=\left\{\begin{array}{cc}
-2 x \cos x^2 & , x < 0 \\
2 x \cos x^2 & , x \geq 0
\end{array}\right.
\end{aligned}
$$
Clearly, L(gof )' $(0)=0=\mathrm{R}(\text { gof })^{\prime}(0)$
$\therefore$ gof is differentiable at $x=0$ and also its derivative is continuous at $x=0$
$$
\begin{aligned}
& \text { Now, (gof)" }(\mathrm{x})=\left\{\begin{array}{cc}
-2 \cos \mathrm{x}^2+4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x} < 0 \\
2 \cos \mathrm{x}^2-4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x} \geq 0
\end{array}\right. \\
& \therefore \mathrm{L}(\text { gof })^{\prime \prime}(0)=-2 \text { and } \mathrm{R}(\text { gof })^{\prime \prime}(0)=2 \\
& \therefore \mathrm{L}(\text { gof })^{\prime \prime}(0) \neq \mathrm{R}(\text { gof })^{\prime \prime}(0)
\end{aligned}
$$
$\therefore g o f(x)$ is not twice differentiable at $x=0$.
\begin{aligned}
& f(x)=x|x| \text { and } g(x)=\sin x \\
& g \circ f(x)=\sin (x|x|)=\left\{\begin{array}{cc}
-\sin x^2 & , x < 0 \\
\sin x^2 & , x \geq 0
\end{array}\right. \\
& \therefore \text { (gof })^{\prime}(x)=\left\{\begin{array}{cc}
-2 x \cos x^2 & , x < 0 \\
2 x \cos x^2 & , x \geq 0
\end{array}\right.
\end{aligned}
$$
Clearly, L(gof )' $(0)=0=\mathrm{R}(\text { gof })^{\prime}(0)$
$\therefore$ gof is differentiable at $x=0$ and also its derivative is continuous at $x=0$
$$
\begin{aligned}
& \text { Now, (gof)" }(\mathrm{x})=\left\{\begin{array}{cc}
-2 \cos \mathrm{x}^2+4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x} < 0 \\
2 \cos \mathrm{x}^2-4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x} \geq 0
\end{array}\right. \\
& \therefore \mathrm{L}(\text { gof })^{\prime \prime}(0)=-2 \text { and } \mathrm{R}(\text { gof })^{\prime \prime}(0)=2 \\
& \therefore \mathrm{L}(\text { gof })^{\prime \prime}(0) \neq \mathrm{R}(\text { gof })^{\prime \prime}(0)
\end{aligned}
$$
$\therefore g o f(x)$ is not twice differentiable at $x=0$.
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