Question

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

• A. Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
• B. Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
• C. Statement-1 is true, Statement-2 is false
• D. Statement-1 is false, Statement-2 is true

Answer 1

Answer: (C)

$f \left(x\right) = x \left|x\right|$ and $g\left(x\right) = sin\,x$
$gof \left(x\right)=sin\left(x\left|x\right|\right)=$ $ \begin{cases} -sin^2\,x^2 , & \text{$x<0$} \\[2ex] sin\,x^2 , & \text{$x\ge0$} \end{cases}$
$\therefore (gof)'(x) = \begin{cases} -2x\,cos^2, & \text{$x<0$} \\[2ex] 2x\,cos^2, & \text{$x\ge0$} \end{cases}$
Clearly, $L(gof )'(0) = 0 = R(gof )'(0)$
$∴$ gof is differentiable at $x = 0$ and also its derivative is continuous at $x = 0$
Now, $\therefore (gof)''(x) = \begin{cases} -2x\,cos^2+4x^2\,sin\,x^2, & \text{$x<0$} \\[2ex] 2x\,cos^2-4x^2\,sin\,x^2, & \text{$x\ge0$} \end{cases}$
$∴L(gof )"(0) = −2$ and $R(gof )"(0) = 2$
$∴L(gof )"(0) ≠ R(gof )"(0)$
$∴ gof(x)$ is not twice differentiable at $x = 0.$