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 $g\left(x\right)=sinx$
$gof\left(x\right)=sin\left(x\left|x\right|\right)=$ $\{\begin{array}{ll}-si{n}^2{x}^2,& \text{x<0}\\ sin{x}^2,& \text{x≥0}\end{array}$
$\therefore (gof{)}^{\prime }(x)=\{\begin{array}{ll}-2xco{s}^2,& \text{x<0}\\ 2xco{s}^2,& \text{x≥0}\end{array}$
Clearly, $L(gof{)}^{\prime }(0)=0=R(gof{)}^{\prime }(0)$
$\therefore$ gof is differentiable at $x=0$ and also its derivative is continuous at $x=0$
Now, $\therefore (gof{)}^{\prime \prime }(x)=\{\begin{array}{ll}-2xco{s}^2+4x^{2}sin{x}^2,& \text{x<0}\\ 2xco{s}^2-4x^{2}sin{x}^2,& \text{x≥0}\end{array}$
$\therefore L(gof)"(0)=-2$ and $R(gof)"(0)=2$
$\therefore L(gof)"(0)\ne R(gof)"(0)$
$\therefore gof(x)$ is not twice differentiable at $x=0.$