Question

Let $f\left(x\right)=x|x|$ and $g\left(x\right)=sinx$
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: (A)

Statement 1 is true, Statement 2 is true,Statement 2 is a correct explanation for Statement 1

$g\left(f\right(x\left)\right)=\sin \left(f\right(x\left)\right)=\{$ $\sin x^2,x\ge 0$
$-\sin x^2,x<0$

$\left(g\right(f\left(x\right)){)}^{\prime }=\{\begin{array}{cc}2x\cos x^2,& x\ge 0\\ -2x\cos x^2,& x<0\end{array}$
$gof$ is differentiable at $x=0$ and its derivative is continuous at that point.
R.H.D. of $(g\left(f\right(0\left)\right){)}^{\prime }=\underset{h\to 0^{+}}{lim}\frac{2h{\cosh }^2}{h}=2$
L.H.D. of $(g\left(f\right(0\left)\right){)}^{\prime }=\underset{h\to 0^{-}}{lim}\frac{2(-h){\cosh }^2}{-h}=2$
Clearly $gof$ is twice differentiable at $x=0$.
Therefore, Statement 1 is true, Statement 2 is true,Statement 2 is a correct explanation for Statement 1
Hence, option 'A' is correct.