Question

Assertion :f(x) = $\left|x\right|.\sin x$ is differentiable at $x=0$ Reason: If f(x) is not differentiable and g(x) is differentiable at $x=a$ then f(x). g(x) will be differentiable at $x=a$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

$f\left(x\right)=|x|\sin x=\{\begin{array}{c}-x\sin x,x<0\\ x\sin x,x\ge 0\end{array}$
$f^{\prime }\left(x\right)=\{\begin{array}{c}-\sin x-x\cos x,x<0\\ \sin x+x\cos x,x\ge 0\end{array}$
Clearly at $x=0$, L.H.D $=0=$ R.H.D. Hence $f\left(x\right)$ is differentiable at $x=0$
Reason is not correct. Example $f\left(x\right)=[x-1],g\left(x\right)=x^2$ where $[\ast ]$ represents GIF.