Question

If $f\left(x\right)=|\log 2-\sin x|$ and $g\left(x\right)=f\left(f\right(x\left)\right),$ where $x\in R$, then

• A. $g\left(x\right)$ is not differentiable at $x=0$
• B. $g\left(x\right)$ is differentiable at $x=0$ and $g^{\prime }\left(0\right)=\cos \left(\log 2\right)$
• C. $g\left(x\right)$ is differentiable at $x=0$ and $g^{\prime }\left(0\right)=-\cos \left(\log 2\right)$
• D. $g\left(x\right)$ is differentiable at $x=0$ and $g^{\prime }\left(0\right)=-\sin \left(\log 2\right)$

Answer 1

The correct option is B $g\left(x\right)$ is differentiable at $x=0$ and $g^{\prime }\left(0\right)=\cos \left(\log 2\right)$

Given : $f\left(x\right)=|\log 2-\sin x|$ and $g\left(x\right)=f\left(f\right(x\left)\right)$
In the neighorhood of $x=0$, $\sin x<\log 2$, so
$f\left(x\right)=\log 2-\sin x\Rightarrow g\left(x\right)=\log 2-\sin f\left(x\right)\Rightarrow g\left(x\right)=\log 2-\sin (\log 2-\sin x)$
So, $g\left(x\right)$ is differentiable at $x=0$
Now,
$g^{\prime }\left(x\right)=f^{\prime }\left(f\right(x\left)\right)\times f^{\prime }\left(x\right)\Rightarrow g^{\prime }\left(0\right)=f^{\prime }\left(f\right(0\left)\right)\times f^{\prime }\left(0\right)\Rightarrow g^{\prime }\left(0\right)=-\cos \left(\log 2\right)\times (-\cos 0)\therefore g^{\prime }\left(0\right)=\cos \left(\log 2\right)$