For $x \in R, f(x)=|\log 2-\sin x|$ and $g(x)=f(f(x))$, then

For $x \in R, f(x)=|\log 2-\sin x|$ and $g(x)=f(f(x))$, then
  1. $g^{\prime}(0)=-\cos (\log 2)$
  2. $g$ is not differentiable at $x=0$.
  3. $g^{\prime}(0)=\cos (\log 2)$
  4. $g$ is differentiable at $x=0$ and $g^{\prime}(0)=-\sin (\log 2)$.

Solution

for $x \rightarrow 0^{-}$ $g(x)=\log 2-\sin (\log 2-\sin x)$ for $x \rightarrow 0^{+}$ $g(x)=\log 2-\sin (\log 2-\sin x)$ [as $\log 2>\sin 0$ and $x>\sin x$ ] and $g(x)$ is continuous at $x=0$ $\begin{aligned} & \text { now } g^{\prime}(x)=0-\cos (\log 2-\sin x)(0-\cos x) \\ & \Rightarrow g^{\prime}(0)=\cos (\log 2)\end{aligned}$

Asked in: MHT CET 2022 (10 Aug Shift 2)

Practice more Differentiation questions on Aicharya