For xR, f(x)=|2- x| and g(x)=f(f(x)), then

Question

For $x\in\mathbb{R}$, $f(x)=|\log2-\sin x|$ and $g(x)=f(f(x))$, then

Accepted Answer

$$g\left( x \right) = f\left( {f\left( x \right)} \right)$$

In the neighbourhood of $x=0,$

$$f\left( x \right) = \left| {\log 2 - \sin \,x} \right| = \left( {\log 2 - \sin x} \right)$$

$$\therefore$$ $$g\left( x \right) = \left| {\log 2 - \sin \left. {\left| {\log 2 - \sin x} \right|} \right|} \right.$$

$$ = \left( {\log 2 - \sin \left( {\log 2 - \sin x} \right)} \right)$$

$$\therefore$$ $g(x)$ is differentiable

and $$g'\left( x \right) =  - \cos \left( {\log 2 - \sin x} \right)\left( { - \cos x} \right)$$

$$ \Rightarrow g'\left( 0 \right) = \cos \left( {\log 2} \right)$$

Solutions · 1