Question

If $f\left(x\right)=2x+|x|,g\left(x\right)=\frac{1}{3}(2x-|x|)$ and $h\left(x\right)=f\left(g\right(x\left)\right)$, then domain of ${\sin }^{-1}\underset{\text{n times}}{\underset{}{\left(h\right(h\left(h\right(h.....h\left(x\right).....\left)\right)\left)\right)}}$ is

• A. $[-1,1]$
• B. $[-1,-\frac{1}{2}]\cup [\frac{1}{2},1]$
• C. $[-1,-\frac{1}{2}]$
• D. $[\frac{1}{2},1]$

Answer 1

The correct option is A $[-1,1]$

Since, $f\left(x\right)=\{\begin{array}{cc}2x+x,& x\ge 0\\ 2x-x,& x<0\end{array}=\{\begin{array}{cc}3x,& x\ge 0\\ x,& x<0\end{array}$
and $g\left(x\right)=\frac{1}{3}\{\begin{array}{cc}2x-x,& x\ge 0\\ 2x+x,& x<0\end{array}=\{\begin{array}{cc}\frac{x}{3},& x\ge 0\\ x,& x<0\end{array}$
$\therefore f\left(g\right(x\left)\right)=\{\begin{array}{cc}3\left(\frac{x}{3}\right)& x\ge 0\\ x& x<0\end{array}$
$\Rightarrow f\left(g\right(x\left)\right)=x,\forall x\in R$
$\therefore h\left(x\right)=x$
$\Rightarrow {\sin }^{-1}\left(h\right(h(h....h(x)....)\left)\right)={\sin }^{-1}x$
$\therefore$ Domain of ${\sin }^{-1}\left(h\right(h\left(h\right(h.....h\left(x\right)....\left)\right)\left)\right)$ is $[-1,1]$
Hence, A is the correct answer.