Question

Let $f\left(x\right)=x^2$ and $g\left(x\right)=\sin x$ for all $x\in R$. Then the set of all $x$ satisfying $\left(fogogof\right)\left(x\right)=\left(gogof\right)\left(x\right)$, where $\left(fog\right)\left(x\right)=f\left(g\right(x\left)\right)$, is

• A. Option a missing
• B. Option c missing
• C. Option b missing
• D. $2n\pi ,n\in \{...,-2,-1,0,1,2,...\}$

Answer 1

The correct option is A Option a missing

We have,
$f\left(x\right)=x^2$ and $g\left(x\right)=\sin x,\forall x$
$\Rightarrow f\left(g\right(g\left(f\right(x\left)\right)\left)\right)=g\left(g\right(f\left(x\right)\left)\right)$
$\Rightarrow g\left(f\right(x\left)\right)=g\left(x^2\right)=\sin x^2$
$\Rightarrow g\left(g\right(f\left(x\right)\left)\right)=g\left(\sin x^2\right)=\sin \left(\sin x^2\right)$
$\Rightarrow f\left(g\right(g\left(f\right(x\left)\right)\left)\right)=\left(\sin \right(\sin x^2){)}^2$
$\Rightarrow (\sin \sin x^2{)}^2=\sin (\sin x^2)$
$\Rightarrow \sin \left(\sin x^2\right)=0$ or $\sin \left(\sin x^2\right)=1$
But $\sin \left(\sin x^2\right)=1$ is not possible hence $sin{x}^2=0$
$\Rightarrow x^2=n\pi$
$\Rightarrow x=\pm \sqrt{n}\pi ,n\in \{0,1,2,\mathrm{3...}\}$