Question

If $f\left(x\right)=x^2,g\left(x\right)=\sin x,\forall x\in R$ Then the set of values of $x$ satisfying $fogogof\left(x\right)=gogof\left(x\right)$ is

• A. $\pm \sqrt{n\pi },n\in W$
• B. $\pm \sqrt{n\pi },n\in N$
• C. $\frac{\pi }{2}+2n\pi ,n\in Z$
• D. $2n\pi ,n\in Z$

Answer 1

The correct option is A $\pm \sqrt{n\pi },n\in W$

Given $f\left(x\right)=x^2,g\left(x\right)=\sin x\Rightarrow gof\left(x\right)=\sin x^2$
$\Rightarrow gogof\left(x\right)=\text{sin}\left(\sin x^2\right)$
and $fogogof\left(x\right)=\text{si}{\text{n}}^2\left(\sin x^2\right)$

$\because fogogof\left(x\right)=gogof\left(x\right)$
$\Rightarrow \text{si}{\text{n}}^2\left(\sin x^2\right)=\sin \left(\sin x^2\right)$
$\Rightarrow \text{sin}\left(\sin x^2\right)\left[\sin \left(\sin x^2\right)-1\right]=0$
$\Rightarrow \sin \left(\sin x^2\right)=0\text{or}\sin \left(\text{sin}{x}^2\right)=1$
$\Rightarrow \text{sin}{x}^2=m\pi ,m\in Z\text{or}\sin x^2=\left(4k+1\right)\frac{\pi }{2},k\in Z$
$\therefore \sin x^2=0\because \sin x^2\in \left[-1,1\right]$
$\Rightarrow x^2=n\pi \Rightarrow x=\pm \sqrt{n\pi },n\in \left\{0,1,2,\dots ..\right\}$