Question

Let $f ( x )= x ^{2}$ and $g ( x )=\sin x$ for all $x \in R$. Then the set of all $x$ satisfying $( f o g o g o f )( x )=( g o g o f )( x )$ where $( f o g )( x )= f ( g ( x ))$, is

• A. $\pm \sqrt{n \pi}, n \in\{0,1,2, \ldots\}$
• B. $\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$
• C. $\frac{\pi}{2}+2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots .\}$
• D. $2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$

Answer 1

Answer: (A)

$($ fogogof $)(x)=\sin ^{2}\left(\sin x^{2}\right)$
$($ gogof $)(x)=\sin \left(\sin x^{2}\right)$
$\therefore \sin ^{2}\left(\sin x^{2}\right)=\sin \left(\sin x^{2}\right)$
$\Rightarrow \sin \left(\sin x^{2}\right)\left[\sin \left(\sin x^{2}\right)-1\right]=0$
$\Rightarrow \sin \left(\sin x^{2}\right)=0$ or 1
$\Rightarrow \sin x^{2}=n \pi$ or $2 m \pi+\pi / 2$, where $m, n \in I$
$\Rightarrow \sin x^{2}=0$
$\Rightarrow x^{2}=n \pi \Rightarrow x=\pm \sqrt{n \pi}, n \in\{0,1,2, \ldots\}$