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Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then, the set of all $x$ satisfying $($ fogogof $)(x)=(\operatorname{gogof})(x)$, where $(f \circ g)(x)=f(g(x))$ is
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$\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$
$\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$
$f(x)=x^2, g(x)=\sin x$
$(g \circ f)(x)=\sin x^2$
go(gof) $(x)=\sin \left(\sin x^2\right)$
$(f \circ g \circ g \circ f)(x)=\left(\sin \left(\sin x^2\right)\right)^2$
Again, $(g \circ f)(x)=\sin x^2$
(gogof) $(x)=\sin \left(\sin x^2\right)$
Given, $($ fogogof $)(x)=($ gogof $)(x)$
$\Rightarrow \quad\left(\sin \left(\sin x^2\right)\right)^2=\sin \left(\sin x^2\right)$
$\Rightarrow \sin \left(\sin x^2\right)\left\{\sin \left(\sin x^2\right)-1\right\}=0$
$\Rightarrow \sin \left(\sin x^2\right)=0$ or $\sin \left(\sin x^2\right)=1$
$\Rightarrow \sin x^2=0$ or $\sin x^2=\frac{\pi}{2}$
$\therefore \quad x^2=n \pi$
(i.e. not possible as $-1 \leq \sin \theta \leq 1$ )
$(g \circ f)(x)=\sin x^2$
go(gof) $(x)=\sin \left(\sin x^2\right)$
$(f \circ g \circ g \circ f)(x)=\left(\sin \left(\sin x^2\right)\right)^2$
Again, $(g \circ f)(x)=\sin x^2$
(gogof) $(x)=\sin \left(\sin x^2\right)$
Given, $($ fogogof $)(x)=($ gogof $)(x)$
$\Rightarrow \quad\left(\sin \left(\sin x^2\right)\right)^2=\sin \left(\sin x^2\right)$
$\Rightarrow \sin \left(\sin x^2\right)\left\{\sin \left(\sin x^2\right)-1\right\}=0$
$\Rightarrow \sin \left(\sin x^2\right)=0$ or $\sin \left(\sin x^2\right)=1$
$\Rightarrow \sin x^2=0$ or $\sin x^2=\frac{\pi}{2}$
$\therefore \quad x^2=n \pi$
(i.e. not possible as $-1 \leq \sin \theta \leq 1$ )
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