Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then, the set of all $x$ satisfying $($ fogogof…

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
  1. $\pm \sqrt{n \pi}, n \in\{0,1,2, \ldots\}$
  2. $\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$
  3. $\frac{\pi}{2}+2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$
  4. $2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$

Solution

$f(x)=x^2, g(x)=\sin x$ $(g \circ f)(x)=\sin x^2$ 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 $-1 \leq \sin \theta \leq 1$ )

Asked in: JEE Advanced 2011 (Paper 2)

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