General Questions | JEE 2011 | Discussion Page

Let f(x)= x² and g(x)= $\sin x $ for all x $\epsilon$ R. Then, the set of all x satisfying (fogogof)(x) = (gogof)(x), where (fog)(x) = f(g(x))is

A) $\pm \sqrt{n \pi},n\epsilon\left\{0,1,2....\right\}$

B) $\pm \sqrt{n \pi},n\epsilon\left\{1,2....\right\}$

C) $\frac{\pi}{2}+2 n\pi ,n\epsilon\left\{.....-2,-1,0,1,2....\right\}$

D) $2 n\pi ,n\epsilon\left\{.....-2,-1,0,1,2....\right\}$

Option B

$f(x)=x^{2},g(x)= \sin x$

$(g o f)(x)= \sin x^{2}$

$g o(gof)(x)=\sin(\sin x^{2})$

$(f ogogof)(x)=(\sin(\sin x^{2}))^{2}$.......(i)

Again , $(gof)(x)=\sin x^{2}$

$(gogof)(x)=\sin (\sin x^{2})$.....(ii)

Given, $(fogogof)(x)=(gogof)(x)$

$\Rightarrow$ $(\sin (\sin x^{2}))^{2}= \sin (\sin x^{2})$

$\Rightarrow$ $\sin (\sin x^{2}) ( \sin (\sin x^{2})-1)=0$

$\Rightarrow$ $\sin (\sin x^{2})=0 $ or $\sin (\sin x^{2})=1$

$\Rightarrow$ $\sin x^{2}=0$ or $\sin x^{2}=\frac{\pi}{2}$

$\therefore$ $x^{2}=n \pi$

(i.e, not possible as $ -1 \leq \sin \theta \leq1)$

$ x=\pm \sqrt{n \pi}$

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