Discussion Forum



1)

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

Answer:

Option B

Explanation:

 $f(x)=x^{2},g(x)= \sin x$

 $(g o f)(x)= \sin x^{2}$

 $g o(gof)(x)=\sin(\sin x^{2})$

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