Question

Let $f\left(x\right)=x^2$ and $g\left(x\right)=\sin x$ for all $x\in R$. Then the set of all $x$ satifying $(f\circ g\circ g\circ f)\left(x\right)=(g\circ g\circ f)\left(x\right)$, where $(f\circ g)\left(x\right)=f\left(g\right(x\left)\right)$, is

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

Answer 1

The correct option is A

$\pm \sqrt{n\pi ,}n\in 0,1,2,...$

Given that $f\left(x\right)=x^2$ and $g\left(x\right)=\sin x,\forall x\in R$
Then $(g\circ f)\left(x\right)=\sin x^2$
$\Rightarrow (g\circ g\circ f)\left(x\right)=\sin \left(\sin x^2\right)$
As given that $(f\circ g\circ g\circ f)\left(x\right)=(g\circ g\circ f)\left(x\right)$
$\Rightarrow {\sin }^2\left(\sin x^2\right)=\sin \left(\sin x^2\right)\Rightarrow \sin \left(\sin x^2\right)=0,1$
$\Rightarrow \sin x^2=n\pi$ or $(4n+1)\frac{\pi }{2}$ where $n\in Z$
$\Rightarrow \sin x^2=0\because \sin x^2\in [-1,1]\Rightarrow x^2=n\pi$
$\Rightarrow x=\pm \sqrt{n\pi }$ where $n\in W$