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 $x$ satisfying $(f\circ g\circ g\circ f)\left(x\right)=(g\circ g\circ f)\left(x\right)$ , where $(f\circ g)\left(x\right)=f\left(g\right(x\left)\right)$ , is

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

Answer 1

The correct option is A $\pm \sqrt{n\pi },n\in \{0,1,2,\dots \}$

Given, $f\left(x\right)=x^2$ for all $x\in R$

$g\left(x\right)=\sin x$ for all $x\in R$

Now, $\left(fogogof\right)\left(x\right)=f\left(g\right(g\left(f\right(x\left)\right))$

$=f\left(g\right(g\left(x^2\right)\left)\right)$

$=f\left(g\right(\sin x^2\left)\right))$

$=f\left(\sin \sin x^2\right)$

$\left(fogogof\right)\left(x\right)={\left(\sin \sin x^2\right)}^2$

Now, $\left(gogof\right)\left(x\right)=g\left(g\right(f\left(x\right))$

$=g\left(g\right(x^2\left)\right)$

$=g\left(\sin x^2\right)$

$\left(gogof\right)\left(x\right)=\sin \sin x^2$

Given ,$\left(fogogof\right)\left(x\right)=\left(gogof\right)\left(x\right)$

So, ${\left(\sin \sin x^2\right)}^2=\sin \sin x^2$

${\left(\sin \sin x^2\right)}^2-\sin \sin x^2=0$

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

So, $\sin \sin x^2=0$ or $\sin \sin x^2=1\Rightarrow \sin x^2=\frac{\pi }{2}$ it is not possible

$\sin x^2=0$ implies $x^2=n\pi$ or $x=\pm \sqrt{n\pi },n\in \{0,1,2,\mathrm{3.....}\}$

$x=\pm \sqrt{n\pi }n\in \{0,1,2,\mathrm{3.....}\}$