Question

If $f\left(x\right)=x^2$ and $g\left(x\right)=\sin x,\forall x\in R$, then the set of all $x$ satisfying $\left(f\circ g\circ g\circ f\right)\left(x\right)=\left(g\circ g\circ f\right)\left(x\right)$, where $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ is

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

Answer 1

The correct option is A

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

Explanation for the correct option.

Step 1. Find the value of $\left(f\circ g\circ g\circ f\right)\left(x\right)$ and $\left(g\circ g\circ f\right)\left(x\right)$.

For functions $f\left(x\right)=x^2$ and $g\left(x\right)=\sin x$, the composite function $\left(f\circ g\circ g\circ f\right)\left(x\right)$ is given as:

$\begin{array}{rcl}\left(f\circ g\circ g\circ f\right)\left(x\right)& =& f\left(g\left(g\left(f\left(x\right)\right)\right)\right)\\ & =& f\left(g\left(g\left(x^2\right)\right)\right)\\ & =& f\left(g\left(\sin x^2\right)\right)\\ & =& f\left(\sin \left(\sin x^2\right)\right)\\ & =& {\sin }^2\left(\sin x^2\right)\end{array}$

And the function $\left(g\circ g\circ f\right)\left(x\right)$ is given as:

$\begin{array}{rcl}\left(g\circ g\circ f\right)\left(x\right)& =& g\left(g\left(f\left(x\right)\right)\right)\\ & =& g\left(g\left(x^2\right)\right)\\ & =& g\left(\sin x^2\right)\\ & =& \sin \left(\sin x^2\right)\end{array}$

Step 2. Find the set of all $x$ satisfying $\left(f\circ g\circ g\circ f\right)\left(x\right)=\left(g\circ g\circ f\right)\left(x\right)$.

The equation $\left(f\circ g\circ g\circ f\right)\left(x\right)=\left(g\circ g\circ f\right)\left(x\right)$ can be written as:

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

So either $\sin \left(\sin x^2\right)=0$ or $\sin \left(\sin x^2\right)=1$.

For $\sin \left(\sin x^2\right)=0$, and so $\sin x^2=n\mathrm{\pi }$ where $n\in I$. But $\sin x^2$ cannot be greater than $1$, so $\sin x^2=0$ is only accepted.

For $\sin x^2=0$, $x^2=n\mathrm{\pi }$, where $n\in I^{+}$ and thus $x=\pm \sqrt{n\mathrm{\pi }},n\in \left\{0,1,2,...\right\}$.

Again for $\sin \left(\sin x^2\right)=1$, and so $\sin x^2=\frac{n\mathrm{\pi }}{2}$ . But $\sin x^2$ cannot be greater than $1$, and $\frac{n\mathrm{\pi }}{2}>1$ is always true, so this has no accpeted solution.

Thus, for $x=\pm \sqrt{n\mathrm{\pi }},n\in \left\{0,1,2,...\right\}$, the equation $\left(f\circ g\circ g\circ f\right)\left(x\right)=\left(g\circ g\circ f\right)\left(x\right)$ is true.

Hence, the correct option is A.