Question

Let $f\left( x \right)={{x}^{2}}$ and $g\left( x \right)=\sin x$ for all $x\in \mathbb{R}$. 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 $\left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right)$.
A. $\pm \sqrt{n\pi },n\in \left\{ 0,1,2,3,... \right\}$
B. $\pm \sqrt{n\pi },n\in \left\{ 1,2,3,... \right\}$
C. $\dfrac{\pi }{2}+2n\pi ,n\in \left\{ .....,-2,-1,0,1,2,... \right\}$
D. $2n\pi ,n\in \left\{ .....,-2,-1,0,1,2,... \right\}$

Answer 1

Hint: We first simplify the equation and use the concept of composite function to find the actual equation of $\left( f\circ g\circ g\circ f \right)\left( x \right)=\left( g\circ g\circ f \right)\left( x \right)$. We find the range of the function and try to find the right option for the domain.

Complete step by step solution:
It’s given that $f\left( x \right)={{x}^{2}}$ and $g\left( x \right)=\sin x$, $\left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right)$.
Similarly, $\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)$.
Now putting the different function, we get $\left( f\circ f \right)\left( x \right)=f\left( f\left( x \right) \right)$ and $\left( g\circ g \right)\left( x \right)=g\left( g\left( x \right) \right)$.
Putting the values of the functions we get $\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)=\sin {{x}^{2}}$.
Now, $\left( g\circ g\circ f \right)\left( x \right)=g\left( g\left( f\left( x \right) \right) \right)=\sin \left( \sin {{x}^{2}} \right)$.
$\left( f\circ g\circ g\circ f \right)\left( x \right)=f\left[ g\left\{ g\left( f\left( x \right) \right) \right\} \right]={{\sin }^{2}}\left( \sin {{x}^{2}} \right)$.
This relation $\left( f\circ g\circ g\circ f \right)\left( x \right)=\left( g\circ g\circ f \right)\left( x \right)$ gives
${{\sin }^{2}}\left( \sin {{x}^{2}} \right)=\sin \left( \sin {{x}^{2}} \right)$.
If we take ${{\sin }^{2}}\left( \sin {{x}^{2}} \right)$, the value will be in between $\left[ 0,1 \right]$.
We can see that for any value of $x=\pm \sqrt{n\pi },n\in \left\{ 0,1,2,3,... \right\}$, the equation satisfies the relation.
The inner bracket part becomes 0 for any value of $n\in \left\{ 0,1,2,3,... \right\}$.
Using the same theory of the value being equal to 0, the second and fourth options also satisfy the equation as the sin curve becomes 0 for any integer multiple of $\pi $.
Mathematically we can express it this way that for any value of $x=n\pi ,n\in \mathbb{Z}$, the value of $\sin x=0$.
We can say that $\pm \sqrt{n\pi },n\in \left\{ 1,2,3,... \right\}$ is a subset of $\pm \sqrt{n\pi },n\in \left\{ 0,1,2,3,... \right\}$.
Again, the given domain of $2n\pi ,n\in \left\{ .....,-2,-1,0,1,2,... \right\}$ satisfies the equation of ${{\sin }^{2}}\left( \sin {{x}^{2}} \right)=\sin \left( \sin {{x}^{2}} \right)$.
Therefore, the correct options are A, B, D.

Note: A composite function is possible (or exists) if the range of the first function is a subset of the domain of the second function. If this is not the case then it is obvious from the flow chart above that the link between the two functions will be broken.