Question

If $f(x) = {x^2}$ and $g(x) = \sin x$ $x \in \mathbb{R}$. Then the set of all $x$ satisfying $\left( {fogogof} \right)\left( x \right) = \left( {gogof} \right)\left( x \right)$ where $fog(x) = f(g(x))$ is
A. \pm \sqrt {n\pi } ,n \in \left\\{ {0,1,2...} \right\\}
B. \pm \sqrt {n\pi } ,n \in \left\\{ {1,2...} \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

Find the required composition of the functions correctly by solving for the Left Hand Side and the Right Hand Side separately. Equate both the sides and find the solution set for the values of x accordingly. This question requires us to have the knowledge of basic and simple algebraic rules and operations such as substitution, addition, multiplication, subtraction and many more like these.

Complete answer: The composition of a function is an operation where two functions say $f$and $g$generate a new function say $h$in such a way that $h(x) = f(g(x))$ . It means here the function $g$ is applied to the function $f$ of$x$. So, basically, a function is applied to the result of another function.
The order of function is an important thing while dealing with the composition of functions since$fog(x) \ne gof(x)$.
Symbol: It is also denoted as $gof(x)$where $o$is a small circle symbol. We cannot replace $o$ with a dot (.) because it will show as the product of two functions, such as$f.g(x)$.
Domain: $f(g(x))$is read as $f$ of $g$ of $x$ . In the composition of $fog(x)$ the domain of function $f$ becomes $g(x)$ . The domain is a set of all values which go into the function.
The function composition of one-to-one function is always one to one.
The function composition of two onto function is always onto
The inverse of the composition of two functions $f$and $g$is equal to the composition of the inverse of both the functions.

We are given $f(x) = {x^2}$
$g(x) = \sin x$
Now we have $\left( {fogogof} \right)(x) = f(g(g(f(x))))$
Putting value of function $f$
$= f(g(g({x^2})))$
Putting value of function $g$
$= f(g(\sin {x^2}))$
Putting value of function $g$
$= f(\sin (\sin {x^2}))$
Putting value of function $f$
$= {\sin ^2}(\sin {x^2})$
Similarly we have $\left( {gogof} \right)(x) = g(g(f(x)))$
Putting value of function $f$
$= g(g({x^2}))$
Putting value of function $g$
$= g(\sin {x^2})$
Putting value of function $g$
$= \sin (\sin {x^2})$
Therefore we have $\left( {fogogof} \right)\left( x \right) = {\sin ^2}\left( {\sin {x^2}} \right)$ and $\left( {gogof} \right)\left( x \right) = \sin \left( {\sin {x^2}} \right)$
We are given that $\left( {fogogof} \right)\left( x \right) = \left( {gogof} \right)\left( x \right)$
Therefore we get ${\sin ^2}\left( {\sin {x^2}} \right) = \sin \left( {\sin {x^2}} \right)$
Taking all the terms on one side we get ${\sin ^2}\left( {\sin {x^2}} \right) - \sin \left( {\sin {x^2}} \right) = 0$
$\sin \left( {\sin {x^2}} \right)\left[ {\sin \left( {\sin {x^2}} \right) - 1} \right] = 0$
Therefore we get $\sin \left( {\sin {x^2}} \right) = 0$ or $\sin \left( {\sin {x^2}} \right) = 1$
Therefore we get $\sin {x^2} = n\pi$ or $\sin {x^2} = 2m\pi + \dfrac{\pi }{2}$
Where $m,n \in \mathbb{Z}$
Therefore $\sin {x^2} = 0$
${x^2} = n\pi$ and hence x = \pm \sqrt {n\pi } ,n \in \left\\{ {0,1,2...} \right\\}
Therefore option (A) is the correct answer.

Note:
Composition of a function is a function that is applied to the result of another function. The order of function is an important thing while dealing with the composition of functions since $fog(x) \ne gof(x)$. We cannot replace $o$ with a dot (.) because it will show as the product of two functions, such as $f.g(x)$.