Question

If \[f:R\to R,g:R\to R\] and \[h:R\to R\] are such that $f(x)={{x}^{2}},~g(x)=\tan x$ and $h(x)=\log x$ then the value of $(ho(gof))(x)$if $x=\sqrt{\dfrac{\pi }{4}}$ will be
A. $0$
B. $1$
C. $-1$
D. $\pi $

Answer 1

Hint: To solve this question, we will use the formula of composition of functions. We will rewrite the given function using composition of function and substitute the value of the functions and derive an equation. We will then put the value of $x$ in that equation and derive the result.

Formula Used: If there are two functions $f$and $g$, then the composite function $gof(x)$ can be written as $gof(x)=g(f(x))$.

Complete step by step solution: We are given three functions $f(x)={{x}^{2}},~g(x)=\tan x$, $h(x)=\log x$ and we have to determine the value of $(ho(gof))(x)$ if $x=\sqrt{\dfrac{\pi }{4}}$.
We will rewrite the given equation,
$(ho(gof))(x)=h(g(f(x)))$
We will now substitute the value of the function $f(x)={{x}^{2}}$ in the above equation.
$\begin{align}
  & (ho(gof))(x)=h(g(f(x))) \\
 & =h(g({{x}^{2}}))
\end{align}$
We will now substitute the value of the function$g(x)=\tan x$ in place of $x$ in the above equation.
$\begin{align}
  & (ho(gof))(x)=h(g({{x}^{2}})) \\
 & =h(\tan {{x}^{2}})
\end{align}$
Now we will substitute the value of the function $h(x)=\log x$ in the above equation.
$\begin{align}
  & (ho(gof))(x)=h(\tan {{x}^{2}}) \\
 & =\log (\tan {{x}^{2}})
\end{align}$
We will now substitute the value $x=\sqrt{\dfrac{\pi }{4}}$ in the above equation.
\[\begin{align}
  & (ho(gof))(x)=\log (\tan {{x}^{2}}) \\
 & =\log \left( \tan {{\left( \sqrt{\dfrac{\pi }{4}} \right)}^{2}} \right)
\end{align}\]
\[\begin{align}
  & (ho(gof))(x)=\log \tan \dfrac{\pi }{4} \\
 & =\log 1 \\
 & =0
\end{align}\]

The value of $(ho(gof))(x)$ is $0$ if \[f:R\to R,g:R\to R\] and \[h:R\to R\] such that $f(x)={{x}^{2}},~g(x)=\tan x$, $h(x)=\log x$ and $x=\sqrt{\dfrac{\pi }{4}}$. Hence the correct option is (A).

Note: The composite functions can be defined as the combination of one or two functions to result in another function. If we have two functions $f(x)$ and $g(x)$, then the function inside the brackets will be the input for the function outside of the brackets. That is,
In composite function $f(g(x))$, $g(x)$ will be the input for the function $f(x)$ and in function $g(f(x))$,~$f(x)$ will be the input for the function $g(x)$.
The function $f(g(x))$ can be also shown as $fog(x)$.