Question

If \[f(x)=2x+|x|\],~\[g(x)=\dfrac{1}{3}(2x-|x|)\] and $h(x)=f(g(x))$ then domain of ${{\sin }^{-1}}\underbrace{(h(h(h(h.....h(x).....))))}_{ntimes}$ is
A. $\left[ -1,1 \right]$
B. \[\left[ -1,-\dfrac{1}{2} \right]\cup \left[ \dfrac{1}{2},1 \right]\]
C. \[\left[ -1,-\dfrac{1}{2} \right]\]
D. \[\left[ \dfrac{1}{2},1 \right]\]

Answer 1

Hint: To solve this question, we will consider both the functions \[f(x)=2x+|x|\] and \[g(x)=\dfrac{1}{3}(2x-|x|)\] and write it as a piecewise function. Then we will derive the value of function $h(x)=f(g(x))$ by using both the functions. After finding the value of $h(x)$, we will substitute its value in ${{\sin }^{-1}}\underbrace{(h(h(h(h.....h(x).....))))}_{ntimes}$ and determine its domain.

Complete step by step solution: We are given functions \[f(x)=2x+|x|\],\[g(x)=\dfrac{1}{3}(2x-|x|)\],$h(x)=f(g(x))$ and we have to determine the domain of ${{\sin }^{-1}}\underbrace{(h(h(h(h.....h(x).....))))}_{ntimes}$.
As we know that absolute function is defined as a piecewise function with sub-functions on different intervals we will define the functions \[f(x)=2x+|x|\] and \[g(x)=\dfrac{1}{3}(2x-|x|)\] as a piecewise function.
We have absolute value $|x|$, so we will write both the functions as \[f(x)=\left\{ \begin{matrix}
   2x+x,x\ge 0 \\
   2x-x,x<0 \\
\end{matrix} \right.=\left\{ \begin{matrix}
   3x,x\ge 0 \\
   x,x<0 \\
\end{matrix} \right.\] and \[g(x)=\dfrac{1}{3}\left\{ \begin{matrix}
   2x-x,x\ge 0 \\
   2x+x,x<0 \\
\end{matrix} \right.=\left\{ \begin{matrix}
   \dfrac{x}{3},x\ge 0 \\
   x,x<0 \\
\end{matrix} \right.\].
So the function $h(x)=f(g(x))$ can be written as,

$\begin{align}
  & h(x)=f(g(x)) \\
 & =f\left( \left\{ \begin{matrix}
   \dfrac{x}{3},x\ge 0 \\
   x,x<0 \\
\end{matrix} \right. \right)
\end{align}$
$\begin{align}
  & h(x)=\left\{ \begin{matrix}
   3\left( \dfrac{x}{3} \right),x\ge 0 \\
   x,x<0 \\
\end{matrix} \right. \\
 & =\left\{ \begin{matrix}
   x,x\ge 0 \\
   x,x<0 \\
\end{matrix} \right.
\end{align}$
So we can say that the function $h(x)=x$ as $f(g(x))=x\forall x\in R$.
We will put the value of function $h(x)$ in ${{\sin }^{-1}}\underbrace{(h(h(h(h.....h(x).....))))}_{ntimes}$. So,
$\begin{align}
  & ={{\sin }^{-1}}\underbrace{(h(h(h(h.....h(x).....))))}_{ntimes} \\
 & ={{\sin }^{-1}}x \\
\end{align}$
Now we know that the domain is the set of all the input numbers which is accepted by that function and the domain of the trigonometric function sine inverse is $\left[ -1,1 \right]$.
So the domain of ${{\sin }^{-1}}\underbrace{(h(h(h(h.....h(x).....))))}_{ntimes}$ will also be $\left[ -1,1 \right]$.
The domain of ${{\sin }^{-1}}\underbrace{(h(h(h(h.....h(x).....))))}_{ntimes}$ is $\left[ -1,1 \right]$ when \[f(x)=2x+|x|\],\[g(x)=\dfrac{1}{3}(2x-|x|)\] and $h(x)=f(g(x))$.

So, Option ‘A’ is correct

Note: The modulus function $|x|$ which is also termed as an absolute value function gives the non-negative magnitude of a number. If we have a modulus function $f(x)=|x|$ then if the value of $x$ is negative then the function will give the same magnitude as $x$ that is $f(x)=-x$ and if $x$ is positive then the function will have a similar value as $x$ that is $f(x)=x$.
It is depicted as\[f(x)=|x|=\left\{ \begin{matrix}
   x,x\ge 0 \\
   -x,x<0 \\
\end{matrix} \right.\].