Question

$f\left(x\right)=\{\begin{array}{cc}x-1,& -1\le x<0\\ x^2,& 0\le x\le 1\end{array}$ and $g\left(x\right)=\sin x$. Consider the function $h_1\left(x\right)=f\left(|g\right(x\left)|\right)$ and $h_2\left(x\right)=|f\left(g\right(x\left)\right)|$
For when $h_1\left(x\right)$ and $h_2\left(x\right)$ are identical functions, then which of the following is not true?

• A. Domain of $h_1\left(x\right)$ and $h_2\left(x\right)$ is $[2n\pi ,(2n+1\left)\pi \right],n\in Z$
• B. Range of $h_1\left(x\right)$ and $h_2\left(x\right)$ is $[0,1]$
• C. Period of $h_1\left(x\right)$ and $h_2\left(x\right)$ is $\pi$
• D. None of these

Answer 1

The correct option is C Period of $h_1\left(x\right)$ and $h_2\left(x\right)$ is $\pi$

$|g\left(x\right)|=|\sin x|,\forall x\in R$
$f\left(|g\right(x\left)|\right)=\{\begin{array}{cc}|\sin x|-1,& -1\le |\sin x|<0\\ (|\sin x|{)}^2,& 0\le |\sin x|\le 1\end{array}\Rightarrow h_1\left(x\right)=f\left(|g\right(x\left)|\right)={\sin }^{2}x$

$f\left(g\right(x\left)\right)=\{\begin{array}{cc}\sin x-1,& -1\le \sin x<0\\ {\sin }^{2}x,& 0\le \sin x\le 1\end{array}\Rightarrow f\left(g\right(x\left)\right)=\{\begin{array}{cc}\sin x-1,& (2n-1)\pi <x<2n\pi \\ {\sin }^{2}x,& 2n\pi \le x\le (2n+1)\pi \end{array},n\in Z$
$\Rightarrow h_2\left(x\right)=|f\left(g\right(x\left)\right)|=\{\begin{array}{cc}1-\sin x,& (2n-1)\pi <x<2n\pi \\ {\sin }^{2}x,& 2n\pi \le x\le (2n+1)\pi \end{array},n\in Z$


For $h_1\left(x\right)=h_2\left(x\right)={\sin }^{2}x,x\in [2n\pi ,(2n+1\left)\pi \right],n\in Z$, and has range $[0,1]$ for the common domain.
Also, from the graph the period is $2\pi$