Question

If $f\left(x\right)=\sin \left\{\begin{array}{c}[x+5]+\left\{\begin{array}{c}x-\left\{\begin{array}{c}x-\left\{\begin{array}{c}x\end{array}\right\}\end{array}\right\}\end{array}\right\}\end{array}\right\}$ for $x\in \left(\begin{array}{c}0,\frac{\pi }{4}\end{array}\right)$ is invertible, where $\left\{\begin{array}{c}.\end{array}\right\}$ and $[.]$ represent fractional part and greatest integer functions respectively, then $f^{-1}\left(x\right)$ is

• A. ${\sin }^{-1}x$
• B. $\frac{\pi }{2}-{\cos }^{-1}x$
• C. ${\sin }^{-1}\left\{\begin{array}{c}x\end{array}\right\}$
• D. ${\cos }^{-1}\left\{\begin{array}{c}x\end{array}\right\}$

Answer 1

Answer: (A), (B), (C)

$\frac{\pi }{2}-{\cos }^{-1}x$
${\sin }^{-1}\left\{\begin{array}{c}x\end{array}\right\}$
${\sin }^{-1}x$

$x-\left\{\begin{array}{c}x\end{array}\right\}=\left[x\right]$

$\Rightarrow \left\{\begin{array}{c}x-\left\{\begin{array}{c}x\end{array}\right\}\end{array}\right\}=0$
$\Rightarrow \left\{\begin{array}{c}x-\left\{\begin{array}{c}x-\left\{\begin{array}{c}x\end{array}\right\}\end{array}\right\}\end{array}\right\}=\left\{\begin{array}{c}x\end{array}\right\}$
$\Rightarrow \sin \left\{\begin{array}{c}[x+5]+\left\{\begin{array}{c}x-\left\{\begin{array}{c}x-\left\{\begin{array}{c}x\end{array}\right\}\end{array}\right\}\end{array}\right\}\end{array}\right\}=\sin \left\{\right[x]+5+\{x\left\}\right\}$
$\therefore f\left(x\right)=\sin \left\{x\right\}=\sin x,\because x\in (0,\frac{\pi }{4})$
$\therefore f^{-1}\left(x\right)={\sin }^{-1}x=\frac{\pi }{2}-{\cos }^{-1}x={\sin }^{-1}\left\{\begin{array}{c}x\end{array}\right\}$