question_answer

The area of the portion common to \[y={{\sin }^{-\,1}}(\sin x)\] and \[y=[{{\sin }^{-1}}(\sin x)]\] in \[[0,\pi ],\]here \[[\cdot ]\]denotes the greatest integer function is

A) \[{{(\pi -2)}^{2}}\]                  

B) \[\frac{{{(\pi -2)}^{2}}}{2}\]

C) \[\frac{{{(\pi -2)}^{2}}}{4}\]   

D) None of these

Correct Answer: C

Solution :

[c]

We have, \[y={{\sin }^{-\,1}}(\sin x)=x\sin [0,\pi /2]\]\[=\pi -x\sin [\pi /2,\pi ]\]

\[y=[{{\sin }^{-\,1}}(\sin x)=0\]in \[[0,1)\]\[=1\]in \[[1,\pi -1)\]\[=0\]in \[[\pi -1,\pi ]\]

Required area \[=\frac{1}{2}(\pi -2)\left( \frac{\pi }{2}-1 \right)\]\[=\frac{{{(\pi -2)}^{2}}}{4}\]