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