A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[{{\sin }^{-1}}\left( \frac{2}{\pi } \right)\]
D) \[{{\cos }^{-1}}\left( \frac{2}{\pi } \right)\]
Correct Answer: C
Solution :
We know that \[f'(c)=\frac{f(b)-f(a)}{b-a}\] \[\Rightarrow f'(c)=\frac{0-1}{\pi /2}=-\frac{2}{\pi }\] .....(i) But \[f'(x)=-\sin x\Rightarrow f'(c)=-\sin c\] ....(ii) From (i) and (ii), we get \[-\sin c=-\frac{2}{\pi }\Rightarrow c={{\sin }^{-1}}\left( \frac{2}{\pi } \right)\].
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