If f(x) = cos x,0 le x le {pi over 2}, then t | vedclass

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Question

(c) We know that $f'(c) = \frac{{f(b) - f(a)}}{{b - a}}$

$ \Rightarrow f'(c) = \frac{{0 - 1}}{{\pi /2}} = - \frac{2}{\pi }$.....$(i)$

Bu

Vedclass · Accepted Answer

${\sin ^{ - 1}}\left( {{2 \over \pi }} \right)$

Vedclass · Answer

(c) 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)$.

If $f(x) = \cos x,0 \le x \le {\pi \over 2}$, then the real number $ ‘c’ $ of the mean value theorem is

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