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

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5. Continuity and Differentiation

easy

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

A

${\pi \over 6}$

B

${\pi \over 4}$

C

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

D

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

Solution

(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)$.

Standard 12

Mathematics

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