Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)=\cos x, 0 \leq x \leq \frac{\pi}{2}$, then the real number ' $c$ ' of the mean value theorem is
Options:
Solution:
2587 Upvotes
Verified Answer
The correct answer is:
$\sin ^{-1}\left(\frac{2}{\pi}\right)$
We know that
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
$\Rightarrow f^{\prime}(c)=\frac{0-1}{\pi / 2}=-\frac{2}{\pi}$
But $f^{\prime}(x)=-\sin x \Rightarrow f^{\prime}(c)=-\sin c$
From (i) and (ii), we get
$-\sin c=-\frac{2}{\pi} \Rightarrow c=\sin ^{-1}\left(\frac{2}{\pi}\right)$
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
$\Rightarrow f^{\prime}(c)=\frac{0-1}{\pi / 2}=-\frac{2}{\pi}$
But $f^{\prime}(x)=-\sin x \Rightarrow f^{\prime}(c)=-\sin c$
From (i) and (ii), we get
$-\sin c=-\frac{2}{\pi} \Rightarrow c=\sin ^{-1}\left(\frac{2}{\pi}\right)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.