Question

Find the values of $c$ that satisfy the MVT for integrals on $[\frac{3\pi }{4},\pi ]$.
$f\left(x\right)=\cos (2x-\pi )$

• A. $c=\frac{5\pi }{2}-\frac{1}{2}{\cos }^{-1}(-\frac{2}{\pi })$
• B. $c=\pi -\frac{1}{2}{\cos }^{-1}(-\frac{2}{\pi })$
• C. $c=\frac{\pi }{2}+\frac{1}{2}{\sin }^{-1}(-\frac{2}{\pi })$
• D. $c=\frac{\pi }{2}+\frac{1}{2}{\sin }^{-1}\left(\frac{2}{\pi }\right)$

Answer 1

The correct option is D $c=\frac{\pi }{2}+\frac{1}{2}{\sin }^{-1}\left(\frac{2}{\pi }\right)$

Given $f\left(x\right)=cos(2x-\pi )$ and $a=\frac{3\pi }{4}$ and $b=\pi$.

So $f\left(a\right)=0$ and $f\left(b\right)=-1$.

Now by MVT:

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

$=>-sin(2c-\pi )2=\frac{-1}{\pi /4}$

$=>sin(2c-\pi )=2/\pi$

hence, $c=\frac{1}{2}si{n}^{-1}\frac{2}{\pi }+\frac{\pi }{2}$