Question

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

• A. $\frac{\mathrm{\pi }}{6}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. ${\sin }^{-1}\left(\frac{2}{\mathrm{\pi }}\right)$
• D. ${\cos }^{-1}\left(\frac{2}{\mathrm{\pi }}\right)$

Answer 1

The correct option is C

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

Explanation for the correct option.

Find the value of $c$.

The mean value theorem states that if a function $f\left(x\right)$ is differentiable in the interval $\left[a,b\right]$ then there exists a real number $c$ in that interval such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

For the function $f\left(x\right)=\cos x$, the derivative is given as $f\text{'}\left(x\right)=-\sin x$.

Now in the interval $0\le x\le \frac{\mathrm{\pi }}{2}$ the mean value theorem is given as:

$$\begin{gathered} f\text{'}\left(c\right)=\frac{f\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)-f\left(0\right)}{{\displaystyle \frac{\mathrm{\pi }}{2}}-0} \\ \Rightarrow -\sin c=\frac{\cos {\displaystyle \frac{\mathrm{\pi }}{2}}-\cos 0}{{\displaystyle \frac{\mathrm{\pi }}{2}}} \\ \Rightarrow -\sin c=\frac{2}{\mathrm{\pi }}\left(0-1\right) \\ \Rightarrow -\sin c=-\frac{2}{\mathrm{\pi }} \\ \Rightarrow \sin c=\frac{2}{\mathrm{\pi }} \\ \Rightarrow c={\sin }^{-1}\left(\frac{2}{\mathrm{\pi }}\right) \end{gathered}$$

Hence, the correct option is C.