#### Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (xi)

$c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$

Hint:

$f(0)=f\left(\frac{\pi}{2}\right)$, so there exists at c belongs to$\left(0, \frac{\pi}{2}\right)$

Given:

$f(x)=\sin x+\cos x$ on$\left[0, \frac{\pi}{2}\right]$

Explanation:

We have

$f(x)=\sin x+\cos x$ on$\left[0, \frac{\pi}{2}\right]$

We know that cosine and sine functions are continuous and differentiable on R.

Let’s find the values of the function at an extreme

$\begin{array}{ll}\Rightarrow & f(0)=\sin (0)+\cos (0) \\ \Rightarrow & f(0)=0+1 \\ \therefore & f(0)=1\end{array} \\\\ \Rightarrow \quad f\left(\frac{\pi}{2}\right)=\sin \left(\frac{\pi}{2}\right)+\cos \left(\frac{\pi}{2}\right) \\\\ \Rightarrow \quad f\left(\frac{\pi}{2}\right)=1+0 \\\\ \therefore f\left(\frac{\pi}{2}\right)=1$

We have$f(0)=f\left(\frac{\pi}{2}\right)$ , so there exists at$c \in\left(0, \frac{\pi}{2}\right)$ , such that$f^{\prime}(c)=0$

Let’s find the derivative of f(x)

$f^{\prime}(x)=\frac{d(\sin x+\cos x)}{d x}\\\\ \Rightarrow \quad f^{\prime}(x)=\cos x-\sin x \left[\begin{array}{l}\because \frac{d(\cos x)}{d x}=-\sin x \\\\ \because \frac{d(\sin x)}{d x}=\cos x\end{array}\right]$

We have$f^{\prime}(c)=0$

$\\ \Rightarrow \quad \cos c-\sin c=0 \\\\ \Rightarrow \quad \frac{1}{\sqrt{2}} \cos c-\frac{1}{\sqrt{2}} \sin c=0 \quad$                                          [Multiplying both the sides by$\frac{1}{\sqrt2}$]$\\ \Rightarrow \sin \left(\frac{\pi}{4}\right) \cos c-\cos \left(\frac{\pi}{4}\right) \sin c=0\\\\ \Rightarrow \sin \left(\frac{\pi}{4}-c\right)=0 \ \ \ \ \quad[\because \sin (A-B)=\sin A \cos B-\cos A \sin B] \\\\ \Rightarrow \quad \frac{\pi}{4}-c=0 \\\\ \Rightarrow \quad c=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)$

Hence, Rolle’s Theorem is verified.