#### Provide solution RD Sharma maths class 12 chapter Maxima and Minima exercise 17.2 question 11 maths textbook solution

$x= \frac{\pi}{3}$ and $x= -\frac{\pi}{3}$is the point of local maxima an local minima respectively. The value of the local

maxima and local minima is $\sqrt{3} -\frac{\pi}{3}$ and  -$-\sqrt{3} +\frac{\pi}{3}$ respectively

Hint

Use first derivative test to find the value and point of local maxima and local minima

Given

$f(x)=2 \sin x-x,-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$

Solution:

$f(x)=2 \sin x-x$

Differentially $f(x) w \cdot r, t^{\prime} x^{\prime} \text { then }$

\begin{aligned} &\begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}(2 \operatorname{Sin} x-x)=\frac{d}{d x}(2 \operatorname{Sin} x)-\frac{d}{d x}(x) \\ \end{aligned} \end{aligned}          $\left[\because \frac{d}{d x}\{h(x) \pm g(x)\}=\frac{d}{d x}\{h(x)\} \pm \frac{d}{d x}\{g(x)\}\right] \\$

\begin{aligned} &\begin{aligned} \\ &=2 \frac{d}{d x} \sin x-\frac{d}{d x}(x) & \end{aligned} \end{aligned}                                                      $\left[\because \frac{d}{d x} a f(x)=a \frac{d}{d x}\{f(x)\}\right]$

$\therefore f^{\prime}(x)=-(1-2 \operatorname{Cos} x)$

$\Rightarrow f^{\prime}(x)=-(1-2 \operatorname{Cos} x)$                                                         $\left[\because \frac{d}{d x} \operatorname{Sin} x=\operatorname{Cos} x, \frac{d}{d x}(x)=1\right]$

By first derivative test, for local maxima and local minima, we have

\inline \begin{aligned} &f^{\prime}(x)=0 \quad \Rightarrow-(1-2 \operatorname{Cos} x)=0 \Rightarrow 1-2 \operatorname{Cos} x=0 \\ &\Rightarrow 2 \cos x-1=0 \Rightarrow 2 \cos x=1 \\ &\Rightarrow \cos x=\frac{1}{2} \quad \Rightarrow \cos x=\cos \frac{\pi}{3} \quad\left[\because \frac{1}{2}=\cos \frac{\pi}{3}\right] \\ &\Rightarrow x=2 n \pi \pm \frac{\pi}{3} \quad, n \in \mathbb{Z}[\because \operatorname{Cos} \theta=\operatorname{Cos} \alpha \Rightarrow \theta=2 n \pi \pm \propto, n \in \mathbb{Z}] \\ &x=\pm \frac{\pi}{3}, 2 \pi \pm \frac{\pi}{3},-2 \pi \pm \frac{\pi}{3} \\ &x=\frac{\pi}{3},-\frac{\pi}{3}\left[\text { Neglecting other values of } x \text { since }-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\right] \end{aligned}

-               +                      -

-∞                       $-\frac{\pi}{3}$                  $\frac{\pi}{3}$                           ∞

since ${f}'\left ( x \right )$ changes from +ve to -ve when $x$ increases through $\frac{\pi}{3}$                    .

So, $\inline x=\frac{\pi}{3}$    is the point of local maxima

The value of local maxima of $f\left ( x \right )$ at $x=\frac{\pi}{3}$  is

\begin{aligned} f\left(\frac{\pi}{3}\right) &=2 \sin \frac{\pi}{3}-\frac{\pi}{3} \quad\left[\because \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right] \\ &=2 \frac{\sqrt{3}}{2}-\frac{\pi}{3}=\sqrt{3}-\frac{\pi}{3} \end{aligned}

Again since ${f}'\left ( x \right )$ changes from -ve to +ve    when $x$ increases through $\frac{-\pi}{3}$ .

So, $x=\frac{-\pi}{3}$  is the point of local minima.

The value of local minima of $f\left ( x \right )$ at $x=\frac{-\pi}{3}$is

\begin{aligned} f\left(\frac{-\pi}{3}\right) &=2 \sin \left(-\frac{\pi}{3}\right)-\left(-\frac{\pi}{3}\right) \\ &=-2 \operatorname{Sin} \frac{\pi}{3}+\frac{\pi}{3} \quad[\because \operatorname{Sin}(-\theta)=-\operatorname{Sin} \theta] \end{aligned}

\begin{aligned} &=-2 \cdot \frac{\sqrt{3}}{2}+\frac{\pi}{3}\left[\because \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right] \\ &=-\sqrt{3}+\frac{\pi}{3} \end{aligned}