#### Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 8

$x=\frac{3 \pi}{4}$ and $x=\frac{7\pi}{4}$is the point of  local maxima and  local minima respectively. The value of   local maxima and  local minima is $\sqrt{2}$and $-\sqrt{2}$ respectively.

Hint:

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

Given:

$f(x)=\sin x-\cos x, 0

Solution:

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

Differentiating $f\left ( x \right )$ with respect to ‘x’ then,

\begin{aligned} &\frac{d(f(x))}{d x}=\frac{d}{d x}(\sin x-\cos x) \\ &=\frac{d}{d x}(\sin x)-\frac{d}{d x}(\cos x) \quad\left[\because \frac{d}{d x}(g(x)-f(x))=\frac{d}{d x}(g(x))-\frac{d}{d x}(f(x))\right] \\ &=\cos x-(-\sin x) \quad\left[\because \frac{d}{d x}(\sin x)=\cos x, \frac{d}{d x}(\cos x)=-\sin x\right] \\ &\therefore f^{\prime}(x)=\cos x+\sin x \end{aligned}

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

$f^{\prime}(x)$

\begin{aligned} &\Rightarrow \cos x+\sin x=0 \Rightarrow-\cos x=+\sin x\\ &\Rightarrow-1=\frac{\sin x}{\cos x} \Rightarrow \tan x=-1 \quad\left[\because \frac{\sin \theta}{\cos \theta}=\tan \theta\right]\\ &\Rightarrow \tan x=-\left(\tan \frac{\pi}{4}\right) \quad\left[\because 1=\tan \frac{\pi}{4}\right]\\ &\Rightarrow \tan x=\tan \left(-\frac{\pi}{4}\right) \quad[\because-\tan \theta=\tan (-\theta)]\\ &\Rightarrow x=n \pi+\left(-\frac{\pi}{4}\right) ; n \in z \quad[\tan \theta=\tan \alpha \Rightarrow \theta=n \pi+\alpha ; n \in z]\\ \end{aligned}

$\Rightarrow x=n \pi-\frac{\pi}{4}\\$

$x=-\frac{\pi}{4},\left(\pi-\frac{\pi}{4}\right),\left(2 \pi-\frac{\pi}{4}\right),\left(-\pi-\frac{\pi}{4}\right)\left(-2 \pi-\frac{\pi}{4}\right)\\$

$\Rightarrow x=-\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{7 \pi}{4}, \frac{-5 \pi}{4}, \frac{-9 \pi}{4}\\$

$\Rightarrow x=\frac{3 \pi}{4}, \frac{7 \pi}{4}[\text { since } 0

+                -                       +

-∞                            $\frac{3 \pi}{4}$                     $\frac{7\pi}{4}$

since ${f}'\left ( x \right )$ changes from +ve  to –ve when $x$ increases through $\frac{3\pi}{4}$                   .

So, $x=\frac{3\pi}{4}$    is the point of local maxima.

The value of local maxima of $f\left ( x \right )$ at $x=\frac{3\pi}{4}$   is

\begin{aligned} &f\left(\frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{4}-\cos \frac{3 \pi}{4} \\ \end{aligned}

$=\operatorname{Sin}\left(\pi-\frac{\pi}{4}\right)-\cos \left(\pi-\frac{\pi}{4}\right) \\$

$\quad=\operatorname{Sin} \frac{\pi}{4}-\left(-\operatorname{Cos} \frac{\pi}{4}\right) \quad[\because \sin (\pi-\theta)=\sin \theta \& \cos (\pi-\theta)=-\operatorname{Cos} \theta] \\$

$=\operatorname{Sin} \frac{\pi}{4}+\operatorname{Cos} \frac{\pi}{4}$

$\\ \quad=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \quad\left[\because \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}\right] \\ \quad=\frac{2}{\sqrt{2}}=\sqrt{2}$

Again since ${f}'\left ( x \right )$ changes from –ve  to +ve  when $x$ increases through $\frac{7\pi}{4}$ .

So,$x=\frac{7\pi}{4}$  is the point of local minima

The value of local minima of $f\left ( x \right )$ at $x=\frac{7\pi}{4}$is

\begin{aligned} &f\left(\frac{7 \pi}{4}\right)=\sin \frac{7 \pi}{4}-\cos \frac{7 \pi}{4} \\ &=\sin \left(2 \pi-\frac{\pi}{4}\right)-\cos \left(2 \pi-\frac{\pi}{4}\right) \\ &=-\sin \frac{\pi}{4}-\cos \frac{\pi}{4} \\ &=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \quad\left[\begin{array}{l} \therefore \sin (2 \pi-\theta)=-\sin \theta \\ \cos (2 \pi-\theta)=\cos \theta \end{array}\right] \\ &=-\frac{2}{\sqrt{2}} \\ &=-\sqrt{2} \end{aligned}