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

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 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.


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


f(x)=\sin x-\cos x, 0<x<2 \pi


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


\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<x<2 \pi, \text { so neglecting other values }]

                                                             +                -                       +

                                   -∞                            \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}

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