RD Sharma Class 12 Exercise 17.2 Maxima And Minima Solutions Maths - Download PDF Free Online

Maxima and Minima Excercise: 17.2

Minima and Maxima Exercise 17.2 Question 1
Answer:

Maxima and Minima Exercise 17.2 question 2

Maxima and Minima Exercise 17.2 Question 3

Maxima and Minima Exercise 17.2 Question 4

Maxima and Minima Exercise 17.2 Question 5

Maxima and Minima Exercise 17.2 Question 6

Maxima and Minima Exercise 17.2 Question 7

Maxima and Minima Exercise 17.2 Question 8
Answer:

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

$f^{\prime }(x)$

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

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

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

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

$\Rightarrow x=\frac{3\pi }{4},\frac{7\pi }{4}[\text{ since }0<x<2\pi ,\text{ so neglecting other values }]$

+ - +

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

since $f^{\prime }\left(x\right)$ 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

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

$=\mathrm{Sin}(\pi -\frac{\pi }{4})-\cos (\pi -\frac{\pi }{4})$

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

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

$$\begin{gathered} =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}[\because \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}] \\ =\frac{2}{\sqrt{2}}=\sqrt{2} \end{gathered}$$

Again since $f^{\prime }\left(x\right)$ 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
$\begin{array}{rl}& f\left(\frac{7\pi }{4}\right)=\sin \frac{7\pi }{4}-\cos \frac{7\pi }{4}\\ & =\sin (2\pi -\frac{\pi }{4})-\cos (2\pi -\frac{\pi }{4})\\ & =-\sin \frac{\pi }{4}-\cos \frac{\pi }{4}\\ & =-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\left[\begin{array}{c}\therefore \sin (2\pi -\theta )=-\sin \theta \\ \cos (2\pi -\theta )=\cos \theta \end{array}\right]\\ & =-\frac{2}{\sqrt{2}}\\ & =-\sqrt{2}\end{array}$

maxima and minima exercise 17.2 question 9

Answer:

There is no local maxima and local minima of $f\left(x\right)$ at interval (0,π)

Hint:

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

Given:

$f(x)=\mathrm{Cos}x0<x<\pi$

Solution:

$f(x)=\mathrm{Cos}x$

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

$\begin{array}{rl}& \frac{d}{dx}\{f(x)\}=\frac{d}{dx}\mathrm{Cos}x\\ & \because f^{\prime }(x)=-\sin x[\because \frac{d(\cos x)}{dx}=-\sin x]\end{array}$

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

$\begin{array}{rl}& f^{\prime }(x)=0\\ & \Rightarrow -\sin x=0\Rightarrow \mathrm{Sin}x=0\\ & \Rightarrow x=n\pi ;\mathrm{n}\in \mathbb{Z}\\ & \Rightarrow \mathrm{x}=0,\pi ,-\pi ,2\pi ,-2\pi \dots .\end{array}$

But these points of x lies outside the interval (0,π)

So there is no local maxima and minima will exist in the interval (0,π)

Maxima and Minima exercise 17.2 question 10
Answer:

$x=\frac{\pi }{6}$ and $x=-\frac{\pi }{6}$is the point of local maxima and local minima respectively. The value of

local maxima and local minima is $\frac{\sqrt{3}}{2}-\frac{\pi }{6}\text{ and }\frac{-\sqrt{3}}{2}+\frac{\pi }{6}$ respectively.

Hint:

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

Given:

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

Solution:

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

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

$\begin{array}{rl}\frac{d}{dx}\{f^{\prime }(x)\}& =\frac{d}{dx}(\mathrm{Sin}2x-x)=\frac{d}{dx}\mathrm{Sin}2x-\frac{d}{dx}(x)[\because \frac{d}{dx}(h(x)\pm g(x))=\frac{d}{dx}h(x)\pm \frac{d}{dx}g(x)]\\ & =\mathrm{Cos}2x.2-1[\because \frac{d}{dx}(\mathrm{Sin}ax)=\mathrm{Cos}ax\cdot a]\\ f^{\prime }(x)& =2\mathrm{Cos}2x-1[\because \frac{d}{dx}(x)=1]\\ \Rightarrow f^{\prime }(x)& =-(1-2\mathrm{Cos}2x)\end{array}$

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

$\begin{array}{rl}& f^{\prime }(x)=0\\ & \Rightarrow -(1-2\mathrm{Cos}2x)=0\Rightarrow 1-2\mathrm{Cos}2x=0\\ & \Rightarrow 2\mathrm{Cos}2x-1=0\Rightarrow 2\mathrm{Cos}2x=1\\ & \Rightarrow \mathrm{Cos}2x=\frac{1}{2}\Rightarrow \mathrm{Cos}2x=\mathrm{Cos}\frac{\pi }{3}\\ & \Rightarrow 2x=2n\pi \pm \frac{\pi }{3};\mathrm{n}\in \mathbb{Z}[\cos \theta =\cos \alpha \Rightarrow \theta =2n\pi \pm \alpha ,n\in \mathbb{Z}]\end{array}$

$\Rightarrow \mathrm{x}=\pm \frac{\pi }{6};\mathrm{n}\in \mathbb{Z}$

$\Rightarrow x=\frac{\pi }{6},\pi \pm \frac{\pi }{6}\cdots$

$\Rightarrow \mathrm{x}=\frac{\pi }{6},-\frac{\pi }{6}[\text{ neglecting other values of }\mathrm{x}\text{ since }-\frac{\pi }{2}\le \mathrm{x}\le \frac{\pi }{2}]$

- + -

-∞ $\frac{-\pi }{6}$ $\frac{\pi }{6}$ ∞

since $f^{\prime }\left(x\right)$ changes from +ve to –ve when $x$ increases through $\frac{\pi }{6}$ .

So,$x=\frac{\pi }{6}$ is the point of local maxima

The value of local maxima of $f\left(x\right)$ at $x=\frac{\pi }{6}$ is

$\begin{array}{rl}& f\left(\frac{\pi }{6}\right)=\sin 2\frac{\pi }{6}-\frac{\pi }{6}\\ & =\sin \frac{\pi }{3}-\frac{\pi }{6}\\ & =\frac{\sqrt{3}}{2}-\frac{\pi }{6}\end{array}$

Again since$f^{\prime }\left(x\right)$ changes from –ve to +ve when $x$ increases through $\frac{-\pi }{6}$ .

So,$x=\frac{-\pi }{6}$ is the point of local minima

The value of local minima of $f\left(x\right)$ at $x=\frac{-\pi }{6}$is

$\begin{array}{rl}& f(-\frac{\pi }{6})=\sin 2\times (-\frac{\pi }{6})-(-\frac{\pi }{6})\\ & =-\sin \frac{\pi }{3}+\frac{\pi }{6}[\because \sin (-\theta )=-\sin \theta ]\\ & =-\frac{\sqrt{3}}{2}+\frac{\pi }{6}[\because \sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}]\end{array}$

Maxima and Minima exercise 17.2 question 11