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

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

$f^{\prime}(x)$

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

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

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

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

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

+ - +

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

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

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:

$f(x)=\operatorname{Cos} x \quad 0<x<\pi$

Solution:

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

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

$\begin{aligned} &\frac{d}{d x}\{f(x)\}=\frac{d}{d x} \operatorname{Cos} x \\ &\because f^{\prime}(x)=-\sin x\left[\because \frac{d(\cos x)}{d x}=-\sin x\right] \end{aligned}$

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

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

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:

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

Solution:

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

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

$\begin{aligned} \frac{d}{d x}\left\{f^{\prime}(x)\right\} &=\frac{d}{d x}(\operatorname{Sin} 2 x-x)=\frac{d}{d x} \operatorname{Sin} 2 x-\frac{d}{d x}(x)\left[\because \frac{d}{d x}(h(x) \pm g(x))=\frac{d}{d x} h(x) \pm \frac{d}{d x} g(x)\right] \\ &=\operatorname{Cos} 2 x .2-1 \quad\left[\because \frac{d}{d x}(\operatorname{Sin} a x)=\operatorname{Cos} a x \cdot a\right] \\ f^{\prime}(x) &=2 \operatorname{Cos} 2 x-1 \quad\left[\because \frac{d}{d x}(x)=1\right] \\ \Rightarrow f^{\prime}(x) &=-(1-2 \operatorname{Cos} 2 x) \end{aligned}$

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

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

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

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

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

- + -

-∞ $\frac{-\pi}{6}$ $\frac{\pi}{6}$ ∞

since ${f}'\left ( x \right )$ 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

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

Again since${f}'\left ( x \right )$ 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

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

Maxima and Minima exercise 17.2 question 11

maxima and minima exercise 17.2 question 12

Answer

$$$ x=2 / 3 \text { is the point of local maxima and the value of local maxima is } \frac{2 \sqrt{3}}{9}$

Hint:

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

Given:

$$$ f(x)=x \sqrt{1-x} \quad, x>0$

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

$$$ \begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}(x \sqrt{1-x}) \\ &=x \frac{d}{d x}(\sqrt{1-x})+\sqrt{1-x} \frac{d}{d x}(x) \quad\left[\because \frac{d}{d x}(y x)=y \frac{d}{d x} x+x \frac{d}{d x} y\right] \\ &=x \frac{d}{d x}(\sqrt{1-x})+(\sqrt{1-x}) \frac{d}{d x}(x) \\ &=x \frac{1}{2}(1-x)^{\frac{1}{2}-1} \frac{d}{d x}(1-x)+\sqrt{1-x} \cdot 1\left[\because \frac{d}{d x}(x+a)^{n}=n(x+a)^{n-1} \frac{d}{d x}(x+a)\right] \\ \end{aligned}$

$=x \cdot \frac{1}{2}(1-x)^{\frac{1-2}{1}}\left\{\frac{d}{d x}(1)-\frac{d}{d x}(x)\right\}+\sqrt{1-x} \quad\left[\because \frac{d}{d x}(x+a)=\frac{d}{d x}(x)+\frac{d}{d x}(a)\right] \\$

$=\frac{x}{2}(1-x)^{-1 / 2}\{0-1\}+\sqrt{1-x} \quad\left[\because \frac{d}{d x} \text { Constant }=0, \frac{d}{d x}(x)=1\right] \\$

$=\frac{x}{2 \sqrt{1-x}}(0-1)+\sqrt{1-x} \\$

$=\frac{-x}{2 \sqrt{1-x}}+\sqrt{1-x} \\$

$=\frac{-x+2(1-x)}{2 \sqrt{1-x}} \\$

$=\frac{-x+2-2 x}{2 \sqrt{1-x}} \\$

$=\frac{2-3 x}{2 \sqrt{1-x}}$

${f}'\left ( x \right )=\frac{-\left ( 3x-2 \right )}{2 \sqrt{1-x}}$

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

$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{-(3 x-2)}{2 \sqrt{1-x}}=0 \quad \Rightarrow \frac{3 x-2}{2 \sqrt{1-x}}=0 \Rightarrow \frac{3 x-2}{\sqrt{1-x}}=0 \quad\left[\because \frac{1}{2} \neq 0\right] \\ &\Rightarrow 3 x-2=0 \quad \Rightarrow \quad 3 x=2 \quad x=2 / 3 \end{aligned}$

$\text { Since, } f^{\prime}(x) \text { changes } f \text { rom }+\text { ve to }-\text { ve when } x \text { increases through } 2 / 3 \text { is the point of local minima }$

$\begin{aligned} &\therefore \text { the value of the local maxima of } f(x) \text { at } x=2 / 3 \text { is }\\ &f\left(\frac{2}{3}\right)=\frac{2}{3} \sqrt{1-\frac{2}{3}}=\frac{2}{3} \sqrt{\frac{3-2}{3}}=\frac{2}{3} \sqrt{\frac{1}{3}}\\ &=\frac{2}{3} \cdot \frac{1}{\sqrt{3}}=\frac{2}{3 \sqrt{3}}\\ &=\frac{2}{3 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2 \sqrt{3}}{3 \times 3}=\frac{2 \sqrt{3}}{9} \end{aligned}$

Maxima and Minima Exercise 17.2 Question 13

Answer

$x=\frac{1}{4}$ is the point of local minima and the values of local minima at $x=\frac{1}{4} \ \ \ is \ \ \ \ \frac{-1}{512}$

Hint

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

Given:

$f(x)=x^{3}(2 x-1)^{3}$

Solution: -

$f(x)=x^{3}(2 x-1)^{3}$
Differentiating $f(x) w \cdot r . t^{\prime} x^{\prime} then$
$\begin{aligned} f^{\prime}(x) &=\frac{d}{d x}\left\{x^{3}(2 x-1)^{3}\right\} \\ &=x^{3} \frac{d}{d x}(2 x-1)^{3}+(2 x-1)^{3} \frac{d}{d x} x^{3} \quad\left\{\because \frac{d}{d x}(y x)=y \frac{d}{d x} x+x \frac{d}{d x} y\right\} \\ &=x^{3} 3(2 x-1)^{3-1} \frac{d}{d x}(2 x-1)+(2 x-1)^{3} 3 x^{3-1} \quad\left\{\because \frac{d}{d x} x^{n}=n x^{n-1} \& \frac{d}{d x}(a x+b)^{n}=n(a x+b)^{n-1} \frac{d}{d x}(a x+b)\right\} \\ &=x^{3} 3(2 x-1)^{2}\left\{\frac{d}{d x}(2 x)-\frac{d}{d x}(1)^{2}\right\}+(2 x-1)^{3} 3 x^{2} \quad\left\{\because \frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)\right\} \end{aligned}$
$=x^{3} 3(2 x-1)^{2}\left\{2 \frac{d}{d x}(x)-0\right\}+(2 x-1)^{3} 3 x^{2} \quad\left\{\because \frac{d}{d x}(a x)=a \frac{d}{d x}(x), \frac{d}{d x}(x)=1, \frac{d}{d x} \text { constant }=0\right\}$
$\\ =3 x^{3}(2 x-1)^{2}(2 \cdot 1)+3 x^{2}(2 x-1)^{3} \\ =6 x^{3}(2 x-1)^{2}+3 x^{2}(2 x-1)^{3} \\ =3 x^{2}(2 x-1)^{2}(2 x+2 x-1) \\$
$f^{\prime}(x) =3 x^{2}(2 x-1)^{2}(4 x-1)$
By first derivative test, for local maxima and local minima, we have
$\Rightarrow f^{\prime}(x)=0$
$\Rightarrow 3 x^{2}(2 x-1)^{2}(4 x-1)=0$
$\Rightarrow x^{2}(2 x-1)^{2}(4 x-1)=0 \quad[\because 3 \neq 0]$
$\Rightarrow x^{2}=0 \quad$ or $(2 x-1)^{2}=0 \Rightarrow(4 x-1)=0$
$\Rightarrow x=0 \quad$ or $(2 x-1)=0 \Rightarrow 2 x=1$ or $\Rightarrow 4 x=1 \Rightarrow x=1 / 4$
$\Rightarrow x=1 / 2$
$\therefore x=0, \frac{1}{2}, \frac{1}{4}$

- - + +

-∞ 0 1/4 1/2 ∞

Since f(x) changes from –ve to +ve when x increases through $\frac{1}{4}$ so $x=\frac{1}{4}$ is the point of local minima

The value of the local minima of $f\left ( x \right )$ at $x=\frac{1}{4}$

$\begin{aligned} f\left(\frac{1}{4}\right) &=\left(\frac{1}{4}\right)^{3}\left(2 \cdot \frac{1}{4}-1\right)^{3} \\ &=\frac{1}{4^{3}}\left(\frac{1}{2}-1\right)^{3}=\frac{1}{64}\left(\frac{1-2}{2}\right)^{3}=\frac{1}{64}\left(\frac{-1}{2}\right)^{3} \\ &=\frac{1}{64} \cdot \frac{-1}{8}=\frac{-1}{512} \end{aligned}$

Maxima and Minima Exercise 17.2 Question 14

Answer:

$x=2$ is the point of local minima and the value of local minima is 2

Hint:

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

Given:

$f(x)=\frac{x}{2}+\frac{2}{x}, x>0$

Solution:-

$f(x)=\frac{x}{2}+\frac{2}{x}$

Differentiating $f(x) w \cdot r . t^{\prime} x^{\prime}then$

$\begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}\left\{\frac{2}{x}+\frac{x}{2}\right\} \\ &=\frac{d}{d x}\left\{\frac{2}{x}\right\}+\frac{d}{d x}\left\{\frac{x}{2}\right\} \quad\left[\frac{d}{d x}\{f(x)+h(x)\}=\frac{d}{d x}\{f(x)\}+\frac{d}{d x}\{h(x)\}\right] \\ &=2 \frac{d}{d x}\left(x^{-1}\right)+\frac{1}{2} \frac{d}{d x}(x) \quad\left[\frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \\ &=2(-1) x^{-1-1}+\frac{1}{2} x^{1-1} \quad\left[\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right] \\ \end{aligned}$

$=-2 x^{-2}+\frac{1}{2} x^{0} \\$

$=\frac{-2}{x^{2}}+\frac{1}{2} \\ f^{\prime}(x) =\frac{1}{2}-\frac{2}{x^{2}}$

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

$f^{\prime}(x)=0$

$\Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0 \Rightarrow \frac{1}{2}=\frac{2}{x^{2}}$

$\Rightarrow \quad x^{2}=4 \quad \Rightarrow x=\pm 2$

$\Rightarrow x=2$ $[x=-2$ is not possible since $x>0]$

- +

-∞ 2 +∞

Since ${f}'\left ( x \right )$ changes from $-v e$ to $+v e$ when $x$ increases through 2. So $x=2$ is the point of local minima

The value of local minima of $f\left ( x \right )$ at $x=2$ is

$\begin{aligned} f(2) &=\frac{2}{2}+\frac{2}{2}=1+1 \\ &=2 \end{aligned}$