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

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

Edited By Satyajeet Kumar | Updated on Jan 21, 2022 02:25 PM IST

RD Sharma is well known by many students as it is one of the top-selling books in the country. RD Sharma's books are known for their extremely detailed and informative answers which are solved step-by-step to make it easier for students to understand. Class 12 RD Sharma chapter 17 exercise 17.2 solution is titled 'Maxima and Minima.' The book will have many examples based on the chapter and students can practice these solutions to develop their skills on the maths subject. However, solving a complex chapter like Maxima and Minima is a tricky and tedious task. The RD Sharma class 12th exercise 17.2 solution will then come to the rescue of students and help them score well.

## Maxima and Minima Excercise: 17.2

### Minima and Maxima Exercise 17.2 Question 1Answer:

$x=5$ is the point of local minima and the value of local minimum of $f\left ( x \right )$ is 0.
Hint:
Use first derivative test to find the point and values of local maximum or local minimum.
Given:
$f\left ( x \right )=\left ( x-5 \right )^{4}$
Solution:
$f\left ( x \right )=\left ( x-5 \right )^{4}$
Differentiating $f\left ( x \right )$ with respect to ‘x’ then
\begin{aligned} \frac{d}{d x}(f(x)) &=\frac{d}{d x}(x-5)^{4} \\ &=4(x-5)^{4-1} \frac{d}{d x}(x-5) \\ \end{aligned}
$=4(x-5)^{3}\left(\frac{d x}{d x}-\frac{d 5}{d x}\right) \\$ $\left[\therefore \frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \quad \& \frac{d}{d x} \operatorname{constan} t=0\right]$
$=4(x-5)^{3}(1-0) \\$
$=4(x-5)^{3} \\$
$\therefore f^{\prime}(x)=+4(x-5)^{3}$
By first derivative test, for local maxima or local minima, we have
\begin{aligned} &\Rightarrow 4(x-5)^{3}=0 \\ &\Rightarrow \quad(x-5)^{3}=0 \\ &\Rightarrow \quad(x-5)=0 \\ &\Rightarrow \quad x=5 \end{aligned}

-∞ - 5 + +∞

Since ${f}'\left ( x \right )$ changes from -ve to +ve when $x$ increases through 5.
so, $x$ = 5 is the point of local minima.
The value of local minima of $f\left ( x \right )$ at x = 5 is
$f\left ( 5 \right )=\left ( 5-5 \right )^{4}=0$

Maxima and Minima Exercise 17.2 question 2

X=1 and x =-1 is the point of local minima and local maxima respectively and -2 and 2 is the value of local minima and local maxima respectively.
Hint:
Use first derivative test to find the point and value of local maxima or local minima.
Given:
$f\left ( x \right )=x^{3}-3x$
Solution:
$f\left ( x \right )=x^{3}-3x$
Differentiating $f\left ( x \right )$ with respect to ‘x’ then
$\frac{d(f(x))}{d x}=\frac{d}{d x}\left(x^{3}-3 x\right)=\frac{d\left(x^{3}\right)}{d x}-\frac{d(3 x)}{d x} \quad$ $\left[\because \frac{d}{d x}(a x+b)=\frac{d(a x)}{d x}+\frac{d(b)}{d x}\right]$
$=\frac{d\left(x^{3}\right)}{d x}-3 \frac{d(x)}{d x} \quad$ $\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d\left(x^{n}\right)}{d x}\right]$
$=3 x^{3-1}-3 x^{1-1} \quad$ $\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$
$=3x^{2}-3x^{0}$
$=3x^{2}-3$ $\left [ \because x^{0}=1 \right ]$
$\because {f}'\left ( x \right )=+3\left ( x^{2}-1 \right )$
By first derivation test, for local maxima or local minima ,we have
$\because {f}'\left ( x \right )=0$
$\begin{array}{ll} \Rightarrow 3\left(x^{2}-1\right)=0 & \Rightarrow\left(x^{2}-1\right)=0 \quad[\because 3 \neq 0] \\ \Rightarrow x^{2}=1 & \Rightarrow x=\pm 1 \end{array}$

− +
-∞ -1 +1 ∞
Since ${f}'\left ( x \right )$ changes from –ve to +ve as $x$ increases through 1.
So, $x$ = 1 is the point of local minima.
The value of local minima of $f\left ( x \right )$ at x=1 is
$f\left ( 1\right )=1^{3}-3\cdot 1=3-1=-2$
Again, since ${f}'\left ( x \right )$ changes from +ve to -ve as $x$ increases through -1.
So, $x$ = -1 is the point of local maxima.
The value of local maxima of $f\left ( x \right )$ at x=-1 is
$f\left (-1\right )=\left ( -1 \right )^{3}-3\left ( -1 \right )$
$-1+3=2$

Maxima and Minima Exercise 17.2 Question 3

$x= 1$,$x= \frac{3}{5}$ is the point of local minima and local maxima respectively. The value of local minima and local maxima is 0 and $\frac{108}{3125}$respectively.
Hint:
Use first derivative test to find the value and point of local maxima or local minima.
Given:
$f(x)=x^{3}(x-1)^{2}$
Solution:
$f(x)=x^{3}(x-1)^{2}$
Differentiating f(x) with respect to ‘x’ then,
$\frac{d}{d x}(f(x))=\frac{d}{d x}\left\{x^{3}(x-1)^{2}\right\}$
$=x^{3} \frac{d}{d x}(x-1)^{2}+(x-1)^{2} \frac{d}{d x}\left(x^{3}\right) \quad\left[\because \frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$
$=x^{3} 2(x-1)^{2-1} \frac{d}{d x}(x-1)+(x-1)^{2} .3 x^{3-1}\left[\because \frac{d}{d x}(x+a)^{n}=n(x+a)^{n-1} \frac{d}{d x}(x+a), \frac{d x^{n}}{d x}=n x^{n-1}\right]$
$=x^{3} \cdot 2(x-1)\left\{\frac{d x}{d x}-\frac{d(1)}{d x}\right\}+(x-1)^{2} \cdot 3 x^{2} \quad\left[\because \frac{d}{d x}(a x+b)=\frac{d(a x)}{d x}+\frac{d(b)}{d x}\right]$
\begin{aligned} &=2 x^{3}(x-1)+3 x^{2}(x-1)^{2} \\ &=x^{2}(x-1)\{2 x+3(x-1)\}=x^{2}(x-1)(2 x+3 x-3) \\ &\therefore f^{\prime}(x)=+x^{2}(x-1)(5 x-3) \end{aligned}
By first derivative test, for local maxima and local minima ,we have
$f^{\prime}(x)$
\begin{aligned} &\Rightarrow x^{2}(x-1)(5 x-3)=0 \\ &\Rightarrow x^{2}=0 \quad \text { or } \quad(x-1)=0 \quad \text { or } 5 x-3=0 \\ &\Rightarrow x=0 \quad \text { or } \quad x=1 \quad \text { or } \quad 5 x=3 \Rightarrow x=\frac{3}{5} \\ &\quad \therefore x=0,1, \frac{3}{5} \end{aligned}
− + - +
-∞ 0 3/5 1 ∞
Since $f^{\prime}(x)$ changes from negative to positive when $x$ increases through 1.
So, $x= 1$ is the point of local minima.
The value of local minima of $f\left ( x \right )$ at x=1 is
$f\left ( 1 \right )= 1^{3}\left ( 1-1 \right )=0$
Again, since $f^{\prime}(x)$ changes from +ve to –ve when $x$ increases through 3/5.
So, $x= \frac{3}{5}$ is the point of local maxima.
The value of local maxima of $f\left ( x \right )$ at $x= \frac{3}{5}$ is
\begin{aligned} f\left(\frac{3}{5}\right) &=\left(\frac{3}{5}\right)^{3}\left(\frac{3}{5}-1\right)^{2}=\frac{27}{125}\left(\frac{3-5}{5}\right)^{2} \\ &=\frac{27}{125}\left(\frac{-2}{5}\right)^{2}=\frac{27}{125} \times \frac{4}{25} \\ &=\frac{108}{3125} \end{aligned}
NOTE: Answer given in the book is incorrect.

Maxima and Minima Exercise 17.2 Question 4

X= 0 and x= -2 is the point of local minima and local maxima respectively.
The value of the local maxima and local minima is 0 and -4 respectively.
Hint:
Use first derivative test to find the point and value of the local maxima and local minima.
Given:
$f\left ( x \right )= \left ( x-1 \right )\left ( x+2 \right )^{2}$
Solution:
$f\left ( x \right )= \left ( x-1 \right )\left ( x+2 \right )^{2}$
Differentiating f(x) with respect to ‘x’ then
\begin{aligned} &\frac{d}{d x}\{f(x)\}=\frac{d}{d x}\left\{(x-1)(x+2)^{2}\right\} \\ &=(x-1) \frac{d}{d x}(x+2)^{2}+(x+2)^{2} \frac{d}{d x}(x-1) \\ &=(x-1) \cdot 2(x+2)+(x+2)^{2} \\ &=(x+2)(2 x-2+x+2) \\ &=(x+2)(3 x) \end{aligned}
By first derivative test,for local maxima or local minima ,we have
\begin{aligned} &\frac{d}{d x}\{f(x)\}=\frac{d}{d x}\left\{(x-1)(x+2)^{2}\right\} \\ &=(x-1) \frac{d}{d x}(x+2)^{2}+(x+2)^{2} \frac{d}{d x}(x-1) \\ &=(x-1) \cdot 2(x+2)+(x+2)^{2} \\ &=(x+2)(2 x-2+x+2) \\ &=(x+2)(3 x) \end{aligned}
+ - +
-∞ -2 0 ∞
Since, ${f}'\left ( x \right )$ changes from –ve to +ve when $x$ increases through 0. So $x=0$ is the point of local minima.
The value of the local minima of $f\left ( x \right )$ at $x=0$is

Again, Since ${f}'\left ( x \right )$ changes from +ve to -ve when $x$ increases through -2.
So, $x=-2$ is the point of local maxima.
The value of local maxima of $f\left ( x \right )$ at $x=-2$ is
$f\left ( -2 \right )=\left ( -2-1 \right )\left ( -2+2 \right )^{2}=-3\times 0=0$

Maxima and Minima Exercise 17.2 Question 5

$x=0$ is the point of local maxima and the value of local maxima is $\frac{1}{2}$
Hint:
Use first derivative test to find the value and point of local maxima and local minima.
Given:
$f\left ( x \right )=\frac{1}{x^{2}+2}$
Solution:
$f\left ( x \right )=\frac{1}{x^{2}+2}$
Differentiating $f\left ( x \right )$ with respect to ‘x’ then,
$\frac{d}{d x}\{f(x)\}=\frac{d\left(\frac{1}{x^{2}+2}\right)}{d x}=\frac{d\left(x^{2}+2\right)-1}{d x} \quad\left[\because \frac{d}{d x}\left(x^{n}+a\right)^{n}=n\left(x^{n}+a\right)^{n-1} \frac{d}{d x}\left(x^{n}+a\right)\right]$
$=-1\left(x^{2}+2\right)-1-1 \cdot \frac{d}{d x}\left(x^{2}+2\right)$
$=-1\left(x^{2}+2\right)^{-2}\left\{\frac{d\left(x^{2}\right)}{d x}+\frac{d(2)}{d x}\right\} \ \ \ \ \ \ \ \ \left[\because \frac{d}{d x}\left(x^{n}+a\right)=\frac{d\left(x^{n}\right)}{d x}+\frac{d(a)}{d x}\right]$
\begin{aligned} &=\frac{-1}{\left(x^{2}+2\right)^{2}}\left[2 x^{2-1}+0\right] \quad\left[\because \frac{d x^{n}}{d x}=n x^{n-1}, \frac{d}{d x} \text { constant }=0\right. \\ &\therefore f^{i}(x)=\frac{-1}{\left(x^{2}+2\right)^{2}} 2 x=\frac{-2 x}{\left(x^{2}+2\right)^{2}} \end{aligned}
By first derivative test, for local maxima or local minima, we have
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{-2 x}{\left(x^{2}+2\right)^{2}}=0 \quad \Rightarrow \quad \frac{x}{\left(x^{2}+2\right)^{2}}=0[\because-2 \neq 0] \\ &\Rightarrow x=0 \end{aligned}

+ -
-∞ 0 ∞
since ${f}'\left ( x \right )$ changes from +ve to -ve when through $x=0$
So, $x=0$ is the point of local maxima.
The value of local maxima of ${f}'\left ( x \right )$ at is$-x=0$
$f(0)=\frac{1}{0^{2}+2}=\frac{1}{2}$

Maxima and Minima Exercise 17.2 Question 6

$x= 3$ and $x= 1$ is the point of local minima and local maxima respectively. The value of local minima and maxima is 15 and 19 respectively.
Hint:
Use first derivative test to find the point and value of local maxima and local minima.
Given:
$f\left ( x \right )=x^{3}-6x^{2}+9x+15$
Solution:
$f\left ( x \right )=x^{3}-6x^{2}+9x+15$
Differentiating f(x) with respect to ‘x’ then,
$\frac{d}{d x}\{f(x)\}=\frac{d}{d x}\left\{x^{3}-6 x^{2}+9 x+15\right\} \quad\left[\because \frac{d}{d x}\left(a x^{3}+b x^{2}+c x+d\right)=\frac{d}{d x}\left(a x^{3}\right)+\frac{d}{d x}\left(b x^{2}\right)+\frac{d}{d x}(c x)+\frac{d}{d x}(d)\right]$
$=\frac{d}{d x}\left(x^{3}\right)-\frac{d}{d x}\left(6 x^{2}\right)+\frac{d}{d x}(9 x)+\frac{d}{d x}(15)$
$=\frac{d}{d x}\left(x^{3}\right)-6 \frac{d}{d x}\left(x^{2}\right)+9 \frac{d}{d x}(x)+\frac{d}{d x}(15)\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right]$
$=3 x^{3-1}-6.2 x^{2-1}+9 x^{1-1}+0 \quad\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}, \frac{d}{d x} \operatorname{constan} t=0\right]$
$\therefore f^{\prime}(x)=3 x^{2}-12 x+9 x=3\left(x^{2}-4 x+3\right)$
By first derivative test, for local maxima or local minima ,we have
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 3\left(x^{2}-4 x+3\right)=0 \Rightarrow\left(x^{2}-4 x+3\right)=0 \quad[\because 3 \neq 0] \\ &\Rightarrow(x-1)(x-3)=0 \quad \Rightarrow x-1=0 \quad \text { or } x-3=0 \\ &\Rightarrow x=1 \quad \text { or } \quad x=3 \end{aligned}
+ - +
-∞ 1 3 ∞

since$f^{\prime}(x)$ changes from -ve to +ve when $x$ increases through 3.
So, $x=3$ is the point of local minima
The value of local minima of $f\left ( x \right )$ at $x=3$ is
\begin{aligned} f(3) &=(3)^{3}-6.3^{2}+9.3+15 \\ &=27-6.9+27+15 \\ &=54-54+15 \\ &=15 \end{aligned}
Again since $f^{\prime}(x)$ changes from +ve to -ve when $x$ increases through 1.
So, $x=1$ is the point of local maxima
The value of local maxima of $f\left ( x \right )$ at $x=1$ is
\begin{aligned} f(1) &=(1)^{3}-6.1^{2}+9.1+15 \\ &=1-6+9+15 \\ &=25-6 \\ &=19 \end{aligned}

Maxima and Minima Exercise 17.2 Question 7

$x=\frac{\pi}{4}$ and $x=\frac{3\pi}{4}$ is the point of local maxima and local minima respectively. The value of local maxima and local minima is 1 and -1 respectively.
Hint:
Use first derivative test to find the point and value of local maxima or local minima.
Given:
$f\left ( x \right )=\sin2x, 0< x< \pi$
Solution:
$f\left ( x \right )=\sin2x$
Differentiating $f\left ( x \right )$ with respect to ‘x’ then,
\begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}(\sin 2 x) \\ &=\cos 2 x \frac{d}{d x}(2 x) \quad\left[\because \frac{d}{d x}(\sin a x)=\cos a x \frac{d}{d x}(a x)\right] \\ &=\cos 2 x .2 \frac{d}{d x}(x)\left[\because \frac{d}{d x}(a x)=a \frac{d}{d x}(x)\right] \\ &=\cos 2 x .2 .1 \quad\left[\because \frac{d}{d x}(x)=1\right] \\ f^{\prime}(x) &=2 \cos 2 x \end{aligned}
By first derivative test, for local maxima or local minima ,we have
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 2 \cos 2 x=0 \\ &\Rightarrow \cos 2 x=0 \quad[\because 2 \neq 0] \\ &\Rightarrow 2 x=(2 n+1) \frac{\pi}{2} \quad ;\left[[\cos \theta=0] \Rightarrow \theta=(2 n+1) \frac{\pi}{2}\right] \\ &\Rightarrow x=(2 n+1) \frac{\pi}{4} ; n \in N \end{aligned}
$\Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4} \quad[\text { neglecting other value of } x \text { since } 0
+ - +
-∞ $\frac{\pi}{4}$ $\frac{3\pi}{4}$

since ${f}'\left ( x \right )$ changes from +ve to -ve when $x$ increases through $\frac{\pi}{4}$ .
So, $x=\frac{\pi}{4}$ is the point of local maxima
The value of local maxima of $f\left ( x \right )$ at $x=\frac{\pi}{4}$ is
$f\left(\frac{\pi}{4}\right)=\sin \left(2 \frac{\pi}{4}\right)=\sin \frac{\pi}{2}=1\left[\because \sin \frac{\pi}{2}=1\right]$
Again 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 minima
The value of local minima of$f\left ( x \right )$ at $x=\frac{3\pi}{4}$is
\begin{aligned} f\left(\frac{3 \pi}{4}\right) &=\sin \left(2 \frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{2} \\ &=\sin \left(\pi+\frac{\pi}{2}\right)=-\sin \frac{\pi}{2} \quad[\because \sin (\pi+\theta)=-\sin \theta] \\ &=-1 \quad\left[\because \sin \frac{\pi}{2}=1\right] \end{aligned}
$x=\frac{3 \pi}{4}$ 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}

maxima and minima exercise 17.2 question 9

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

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)

$x=\frac{\pi}{6}$ 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

$x= \frac{\pi}{3}$ 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}

maxima and minima exercise 17.2 question 12

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

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}

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

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

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

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

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

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

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

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

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

\inline \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}

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

\inline \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

$\inline x=\frac{1}{4}$ 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:

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

Solution: -

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

- - + +

- 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 $x=\frac{1}{4}$

\inline \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

$x=2$ 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}

The RD Sharma class 12 solution of Minima and maxima exercise 17.2 consists of 14 questions including subparts, that cover up almost the majority of all the topics of the chapter maxima and minima. The concept covered in this chapter are-

• Higher-order derivative test for local maxima and minima

• Theorem based on higher derivative test

• Point of inflection

• Point of inflection

• Properties of maxima and minima

• Maximum and minimum values in a closed interval

• Applied problems on maxima and minima

The RD Sharma class 12 solutions chapter 17 exercise 17.2 is given into two levels that divide the questions into easy, moderate, and tough categories, these two-level questions are designed by maths experts with helpful tips to prepare well for the 12th board exams. The solutions provided in the RD Sharma class 12th exercise 17.2 are best for revision and also help in solving homework as it contains solved questions for reference.

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