RD Sharma Class 12 Exercise 17.2 Maxima And Minima Solutions Maths

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.

Latest: JEE Main: high scoring chapters | Past 10 year's papers

Don't Miss: Most scoring concepts for NEET | NEET papers with solutions

New: Aakash iACST Scholarship Test. Up to 90% Scholarship. Register Now

This Story also Contains

RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise
Maxima and Minima Excercise: 17.2
RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise

Maxima and Minima Excercise: 17.2

Minima and Maxima Exercise 17.2 Question 1
Answer:

$x=5$ 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

Answer:
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

Answer:
$x= 1$ , $x= \frac{3}{5}$ 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

Answer:
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

Answer:
$x=0$ 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

Answer:
$x= 3$ 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

Answer:
$x=\frac{\pi}{4}$ 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
$\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 2 \cos 2 x=0 \\ &\Rightarrow \cos 2 x=0 \quad[\because 2 \neq 0] \\ &\Rightarrow 2 x=(2 n+1) \frac{\pi}{2} \quad ;\left[[\cos \theta=0] \Rightarrow \theta=(2 n+1) \frac{\pi}{2}\right] \\ &\Rightarrow x=(2 n+1) \frac{\pi}{4} ; n \in N \end{aligned}$
$\Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4} \quad[\text { neglecting other value of } x \text { since } 0<x<\pi]$
+ - +
-∞ $\frac{\pi}{4}$ $\frac{3\pi}{4}$ ∞

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
$\begin{aligned} f\left(\frac{3 \pi}{4}\right) &=\sin \left(2 \frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{2} \\ &=\sin \left(\pi+\frac{\pi}{2}\right)=-\sin \frac{\pi}{2} \quad[\because \sin (\pi+\theta)=-\sin \theta] \\ &=-1 \quad\left[\because \sin \frac{\pi}{2}=1\right] \end{aligned}$

Maxima and Minima Exercise 17.2 Question 8
Answer:

$x=\frac{3 \pi}{4}$ 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:
$f(x)=\sin x-\cos x, 0<x<2 \pi$
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

$f^{\prime}(x)$

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

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

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

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

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

+ - +

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

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

Answer
$x= \frac{\pi}{3}$ 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
$\begin{aligned} &f^{\prime}(x)=0 \quad \Rightarrow-(1-2 \operatorname{Cos} x)=0 \Rightarrow 1-2 \operatorname{Cos} x=0 \\ &\Rightarrow 2 \cos x-1=0 \Rightarrow 2 \cos x=1 \\ &\Rightarrow \cos x=\frac{1}{2} \quad \Rightarrow \cos x=\cos \frac{\pi}{3} \quad\left[\because \frac{1}{2}=\cos \frac{\pi}{3}\right] \\ &\Rightarrow x=2 n \pi \pm \frac{\pi}{3} \quad, n \in \mathbb{Z}[\because \operatorname{Cos} \theta=\operatorname{Cos} \alpha \Rightarrow \theta=2 n \pi \pm \propto, n \in \mathbb{Z}] \\ &x=\pm \frac{\pi}{3}, 2 \pi \pm \frac{\pi}{3},-2 \pi \pm \frac{\pi}{3} \\ &x=\frac{\pi}{3},-\frac{\pi}{3}\left[\text { Neglecting other values of } x \text { since }-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\right] \end{aligned}$
- + -
-∞ $-\frac{\pi}{3}$ $\frac{\pi}{3}$ ∞

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

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

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

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.

The RD Sharma class 12th exercise 17.2 is trusted by a thousand of students for its originality and basic concepts that is easy to understand because the RD Sharma class 12th exercise 17.2 is prepared by hand-picked experts in maths from around the country and in addition they also provide tips to guide students through the process of understanding maths in an easy alternate way.

The study material for the RD Sharma class 12th exercise 17.2 can be accessed online through the website of Careers360 for free of cost. Even all the materials of RD Sharma solutions are available on the website of career360 which can be downloaded online without paying any charge for it.

RD Sharma Chapter-wise Solutions

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

Frequently Asked Question (FAQs)

1. Do the RD Sharma books contain all the latest updates?

Yes, the RD Sharma books are updated from time to time with any change in the CBSE syllabus or question pattern.

2. What is the price of RD Sharma class 12th exercise 17.2 solutions?

RD Sharma class 12 chapter 17 ex 17.2 pdf does not require students to purchase it. You can get the free E-book through the Career360 website.

3. What is the total number of questions in RD Sharma solution of chapter 12 ex 17.2?

The RD Sharma solution of chapter 12 ex 17.2 consists of 14 questions that cover the entire chapter, making it efficient enough to follow and study.

4. Which secondary material is helpful in solving homework?

The RD Sharma solution is super helpful to solve homework. Many teachers use this book to give all the homework to their students. Solving homework with the help of RD Sharma class 12 chapter 17 exercise 17.2 solutions can help solve questions accurately.

5. Which is a trusted source to download RD Sharma solution pdfs?

Students can simply download the online free PDF of any RD Sharma solutions book from the official website of Career360.