RD Sharma Solutions Class 12 Mathematics Chapter 17 FBQ

Edited By Satyajeet Kumar | Updated on Jan 21, 2022 02:52 PM IST

The Class 12 RD Sharma chapter 17 exercise FBQ solution is one of the most prominent solution materials for the chapter of Maxima and Minima. RD Sharma class 12th exercise FBQ gives answers to all the necessary queries that a student might come across while solving the questions. The CBSE students have conquered differences in their performance after practicing from the RD Sharma class 12th solution of Maxima and Minima exercise FBQ.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

This Story also Contains

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

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

Maxima and Minima Excercise: FBQ

Maxima and Minima Exercise Fill in the blanks Question 1

Answer:
$\frac{4}{3}$
Hint:
For maxima or minima we must have ${f}'(x)$
Given:
$f(x)=\frac{1}{4 x^{2}+2 x+1}$
Solution:
We have
$f(x)=\frac{1}{4 x^{2}+2 x+1}$
${f}'(x)=\frac{-\left ( 8x+2 \right )}{\left ( 4 x^{2}+2 x+1 \right )^{2}}$
Put ${f}'(x)=0$
$8x+2=0$
$x=-\frac{1}{4}$
$f^{\prime \prime}(x)=\frac{\left[-\left(4 x^{2}+2 x+1\right)^{2} \times 8-(8 x+2) \times 2 \times\left(4 x^{2}+2 x+1\right)(8 x+2)\right]}{\left[4 x^{2}+2 x+1\right]^{4}}$
Now $f^{\prime \prime}(\frac{-1}{4})$ is negative (point of maxima)
So
$\begin{aligned} &f\left(\frac{-1}{4}\right)=\frac{1}{\left(4 \times\left(\frac{1}{16}\right)-2 \times\left(\frac{1}{4}\right)+1\right)} \\ &f\left(\frac{-1}{4}\right)=\frac{4}{3} \end{aligned}$

Maxima and Minima Exercise Fill in the blanks Question 2

Answer:
-1
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given:
$f(x)=\sin x\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ $f(x)=\sin x\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Solution:
$\begin{aligned} &f(x)=\sin x \\ &f^{\prime}(x)=\cos x \end{aligned}$
For maxima and minima
$\begin{aligned} &f^{\prime}(x)=0 \\ &\cos x=0, x=\frac{\pi}{2},-\frac{\pi}{2} \\ &f^{\prime \prime}(x)=-\sin x>0 \end{aligned}$ When $x=\frac{\pi}{2}$
Now
$f^{\prime \prime}\left ( \frac{\pi}{2} \right )=\sin \left ( -\frac{\pi}{2} \right )$
$=-\sin\frac{\pi}{2}$
$= -1$
Hence Minimum value of
$f(x)=\sin x \quad \text { in }-\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \text { is }-1$

Maxima and Minima Exercise Fill in the blanks Question 3
Answer:

$\sqrt{2}$
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given:
$f\left ( x \right )=\sin x+\cos x$
Solution:
$\begin{aligned} &f(x)=\sin x+\cos x \\ &\frac{d}{d x}=(f(x))=\cos x-\sin x \end{aligned}$ [ $\frac{d}{d x}(\sin x)=\cos x ; \frac{d}{d x}(\cos x)=-\sin x$ ]
${f}'\left ( \cos x-\sin x \right )\rightarrow {f}'\left ( x \right )=0$ for Maxima
${{f}'}'\left ( x \right )=-\left ( \sin x+\cos x \right )$
Therefore
$\sin x=\cos x$ and $\tan x=1$
$\begin{aligned} &x=\frac{\pi}{4} \\ &f(x) \Rightarrow \max \text { at } \frac{\pi}{4} \\ &f(x)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ &f(x)=\sqrt{2} \end{aligned}$

Maxima and Minima Exercise Fill in the blanks Question 4

Answer:
Local minimum
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Solution:
Let f have second derivative at c such that ${f}'\left ( c \right )=0$ and ${{f}'}'\left ( c \right )> 0$ ,then c is a point of local minimum.

Maxima and Minima Exercise Fill in the blanks Question 5

Answer:
Local maximum
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Solution:
Let c is called a point of local maxima. If there is an h > 0 such that $f\left ( c \right )\geq f\left ( x \right )$
for all x in $\left ( c-h,c+h \right )$
The value $f\left ( c \right )$ is called the local maximum value of $f$

Maxima and Minima Exercise Fill in the blanks Question 6

Answer:

Answer:
Local minimum
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Solution:
Let C is called a point of local minimum. If there is an $h< 0$ , such that $f\left ( c \right )\leq f\left ( x \right )$
for all x in $\left ( c-h,c+h \right )$
The value $f\left ( c \right )$ is called the local minimum of $f$

Maxima and Minima Exercise Fill in the blanks Question 7

Answer:
1
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Solution:
$\begin{aligned} &f(x)=x+\frac{1}{x}\\ &f^{\prime}(x)=1-\left(\frac{1}{x^{2}}\right)\\ &\text { Put } f^{\prime}(x)=0\\ &1-\left(\frac{1}{x^{2}}\right)=0\\ &x=\pm 1\\ &f^{\prime \prime}(x)=\frac{2}{x^{3}}\\ \end{aligned}$
$\text { At } x=1, f^{\prime \prime}(x)=\text { Positive }\\$
$\text { At } x=-1, f^{\prime \prime}(x)=\text { Negative }\\$
$f(x) \text { Attains minimum value at } x=1$

Maxima and Minima Exercise Fill in the blanks Question 8

Answer:
$\frac{1}{\sqrt{3}}$
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Solution:
Let the number be $x$ , then the cube of the number is $x^{3}$
$f\left ( x \right )=x-x^{3}$
${f}'\left ( x \right )=1-3x^{2}$
$\begin{aligned} &\text { Put } f^{\prime}(x)=0 \\ &1-3 x^{2}=0 \\ &x=\pm \frac{1}{\sqrt{3}} \\ &f^{\prime \prime}(x)=-6 x[\text { Negative }] \\ &x=\frac{1}{\sqrt{3}} \end{aligned}$

Maxima and Minima Exercise Fill in the blanks Question 9

Answer:
$\sqrt{\frac{b}{a}}$
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given:
$\begin{aligned} &f(x)=a x+\frac{b}{x} \\ \end{aligned}$
Solution:
$\begin{aligned} &f(x)=a x+\frac{b}{x} \\ &f^{\prime}(x)=a-\frac{b}{x^{2}} \\ &f^{\prime}(x)=0 \\ &a=\frac{b}{x^{2}} \text { or } x=\sqrt{\frac{b}{a}} \end{aligned}$
Now
$f^{\prime \prime}(x)=\frac{2 b}{x^{3}}>0 \text { for } x=\sqrt{\frac{b}{a}}$
Thus least value of $f\left ( x \right )$ is $x=\sqrt{\frac{b}{a}}$

Maxima and Minima Exercise Fill in the blanks Question 10

Answer:

2
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given:
$f(x)=a \sin x+\frac{1}{3} \sin 3 x$
Solution:
$\begin{aligned} &f(x)=a \sin x+\frac{1}{3} \sin 3 x \\ &f^{\prime}(x)=a \cos x+\cos 3 x \\ &f^{\prime}\left(\frac{\pi}{3}\right)=0 \text { (At extremum) } f^{\prime}(x)=0 \\ &a\left(\frac{1}{2}\right)-1=0 \\ &a=2 \end{aligned}$

Maxima and Minima Exercise Fill in the blanks Question 12

Answer:
$\frac{1}{e}$
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given: $f\left ( x \right )=y=xe^{-x}$
Solution:
Let $y=xe^{-x}$
Now differentiate it
$\frac{dy}{dx}=e^{-x}-xe^{-x}$ (product rule)
$\frac{dy}{dx}=e^{-x}\left ( 1-x \right )$ For max/min
X=1
Hence maximum value of given expression occurs at x=1
$\begin{aligned} &y_{\max }=(1) e^{-1}=\frac{1}{e} \\ &y_{\max }=\frac{1}{e} \end{aligned}$

Maxima and Minima Exercise Fill in the blanks Question 14

Answer:

1
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given:
Sum of two non-zero numbers is 4
Solution:
Let us consider
$x+y=4$
Or $y=4-x$
That is
$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$
Or $f\left ( x \right )=\frac{4}{xy}=\frac{4}{x\left ( 4-x \right )}$
Now
$\begin{aligned} &f(x)=\frac{4}{\left(4 x-x^{2}\right)} \\ &f^{\prime}(x)=\left(\frac{-4}{\left(4 x-x^{2}\right)^{2}}\right)(4-2 x) \end{aligned}$
Substituting ${f}'\left ( x \right )$ we get $4-2x=0$
${f}'\left ( x\therefore x=2,y=2 \right )$
Therefore
$\begin{aligned} &\min \left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{2}+\frac{1}{2} \\ &\min \left(\frac{1}{x}+\frac{1}{y}\right)=1 \end{aligned}$

Maxima and Minima Exercise Fill in the blanks Question 15

Answer:

2
Hint:
For maxima or minima, we must have ${f}'\left ( x \right )=0$
Given:
x & y are two real numbers such that x >0 and $xy=1$
Solution:
Let $f\left ( x \right )=x+y$ where $xy=1$
$\begin{aligned} &f(x)=x+\frac{1}{x} \\ &f^{\prime}(x)=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}} \\ &f^{\prime \prime}(x)=\frac{2}{x^{3}} \end{aligned}$
On putting ${f}'\left ( x \right )=0$ we get
$x=\underline{+}1$ but $x> 0$ (Neglecting $x=-1$ )
${{f}'}'\left ( x \right )> 0$ for $x=1$
Hence $f\left ( x \right )$ attains minimum at $x=1,y=1$
$\left ( x+y \right )$ has minimum value 2.

Maxima and Minima Exercise Fill in the blanks Question 16

Answer:
$\frac{1}{2}$
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Solution:
Let the number be x, then
$\begin{aligned} &y=x-x^{2} \\ &\frac{d y}{d x}=1-2 x \text { and } \frac{d^{2} y}{d x^{2}}=-2(<0) \\ &1-2 x=0 \\ &x=\frac{1}{2} \end{aligned}$

Maxima and Minima Exercise Fill in the blanks Question 17

Answer:

$\left ( 3,19 \right )$
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given:
$f\left ( x \right )=\left ( x-1 \right )^{2}$
Solution:
$f\left ( x \right )$ is the sum of a square and a constant term. It will acquire minimum value when the square term becomes zero
Hence
$f\left ( x \right )$ is max at $x=-3$
$f\left ( -3 \right )=19$
Or
$\begin{aligned} &f^{\prime}(x)=(x-1)^{2}+1 \\ &f^{\prime}(x)=2(x-1)=0 \\ &x \in[-3,1] \end{aligned}$
$\begin{aligned} &f^{\prime}(x)=2> 0 \end{aligned}$
$x=1$ is local minimum
$\begin{aligned} &f^{\prime \prime}(x)=2>0\\ &x=1 \text { is local minimum }\\ &f(-3)=(-3-1)^{2}+3\\ &=19(m, M)\\ &=(3,19)\\ &f(1)=3 \rightarrow \min =m \end{aligned}$

Maxima and Minima Exercise Fill in the blanks Question 18

Answer:
75
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given:
$\begin{aligned} &f^{\prime}(x)=\left(x^{2}-\frac{250}{x^{2}}\right) \\ \end{aligned}$
Solution:
Let
$\begin{aligned} &f^{\prime}(x)=\left(x^{2}-\frac{250}{x^{2}}\right) \\ \end{aligned}$
Then
$\begin{aligned} &f^{\prime}(x)=\left(2 x-\frac{250}{x^{2}}\right) \\ &f^{\prime \prime}(x)=\left(2+\frac{500}{x^{3}}\right) \\ &f^{\prime}(x)=0 \Rightarrow 2 x^{3}-250=0 \\ &2 x^{3}=250 \\ &x^{3}=125 \\ &x=5 \end{aligned}$
$\begin{aligned} &{{f}'}'(5)=\left(2+\frac{500}{125}\right)=6> 0 \\ \end{aligned}$
$\therefore f\left ( x \right )$ is minimum at x=5 and minimum value
$=\left ( 25+\frac{250}{x} \right )=75$

Maxima and Minima Exercise Fill in the blanks Question 19
Answer:

12
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given:
$y=-x^{3}+3 x^{2}+9 x-27$
Solution: we have, $y=-x^{3}+3 x^{2}+9 x-27$
To find the slope we differentiate the function once
${y}'=-3x^{2}+6x+9$
To find the extremum points we again differentiate at equate it to zero
${{y}'}'=-6x+6$
${{y}'}'=0$
$-6x+6=0$
$x=1$
Now to find whether at the critical points we find a maxima or minima, we use the second derivative test
${{{y}'}'}'=-6$
${{{y}'}'}'\left ( 1 \right )=-6$
${{{y}'}'}'\left ( 1 \right )< 0$
Hence we find maxima at $x=1$ in the equation of the slope. Hence the max value of the slope is ${y}'\left ( 1 \right )$ which is
${y}'\left ( 1 \right )=-3+6+9$
${y}'\left ( 1 \right )=12$

Maxima and Minima Exercise Fill in the blanks Question 20

Answer:

2
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given: $f(x)=\frac{x}{2}+\frac{2}{x}$
Solution:
$\begin{aligned} &f(x)=\frac{x}{2}+\frac{2}{x} \\ &f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}} \end{aligned}$
For extremum ${f}'\left ( x \right )=0$
$\frac{1}{2}-\frac{2}{x^{2}}=0$
$x=\underline{+}2$
${{f}'}'\left ( x \right )=\frac{4}{x^{3}}> 0$ for $x=2$
$\therefore x=2$ is the point of minima
$\therefore$ Given function has local minima at $x=2$

Maxima and Minima Exercise Fill in the blanks Question 21
Answer:

$2\sqrt{ab}$
Hint:
For maxima or minima we must have ${f}'\left ( x \right )=0$
Given:
$f\left ( x \right )=ax+\frac{b}{x}$ $a> 0,b> 0,x> 0$
Solution:
$f\left ( x \right )=ax+\frac{b}{x}$
${f}'\left ( x \right )=a-\frac{b}{x^{2}}$
For critical points
${f}'\left ( x \right )=0$
$a-\frac{b}{x^{2}}=0$
$x^{2}=\frac{b}{a}$
$x=\underline{+}\sqrt{\frac{b}{a}}$
But $x> 0$
$x=\sqrt{\frac{b}{a}}$
${{f}'}'\left ( x \right )=-b\left ( \frac{-2}{x^{3}} \right )$
${{f}'}'\left ( x \right )=\frac{2b}{x^{3}}$
${{f}'}'\ \left ( \sqrt{\frac{b}{a}} \right ) =\frac{2b}{\left ( \sqrt{\frac{a}{b}} \right )^{3}}> 0$ as $a> 0,b> 0$
At $x=\sqrt{\frac{b}{a}}$ we will get minimum value
$f\left ( \sqrt{\frac{b}{a}} \right )=a\cdot \sqrt{\frac{b}{a}}+b\cdot \sqrt{\frac{a}{b}}$
$=\sqrt{ab}+\sqrt{ab}$
$=2\sqrt{ab}$
$\therefore$ The least value of $f\left ( x \right )$ is $=2\sqrt{ab}$

The RD Sharma chapter 12th exercise FBQ solution includes the questions from Maxima and minima, which covers the exercise FBQ consisting of only 21 questions. The FBQ section is pretty important to study as it covers the entire chapter’s concepts and includes questions from all aspects. Some of them are mentioned below:-

Maximum and minimum values of a function
First derivative test for maxima and minima
How to find Local maxima and minima
Theorem and algorithm based on higher-order derivative test
Derivatives of the function

Here are a few reasons why RD Sharma class 12 solutions chapter 17 exercise FBQ is arguably the best guide for students:-

The RD Sharma class 12th exercise FBQ is trusted by students across the country as it has been found that the questions asked in the board exams are similar to the questions provided in the textbook.
RD Sharma class 12th exercise FBQ solutions are designed by experts in maths for the better performance of students.
A student can also take help from the solutions while solving homework, as it gives a brief about the concept and saves time as well.
RD Sharma class 12th exercise FBQ solution will always have questions from the latest and updated versions of the NCERT textbooks.
You can download the RD Sharma class 12 chapter 17 exercise FBQ from the Career360 website and access all the solutions for every chapter.
The Career360 website provides online PDFs of RD sharma solutions for free of cost, so everyone can own the solutions without spending any money.

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

RD Sharma Chapter-wise Solutions

Frequently Asked Questions (FAQs)

1. Where to download the class 12 RD Sharma chapter 17 ex FBQ solution?

Students need to navigate to the Career360 website to download the free copy of the class 12 chapter 17 solution. They can be availed for free of cost.

2. Is It okay to use RD Sharma class 12 ex FBQ for JEE mains preparation?

It is absolutely okay for students to use RD Sharma class 12 chapter 17 ex FBQ for JEE Mains preparation.

3. Does RD Sharma's class 12 exercise FBQ solution covers all the latest syllabus?

Yes, RD Sharma class 12 FBQ solution is designed to cover the latest syllabus and even receives updates to stay relevant. The latest version can be obtained from the Career360 website.

4. Can the RD Sharma class 12 exercise be used for solving homework?

Since school teachers like to give questions from NCERT books, students can use RD Sharma Solutions to find and check their homework answers.

5. How to make use of the class 12 RD Sharma chapter 17 exercise FBQ for self-practice?

Students can solve questions from their NCERT textbooks and compare answers with class 12 RD Sharma chapter 17 exercise FBQ solution to mark their performance on paper.