RD Sharma Class 12 Exercise 9.2 Differentiability Solutions Maths

RD Sharma Class 12 Exercise 9.2 Differentiability Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 04:11 PM IST

Class 12 RD Sharma chapter 9 exercise 9.2 solution book is now available for the welfare of the class 12 students. The expert provided solution book is a boon for the class 12 students to learn mathematical concepts easily. Therefore, the RD Sharma Class 12 Solutions Differentiability Ex 9.2 serves an essential purpose for the students.

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

Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 9 Differentiability - Other Exercise

Differentiability Excercise:9.2

Differentiability Exercise 9.2 Question 1

Answer: 4
Hint:Differentiate f(x) w.r.t x. Then, put x = 2 in f `(x).
Given: $f\left ( x \right )=x^{2}$
Solution:
Differentiating f(x) w.r.t x then,
$\Rightarrow \frac{d}{dx}\left \{ f\left ( x \right ) \right \}=\frac{d}{dx}\left ( x^{2} \right )=2x^{2-1}$ $\left [ \because \frac{d}{dx}\left ( x^{n} \right )=nx^{n-1} \right ]$
$\therefore f\left ( x \right )=2x$ $\left [ we\: \: can \: \: write\: f'\left ( x \right )=\frac{d}{dx}\left \{ f\left ( x \right ) \right \} \right ]$

Now put x = 2 in f `(x), then

$f'\left ( 2 \right )=2\left ( 2 \right )=4$

Differentiability Exercise 9.2 Question 3

Answer: f ` (1) = f ` (2) $\Rightarrow$ L.H.S = R.H.S
Hint: Differentiate f(x) to find f `(x). Put x = 1 and x = 2 in f `(x). Then, we will check L.H.S and R.H.S are equal or not.
Given: $f\left ( x \right )=2x^{3}-9x^{2}+12x+9$
Solution:
Differentiating f(x) w.r.t x then,
$\begin{aligned} &\Rightarrow \frac{d}{d x}\{\mathrm{f}(\mathrm{x})\}=\frac{d}{d x}\left(2 \mathrm{x}^{3}-9 \mathrm{x}^{2}+12 \mathrm{x}+9\right) \\ &=\frac{d}{d x}\left(2 \mathrm{x}^{3}\right)+\frac{d}{d x}\left(-9 \mathrm{x}^{2}\right)+\frac{d}{d x}(12 \mathrm{x})+\frac{d}{d x}(9) \end{aligned}$
$\left[\because \frac{d}{d x}\left(a x^{3}+b x^{2}+c x+d\right)=\frac{d}{d x}\left(a x^{3}\right)+\frac{d}{d x}\left(b x^{2}\right)+\frac{d}{d x}(c x)+\frac{d}{d x}(d)\right]$
$\begin{aligned} &=2 \frac{d}{d x}\left(x^{3}\right)-9 \frac{d}{d x}\left(x^{2}\right)+12 \frac{d}{d x}(x)+0 \\ &=2\left(3 x^{3-1}\right)-9\left(2 x^{2-1}\right)+12\left(1 x^{1-1}\right) \end{aligned}$ $\left [ \because \frac{d}{dx}\left ( constant \right )=0\left [ \because \frac{d}{dx}\left ( ax^{n}=a\frac{d}{dx}\left ( x^{n} \right ) \right ) \right ] \right ]$
$=6x^{2}-18x+12$ $\left [ \because \frac{d}{dx}\left ( x^{n} \right )=nx^{n-1} \right ]$
Now, $f'\left ( 1 \right )=6\left ( 1\right )^{2}-18\left ( 1 \right )+12$
$=6-18+12$
$=18-18$
$=0$
$f'\left ( 2 \right )=6\left ( 2 \right )^{2}-18\left ( 2 \right )+12$
$=24-36+12$
$=36-36$
$=0$
$\therefore f'\left ( 1 \right )=f'\left ( 2 \right )=0$ [Equal]

Differentiability Exercise 9.2 Question 4

Answer: $\Phi =9$
Hint: Differentiate f(x) to find f `(x). Put x = 5 in f `(x). To find the value of $\Phi$ , equate R.H.S of f ` (5) from given and calculate.
Given: $f\left ( x \right )=\Phi x^{2}+7x-4$
Solution:
Differentiating f(x) w.r.t x then,
$\Rightarrow \frac{d}{d x}\{\mathrm{f}(\mathrm{x})\}=\frac{d}{d x}\left(\Phi \mathrm{x}^{2}+7 \mathrm{x}-4\right) \; \; \; \; \; \; \; \; \; \quad\left[\because \frac{d}{d x}\left(a x^{2}+b x+c\right)=\frac{d}{d x}\left(a x^{2}\right)+\frac{d}{d x}(b x)+\frac{d}{d x}(c)\right]$
$\begin{array}{ll} =\frac{d}{d x}\left(\Phi \mathrm{x}^{2}\right)+\frac{d}{d x}(7 \mathrm{x})+\frac{d}{d x}(-4) & {\left[\because \frac{d}{d x}(\text { constant })=0\right]} \; \; \; \\ =\Phi \frac{d}{d x}\left(\mathrm{x}^{2}\right)+7 \frac{d}{d x}(\mathrm{x})+0 & {\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right]} \\ =\Phi\left(2 \mathrm{x}^{2-1}\right)+7\left(1 \mathrm{x}^{1-1}\right)+0 & {\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]} \end{array}$
$=2\Phi x+7$
$\therefore f'\left ( x \right )=2\Phi\; x+7$
We Know that
$f'\left ( 5 \right )=97$
$2\: \Phi \left ( 5 \right )+7=97$
$10\: \Phi =97-7$
$10\: \Phi =90$
$\Phi =9$

Differentiability Exercise 9.2 Question 5

Answer: 112
Hint: Differentiate f(x) to find f `(x). Put x = 4 in f `(x).
Given: $f\left ( x \right )=x^{3}+7x^{2}+8x-9$
Solution:
Differentiating f(x) w.r.t x then,
$\begin{aligned} \Rightarrow \frac{d}{d x}\{\mathrm{f}(\mathrm{x})\} &=\frac{d}{d x}\left(\mathrm{x}^{3}+7 \mathrm{x}^{2}+8 \mathrm{x}-9\right) \\ &=\frac{d}{d x}\left(\mathrm{x}^{3}\right)+\frac{d}{d x}\left(7 \mathrm{x}^{2}\right)+\frac{d}{d x}(8 \mathrm{x})+\frac{d}{d x}(-9) \end{aligned}$
$\left[\because \frac{d}{d x}\left(a x^{2}+b x+c\right)=\frac{d}{d x}\left(a x^{2}\right)+\frac{d}{d x}(b x)+\frac{d}{d x}(c)\right]$
$=\frac{d}{d x}\left(\mathrm{x}^{3}\right)+7 \frac{d}{d x}\left(\mathrm{x}^{2}\right)+8 \frac{d}{d x}(\mathrm{x})+0\; \; \; \; \; \; \quad\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right]$
$=3\: x^{3-1}+7\left ( 2x^{2-1} \right )+8\left ( 1x^{1-1} \right )$
$\left [ \because \frac{d}{dx}\left ( constant \right ) =0,\left [ \because \frac{d}{dx}\left ( x^{n} \right )=nx^{n-1} \right ]\right ]$
$\therefore f'\left ( x \right )=3x^{2}+14x+8$
Now, for f ` (4), we will put x = 4 in f ` (x), then:

$f'\left ( 4 \right )=3\left ( 4 \right )^{2}+14\left ( 4 \right )+8$

$=48+56+8$

$=112$

Differentiability Exercise 9.2 Question 6

Answer: $f'\left ( 0 \right )=m$
Hint: Differentiate f(x) to find f `(x). Put x = 0 in f `(x).
Given: $f\left ( x \right )=mx+c$
Solution:
Differentiating f(x) w.r.t x then,
$\Rightarrow \frac{d}{dx}\left \{ f\left ( x \right ) \right \}=\frac{d}{dx}\left ( mx+c \right )$
$=\frac{d}{dx}\left ( mx \right )+\frac{d}{dx}\left ( c \right )$
$\begin{aligned} &{\left[\because \frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)\right]} \\ &=m \frac{d}{d x}(\mathrm{x})+\frac{d}{d x}(\mathrm{c}) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \end{aligned}$
$m=\left ( 1x^{1-1} \right )+0$ $\left[\because \frac{d}{d x} \text { (constant) }=0,\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]\right]$
$\therefore$ f `(x) = m
For f `(0), put x = 0 in f `(x), then
f `(x) = m
$\Rightarrow$ f ` (x) is not affected at x = 0, since in f `(x) there is no term of x.

Differentiability Exercise 9.2 Question 7

Answer: Function is differentiable at x = 0 & not differentiable at x = -2
Hint: Function is differentiable on x $\epsilon$ [-3, 2], x $\epsilon$ (-2, 0) and x $\epsilon$ [0, 1]. So, we need to check differentiability of f(x) at x = - 2 and x = 0.
Given: $f\left ( x \right )=$ $\left\{\begin{array}{c} 2 x+3, \text { if }-3 \leq \mathrm{x}<-2 \\ x+1, \text { if }-2<x<0 \\ x+2, \text { if } 0 \leq \mathrm{x} \leq 1 \end{array}\right.$
Solution:
To check differentiability at x = -2,
$(\mathrm{LHD} \text { at } \mathrm{x}=-2)=\lim _{x \rightarrow-2^{-}} \frac{f(x)-f(-2)}{x-(-2)}$
$=\lim _{x \rightarrow-2^{-}} \frac{2x+3-\left ( 2\left ( -2 \right ) +3\right )}{x+2}$ $\left [ \because f\left ( x \right )=2x+3,x< -2 \right ]$
$\begin{aligned} &=\lim _{x \rightarrow-2^{-}} \frac{2 x+3-(-4+3)}{x+2} \\ &=\lim _{x \rightarrow-2^{-}} \frac{2 x+3-(-1)}{x+2} \\ &=\lim _{x \rightarrow-2^{-}} \frac{2 x+4}{x+2} \\ &=\lim _{x \rightarrow-2^{-}} \frac{2(x+2)}{x+2} \\ &=2 \end{aligned}$
$(\mathrm{RHD} \text { at } \mathrm{x}=-2)=\lim _{x \rightarrow-2^{+}} \frac{f(x)-f(-2)}{x-(-2)}$
$=\lim _{x \rightarrow-2^{+}} \frac{x+1-\left ( -2+1 \right )}{x-\left ( -2 \right )}$ $\left [ \because f\left ( x \right )=x+1,x> -2 \right ]$
$=\lim _{x \rightarrow-2^{+}} \frac{x+1+1}{x+2}$
$=\lim _{x \rightarrow-2^{+}} \frac{x+2}{x+2}$
$=1$
$\because$ (LHD at x = -2) $\neq$ (RHD at x = -2)
Thus, f(x) is not differentiable at x = -2.
Now, we need to check differentiability at x = 0.
$\begin{aligned} (\mathrm{LHD} \text { at } \mathrm{x}=0) &=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{x \rightarrow 0^{-}} \frac{x+1-(0+1)}{x-0} \end{aligned}$ $\left [ \because f\left ( x \right ) =x+1,x< 0\right ]$
$=\lim _{x \rightarrow 0^{-}} \frac{x}{x}$
$=1$
$\begin{aligned} (\mathrm{RHD} \text { at } \mathrm{x}=0) &=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{x \rightarrow 0^{+}} \frac{x+2-(0+2)}{x-0} \end{aligned}$ $\left [ \because f\left ( x \right ) =x+2,x< 0\right ]$
$=\lim _{x \rightarrow 0^{+}} \frac{x}{x}$
$=1$
$\because$ (LHD at x = 0) = (RHD at x = 0)
Thus, f(x) is differentiable at x = 0.

Differentiability Exercise 9.2 Question 8

Answer: $f\left ( x \right )=|x-2|+|x-3|+|x-4|+|x-5|+|x-6|$
Hint: Modulus function f(x) = $|x|$ is continuous but not differentiable at x = 2,3,4,5,6
Given: To write a function which is everywhere continuous but fails differentiability exactly at five points.
Solution:
We know that, polynomial functions are always continuous and differentiable.
Let us consider, $f\left ( x \right )=|x|$
Modulus functions are always continuous.
So, $|x|$ is continuous.
Now, there is more than one slope possible. So it is not differentiable as we know that
Using mod function, we can write the function which is non-derivable at exactly 5 points and continuous always is
$f\left ( x \right )=|x-2|+|x-3|+|x-4|+|x-5|+|x-6|$

Differentiability Exercise 9.2 Question 9

Answer: Function is differentiable at x = 0 & not differentiable at x = -2
Hint: Using graphical approach, we can observe that modulus function has more than one slope possible.
Given: $f\left ( x \right )=|log|x||$
Solution:
The given function is absolute function. So it is continuous. Consider the graph of the given function.

Using graphs of the function, we can observe that the function f(x) is not differentiable at x= --1 and x = 1 but continuous for all x expect 0

Differentiability Exercise 9.2 Question 10

Answer: Not differentiable at x = 0
Hint: Using graphical approach, we can observe that modulus function has more than one slope possible.
Given: $f\left ( x \right )=e^{|x|}$
Solution:
We know that exponential function is always continuous. So, the given function is continuous. Consider the graph of the given function.

Using graph of the function, it is clear that f(x) is not differentiable at x = 0 but continuous for all x.

Differentiability Exercise 9.2 Question 11

Answer: Function is continuous but not differentiable at x = c
Hint: To check whether the function f(x) is differentiable at point x=c, we need to check that $f'\left ( c^{-} \right )=f'\left ( c^{+} \right )$
Given: $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} (x-c) \cos \left(\frac{1}{x-c}\right), \mathrm{x} \neq \mathrm{c} \\ 0, x=c \end{array}\right.$
Solution:
It is given that, f(c) = 0.
Consider,
$\lim _{x \rightarrow c}(x-c) \cos \left(\frac{1}{x-c}\right)$
Substituting x – c = y, we get:
$\begin{aligned} \lim _{y \rightarrow 0} y \cos \left(\frac{1}{y}\right) &=\lim _{y \rightarrow 0} y \cdot \lim _{y \rightarrow 0} \cos \left(\frac{1}{y}\right) \\ &=0 \cdot \lim _{y \rightarrow 0} \cos \left(\frac{1}{y}\right) \end{aligned}$
$=0$
So, $\lim_{x\rightarrow c}f\left ( x \right )=f\left ( c \right )=0$
Thus, f(x) is continuous at x = c.
Now, we need to check differentiability at x=c.
Using the formula, we have:
$f'\left ( c \right )=\lim_{x\rightarrow c}\frac{f\left ( x \right )-f(c)}{x-c}$
So, LHD is given by,
$f'\left ( c \right )=\lim_{x\rightarrow c}\frac{f\left ( x \right )-f(c)}{x-c}$
$=\lim_{x\rightarrow c}\frac{\left ( x-c \right )\cos \left ( \frac{1}{x-c} \right )-0}{x-c}$
$=\lim_{x\rightarrow c}\cos \left ( \frac{1}{x-c} \right )$
Put x = c – h

$f'\left ( c \right )=\lim_{x\rightarrow c}\cos \left ( \frac{1}{c-h-c} \right )$

$=\lim_{x\rightarrow c}\cos \frac{1}{h}$ $\left [ \cos \left ( \Theta \right )=\cos \Theta \right ]$
Since the value of cos is going to infinity, its limit will oscillate between -1 and 1.
Now, R.H.D is given by,
$\begin{aligned} \mathrm{f}^{\prime}\left(\mathrm{c}^{+}\right) &=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c} \\ &=\lim _{x \rightarrow c^{+}} \frac{(x-c) \cos \left(\frac{1}{x-c}\right)-0}{x-c} \\ &=\lim _{h \rightarrow 0} \cos \left(\frac{1}{c-h-c}\right) \\ &=\lim _{h \rightarrow 0} \cos \left(\frac{1}{h}\right) \end{aligned}$
Since the value of cos is going to infinity, its limit will oscillate between -1 and 1. As the value of limit is not a finite value, the function is not differentiable.

Differentiability Exercise 9.2 Question 12

Answer: $|\sin x|$ is not differentiable at $x=n\pi$ and $|\cos x|$ is differentiable everywhere.
Hint:
Check LHD at $x=n\pi$ is equal to RHD at $x=n\pi$ or not for $n$ is even and $n$ is odd respect for the function $|\sin x|$
Given:
Given functions are $|\sin x|$ and $|\cos x|$ .
Solution:
Let $f\left ( x \right )=|\sin x|$
$= \begin{cases}-\sin x & , x<n \pi \\ \sin x & , x \geq n \pi\end{cases}$
Case 1
For $x=n\pi$ (Where $n$ is even)
LHD at $x=n\pi$
$\begin{aligned} &=\lim _{x \rightarrow n-1} \frac{f(x)-f(n \pi)}{x-n \pi} \\ &=\lim _{h \rightarrow 0} \frac{-\sin (n \pi-h)-\sin n \pi}{n \pi-h-n \pi} \\ &=\lim _{h \rightarrow 0} \frac{\sinh -0}{-h} \end{aligned}$
LHD at $x=n\pi =-1$ … (i)
RHD at $x=n\pi,$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\sin (n \pi+h)-\sin n \pi}{h} \\ &=\lim _{h \rightarrow 0} \frac{\sinh }{h} \end{aligned}$
RHD at $x=n\pi =1$ ....(iii)
From(i) and (ii)
LHD at $x=n\pi \neq RHD \: at\: x=n\pi$
Case II
For $x=n\pi$ (Where $n$ is odd)
LHD at $x=n\pi$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\sin (n \pi-h)-\sin n \pi}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sinh }{-h} \end{aligned}$
RHD at $x=n\pi =-1$ .....(iv)
From (iii) and (iv)
LHD at $x=n\pi \neq RHD \: at\: x=n\pi$
Thus, $f\left ( x \right )=|\sin x|$ is not differentiable at $x=n\pi$
Now we need to check $\cos |x|$ is differentiable or not
Let,
$g\left ( x \right )=\cos |x|$
$\cos \left ( -x \right )=\cos x$
$g\left ( x \right )=\cos \: x$
$g\left ( x \right )=\cos |x|$ is differentiable everywhere.

Differentiability Exercise 9.2 Question 6

Answer: f(x) is not differentiable at x = 1 but differentiable at x = 2
Hint: The function y = f(x) is said to be differentiable in the closed interval [a, b] if R f ` (a) and L f ` (b) exist and f `(x) exist for every point of (a, b).
If the left hand limit, right hand limit and the value of the function at x = c exist and are equal to each other, then f is said to be continuous at x = c.
Given: $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} x, x \leq 1 \\ 2-x, 1 \leq \mathrm{x} \leq 2 \\ -2+3 x-x^{2}, x>2 \end{array}\right.$
Solution:
Differentiability at x = 1:
$\begin{aligned} &\mathrm{LHD} \text { at } \mathrm{x}=0: \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{(0-h)-0} \\ &=\lim _{h \rightarrow 0} \frac{(0-h)^{m} \sin \left(\frac{1}{-h}\right)}{-h} \\ &=\lim _{h \rightarrow 0}(-h)^{m-1} \sin \left(\frac{1}{-h}\right) \end{aligned}$ $\left \{ \frac{a^{m}}{a^{n}}\Rightarrow a^{m-n} \right \}$
$=\lim _{h \rightarrow 0}(-h)^{m-1} \sin \left(\frac{1}{h}\right)$
$\Rightarrow \left \{ when-1\leq \frac{1}{h} \leq 1\right \}$
$\Rightarrow 0\: x\: k$
$\Rightarrow 0$
$\begin{aligned} &\text { RHD at } x=0: \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{(0+h)-0} \\ &=\lim _{h \rightarrow 0} \frac{h^{m} \sin \left(\frac{1}{h}\right)}{h} \end{aligned}$
$\Rightarrow \left \{ when-1\leq k\leq 1 \right \}$ $\left\{\begin{array}{c} (0)^{m-1}=0 \\ \sin \left(\frac{1}{0}\right)=\sin (\infty) \end{array}\right.$
$\Rightarrow 0\; x\: k$
$\Rightarrow 0$
(LHD at x = 0)= (RHD at x = 0)
Hence, f(x) is differentiable at x = 0.

The chapter 9 in mathematics for class 12 consists of two exercises, ex 9.1 and 9.2. the second or the last exercise, ex 9.2 consists of 12 questions to be solved. It covers the topics like differentiability at a point, derivative of the capacity and differentiability in a set. All the solutions for these questions can be found at the RD Sharma Class 12 Solutions Chapter 9 exercise 9.2 reference book.

The Class 12 RD Sharma chapter 9 exercise 9.2 reference material contains additional questions and sample question papers. This allows the students to try various mock tests before facing a class test or exam. Therefore, RD Sharma Class 12 Solutions Chapter 9 Exercise 9.2 makes the students exam-ready.

Continuous practice with the RD Sharma Class 12th Exercise 9.2 Chapter 9 Differentiability solution book will increase their speed in finding the answers. As a result, they would have a lot of time to recheck their answers after writing the exam.

Here are some of the benefits that the students who use the RD Sharma solution books would reap:

The solution books are available at the Career 360 website where not even a single rupee is required to be paid.
The RD Sharma Class 12 Solutions Chapter 9 ex 9.2 can be downloaded as a PDF file at an instance.
Formulae are provided separately for convenience.
The sums are solved in many methods.
Students cross their benchmark scores effortlessly due to proper practice.
All the sums are solved in the same order as present in the textbook to avoid confusion.

RD Sharma Chapter-wise Solutions

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book

Frequently Asked Questions (FAQs)

1. What is the proper way to attain the RD Sharma Class 12th exercise 9.2 Solutions on the Career 360 website?

Not every website provides the authorized set of books, therefore, download the RD Sharma Class 12th Chapter 9 Exercise 9.2 material from the Career 360 website.

2. How many questions are asked in ex 9.2 of class 12 mathematics?

This exercise consists of 12 questions that are solved in the RD Sharma Class 12 Chapter 9 Exercise 9.2 book.

3. Can everyone utilize the RD Sharma Class 12 Mathematics chapter 9 solutions book?

Yes, there are no restrictions or specific access given in use the RD Sharma books in the career 360 website. Therefore, everyone can access it easily.

4. What makes the Class 12 RD Sharma Chapter 9 Exercise 9.2 solution book the best option to select?

Some of the questions for the public exams in the previous years were asked from the Class 12 RD Sharma Chapter 9 exercise 9.2 book. therefore, it is a wise man’s option to select this book.

5. Are the solutions provided in the RD Sharma solutions accurate?

Distinct experts have contributed their knowledge and skill in developing this set of solution books. Therefore, the students are assured about the accuracy of the solutions.