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

# 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.

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

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

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
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
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.

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• All the sums are solved in the same order as present in the textbook to avoid confusion.

## RD Sharma Chapter-wise Solutions

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