Which Solution book can I use to clarify my doubts regarding the Very Short Answers in the Differentiability concept?

The best solution book to guide you with the VSA part in the Differentiability Chapter is the RD Sharma Class 12th VSA book.

Can a slow-leaner student adapt to the solutions given in the RD Sharma solution books?

All solutions are provided in various methods to arrive at an accurate answer. Each student can select and adapt the way that they find easy. Hence, all types of students, including the toppers, average learners, and slow-learner students, can utilize this material.

Where can I find the RD Sharma books for online study?

You can visit the Career 360 website to access the RD Sharma solution books for any subject.

Are there any different solution materials to refer to the VSAs in the RD Sharma book?

You can use the RD Sharma Class 12th VSA book for the required chapters to clarify your doubts in this portion.

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The RD Sharma books are available for free at the Career 360 website. You can also download it for easy reference.

RD Sharma Solutions Class 12 Mathematics Chapter 9 VSA

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RD Sharma Solutions Class 12 Mathematics Chapter 9 VSA

RD Sharma Solutions Class 12 Mathematics Chapter 9 VSA

Updated on 20 Jan 2022, 04:02 PM IST

The majority of the CBSE Board students use the RD Sharma Solution books to clear their doubts in their homework and score good marks in tests and public examinations. Class 12 mathematics has many portions that the students face challenges in solving. Shortcut methods, tricks, and formulas play a prominent role when it comes to the Very Short Answers (VSA) part. Here is where the RD Sharma Solution books come to help. The RD Sharma Class 12th VSA solution material can be used for clarifying doubts.

This Story also Contains

RD Sharma Class 12 Solutions Chapter 9 VSA Differentiability - Other Exercise
Differentiability Excercise: VSA
RD Sharma Chapter wise Solutions

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

RD Sharma Class 12 Solutions Chapter 9 VSA Differentiability - Other Exercise

Differentiability Excercise: VSA

Differentiability exercise Very short answer type question 1

Answer:

Differentiability of a function at a point:
let$f(x)$ be a real valued function defined on an open interval (a,b) and $c \in(a, b)$
then f(x) is said to be differentiable at a point if $\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}$exists finitely.

Differentiability exercise Very short answer type question 2

Answer: Yes
Hint: With the help of definition of differentiability we will form the condition of continuity.
Given: Function is differentiable
Explanation: Let a function is differentiable at $x=c$, then
$f^{\prime}(c)=\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}$ ............(i)
To prove a function is continuous, we have to show
$\begin{aligned} &\lim _{x \rightarrow c} f(x)=f(c) \\\\ &\text { Now, } \lim _{x \rightarrow c}[f(x)-f(c)]=\lim _{x \rightarrow c}\left[\frac{f(x)-f(c)}{x-c}(x-c)\right] \end{aligned}$[Divide and multiply by$(x-c)$]
$\lim _{x \rightarrow c}[f(x)-f(c)]=\lim _{x \rightarrow c}\left[\frac{f(x)-f(c)}{x-c}\right] \lim _{x \rightarrow c}(x-c)$
$\lim _{x \rightarrow c}[f(x)-f(c)]=f^{\prime}(c) \lim _{x \rightarrow c}(x-c)$ [From equation (i)]
$\begin{aligned} &\lim _{x \rightarrow c}[f(x)-f(c)]=f^{\prime}(c) \times 0 \\\\ &\lim _{x \rightarrow c}[f(x)-f(c)]=0 \\\\ &\lim _{x \rightarrow c} f(x)=f(c) \end{aligned}$
Hence the given function is continuous.
So, every differentiable function is continuous.

Differentiability exercise Very short answer type question 3

Answer: No
Explanation: Every continuous function is not differentiable.
For example,$f(x)=|x|=\left\{\begin{array}{ll} x & x>0 \\ -x & x<0 \end{array}\right\}$is continuous at x=0.
But at x=0,
$\begin{aligned} L H D &=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\\\ &=\lim _{x \rightarrow 0^{-}} \frac{-x-0}{x-0}=\lim _{x \rightarrow 0}-1 \\\\ &=-1 \end{aligned}$
$\begin{aligned} R H D &=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\\\ &=\lim _{x \rightarrow 0^{-}} \frac{x-0}{x-0}=\lim _{x \rightarrow 0} 1 \\\\ &=1 \end{aligned}$
As $L H D \neq R H D$
Therefore, $\left | x \right |$ is not differentiable at x=0.

Differentiability exercise Very short answer type question 4

Answer: $\mathrm{f}(\mathrm{x})=|x| \text { at } \mathrm{x}=0$

Explanation: let $f(x)=|x|=\left\{\begin{array}{ll} x & x>0 \\ -x & x<0 \end{array}\right\}$
Now $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}-x=0 \text { and } \lim _{x \rightarrow 0^{+}} f(x)=0$

Differentiability exercise Very short answer type question 5

Answer:$f(c)$
Given:$f(x)$ is differentiable at $x=c$
Explanation: as $f(x)$ is differentiable at $x=c$ then it is continuous also. As every differentiable function is continuous.
So, $\lim _{x \rightarrow c} f(x)=f(c)$

Differentiability exercise Very short answer type question 6

Answer: Does not exist
Hint: if $f(x)$ is differentiable at $x=2$ then $\lim _{x \rightarrow c} \frac{f(x)-f(2)}{x-2}=f^{\prime}(2)$
Given:$f(x)=|x-2|$
Explanation:$f(x)=\left(\begin{array}{ll} x-2 & x \geq 2 \\ -(x-2) & x<2 \end{array}\right)$
Now, LHD at x=2
$\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{h \rightarrow 0} \frac{-(2-h-2)-0}{-h}$
$\begin{aligned} &=-1 \\\\ &\text { RHD at\; x }=2 \\\\ &\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2} \frac{(2+h-2)-0}{h} \\\\ &\text { As } L H D \neq R H D \end{aligned}$
$f(x)$ is not differentiable so $f^{'}(2)$ does not exist.

Differentiability exercise Very short answer type question 7

Answer: 1
Hint: if $f(x)$ is not differentiable then $L H D \neq R H D$
Given:$f(x)=\left|\log _{e} x\right|$ is not differentiable.
Explanation: $f(x)=\left|\log _{e} x\right|=\left\{\begin{array}{cc} \log _{e} x & x \geq 1 \\ -\log _{e} x & 0<x<1 \end{array}\right\}$
as logarithmic function is differentiable in its domain. So we only have to check at x=1
$\begin{aligned} &L H D=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\\\ &=\lim _{x \rightarrow 1} \frac{-\log _{e} x-0}{x-1} \\\\ &=\lim _{x \rightarrow 1} \frac{-\log x}{x-1} \end{aligned}$
applying L' Hospital rule
$\begin{aligned} &=\lim _{x \rightarrow 1} \frac{-1}{x} \\ &=-1 \end{aligned}$
$\begin{aligned} &R H D \text { atx }=1 \\\\ &\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\\\ &\lim _{x \rightarrow 1^{+}} \frac{\log _{e} x-\log 1}{x-1} \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 1} \frac{\log _{e} x-0}{x-1} \\\\ &=\lim _{x \rightarrow 1} \frac{\log x}{x-1} \end{aligned}$
applying L' Hospital rule
$\begin{aligned} &=\lim _{x \rightarrow 1} \frac{1}{x} \\\\ &=1 \\\\ &\text { As } L H D \neq R H D_{\text {at }} \mathrm{x}=1 \end{aligned}$
$\left|\log _{e} x\right|$ is not differentiable at $x=1$

Differentiability exercise Very short answer type question 8

Answer:$\pm 1$
Hint: $\log x$ is differentiable in its domain.
Given:$f(x)=|\log | x \|$
Explanation:
$|x|= \begin{cases}-x & , \infty<x<-1 \\ -x & ,-1<x<0 \\ x & , 0<x<1 \\ x & , 1<x<\infty\end{cases}$
$\log |x|= \begin{cases}\log (-x),-\infty<x<-1 \\ \log (-x),-1<x<0 \\ \log x, & , 0<x<1 \\ \log x \quad, 1<x<\infty\end{cases}$
$|\log | x \|=\left\{\begin{array}{c} -\log (-x),-0<x<-1 \\ -\log (-x),-1<x<0 \\ \log x \quad, 0<x<1 \\ \log x, 1<x<\infty \end{array}\right.$
We have to check differentiability at$\pm 1$
$\begin{aligned} &\text { LHD at } x=1\\ &\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\log (1-h)-\log 1}{-h}\\\\ &=\lim _{h \rightarrow 0} \frac{1}{\frac{1-h}{-1}}=-1 \quad \text { [L-Hospital rule] } \end{aligned}$ ......(i)
$=-1$
$\begin{aligned} \operatorname{RHD} \text { at } x=1 \\ \lim _{x \rightarrow T^{-}} \frac{f(x)-f(1)}{x-1}=& \lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\log (1+h)-\log 1}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{1+h}=1 \quad[\mathrm{~L} \text {-Hospital rule }] \end{aligned}$ ......(ii)
As $LHD \neq RHD$
So the function is not differentiable at $x=1$
$\begin{aligned} &\text { At } x=-1 \\ &\text { LHD at } x=-1 \\ &\begin{aligned} \lim _{x \rightarrow-1^{-}} \frac{f(x)-f(-1)}{x-(-1)} &=\lim _{h \rightarrow 0} \frac{f(-1-h)-f(-1)}{-1-h-(-1)} \\ &=\lim _{h \rightarrow 0} \frac{-\log \{-(-1-h)\}-[-\log \{-(-1)\}]}{-h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{-1}{1+h}-0}{-1}=1 \quad \text { [L-Hospital rule] } \end{aligned} \end{aligned}$ .......(iii)
$=1$
$\begin{aligned} \mathrm{RHD} \text { at } x=-1 \\ \lim _{x \rightarrow-1^{+}} \frac{f(x)-f(-1)}{x-(-1)} &=\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{-1+h+1} \\ &=\lim _{h \rightarrow 0} \frac{-\log \{-(-1+h)\}-[-\log \{-(-1)\}]}{h} \\ &=\lim _{h \rightarrow 0} \frac{-1}{1-h}=-1 \quad[\mathrm{~L}-\mathrm{Hospital} \text { rule }] \end{aligned}$ ......(iv)

As $LHD \neq RHD$
So the function is not differentiable at $x=-1$

Differentiability exercise Very short answer type question 9

Answer 0
Hint : If $f(x)$ is differentiable at $x=0$ then $\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=f^{\prime}(0)$ exists.
Given:$\mathrm{f}(\mathrm{x})=|\mathrm{x}|^{3}$
Explanation:$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc} x^{3} & x \geq 0 \\ -x^{3} & x<0 \end{array}\right.$
$\begin{aligned} &\text { LHD of } \mathrm{x}=0\\ &\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{-x^{3}-0}{x}=0\\\\ &R H D \text { at } x=0\\\\ &\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{x^{3}-0}{x}=0 \end{aligned}$
$\begin{aligned} &\text { As } R H D=L H D\\\\ &f(x) \text { is differentiable at } x=0\\\\ &f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=0 \end{aligned}$

Differentiability exercise Very short answer type question 10

Answer: $x=0,1$
Hint: $|x|= \begin{cases}x & x>0 \\ -x & x<0\end{cases}$
Given:$f(x)=|x|+|x-1|$
Explanation:
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc} x+x-1 & x>1 \\ x-(x-1) & 0 \leq x \leq 1 \\ -x-(x-1) & x<0 \end{array}\right.$
$=\left\{\begin{array}{lr} 2 x-1 & x>1 \\ 1 & 0 \leq x \leq 1 \\ -2 x+1 & x<0 \end{array}\right.$
$f(x)$ is a polynomial function for $x > 1, 0\leq x\leq 1$ and $x < 0$ . so, $f(x)$ is differentiable for all $x>1, 0\leq x\leq 1$ & $x < 0$
We only have to check at $x= 0, 1$
$LHD\; at\; x = 0$
$\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=& \lim _{x \rightarrow 0} \frac{-2 x+1-1}{x-0} \\ &=-2 \end{aligned} \quad \text { [L-Hospital rule ] }$$\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{1-1}{x-0}$

$=0$

$\text { As } \mathrm{LHD} \neq \mathrm{RHD} \text { at } \mathrm{x}=0$
$f(x)$ is not differentiable at $x=0$
$\begin{aligned} &\text { At } x=1 \\ &\text { LHD at } x=1 \end{aligned}$

$\begin{aligned} &\text { At } x=1 \\ &\text { LHD at } x=1 \\ &\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 0} \frac{1-1}{x-1}=0 \end{aligned}$

$\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=1 \\ &\begin{aligned} \lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} &=\lim _{x \rightarrow 0} \frac{2 x-1-1}{x-1} \\ &=2 \end{aligned} \quad \text { [L-Hospital rule] } \\ & \end{aligned}$
$\text { As } \mathrm{LHD} \neq \mathrm{RHD} \text { at } \mathrm{x}=1$
$f(x)$ is not differentiable at $x=1$ so that point of non-differentiability
are $0, 1$

Differentiability exercise Very short answer type question 10
Edit Q

Question:10

Differentiability exercise Very short answer type question 10

Answer:

$=0$

Differentiability exercise Very short answer type question 11

Answer:$f(c)$
Hint: If a function is differentiable then it is continuous
Given: $\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}$ exists

Explanation: From above we know if limit exists then it is differentiable and every differentiable function is continuous .
So, from the definition of continuity $\lim _{x \rightarrow c} \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{c})$

Differentiability exercise Very short answer type question 12

Answer: 0
Hint: If f(x) is differentiable at x=2 then $\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}=f^{\prime}(2)$
Given:$\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+|\mathrm{x}-3|$
Explanation:
$f(x)=\left\{\begin{array}{cc} -(x-1)-(x-3) & x<1 \\ x-1-(x-3) & 1<x<3 \\ (x-1)+(x-3) & x>3 \end{array}\right.$
$f(x)=\left\{\begin{array}{lr} -2 x+4 & x<1 \\ 2 & 1<x<3 \\ 2 x-4 & x>3 \end{array}\right.$
Now, at x=2 , f(x)=2 so the function is differentiable at x=2 as it is a polynomial Function.
Hence , from the definition of differentiability
$\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \text { exists. }$
So,
$\begin{aligned} &\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}=f^{\prime} \\\\ &\lim _{x \rightarrow 2} \frac{2-2}{x-2}=f^{\prime} \\\\ &f^{\prime}(2)=0 \end{aligned}$

Differentiability exercise Very short answer type question 13

Answer : $\frac{4}{5}$
Hint: If $f(x)$ is differentiable at $x=4$ then $\lim _{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}=f^{\prime}(4)$
Given: $\mathrm{f}(\mathrm{x})=\sqrt{x^{2}+9}$
Explanation: As$f(x)$ is differentiable at $x=4$
$\begin{aligned} &\lim _{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}=f^{\prime}\\\\ &\lim _{x \rightarrow 4} \frac{\sqrt{x^{2}+9}-\sqrt{4^{2}+9}}{x-4}=f^{\prime}\\\\ &\lim _{x \rightarrow 4} \frac{\sqrt{x^{2}+9}-5}{x-4}=f^{\prime}(4)\\ &\lim _{x \rightarrow 4} \frac{2 x}{2 \sqrt{x^{2}+9}}=f^{\prime} \text { (4) } \quad \text { [L-Hospital Rule] } \end{aligned}$
$\begin{aligned} &\frac{4}{\sqrt{16+9}}=f^{\prime} \\\\ &\frac{4}{5}=f^{\prime}(4) \end{aligned}$

The 9th Chapter in mathematics, Differentiability for class 12, The RD Sharma books are not just papers with solutions; they teach new methods and tricks to make the students understand the concepts deeply. RD Sharma Class 12 Solutions Differentiability VSA has thirteen questions. The concepts include defining differentiability, continuous function, differentiable function and finding values of derivatives. The RD Sharma Class 12 Chapter 9 VSA solution book will help you clarify your doubts regarding it.

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