RD Sharma Solutions Class 12 Mathematics Chapter 8 VSA

Q: Can I download the free RD Sharma class 12th exercise VSA?

Indeed, only a single tick is sufficient to download RD Sharma class 12 solutions VSA PDF from our site Career360.

Q: Where do I get to RD Sharma class 12 solutions chapter 8?

Contenders are urged to check our foundation Career360 to get all pieces of RD Sharma Class 12 Solutions to no end.

Q: How might students utilize the RD Sharma class 12th exercise VSA to prepare for the yearly exam?

Students ought to totally focus on the chapter and practice the exercise inquiries to obtain an unquestionable appreciation of the key thoughts.

Q: What number of questions are there in RD Sharma class 12 Continuity solution exercise VSA?

There are 12 inquiries in RD Sharma Solutions Class 12 RD Sharma chapter 8 exercise VSA.

RD Sharma Solutions Class 12 Mathematics Chapter 8 VSA

Edited By Kuldeep Maurya | Updated on Jan 27, 2022 01:56 PM IST

The chapter Simultaneous direct condition is very trying for a large portion of the students, and the shortfall of suitable assets compounds the situation for them. At Carrer360, you will get free RD Sharma class 12th exercise VSA Chapter 8 Continuity VSAQ. Moreover, rehearsing questions from RD Sharma Mathematics Solutions for Class 12 Chapter 8 Continuity is demonstrated to upgrade your numerical abilities. Hence, to save the students from such a circumstance, the RD Sharma class 12th exercise VSA arrangement book is suggested as specialists handpicked questions in science, giving accommodating tips to the students that they probably won't get in school also.

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 8VSA Continuity - Other Exercise
Continuity Excercise:VSA
RD Sharma Chapter-wise Solutions

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

RD Sharma Class 12 Solutions Chapter 8VSA Continuity - Other Exercise

Continuity Excercise:VSA

8 Continuity exercise Very short answer question 1

Continuity of a function at a point –
A function f(x) is said to be continuous at a point x=a if and only if
$\lim _{x \rightarrow a} f(x)=f(a)_{\text {i.e. }} \lim _{x \rightarrow a^{-}} f(x)=f(a) \text { and } \lim _{x \rightarrow a^{+}} f(x)=f(a)$

Continuity exercise Very short answer question 2

Answer:

Given: $\lim_{x\rightarrow a}f(x)=f(a)$
Explanation:
From the definition of continuity we know that
$\lim_{x\rightarrow a}f(x)=f(a)$
then $f(x)$ is continuous at $x=a$

Continuity exercise Very short answer question 3

Answer: 2
Hint: $\lim _{x \rightarrow 0} f(x)=f(0)$
Given: $f(x)=\frac{x}{1-\sqrt{1-x}}$
Explanation: We have to find $f(0)$ when $f(x)$ becomes continuous at $x = 0$
i.e.,
$\Rightarrow \lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{x}{1-\sqrt{1-x}}=f(0)$
Applying L’Hospital rule as the function is in 0/0 form,
$\begin{aligned} &=>\lim _{x \rightarrow 0} \frac{1}{\frac{-1(-1)}{2 \sqrt{1-x}}}=f(0) \\ &\Rightarrow \lim _{x \rightarrow 0} 2 \sqrt{1-x}=f(0) \\ &\Rightarrow 2 \sqrt{1-0}=f(0) \end{aligned}$
Hence, $f(0)=2$

Continuity exercise Very short answer question 4

Answer: $k=\frac{1}{3}$
Hint: $\lim _{x \rightarrow 0} f(x)=f(0)$
Given: $f(x)=\left\{\begin{array}{l} \frac{x}{\sin 3 x}, x \neq 0 \\ k, x=0 \end{array}\right.$ is continuous at $x = 0$
Explanation:
As $f(x)$ is continuous at $x = 0$
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{x}{3 \sin 3 x}=k \\ &\lim _{x \rightarrow 0} \frac{1}{\frac{\sin 3 x}{x}}=k \\ &\frac{1}{k}=\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \\ &\frac{1}{k}=3 \times 1 \\ &k=\frac{1}{3} \end{aligned}$

Continuity exercise Very short answer question 5

Answer: 10
Hint: $\lim _{x \rightarrow 0} f(x)=f(0)$
Given: $f(x)=\frac{\sin 10 x}{x}, x \neq 0$ is continuous at $x = 0$
Explanation:
As $f(x)$ is continuous at $x = 0$
$\begin{aligned} &=>\lim _{x \rightarrow 0} f(x)=f(0) \\ &=>\lim _{x \rightarrow 0} \frac{\sin 10 x}{x}=f(0) \\ &=>\lim _{x \rightarrow 0} \frac{10 \times \sin 10 x}{10 x}=f(0) \end{aligned}$
$= > 10 \lim _{x \rightarrow 0} \frac{\sin 10 x}{10 x}=f(0) \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\begin{aligned} &=>10 \times 1=f(0) \end{aligned}$
$\begin{aligned} \ = > f(0)=10 \end{aligned}$

Continuity exercise Very short answer question 6

Answer: $k=8$
Hint: $\lim _{x \rightarrow 4} f(x)=f(4)$
Given: $f(x)=\left\{\begin{array}{l} \frac{x^{2}-16}{x-4}, x \neq 4 \\ k, x=4 \end{array}\right.$ is continuous at $x = 4$
Explanation:
As $f(x)$ is continuous at $x = 4$
$\lim _{x \rightarrow 4} f(x)=f(4)$
$=>\lim _{x \rightarrow 4} \frac{\left(x^{2}-16\right)}{x-4}=k \; \; \; \; \; \; \; \; \; \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$
$\begin{aligned} &=>\lim _{x \rightarrow 4} \frac{(x-4)(x+4)}{x-4}=k \\ &=>\lim _{x \rightarrow 4}(x+4)=k \\ &=>k=8 \end{aligned}$

Continuity exercise Very short answer question 7

Answer: Continuous
Hint: If $f(x)$ is continuous at $x = 0$ then $\lim _{x \rightarrow 0} f(x)=f(0)$
Given: $f(x)=\left\{\begin{array}{l} \frac{\sin x^{2}}{x}, x \neq 0 \\ 0, x=0 \end{array}\right.$
Explanation:
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin x^{2}}{x} \text { . } \\ &=\lim _{x \rightarrow 0} \frac{x \times \sin x^{2}}{x^{2}} \\ &=0 \times 1 \\ &=0 \\ &f(0)=0 \end{aligned}$
As $\lim _{x \rightarrow 0} f(x)=f(0)$ , $f(x)$ is continuous at $x = 0$

Continuity exercise Very short answer question 8

Answer: $k=\frac{1}{2}$
Hint: $\lim _{x \rightarrow 0} f(x)=f(0)$
Given: $f(x)=\left\{\begin{array}{l} \frac{1-\cos x}{x^{2}}, x \neq 0 \\ k, x=0 \end{array}\right.$ is continuous at $x = 0$
Explanation:
As $f(x)$ is continuous at $x = 0$
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=k \end{aligned}$
Applying L’Hospital rule as the function is in 0/0 form,
$=>\lim _{x \rightarrow 0} \frac{\sin x}{2 x}=k \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{0}{0} \text { form }\right]$
$=>\frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin x}{x}=k \; \; \; \; \; \; \; \; \; \quad\left[\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$=>\frac{1}{2}=k$

Continuity exercise Very short answer question 9

Answer: 1
Hint: $\lim _{x \rightarrow 0} f(x)=f(0)$
Given: $f(x)=\left\{\begin{array}{l} \frac{\sin ^{-1} x}{x} \\ k, x=0 \end{array}, x \neq 0\right.$ is continuous at $x = 0$
Explanation:
As $f(x)$ is continuous at $x = 0$
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}=k \end{aligned}$
Applying L’Hospital rule as the function is in 0/0 form,
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{1}{\frac{\sqrt{1-x^{2}}}{1}}=k \end{aligned}$ $\left [ \frac{0}{0}form \right ]$
$\begin{aligned} &\frac{1}{\sqrt{1-0}}=k \\ &\therefore k=1 \end{aligned}$

Continuity exercise Very short answer question 10

Answer: b=-1
Hint: $\lim _{x \rightarrow 1} f(x)=f(1)$
Given: $f(x)$ is continuous at $x=1$ and
$f(x)=\left\{\begin{array}{l} 5 x-4,0<x \leq 1 \\ 4 x^{2}+3 b x, 1<x<2 \end{array}\right.$
Explanation:
$f(x)$ is continuous at $x=1$
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=f(1) \\ &\lim _{x \rightarrow 1} 4 x^{2}+3 b x=5 \times 1-4 \\ &=>4+3 b=1 \\ &=>3 b=-3 \\ &=>b=-1 \end{aligned}$

Continuity exercise Very short answer question 11

Answer: -3
Hint: $\lim _{x \rightarrow 0} f(x)=f(0)$
Given: $f(x)$ is continuous at $x=0$ and
Where $f(x)=\left\{\begin{array}{l} \frac{k x}{|x|}, x<0 \\ 3, x \geq 0 \end{array}\right.$
Explanation:
As $f(x)$ is continuous at $x=0$ and
$\lim _{x \rightarrow 0^{-}} f(x)=f(0)$
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} \frac{k x}{|x|}=3 \\ &\text { As }|x|=-x \text { for } x<0, \\ &\Rightarrow \lim _{x \rightarrow 0} \frac{k x}{-x}=3 \\ &\Rightarrow-k=3 \\ &\Rightarrow k=-3 \end{aligned}$

Continuity exercise Very short answer question 12

Answer: 7
Hint: $\lim _{x \rightarrow 2} f(x)=f(2)$
Given: $f(x)=\left\{\begin{array}{l} \frac{x^{2}+3 x-10}{x-2}, x \neq 2 \\ k, x=2 \end{array}\right.$ is continuous at $x = 2$
Explanation:
As $f(x)$ is continuous at $x = 2$ and
$\begin{aligned} &\lim _{x \rightarrow 2} f(x)=f(2) \\ &\lim _{x \rightarrow 2} \frac{x^{2}+3 x-10}{x-2}=k \\ &\lim _{x \rightarrow 2} \frac{x^{2}+5 x-2 x-10}{x-2}=k \\ &\lim _{x \rightarrow 2} \frac{(x-2)(x+5)}{x-2}=k \\ &\lim _{x \rightarrow 2} x+5=k \\ &\therefore k=7 \end{aligned}$
There are just 12 questions in the RD Sharma class 12th exercise VSA. Not exclusively is this book suggested by students, however by every one of the instructors too. Therefore, many students trust the RD Sharma class 12 solution VSA arrangement. The students may know the cut-off and stunts for addressing questions rapidly through the RD Sharma class 12 solution VSA of Simultaneous Linear Equation VSA. This book addresses schoolwork for students, as most educators usually like the RD Sharma class 12 VSA book for allotting schoolwork to the students.

Advantages of taking Career360:

Unique yet organized a demonstration of the topics
A detailed explanation of thoughts and formulae
It helps students to practice with no issue
One book covers all chapters and is sufficient for scoring extraordinary engravings

The main benefit is that the students won't need to pay a solitary penny to possess this book. They can download the free E-books and PDF of the RD Sharma class 12 solutions VSA answer free of charge from the Career360 site. Not just students, instructors too can allude to the Career360 site to download any book of any subject.

Features of RD Sharma class 12 solutions VSA (Continuity and Differentiability)

The significance of consistent capacities is clarified in this chapter regarding diagrams.
By rehearsing the solutions given in this chapter, one can comprehend the confirmations of various hypotheses and conduct of nonstop capacities when these are subjected to mathematical computations.
Students can concentrate on a few culminations extricated from hypotheses and demonstrate them.

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book