RD Sharma Solutions Class 12 Mathematics Chapter 8 VSA

# RD Sharma Solutions Class 12 Mathematics Chapter 8 VSA

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

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

## 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

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

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

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}

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,
$=>\lim _{x \rightarrow 0} \frac{\sin x}{2 x}=k \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{0}{0} \text { form }\right]$
$=>\frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin x}{x}=k \; \; \; \; \; \; \; \; \; \quad\left[\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$=>\frac{1}{2}=k$

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}
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
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

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

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}
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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.

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## RD Sharma Chapter-wise Solutions

1. 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.

2. Where do I get to RD Sharma class 12 solutions chapter 8?

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3. 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.

4. 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.

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