RD Sharma Solutions Class 12 Mathematics Chapter 8 VSA

Kuldeep MauryaUpdated on 27 Jan 2022, 01:56 PM IST

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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
$\begin{aligned} &\lim _{x \rightarrow 2} f(x)=f(2) \\ &\lim _{x \rightarrow 2} \frac{x^{2}+3 x-10}{x-2}=k \\ &\lim _{x \rightarrow 2} \frac{x^{2}+5 x-2 x-10}{x-2}=k \\ &\lim _{x \rightarrow 2} \frac{(x-2)(x+5)}{x-2}=k \\ &\lim _{x \rightarrow 2} x+5=k \\ &\therefore k=7 \end{aligned}$
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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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