RD Sharma Solutions Class 12 Mathematics Chapter 8 VSA
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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: 2Hint: $\lim _{x \rightarrow 0} f(x)=f(0)$
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: 10Hint: $\lim _{x \rightarrow 0} f(x)=f(0)$
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 $x = 4$
$\lim _{x \rightarrow 4} f(x)=f(4)$
$=>\lim _{x \rightarrow 4} \frac{\left(x^{2}-16\right)}{x-4}=k \; \; \; \; \; \; \; \; \; \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$
$\begin{aligned} &=>\lim _{x \rightarrow 4} \frac{(x-4)(x+4)}{x-4}=k \\ &=>\lim _{x \rightarrow 4}(x+4)=k \\ &=>k=8 \end{aligned}$
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 $x = 0$
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,
$=>\lim _{x \rightarrow 0} \frac{\sin x}{2 x}=k \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{0}{0} \text { form }\right]$
$=>\frac{1}{2} \lim _{x \rightarrow 0} \frac{\sin x}{x}=k \; \; \; \; \; \; \; \; \; \quad\left[\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$=>\frac{1}{2}=k$
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,
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{1}{\frac{\sqrt{1-x^{2}}}{1}}=k \end{aligned}$ $\left [ \frac{0}{0}form \right ]$
$\begin{aligned} &\frac{1}{\sqrt{1-0}}=k \\ &\therefore k=1 \end{aligned}$
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,
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{1}{\frac{\sqrt{1-x^{2}}}{1}}=k \end{aligned}$ $\left [ \frac{0}{0}form \right ]$
$\begin{aligned} &\frac{1}{\sqrt{1-0}}=k \\ &\therefore k=1 \end{aligned}$
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 $x=1$
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=f(1) \\ &\lim _{x \rightarrow 1} 4 x^{2}+3 b x=5 \times 1-4 \\ &=>4+3 b=1 \\ &=>3 b=-3 \\ &=>b=-1 \end{aligned}$
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: -3Hint: $\lim _{x \rightarrow 0} f(x)=f(0)$
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: 7Hint: $\lim _{x \rightarrow 2} f(x)=f(2)$
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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Download E-bookRD Sharma Chapter-wise Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
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