RD Sharma Class 12 Exercise 8.1 Continuity Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 27, 2022 02:14 PM IST

Students of class 12 carry a big burden in the form of public exams. Best publications like the RD Sharma books, cater the needs of the students to make them understand every concept clearly. RD Sharma Solution Mathematics is a complicated subject for the class 12 students. The Class 12 RD Sharma Chapter Exercise 8.1 solution plays a key role in making the students understand those concepts in-depth.

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RD Sharma Class 12 Solutions Chapter 8 Continuity - Other Exercise
Continuity Excercise: 8.1
Continuity Exercise 8.1 Question 36 (vii)
RD Sharma Chapter wise Solutions

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

RD Sharma Class 12 Solutions Chapter 8 Continuity - Other Exercise

Continuity Excercise: 8.1

Continuity Excercise 8.1 Question 1

Answer:
Discontinuous
Hint:
The discontinuity occurs when the LHL and RHL are not equal at a point
Solution:
Given
$f(x)=\left(\begin{array}{l} \frac{x}{|x|}, x \neq 0 \\\\ 1, x=0 \end{array}\right)$
We observe,
[LHL at $x=0$ ]
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h) \\ &\lim _{h \rightarrow 0} \frac{-h}{|-h|}=\lim _{h \rightarrow 0} \frac{-h}{h}=\lim _{h \rightarrow 0}(-1)=-1 \end{aligned}$
[RHL at $x=0$ ]
$\therefore$ $\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\ &\lim _{h \rightarrow 0} \frac{h}{|h|}=\lim _{h \rightarrow 0} \frac{h}{h}=\lim _{h \rightarrow 0}(1)=1 \\ &\lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \end{aligned}$
Hence, $f\left ( x \right )$ is discontinuous at the $x=0$

Continuity Excercise 8.1 Question 2

Answer:
$x=3$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{cc} \frac{x^{2}-x-6}{x-3} & , x \neq 3 \\\\ 5 & , x=3 \end{array}\right)$
We observe,
[LHL at $x=3$ ]
$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h)$
$\lim _{h \rightarrow 0} \frac{(3-h)^{2}-(3-h)-6}{(3-h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}-6 h-3+h-6}{-h}$ $\left[\because(a-b)^{2}=\left(a^{2}+b^{2}-2 a b\right)\right]$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h^{2}+5 h}{h} \\\\ &=\lim _{h \rightarrow 0}(5+h)=5 \end{aligned}$
[RHL at $x=3$ ]
$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)$
$\lim _{h \rightarrow 0} \frac{(3+h)^{2}-(3+h)-6}{(3+h)-3}=\lim _{h \rightarrow 0} \frac{9+h^{2}+6 h-3-h-6}{h}$ $\left [ \because \left ( a+b \right )^{2}=\left ( a^{2}+b^{2}+2ab \right ) \right ]$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h^{2}+5 h}{h} \\\\ &=\lim _{h \rightarrow 0}(5+h)=5 \end{aligned}$
Also, $f\left ( 3 \right )=5$
$\therefore$ $\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{-}} f(x)=f(3)$
Hence, $f\left ( x \right )$ is continuous at $x=3$ .

Continuity Excercise 8.1 Question 3

Answer:
$x=3$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{c} \frac{x^{2}-9}{x-3}, \text { if } x \neq 3 \\\\ 6, \text { if } x=3 \end{array}\right)$
We observe,
[LHL at $x=3$ ]
$\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h) \\\\ &\lim _{h \rightarrow 0} \frac{(3-h)^{2}-9}{(3-h)-3}=\lim _{h \rightarrow 0} \frac{3^{2}+h^{2}-6 h-9}{3-h-3} \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{h(h-6)}{-h}$ $\left[\because(a-b)^{2}=\left(a^{2}+b^{2}-2 a b\right)\right]$
$\begin{aligned} &=\lim _{h \rightarrow 0}(6-h) \\ &=6 \end{aligned}$
[RHL at $x=3$ ]
$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)$
$\lim _{h \rightarrow 0} \frac{(3+h)^{2}-9}{(3+h)-3}$
$=\lim _{h \rightarrow 0} \frac{h^{2}+6 h}{h}$ $\left[\because(a+b)^{2}=\left(a^{2}+b^{2}+2 a b\right)\right]$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h(h+6)}{h} \\\\ &=\lim _{h \rightarrow 0}(6+h) \\\\ &=6 \end{aligned}$
Given
$\begin{aligned} &f(3)=6 \\ \\&\therefore \lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{-}} f(x)=f(3) \end{aligned}$
Hence, $f\left ( x \right )$ is continuous at $x=3$ .

Continuity Excercise 8.1 Question 4

Answer:
Continuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{cc} \frac{x^{2}-1}{x-1} & , \text { if } x \neq 1 \\\\ 2 & , \text { if } x=1 \end{array}\right)$
We observe,
[LHL at $x=1$ ]
$\begin{aligned} &\lim _{h \rightarrow 0} \frac{(1-h)^{2}-1}{(1-h)-1} \\\\ &=\lim _{h \rightarrow 0} \frac{1+h^{2}-2 h-1}{1-h-1} \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{h^{2}-2 h}{-h}$ $\left[\because(a-b)^{2}=\left(a^{2}+b^{2}-2 a b\right)\right]$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h(h-2)}{-h} \\\\ &=\lim _{h \rightarrow 0}(2-h) \\\\ &=2 \end{aligned}$
[RHL at $x=1$ ]
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\\\ &\lim _{h \rightarrow 0} \frac{(1+h)^{2}-1}{(1+h)-1} \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{h^{2}+2 h}{h}$ $\left[\because(a+b)^{2}=\left(a^{2}+b^{2}+2 a b\right)\right]$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h(h+2)}{h} \\\\ &=\lim _{h \rightarrow 0}(2+h) \\\\ &=2 \end{aligned}$
Given
$\begin{aligned} &f(1)=2 \\\\ &\therefore \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}$
Hence, $f\left ( x \right )$ is continuous at $x=1$ .

Continuity Excercise 8.1 Question 5

Answer:
$x=0$ (Discontinuous)
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{cc} \frac{\sin 3 x}{x}, \text { if } x \neq 0 \\\\ 1 & , \text { if } x=0 \end{array}\right)$
We observe,
[LHL at $x=0$ ]
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h) \\\\ &\lim _{h \rightarrow 0} \frac{\sin (-3 h)}{-h}=\lim _{h \rightarrow 0} \frac{-\sin (3 h)}{-h}=\lim _{h \rightarrow 0} \frac{3 \sin (3 h)}{3 h}=3 \lim _{h \rightarrow 0} \frac{\sin (3 h)}{3 h}=3 \times 1=3 \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\begin{aligned} &{[\because \sin (-x)=-\sin x]} \\ &{\left[\because \frac{\sin x}{x}=1\right]} \end{aligned}\end{aligned}$
[RHL at $x=0$ ]
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\\\ &\lim _{h \rightarrow 0} \frac{\sin (3 h)}{h}=\lim _{h \rightarrow 0} \frac{3 \sin (3 h)}{3 h}=3 \lim _{h \rightarrow 0} \frac{\sin (3 h)}{3 h}=3 \times 1=3 \end{aligned}$
Given
$f\left ( 0 \right )=1$
It is known that for a function $f\left ( x \right )$ is to be continuous at $x=a$ ,
$\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x) \neq f(a)$
But here
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x) \neq f(0)$
Hence, $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity Exercise 8.1 Question 6

Answer:
$x=0$ (Discontinuous)
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{cc} e^{\frac{1}{x}}, \text { if } x \neq 0 \\\\ 1 & , \text { if } x=0 \end{array}\right)$
We observe,
[LHL at $x=0$ ]
$\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} e^{\frac{-1}{h}}=\lim _{h \rightarrow 0}\left(\frac{1}{e^{\frac{1}{h}}}\right)=\frac{1}{\lim _{h \rightarrow 0} e^{\frac{1}{h}}}=0$ $\left[\because \frac{1}{\infty}=0\right]$
[RHL at $x=0$ ]
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} e^{\frac{1}{h}}=\infty$
Given
$f\left ( 0 \right )=1$
It is known that for a function $f\left ( x \right )$ is to be continuous at $x=a$ ,
$\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x) \neq f(a)$
But here
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x) \neq f(0)$
Hence, $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity Exercise 8.1 Question 7

Answer:
$x=0$ (Discontinuous)
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{c} \frac{1-\cos x}{x^{2}}, \text { if } x \neq 0 \\\\ 1, \text { if } x=0 \end{array}\right)$
Consider,
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}\right)$ $\left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]$
$\begin{aligned} &=\lim _{x \rightarrow 0}\left(\frac{2 \sin \frac{x}{2}}{4\left(\frac{x^{2}}{4}\right)}\right) \\\\ &=\lim _{x \rightarrow 0}\left(\frac{2\left(\sin \frac{x}{2}\right)^{2}}{4\left(\frac{x}{2}\right)^{2}}\right) \end{aligned}$
$=\frac{2}{4} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^{2}$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\lim _{x \rightarrow 0} f(x)=\frac{1}{2} \times 1^{2}=\frac{1}{2}$
Given
$f\left ( 0 \right )=1$
$\lim _{x \rightarrow 0} f(x) \neq f(0)$
Thus, $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity Exercise 8.1 Question 8

Answer:
$x=0$ (Discontinuous)
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left\{\begin{array}{l} \frac{x-x}{2}, x>0 \\\\ \frac{x+x}{2}, x<0 \\\\ 2, x=0 \end{array}\right.$
$f(x)=\left\{\begin{array}{l} 0, x>0 \\ x, x<0 \\ 2, x=0 \end{array}\right.$
We observe,
[LHL at $x=0$ ]
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=0$
[RHL at $x=0$ ]
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=0$
And $f\left ( 0 \right )=2$
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \neq f(0)$
Thus, $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity Exercise 8.1 Question 9

Answer:
$x=a$ (Discontinuous)
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Solution:
Given,
$f(x)=\left(\begin{array}{c} \frac{x-a}{x-a}, \text { if } x>a \\\\ \frac{a-x}{x-a}, \text { if } x<a \\\\ 1 \quad, \text { if } x=a \end{array}\right)$
$f(x)=\left(\begin{array}{ll} 1, & x>a \\ -1, & x<a \\ 1, & x=a \end{array}\right)$
$f(x)=\left(\begin{array}{l} 1, x \geq a \\ -1, x<a \end{array}\right)$
We observe
[LHL at $x=a$ ]
$\lim _{x \rightarrow a^{-}} f(x)=\lim _{h \rightarrow 0} f(a-h)=\lim _{h \rightarrow 0}(-1)=-1$
[RHL at $x=a$ ]
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(x)=\lim _{h \rightarrow 0} f(a+h)=\lim _{h \rightarrow 0}(1)=1 \\\\ &\lim _{x \rightarrow a^{+}} f(x) \neq \lim _{x \rightarrow a^{-}} f(x) \end{aligned}$
Thus, $f\left ( x \right )$ is discontinuous at $x=a$ .

Continuity Exercise 8.1 Question 10 (i)

Continuity Exercise point 1 Question 10 (ii)

Continuity Exercise 8.1 (iii) Question 10

Answer:
Discontinuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{c} \frac{e^{x}-1}{\log (1+2 x)}, \text { if } x \neq 0 \\\\ 7, \text { if } x=0 \end{array}\right)$
We observe,
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{e^{x}-1}{\log (1+2 x)} \\\\ &=\lim _{x \rightarrow 0} \frac{e^{x}-1}{\frac{2 x \log (1+2 x)}{2 x}} \end{aligned}$ [Multiplying and dividing the denominator by $2x$ ]
$\begin{aligned} &=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\left(\frac{e^{x}-1}{x}\right)}{\left(\frac{\log (1+2 x)}{2 x}\right)} \\\\ \end{aligned}$
$=\frac{1}{2} \frac{\left(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\right)}{\left(\lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x}\right)}=\frac{1}{2} \times \frac{1}{1}=\frac{1}{2}$
And $f\left ( 0 \right )=7$
$\lim _{x \rightarrow 0} f(x) \neq f(0)$
Thus, $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity Exercise 8.1 (v) Question 10

Answer:
Discontinuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
Given,
$f(x)=\left(\begin{array}{c} \frac{1-x^{n}}{1-x}, \text { if } x \neq 1 \\\\ n-1, \text { if } x=1 \end{array}\right)$
Here, $f\left ( 1 \right )=n-1$
$\begin{aligned} &\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} \frac{1-x^{n}}{1-x} \\\\ &\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1}\left[(1-x)^{n-1}+n C_{1}(1-x)^{n-2} x+n C_{2}(1-x)^{n-3} x^{2}+\ldots .+n C_{n-1}(1-x)^{0} x^{n-1}\right] \end{aligned}$
$\left[\because(a+b)^{n}={ }^{n} C_{0} a^{n} b^{0}+{ }^{n} C_{1} a^{n-1} b^{1}+{ }^{n} C_{2} d^{n-2} b^{2}+\ldots+{ }^{n} C_{n} a^{0} b^{n}\right]$
$\lim _{x \rightarrow 1} f(x)=0+0+\ldots+(1)^{n-1}=1 \neq f(1)$
Thus, $f\left ( x \right )$ is discontinuous at $x=1$ .

Continuity Excercise 8.1 Question 10 (vi)

Answer:
Continuous at $x=1$
Hint:
For a function to be continuous at a point, its LHL and RHL value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{cc} \left|x^{2}-1\right| & \text { if } x \neq 1 \\\\ x-1 & \text { if } x=1 \end{array}\right)$
We observe
[LHL at $x=1$ ]
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h) \\\\ &=\lim _{h \rightarrow 0} \frac{\left|(1-h)^{2}-1\right|}{(1-h)-1} \\\\ &=\lim _{h \rightarrow 0} \frac{\left|1+h^{2}-2 h-1\right|}{1-h-1} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left|h^{2}-2 h\right|}{h} \\\\ &=\lim _{h \rightarrow 0}-(h-2) \\\\ &=\lim _{h \rightarrow 0}-h+2 \\\\ &=2 \end{aligned}$
[RHL at $x=1$ ]
$\begin{aligned} &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h) \\\\ &=\lim _{h \rightarrow 0} \frac{\left|(1+h)^{2}-1\right|}{(1+h)-1} \end{aligned}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left|1+h^{2}+2 h-1\right|}{1+h-1} \\\\ &=\lim _{h \rightarrow 0} \frac{\left|h^{2}+2 h\right|}{h} \\\\ &=\lim _{h \rightarrow 0}(h+2) \\\\ &=2 \end{aligned}$
RHL=LHL
Therefore, $f\left ( x \right )$ is continuos at $x=1$

Continuity Excercise 8.1 Question 10 (vii)

Answer:
Discontinuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left\{\begin{array}{c} \frac{2|x|+x^{2}}{x}, \text { if } x \neq 0 \\\\ 0, \quad \text { if } x=0 \end{array}\right.$
$f(x)=\left\{\begin{array}{c} \frac{2 x+x^{2}}{x}, \text { if } x>0 \\\\ \frac{-2 x+x^{2}}{x}, \text { if } x<0 \\\\ 0, \quad \text { if } x=0 \end{array}\right.$
$f(x)=\left\{\begin{array}{c} (x+2), \text { if } x>0 \\\\ (x-2), \text { if } x<0 \\\\ 0, \quad \text { if } x=0 \end{array}\right.$
We observe
[LHL at $x=0$ ]
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}(-h-2)=-2$
[RHL at $x=0$ ]
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}(h+2)=2 \\\\ &\lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \end{aligned}$
Thus, $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity Excercise 8.1 Question 10 (viii)

Answer:
Continuous
Hint:
Continuous function must be defined at a point, limit must exist at the point, value of the function at that point must equal the value of the right and left limit at that point.
Solution:
Given,
$f(x)=\left(\begin{array}{c} |x-a| \sin \left(\frac{1}{x-a}\right), \text { if } x \neq a \\\\ 0 \quad, \text { if } x=a \end{array}\right)$
$f(x)=\left(\begin{array}{c} (x-a) \sin \left(\frac{1}{x-a}\right), \text { if } x>a \\\\ (-x+a) \sin \left(\frac{1}{-x+a}\right), \text { if } x<a \\\\ 0, \text { if } x=a \\ \end{array}\right)$
We observe
[LHL at $x=a$ ]
$\lim _{x \rightarrow a^{-}} f(x)=(-a+a) \sin \left(\frac{1}{-a+a}\right)=0$
[RHL at $x=a$ ]
$\begin{aligned} &\lim _{x \rightarrow a^{+}} f(x)=(a-a) \sin \left(\frac{1}{a-a}\right)=0 \\\\ &\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x)=f(a) \end{aligned}$
Thus, $f\left ( x \right )$ is continuous at $x=a$ .

Continuity Exercise 8.1 Question 11

Answer:
Discontinuous
Hint:
Discontinuous occurs when the both number is equal to zero of the numerator and denominator or the function is undefined at its limit.

Solution:
Given,
$f(x)=\left(\begin{array}{ll} 1+x^{2}, & \text { if } 0 \leq x \leq 1 \\\\ 2-x, & \text { if } x>1 \end{array}\right)$
We observe
[LHL at $x=1$ ]
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}\left(1+\left(1-h^{2}\right)\right)=\lim _{h \rightarrow 0}\left(2+h^{2}\right)=2$
[RHL at $x=1$ ]
$\begin{gathered} \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(2-(1+h))=\lim _{h \rightarrow 0}(1-h)=1 \\\\ \lim _{x \rightarrow 1^{+}} f(x) \neq \lim _{x \rightarrow 1^{-}} f(x) \end{gathered}$
Thus, $f\left ( x \right )$ is discontinuous at $x=1$ .

Continuity Exercise 8.1 Question 12

Answer:
Continuous
Hint:
Continuous function must be defined at a point, limit must exist at the point and the value of the function at that point must equal the value of the right hand and left hand limit at that point.
Solution:
Given,
$f(x)=\left(\begin{array}{cl} \frac{\sin 3 x}{\tan 2 x} & , \text { if } x<0 \\\\ \frac{3}{2} & ,\text { if } x=0 \\\\ \frac{\log (1+3 x)}{e^{2 x}-1}&, \text { if } x> 0\\\\ \end{array}\right)$
We observe
[LHL at $x=0$ ]
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h) \\\\ &=\lim _{h \rightarrow 0} f(-h) \\\\ &=\lim _{h \rightarrow 0}\left(\frac{\sin 3(-h)}{\tan 2(-h)}\right) \end{aligned}$
$=\lim _{h \rightarrow 0}\left(\frac{\sin 3 h}{\tan 2 h}\right) \Rightarrow \lim _{h \rightarrow 0}\left(\frac{\frac{3 \sin 3 h}{3h}}{\frac{2 \tan 2 h}{2 h}}\right)$
$\begin{aligned} &=\frac{\lim _{h \rightarrow 0}\left(\frac{3 \sin 3 h}{3 h}\right)}{\lim _{h \rightarrow 0}\left(\frac{2 \tan 2 h}{2 h}\right)} \Rightarrow \frac{3 \lim _{h \rightarrow 0}\left(\frac{\sin 3 h}{3 h}\right)}{2 \lim _{h \rightarrow 0}\left(\frac{\tan 2 h}{2 h}\right)} \\\\ \end{aligned}$ $\left[\begin{array}{l} \because \sin (-x)=-\sin x \\ \because \tan (-x)=-\tan x \end{array}\right]$
$=\frac{3 \times 1}{2 \times 1}=\frac{3}{2}$
$\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
And $f\left ( 0 \right )=\frac{3}{2}$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0)$
Thus, $f\left ( x \right )$ is continuous at $x=0$ .

Continuity exercise 8.1 question 13

Answer:
$a=\frac{1}{2}$
Hint:
Continuous function must be defined at a point, limit must exist at the point and the value of the function at that point must equal to the value of the right hand or left hand limit at that point.
Solution:
Given,
$f(x)=\left(\begin{array}{l} a \sin \frac{\pi}{2}(x+1), \text { if } x \leq 0 \\\\ \frac{\tan x-\sin x}{x^{3}}, \text { if } x>0 \end{array}\right)$
We observe
[LHL at $x=0$ ]
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} a \sin \frac{\pi}{2}(-h+1)=a \sin \frac{\pi}{2}=a$ $\left[\because \sin \frac{\pi}{2}=1\right]$
[RHL at $x=0$ ]
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\\\ &=\lim _{h \rightarrow 0} f(h) \\\\ &=\lim _{h \rightarrow 0} \frac{\tanh -\sinh }{h^{3}} \end{aligned}$ $\left[\begin{array}{l} \because \frac{\sin x}{\cos x}=\tan x \\\\ \because 1-\cos x=2 \sin ^{2} \frac{x}{2} \end{array}\right]$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\frac{\sinh }{\cosh }-\sinh }{h^{3}} \\\\ &=\lim _{h \rightarrow 0} \frac{(1-\cosh ) \tanh }{h^{3}} \end{aligned}$
$=\lim _{h \rightarrow 0} \frac{2\left(\sin ^{2} \frac{h}{2}\right) \tanh }{\frac{4 h^{2}}{4} \times h}$ [Multiplying and dividing the denominator by $4$ ]
$=\frac{2}{4} \lim _{h \rightarrow 0} \frac{\left(\sin ^{2} \frac{h}{2}\right) \tanh }{\frac{h^{2}}{4} \times h}$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$=\frac{1}{2} \lim _{h \rightarrow 0} \frac{\left(\sin \frac{h}{2}\right)}{\frac{h}{2}} \lim _{h \rightarrow 0} \frac{\tanh }{h}$ $\left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right]$
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\frac{1}{2} \times 1 \times 1 \\\\ &\lim _{x \rightarrow 0^{+}} f(x)=\frac{1}{2} \end{aligned}$ $\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\frac{1}{2} \times 1 \times 1 \\ &\lim _{x \rightarrow 0^{+}} f(x)=\frac{1}{2} \end{aligned}$
If $f\left ( x \right )$ is continuous at $x=0$ , then
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x) \\\\ &a=\frac{1}{2} \end{aligned}$

Continuity Exercise 8.1 Question 14

Answer:

Answer:
$f\left ( x \right )$ is discontinuous at the point $x=0$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{ll} 3 x-2, & \text { if } x \leq 0 \\\\ x+1, & \text { if } x>0 \end{array}\right)$ at $x=0$
$f(x)=\left(\begin{array}{l} 3 x-2, \text { if } x<0 \\\\ -2, \text { if } x=0 \\\\ x+1, \quad \text { if } x>0 \end{array}\right)$

We observe
[LHL at $x=0$ ]
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} 3(-h)-2=-2$
[RHL at $x=0$ ]
$\begin{gathered} \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}(h+1)=1 \\\\ \lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \end{gathered}$
Thus, $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity Exercise 8.1 Question 15

Answer:
Discontinuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left\{\begin{array}{l} x, \text { if } x>0 \\\\ 1, \text { if } x=0 \\\\ -x, \text { if } x<0 \end{array}\right.$
We observe
[LHL at $x=0$ ]
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}-(-h)=0$
[RHL at $x=0$ ]
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=0$
And $f\left ( 0 \right )=1$

Thus, $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity exercise 8.1 question 16

Continuity Exercise 8.1 Question 17

Answer:
Discontinuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left\{\begin{array}{l} 2 x-1, \text { if } x<0 \\\\ 2 x+1, \text { if } x \geq 0 \end{array}\right.$
We observe
[LHL at $x=0$ ]
$\lim _{x \rightarrow 0^{-}} f(x)=2(0)-1=-1$
[RHL at $x=0$ ]
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=2(0)+1=1 \\ &\lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \end{aligned}$
Thus, $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity exercise 8.1 question 18

Answer: $k=2$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{c} \frac{x^{2}-1}{x-1}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right)$
If $f\left ( x \right )$ is continuous at $x=1$ , then
$\lim _{x \rightarrow 1} f(x)=f(1)$
$\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{x-1}=k$ $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$
$\begin{aligned} \\\\ &\lim _{x \rightarrow 1}(x+1)=k \\\\ \end{aligned}$
$k=2$

Continuity exercise 8.1 question 19

Answer:
$k=-1$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{c} \frac{x^{2}-3 x+2}{x-1}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right)$
If $f\left ( x \right )$ is continuous at $x=1$ , then
$\begin{aligned} &\lim _{x \rightarrow 1} f(x)=f(1) \\\\ &\lim _{x \rightarrow 1} \frac{x^{2}-3 x+2}{x-1}=k \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow 1} \frac{(x-2)(x-1)}{x-1}=k \\\\ &\lim _{x \rightarrow 1}(x-2)=k \\\\ &k=-1 \end{aligned}$

Continuity exercise 8.1 question 20

Answer:
$k=\frac{5}{3}$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{c} \frac{\sin 5 x}{3 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.$ is continuous at $x=0$
Solution:
$f(x)=\left(\begin{array}{c} \frac{\sin 5 x}{3 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right)$
If $f\left ( x \right )$ is continuous at $x=0$ , then
$\lim _{x \rightarrow 0} f(x)=f(0)$
$\lim _{x \rightarrow 0} \frac{5 \sin 5 x}{3 \times 5 x}=k$ [Multiplying and dividing by $5$ ]
$\frac{5}{3} \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x}=k$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\begin{aligned} &\frac{5}{3} \times 1=k \\ &k=\frac{5}{3} \end{aligned}$

Continuity Exercise 8.1 Question 21

Answer:
$k=\frac{3}{4}$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left(\begin{array}{r} k x^{2}, \text { if } x \leq 2 \\\\ 3, \text { if } x>2 \end{array}\right)$ is continuous at $x=2$
Solution:
$f(x)=\left(\begin{array}{r} k x^{2}, \text { if } x \leq 2 \\\\ 3, \text { if } x>2 \end{array}\right)$
If $f\left ( x \right )$ is continuous at $x=2$ , then
$\begin{aligned} &\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2) \\\\ &\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} k(2-h)^{2}=4 k \end{aligned}$
$\begin{aligned} &f(2)=3 \\ &4 k=3 \\ &k=\frac{3}{4} \end{aligned}$

Continuity Exercise 8.1 Question 22

Answer:
$k=\frac{2}{5}$
Hint:
$f\left ( x \right )$ must be defined. The limit of $f\left ( x \right )$ the approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{l} \frac{\sin 2 x}{5 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.$ is continuous at $x=0$
Solution:
$f(x)=\left\{\begin{array}{l} \frac{\sin 2 x}{5 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.$
If $f\left ( x \right )$ is continuous at $x=0$ , then
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0} \frac{\sin 2 x}{5 x}=k \end{aligned}$
$\lim _{x \rightarrow 0} \frac{2 \sin 2 x}{5 \times 2 x}=k$ [Multiplying and dividing by $2$ ]
$\frac{2}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}=k$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\frac{2}{5}\times 1=k$
$k=\frac{2}{5}$

Continuity Excercise 8.1 Question 23

Answer:
$a=-2$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left(\begin{array}{r} a x+5, \text { if } x \leq 2 \\\\ x-1, \text { if } x>2 \end{array}\right)$
Solution:
$f(x)=\left(\begin{array}{r} a x+5, \text { if } x \leq 2 \\\\ x-1, \text { if } x>2 \end{array}\right)$
We observe
[LHL at $x=2$ ]
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} a(2-h)+5=2 a+5$
[RHL at $x=2$ ]
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}(2+h-1)=1$
$f(2)=a(2)+5=2 a+5$
$f\left ( x \right )$ is continuous at $x=2$ , we have
$\begin{aligned} &\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)=f(2) \\\\ &2 a+5=1 \\\\ &2 a=-4 \\\\ &a=-2 \end{aligned}$

Continuity Excercise 8.1 Question 24

Answer:
Since, $\lim _{x \rightarrow 0^{-}} f(x)$ and $\lim _{x \rightarrow 0^{+}} f(x)$ are not equal, $f\left ( x \right )$ is discontinuous.
Thus, $f\left ( x \right )$ is discontinuous at $x=0$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:

$f(x)=\left\{\begin{array}{c} \frac{x}{|x|+2 x^{2}}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.$

Solution:
$f(x)=\left\{\begin{array}{c} \frac{x}{x+2 x^{2}}, \text { if } x>0 \\\\ \frac{-x}{x-2 x^{2}}, \text { if } x<0 \\\\ k, \text { if } x=0 \end{array}\right.$
$f(x)=\left\{\begin{array}{c} \frac{1}{2 x+1}, \text { if } x>0 \\\\ \frac{1}{2 x-1}, \text { if } x<0 \\\\ k, \text { if } x=0 \end{array}\right.$
We observe
(LHL at $x=0$ )
$\lim _{h \rightarrow 0^{-}} \frac{1}{-2 h-1}=-1$
(RHL at $x=0$ )
$\lim _{h \rightarrow 0^{+}} \frac{1}{2 h+1}=1$
Since, $\lim _{x \rightarrow 0^{-}} f(x)$ and $\lim _{x \rightarrow 0^{+}} f(x)$ are not equal, $f\left ( x \right )$ is discontinuous.
Thus, $f\left ( x \right )$ is discontinuous at $x=0$ , regardless of choice of $k$

Continuity exercise 8.1 question 25

Answer:
$k=6$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $k$ must exist.
Given:

$f(x)=\left\{\begin{array}{c} \frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\\\ 3, \text { if } x=\frac{\pi}{2} \end{array}\right.$

Solution:
$f(x)=\left\{\begin{array}{c} \frac{k \cos x}{\pi-2 x}, \text { if } x \neq \frac{\pi}{2} \\\\ 3, \text { if } x=\frac{\pi}{2} \end{array}\right.$
If $f\left ( x \right )$ is continuous at $x=\frac{\pi}{2}$ , then
$\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\\\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=3 \end{aligned}$ ......(i)

Putting $x=\frac{\pi}{2}-h$
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{k \cos x}{\pi-2 x}=\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}$
From (i)
$\begin{aligned} &\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}=3 \\\\ &\lim _{h \rightarrow 0} \frac{k \sinh }{2 h}=3 \\\\ &\lim _{h \rightarrow 0} \frac{k \sinh }{h}=6 \end{aligned}$
$k \lim _{h \rightarrow 0} \frac{\sinh }{h}=6$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$k\times 1= 6$
$k=6$ , $f\left ( x \right )$ is continuous at $x=\frac{\pi}{2}$ .

Continuity Exercise 8.1 Question 26

Answer:
$a=\frac{-3}{2}, b \neq 0, c=\frac{1}{2} ; b \in R-\{0\}$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{c} \frac{\sin (a+1) x+\sin x}{x}, \text { for } x<0 \\\\ c \quad, \text { for } x=0 \\\\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{\frac{3}{2}}}, \text { for } x>0 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{c} \frac{\sin (a+1) x+\sin x}{x}, \text { for } x<0 \\\\ c \quad, \text { for } x=0 \\\\ \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{\frac{3}{2}}}, \text { for } x>0 \end{array}\right.$
Since $f\left ( x \right )$ is continuous at $x=0$ , we have
$\lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x)$
(LHL at $x=0$ )
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} \frac{\sin (a+1) x+\sin x}{x} \\ &=\lim _{x \rightarrow 0} \frac{(\sin a x \times \cos x)+(\cos a x \times \sin x)+\sin x}{x} \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{(\sin a x \times \cos x)}{x}+\lim _{x \rightarrow 0} \frac{(\cos a x \times \sin x)}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \\ &=\lim _{x \rightarrow 0} \frac{\sin a x}{x} \lim _{x \rightarrow 0} \cos x+\lim _{x \rightarrow 0} \cos a x \lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{\sin a x}{x}(1)+(1) \lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \\ &=\lim _{x \rightarrow 0} \frac{\sin a x}{a x}(a)+\lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \end{aligned}$
$\begin{aligned} &=a \lim _{x \rightarrow 0} \frac{\sin a x}{a x}+\lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x} \\ &=a(1)+1+1 \\ &=a+2 \end{aligned}$ $\left[\begin{array}{l} \because \lim _{x \rightarrow 0} \cos x=1 \\ \because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \end{array}\right]$
RHL at $x=0$
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x \sqrt{x}} \\\\ &=\lim _{x \rightarrow 0} \frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x \sqrt{x}} \times \frac{\sqrt{x+b x^{2}}+\sqrt{x}}{\sqrt{x+b x^{2}}+\sqrt{x}} \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{x+b x^{2}-x}{b x \sqrt{x}\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \\\\ &=\lim _{x \rightarrow 0} \frac{b x^{2}}{b x \sqrt{x}\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{x}{\sqrt{x}\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \\\\ &=\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\left(\sqrt{x+b x^{2}}+\sqrt{x}\right)} \end{aligned}$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{\sqrt{x}}{\sqrt{x}}\right)}{\left(\frac{\sqrt{x+b x^{2}}+\sqrt{x}}{\sqrt{x}}\right)}$
$\begin{aligned} &=\lim _{x \rightarrow 0} \frac{1}{(\sqrt{1+b x}+\sqrt{1})} \\\\ &=\frac{1}{2} \\\\ &f(0)=\lim _{x \rightarrow 0} f(x)=c \end{aligned}$
Since $f\left ( x \right )$ is continuous at $x=0$ , then
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \\\\ &a+2=c=\frac{1}{2} \end{aligned}$
$a=\frac{-3}{2}, c=\frac{1}{2}, b$ is any real number except $0$

Continuity Exercise 8.1 Question 27

Answer:
$k=\pm 1$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{cl} \frac{1-\cos k x}{x \sin x} & , x \neq 0 \\\\ \frac{1}{2} & , x=0 \end{array}\right.$
Solution:
$f\left ( x \right )$ is continuous at $x=0$ , then
$\lim _{x \rightarrow 0} f(x)=f(0)$
Consider,
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos k x}{x \sin x}\right)=\lim _{x \rightarrow 0}\left(\frac{2 \sin ^{2} k \frac{x}{2}}{x \sin x}\right)$ $\left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]$
$\Rightarrow$ $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{2 \sin ^{2} k \frac{x}{2}}{x^{2}\left(\frac{\sin x}{x}\right)}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{2 \frac{k^{2}}{4}\left(\sin k \frac{x}{2}\right)^{2}}{\left(\frac{k x}{2}\right)^{2}\left(\frac{\sin x}{x}\right)}\right)$ [Multiplying and dividing by $\left ( \frac{k}{2} \right )^{2}$ ]
$=\frac{2 k^{2}}{4} \lim _{x \rightarrow 0}\left(\frac{\left(\sin k \frac{x}{2}\right)^{2}}{\left(\frac{k x}{2}\right)^{2}\left(\frac{\sin x}{x}\right)}\right)$
$\lim _{x \rightarrow 0} f(x)=\frac{2 k^{2}}{4}\left(\frac{\lim _{x \rightarrow 0} \frac{\left(\sin k \frac{x}{2}\right)^{2}}{\left(\frac{k x}{2}\right)^{2}}}{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)} \right)$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\lim _{x \rightarrow 0} f(x)=\frac{2 k^{2}}{4} \times 1=\frac{k^{2}}{2}$ … (i)
From (i)
$\begin{aligned} &\frac{k^{2}}{2}=f(0) \\\\ &\frac{k^{2}}{2}=\frac{1}{2}, k=\pm 1 \end{aligned}$

Continuity Exercise 8.1 Question 28

Answer:
$a=1, b=-1$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{r} \frac{x-4}{|x-4|}+a, \text { if } x<4 \\\\ a+b, \text { if } x=4 \\\\ \frac{x-4}{|x-4|}+b, \text { if } x>4 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{r} \frac{x-4}{|x-4|}+a, \text { if } x<4 \\\\ a+b, \text { if } x=4 \\\\ \frac{x-4}{|x-4|}+b, \text { if } x>4 \end{array}\right.$
We observe
(LHL at $x=4$ )
$\begin{aligned} \lim _{x \rightarrow 4^{-}} f(x) &=\lim _{h \rightarrow 0} f(4-h) \\\\ &=\lim _{h \rightarrow 0}\left(\frac{4-h-4}{|4-h-4|}+a\right)=\lim _{h \rightarrow 0}\left(\frac{-h}{|-h|}+a\right)=a-1 \end{aligned}$
(RHL at $x=4$ )
$\begin{aligned} &\lim _{x \rightarrow 4^{+}} f(x)=\lim _{h \rightarrow 0} f(4+h) \\\\ &\quad=\lim _{h \rightarrow 0}\left(\frac{4+h-4}{|4+h-4|}+b\right)=\lim _{h \rightarrow 0}\left(\frac{h}{|h|}+b\right)=b+1 \\\\ &f(4)=a+b \end{aligned}$
If $f\left ( x \right )$ is continuous at $x=4$
$\begin{aligned} &\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{-}} f(x)=f(4) \\ &a-1=b+1=a+b \\ &a-1=a+b, b+1=a+b \\ &b=-1, a=1 \end{aligned}$

Continuity Exercise 8.1 Question 29

Answer:
$k=2$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{c} \frac{\sin 2 x}{x}, x \neq 0 \\\\ k, x=0 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{c} \frac{\sin 2 x}{x}, x \neq 0 \\\\ k, x=0 \end{array}\right.$
If $f\left ( x \right )$ is continuous at $x=0$ ,
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}=k \end{aligned}$
$f\left ( x \right )$ [Multiplying and dividing by $2$ ]
$2 \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}=k \\$ ${\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]}$
$\begin{gathered}\\ 2 \times 1=k \\ k=2 \end{gathered}$

Continuity Exercise 8.1 Question 30

Answer:
$\frac{a+b}{a b}=f(0)$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\frac{\log \left(1+\frac{x}{a}\right)-\log \left(1-\frac{x}{b}\right)}{x}, x \neq 0$
If $f\left ( x \right )$ is continuous at $x=0$ , then
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0}\left(\frac{\log \left(1+\frac{x}{a}\right)-\log \left(1-\frac{x}{b}\right)}{x}\right)=f(0) \end{aligned}$
$\lim _{x \rightarrow 0}\left(\frac{\log \left(1+\frac{x}{a}\right)}{\frac{a x}{a}}-\frac{\log \left(1-\frac{x}{b}\right)}{\frac{b x}{b}}\right)=f(0)$ [Multiplying the denominator of first term by $\frac{a}{a}$ ]
[Multiplying the denominator of second term by $\frac{b}{b}$ ]
$\frac{1}{a} \lim _{x \rightarrow 0}\left(\frac{\log \left(1+\frac{x}{a}\right)}{\frac{x}{a}}\right)-\left(\frac{-1}{b}\right) \lim _{x \rightarrow 0}\left(\frac{\log \left(1-\frac{x}{b}\right)}{\frac{-x}{b}}\right)=f(0)$

$\frac{1}{a} \times 1-\left(\frac{-1}{b}\right) \times 1=f(0)$ $\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$
$\begin{aligned} \\\\ &\frac{1}{a}+\frac{1}{b}=f(0) \\\\ &\frac{a+b}{a b}=f(0) \end{aligned}$

Continuity Exercise 8.1 point 12 Question 31

Answer:
$k=\frac{1}{2}$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{c} \frac{2^{x+2}-16}{4^{x}-16}, \text { if } x \neq 2 \\\\ k, \text { if } x=2 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{c} \frac{2^{x+2}-16}{4^{x}-16}, \text { if } x \neq 2 \\\\ k, \text { if } x=2 \end{array}\right.$
If $f\left ( x \right )$ is continuous at $x=2$ ,
$\begin{aligned} &\lim _{x \rightarrow 2} f(x)=f(2) \\ &\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}=f(2) \end{aligned}$
$\lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}=k$ [Taking $4$ as common] $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$
$\begin{aligned} &\lim _{x \rightarrow 2} \frac{4}{\left(2^{x}+4\right)}=k \\ &\frac{4}{\left(2^{2}+4\right)}=k \\ &k=\frac{4}{8} \\ &k=\frac{1}{2} \end{aligned}$

Continuity Exercise 8.1 Question 32

Answer:
$k=-4$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left(\begin{array}{cc} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} & , x \neq 0 \\\\ k & , x=0 \end{array}\right)$
Solution:
$f(x)=\left(\begin{array}{cc} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1} & , x \neq 0 \\\\ k & , x=0 \end{array}\right)$
If $f\left ( x \right )$ is continuous at $x=0$ , then
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1}=k \end{aligned}$
$\lim _{x \rightarrow 0} \frac{1-\sin ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1}=k$ $\left[\because \cos ^{2} x=1-\sin ^{2} x\right]$
$\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x}{\sqrt{x^{2}+1}-1}=k$
$\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x\left(\sqrt{x^{2}+1}+1\right)}{\left(\sqrt{x^{2}+1}-1\right)\left(\sqrt{x^{2}+1}+1\right)}=k$ [Multiplying and dividing by $\sqrt{x^{2}+1}+1$ ]
$\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x\left(\sqrt{x^{2}+1}+1\right)}{x^{2}}=k$ $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$
$\begin{aligned} &-2 \lim _{x \rightarrow 0} \frac{\left(\sin ^{2} x\right)\left(\sqrt{x^{2}+1}+1\right)}{x^{2}}=k \\\\ &-2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{2} \lim _{x \rightarrow 0}\left(\sqrt{x^{2}+1}+1\right)=k \end{aligned}$
$-2 \times 1 \times 1(1+1)=k$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$k=-4$

Continuity Exercise 8.1 Question 33

Answer:
$f(\pi)=\frac{49}{10}$
Hint:
$f(x)$ must be defined. The limit of the $f(x)$ approaches the value $x$ must exist.
Given:
$f(x)=\frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}, x=\pi$
Solution:
$f(x)=\frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}, x=\pi$
If $f\left ( x \right )$ is continuous at $x=\pi$ , then
$\begin{aligned} &\lim _{x \rightarrow \pi} f(x)=f(\pi) \\ &\lim _{x \rightarrow \pi} \frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}=f(\pi) \end{aligned}$
$\frac{2}{5} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(x-\pi)}{2}\right)}{(x-\pi)^{2}}=f(\pi)$ $\left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]$
$\frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(x-\pi)}{2}\right)}{\frac{49}{4}(x-\pi)^{2}}=f(\pi)$ [Multiplying and dividing by $\frac{49}{4}$ ]
$\frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(x-\pi)}{2}\right)}{\left(\frac{7}{2}(x-\pi)\right)^{2}}=f(\pi)$
$\frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi}\left[\frac{\sin \left(\frac{7(x-\pi)}{2}\right)}{\left(\frac{7}{2}(x-\pi)\right)}\right]^{2}=f(\pi)$
$\frac{2}{5} \times \frac{49}{4} \times 1=f(\pi)$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\begin{aligned} &\frac{1}{5} \times \frac{49}{2} \times 1=f(\pi) \\ &f(\pi)=\frac{49}{10} \end{aligned}$

Continuity Exercise 8.1 Question 34

Answer:
$f\left ( 0 \right )=1$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\frac{2 x+3 \sin x}{3 x+2 \sin x}, x \neq 0$
Solution:
$\begin{aligned} &f(x)=\frac{2 x+3 \sin x}{3 x+2 \sin x}, x \neq 0 \\\\ &\lim _{x \rightarrow 0} f(x)=f(0)\\ \\ &\lim _{x \rightarrow 0} \frac{2 x+3 \sin x}{3 x+2 \sin x}=f(0) \\\\ &\lim _{x \rightarrow 0} \frac{x\left(2+3 \frac{\sin x}{x}\right)}{x\left(3+2 \frac{\sin x}{x}\right)}=f(0) \end{aligned}$
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\left(2+3 \frac{\sin x}{x}\right)}{\left(3+2 \frac{\sin x}{x}\right)}=f(0) \\\\ &\frac{\lim _{x \rightarrow 0}\left(2+3 \frac{\sin x}{x}\right)}{\lim _{x \rightarrow 0}\left(3+2 \frac{\sin x}{x}\right)}=f(0) \end{aligned}$
$\frac{(2+3) \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)}{(3+2) \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)}=f(0)$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\begin{aligned} &\frac{(2+3) 1}{(3+2) 1}=f(0) \\ &\frac{5}{5}=f(0) \\ &f(0)=1 \end{aligned}$

Continuity Exercise 8.1 Question 35

Answer:
$k=1$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left(\begin{array}{c} \frac{1-\cos 4 x}{8 x^{2}}, \text { when } x \neq 0 \\\\ k, w h e n\: x=0 \end{array}\right)$
Solution:
$f(x)=\left(\begin{array}{c} \frac{1-\cos 4 x}{8 x^{2}}, \text { when } x \neq 0 \\\\ k, w h e n\: x=0 \end{array}\right)$
If $f\left ( x \right )$ is continuous at $x=0$ , then
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{8 x^{2}}=f(0) \end{aligned}$
$\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 2 x}{8 x^{2}}=f(0)$ $\left[\because 1-\cos 2 x=2 \sin ^{2} x\right]$
$\frac{2}{2} \lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{4 x^{2}}=f(0)$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$\begin{aligned} &1 \times 1=f(0) \\ &k=1 \end{aligned}$ $[\because f(0)=k]$

Continuity exercise 8.1 question 36 (i)

Answer:
$k=\pm 2$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Given:
$f(x)=\left\{\begin{array}{c} \frac{1-\cos 2 k x}{x^{2}}, \text { if } x \neq 0 \\\\ 8, \text { if } x=0 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{c} \frac{1-\cos 2 k x}{x^{2}}, \text { if } x \neq 0 \\\\ 8, \text { if } x=0 \end{array}\right.$
If $f\left ( x \right )$ is continuous at $x=0$ ,
$\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0} \frac{1-\cos 2 k x}{x^{2}}=8 \end{aligned}$
$\lim _{x \rightarrow 0} \frac{2 k^{2} \sin ^{2} k x}{k^{2} x^{2}}=8$ $(\cos 2x= 1-2\sin ^{2}x)$
$\begin{aligned} &2 k^{2} \lim _{x \rightarrow 0}\left(\frac{\sin k x}{k x}\right)=8 \\\\ &2 k^{2} \times 1=8 \\\\ &k^{2}=4 \\\\ &k=\pm 2 \end{aligned}$

Continuity exercise 8.1 question 36 (ii)

Answer:
$k=\frac{-2}{\pi}$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Given:
$f(x)=\left\{\begin{array}{c} (x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{c} (x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right.$
If $f\left ( x \right )$ is continuous at $x=1$ , then
$\begin{aligned} &\lim _{x \rightarrow 1} f(x)=f(1) \\\\ &\lim _{x \rightarrow 1}(x-1) \tan \frac{\pi x}{2}=k \end{aligned}$
Putting $x-1=y$ , we get
$\begin{aligned} &\lim _{y \rightarrow 0} y \tan \frac{\pi(y+1)}{2}=k \\\\ &\lim _{y \rightarrow 0} y \tan \left(\frac{\pi y}{2}+\frac{\pi}{2}\right)=k \end{aligned}$
$\lim _{y \rightarrow 0} y \tan \left(\frac{\pi}{2}+\frac{\pi y}{2}\right)=k$
$-\lim _{y \rightarrow 0} y \cot \left(\frac{\pi y}{2}\right)=k$ $[\because \tan (90+\theta)=-\cot \theta]$
$-\frac{2}{\pi} \lim _{y \rightarrow 0} \frac{\frac{\pi y}{2} \cos \left(\frac{\pi y}{2}\right)}{\sin \left(\frac{\pi y}{2}\right)}=k$ $\left[\because \cot x=\frac{\cos x}{\sin x}\right]$
$-\frac{2}{\pi} \frac{\lim _{y \rightarrow 0} \cos \left(\frac{\pi y}{2}\right)}{\lim _{y \rightarrow 0}\left(\frac{\sin \left(\frac{\pi y}{2}\right)}{\frac{\pi y}{2}}\right)}=k$
$\begin{aligned} & \\\\ &\frac{-2}{\pi} \times \frac{1}{1}=k \\\\ &k=\frac{-2}{\pi} \end{aligned}$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

Continuity Exercise 8.1 Question 36 (iii)

Answer:
No value of $k$ exists.
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Given:
$f(x)=\left\{\begin{array}{r} k\left(x^{2}-2 x\right), \text { if } x<0 \\\\ \cos x, \text { if } x \geq 0 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{r} k\left(x^{2}-2 x\right), \text { if } x<0 \\\\ \cos x, \text { if } x \geq 0 \end{array}\right.$
We have
(LHL at $x=0$ )
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} k\left(h^{2}+2 h\right)=0$
(RHL at $x=0$ )
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} \cosh =1 \\ &\lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \end{aligned}$
Thus no value of $k$ exists for which $f\left ( x \right )$ is continuous at $x=0$

Continuity Exercise 8.1 Question 36 (iv)

Answer:
$k=\frac{9}{5}$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Given:
$f(x)=\left(\begin{array}{l} k x+1, \text { if } x \leq 5 \\\\ 3 x-5, \text { if } x>5 \end{array}\right)$
Solution:
$f(x)=\left(\begin{array}{l} k x+1, \text { if } x \leq 5 \\\\ 3 x-5, \text { if } x>5 \end{array}\right)$
We have
(LHL at $x=5$ )
$\lim _{x \rightarrow 5^{-}} f(x)=\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0} k(5-h)+1=5 k+1$
(RHL at $x=5$ )
$\lim _{x \rightarrow 5^{+}} f(x)=\lim _{h \rightarrow 0} f(5+h)=\lim _{h \rightarrow 0} 3(5+h)-5=10$
If $f\left ( x \right )$ is continuous at $x=5$ , then
$\begin{aligned} &\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5^{-}} f(x) \\ &5 k+1=10 \\ &k=\frac{9}{5} \end{aligned}$

Continuity Excercise 8.1 Question 36 (vi)

Answer:
$k=10$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left(\begin{array}{cc} \frac{x^{2}-25}{x-5} & , x \neq 5 \\\\ k & , x=5 \end{array}\right)$
Solution:
$f(x)=\left(\begin{array}{cc} \frac{x^{2}-25}{x-5} & , x \neq 5 \\\\ k & , x=5 \end{array}\right)$
$f(x)=\left(\begin{array}{cc} \frac{(x-5)(x+5)}{x-5} \\\\ k, x=5 \end{array}, x \neq 5\right)$ $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$
$f(x)=\left(\begin{array}{cc} (x+5), & x \neq 5 \\\\ k & , x=5 \end{array}\right)$
If $f\left ( x \right )$ is continuous at $x=5$ , then
$\begin{aligned} &\lim _{x \rightarrow 5} f(x)=f(5) \\ &\lim _{x \rightarrow 5}(x+5)=k \\ &k=5+5=10 \\ &k=10 \end{aligned}$

Continuity Exercise 8.1 Question 36 (vii)

Answer:
$k=4$
Hint:
$f(x)$ must be defined. The limit of the $f(x)$ approaches the value $x$ must exist.
Given:
$f(x)=\left(\begin{array}{c} k x^{2}, x \geq 1 \\ 4, x<1 \end{array}\right)$
Solution:
$f(x)=\left(\begin{array}{c} k x^{2}, x \geq 1 \\ 4, x<1 \end{array}\right)$
We have
(LHL at $x=1$ )
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} 4=4$
(RHL at $x=1$ )
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0} k(1+h)^{2}=k$
If $f\left ( x \right )$ is continuous at $x=1$ , then
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\ &k=4 \end{aligned}$

Continuity Exercise 8.1 Question 36 (viii)

Answer:
$k=\frac{1}{2}$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Given:
$f(x)= \begin{cases}k\left(x^{2}+2\right), & x \leq 0 \\ 3 x+1 & , x>0\end{cases}$
Solution:
$f(x)= \begin{cases}k\left(x^{2}+2\right), & x \leq 0 \\ 3 x+1 & , x>0\end{cases}$
We have
(LHL at $x=0$ )
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} k\left((-h)^{2}+2\right)=2 k$
(RHL at $x=0$ )
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} 3 h+1=1$
If $f\left ( x \right )$ is continuous at $x=0$ , then
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x) \\ &2 k=1 \\ &k=\frac{1}{2} \end{aligned}$

Continuity Exercise 8.1 Question 36 (ix)

Answer:
$k=7$
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Given:
$f(x)=\left(\begin{array}{c|c} \frac{x^{3}+x^{2}-16 x+20}{(x-2)^{2}}, & x \neq 2 \\\\ k & , x=2 \end{array}\right)$
Solution:
$f(x)=\left(\begin{array}{c|c} \frac{x^{3}+x^{2}-16 x+20}{(x-2)^{2}}, & x \neq 2 \\\\ k & , x=2 \end{array}\right)$
$f(x)=\left(\begin{array}{cc} \frac{x^{3}+x^{2}-16 x+20}{x^{2}-4 x+4} & , x \neq 2 \\\\ k & , x=2 \end{array}\right)$ $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$f(x)=\left(\begin{array}{c} x+5, x \neq 2 \\ k, x=2 \end{array}\right)$
If $f\left ( x \right )$ is continuous at $x=2$ , then
$\begin{aligned} &\lim _{x \rightarrow 2} f(x)=f(2) \\ &\lim _{x \rightarrow 2}(x+5)=k \\ &2+5=k \\ &k=7 \end{aligned}$

Continuity Exercise 8.1 Question 37

Answer:
$a=3, b=-8$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)= \begin{cases}1 & , \text { if } x \leq 3 \\ a x+b, & \text { if } 3<x<5 \\ 7 & , \text { if } x \geq 5\end{cases}$
Solution:
$f(x)= \begin{cases}1 & , \text { if } x \leq 3 \\ a x+b, & \text { if } 3<x<5 \\ 7 & , \text { if } x \geq 5\end{cases}$
We have
(LHL at $x= 3$ )
$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}(1)=1$
(RHL at $x= 3$ )
$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0} a(3+h)+b=3 a+b$
(LHL at $x= 5$ )
$\lim _{x \rightarrow 5^{-}} f(x)=\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0}[a(5-h)+b]=5 a+b$
(RHL at $x= 5$ )
$\lim _{x \rightarrow 5^{+}} f(x)=\lim _{h \rightarrow 0} f(5+h)=\lim _{h \rightarrow 0} 7=7$
If $f\left ( x \right )$ is continuous at $x= 3$ and $x= 5$ , then
$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{j}} f(x) \text { and } \lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5^{-}} f(x)$
$1=3 a+b$ … (i)
$5 a+b=7$ … (ii)
On solving equation (i) and (ii), we get
$a=3, b=-8$

Continuity Exercise 8.1 Question 39 (i)

Answer:
Continuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Given:
$f(x)=|x|+|x-1|$
Solution:
$f(x)=|x|+|x-1|$
We have
(LHL at $x=0$ )
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}[|0-h|+|0-h-1|]=1$
(RHL at $x=0$ )
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0}[|0+h|+|0+h-1|]=1$
Also
$f(0)=|0|+|0-1|=0+1=1$
Now,
(LHL at $x=1$ )
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[|1-h|+|1-h-1|]=1$
(RHL at $x=1$ )
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[|1+h|+|1+h-1|]=1+0=1$
Also
$\begin{aligned} &f(1)=|1|+|1-1|=1+0=1 \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \text { and } \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}$
Hence $f\left ( x \right )$ is continuous at $x=0$ and $x=1$ .

Continuity Excercise 8.1 Question 39

Answer:
$f\left ( x \right )$ is continuous at $x=-1,1$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=|x-1|+|x+1|$
Solution:
$f(x)=|x-1|+|x+1|$
We have
(LHL at $x=-1$ )
$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{h \rightarrow 0} f(-1-h)=\lim _{h \rightarrow 0}[|-1-h-1|+|-1-h+1|]=2+0=2$
(RHL at $x=-1$ )
$\lim _{x \rightarrow-1^{+}} f(x)=\lim _{h \rightarrow 0} f(-1+h)=\lim _{h \rightarrow 0}[|-1+h-1|+|-1+h+1|]=2+0=2$
Also
$f(-1)=|-1-1|+|-1+1|=|-2|=2$
Now,
(LHL at $x=1$ )
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[|1-h-1|+|1-h+1|]=2$
(RHL at $x=1$ )
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[|1+h-1|+|1+h+1|]=0+2=2$
Also
$\begin{aligned} &f(1)=|1+1|+|1-1|=2+0=2 \\ &\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{-}} f(x)=f(-1) \text { and } \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}$
Hence $f\left ( x \right )$ is continuous at $x=-1$ and $x=1$ .

Continuity Exercise 8.1 Question 40

Answer:
Discontinuous
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{l} \frac{x-x}{x}, \text { if } x>0 \\\\ \frac{x+x}{x}, \text { if } x<0 \\\\ 2, \text { if } x=0 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{l} \frac{x-x}{x}, \text { if } x>0 \\\\ \frac{x+x}{x}, \text { if } x<0 \\\\ 2, \text { if } x=0 \end{array}\right.$
$f(x)=\left\{\begin{array}{l} 0, \text { if } x>0 \\ 2, \text { if } x<0 \\ 2, \text { if } x=0 \end{array}\right.$
We have
(LHL at $x=0$ )
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} 2=2$
(RHL at $x=0$ )
$\begin{gathered} \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} 0=0 \\ \lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \end{gathered}$
Thus $f\left ( x \right )$ is discontinuous at $x=0$ .

Continuity Exercise 8.1 Question 41

Answer:
$k$ can be any real number.
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{l} 2 x^{2}+k, \text { if } x \geq 0 \\ -2 x^{2}+k, \text { if } x<0 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{l} 2 x^{2}+k, \text { if } x \geq 0 \\ -2 x^{2}+k, \text { if } x<0 \end{array}\right.$
We have
(LHL at $x=0$ )
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}- 2-(h)^{2}+k=k$
(RHL at $x=0$ )
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} 2(h)^{2}+k=k$
If $f\left ( x \right )$ is continuous at $x=0$ .
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=k \end{aligned}$
$k$ can be any real number.

Continuity Exercise 8.1 Question 42

Answer:
For any value $\lambda, f(x)$ is Continuous at $x=1$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{l} \lambda\left(x^{2}-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \text { if } x>0 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{l} \lambda\left(x^{2}-2 x\right), \text { if } x \leq 0 \\ 4 x+1, \text { if } x>0 \end{array}\right.$
If $f\left ( x \right )$ is continuous at $x=0$ , then
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0^{-}} \lambda\left(x^{2}-2 x\right) \end{aligned}$
Putting $x=0-h \text { as } x \rightarrow 0^{-}$
When, $h\rightarrow 0$
$\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} \lambda\left((0-h)^{2}-2(0-h)\right) \\\\ &=\lim _{h \rightarrow 0} \lambda\left(h^{2}+2 h\right) \\\\ &=0 \end{aligned}$
At $x=0$ , RHL = $\lim _{x \rightarrow 0^{+}} f(x)$
$=\lim _{x \rightarrow 0^{+}} 4 x+1$
Putting $x=0+h \text { as } x \rightarrow 0^{+}$

When, $h\rightarrow 0$

$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0}[4(0+h)+1] \\\\ &=\lim _{h \rightarrow 0}(4 h+1) \\\\ &=1 \end{aligned}$
LHL $\neq$ RHL. Thus, $f\left ( x \right )$ is discontinuous at $x=0$ for any value $\lambda$ .
At $x=1$ , LHL= $\lim _{x \rightarrow 1^{-}} f(x)$
$=\lim _{x \rightarrow 1^{-}} 4 x+1$
Putting $x=1-\text {has } x \rightarrow 1^{-}$
When, $h\rightarrow 0$
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0}[4(1+h)+1] \\\\ &=\lim _{h \rightarrow 0}(4+4 h+1) \\\\ &=5 \end{aligned}$
At $x=1$ , RHL = $\lim _{x \rightarrow 1^{+}} f(x)$

$=\lim _{x \rightarrow 1^{+}} 4 x+1$
Putting $x=1+h$ as $x\rightarrow 1^{+}$
When, $h\rightarrow 0$
$\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0}[4(1+h)+1] \\\\ &=\lim _{h \rightarrow 0}(4+4 h+1)\\ \\ &=5 \end{aligned}$
LHL $=$ RHL
Therefore, for any value of $\lambda$ for which $f\left ( x \right )$ is continuous at $x=1$

Continuity Excercise 8.1 Question 43

Answer:
$k=2$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{l} 2 x+1, \text { if } x<2 \\ k, \text { if } x=2 \\ 3 x-1, \text { if } x>2 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{l} 2 x+1, \text { if } x<2 \\ k, \text { if } x=2 \\ 3 x-1, \text { if } x>2 \end{array}\right.$
We have
(LHL at $x=2$ )
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}[2(2-h)+1]=[4+1]=5$
(RHL at $x=2$ )
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}[3(2+h)-1]=[6-1]=5$
Also $f\left ( 2 \right )=k$
If $f\left ( x \right )$ is continuous at $x=2$ .
$\begin{aligned} &\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{-}} f(x)=f(2) \\\\ &5=5=k \\\\ &k=5 \end{aligned}$
Since $LHS =RHS ,f\left ( x \right )$ is continuous
Thus, $f\left ( x \right )$ is continuous at $x=2$ .

Continuity exercise 8.1 question 44

Answer:
$a=\frac{1}{2}, b=4$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{l} \frac{1-\sin ^{3} x}{3 \cos ^{2} x}, \text { if } x<\frac{\pi}{2} \\\\ a \quad, \text { if } x=\frac{\pi}{2} \\\\ \left(\frac{6(1-\sin x)}{(\pi-2 x)^{2}}\right), \text { if } x>\frac{\pi}{2} \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{l} \frac{1-\sin ^{3} x}{3 \cos ^{2} x}, \text { if } x<\frac{\pi}{2} \\\\ a \quad, \text { if } x=\frac{\pi}{2} \\\\ \left(\frac{6(1-\sin x)}{(\pi-2 x)^{2}}\right), \text { if } x>\frac{\pi}{2} \end{array}\right.$
We have
(LHL at $x=\frac{\pi}{2}$ )
$\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right)=\lim _{h \rightarrow 0}\left(\frac{1-\sin ^{3}\left(\frac{\pi}{2}-h\right)}{3 \cos ^{2}\left(\frac{\pi}{2}-h\right)}\right)$
$=\lim _{h \rightarrow 0}\left(\frac{1-\cos ^{3} h}{3 \sin ^{2} h}\right)=\frac{1}{3} \lim _{h \rightarrow 0}\left(\frac{(1-\cosh )\left(1+\cos ^{2} h+\cosh \right)}{(1-\cosh )(1+\cosh )}\right)$ $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$
$=\frac{1}{3} \lim _{h \rightarrow 0}\left(\frac{\left(1+\cos ^{2} h+\cosh \right)}{(1+\cosh )}\right)=\frac{1}{3}\left(\frac{1+1+1}{1+1}\right)=\frac{1}{2}$
(RHL at $x=\frac{\pi}{2}$ )
$\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right)=\lim _{h \rightarrow 0}\left[\frac{b\left[1-\sin \left(\frac{\pi}{2}+h\right)\right]}{\left[\pi-2\left(\frac{\pi}{2}+h\right)\right]^{2}}\right]=\lim _{h \rightarrow 0}\left[\frac{b[1-\cosh ]}{[-2 h]^{2}}\right]$
$[\because \sin (90+\theta)=\cos \theta]$
$=\lim _{h \rightarrow 0}\left[\frac{2 b \sin ^{2} \frac{h}{2}}{4 h^{2}}\right]=\lim _{h \rightarrow 0}\left[\frac{2 b \sin ^{2} \frac{h}{2}}{16 \frac{h^{2}}{4}}\right]=\frac{b}{8} \lim _{h \rightarrow 0}\left[\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right]^{2}$ $\left[\begin{array}{l} \because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \because 1-\cos x=2 \sin ^{2} \frac{x}{2} \end{array}\right]$
$\begin{aligned} &=\frac{b}{8} \times 1 \\ &=\frac{b}{8} \end{aligned}$
Also
$f\left(\frac{\pi}{2}\right)=a$
If $f\left ( x \right )$ is continuous at $x=\frac{\pi}{2}$ , then $\begin{aligned} &\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\\\ &\frac{1}{2}=\frac{b}{8}=a \\\\ &a=\frac{1}{2} \text { and } b=4 \end{aligned}$

Continuity exercise 8.1 question 45

Answer:
$k=1$
Hint:
$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.
Given:
$f(x)= \begin{cases}\frac{1-\cos 2 x}{2 x^{2}} & , \text { if } x<0 \\\\ k & , \text { if } x=0 \\\\ \frac{x}{|x|} & , \text { if } x>0\end{cases}$
Solution:
$f(x)= \begin{cases}\frac{1-\cos 2 x}{2 x^{2}} & , \text { if } x<0 \\\\ k & , \text { if } x=0 \\\\ \frac{x}{|x|} & , \text { if } x>0\end{cases}$
$f(x)= \begin{cases}\frac{1-\cos 2 x}{2 x^{2}} & , \text { if } x<0 \\\\ k & \text { , if } x=0 \\\\ 1 & , \text { if } x>0\end{cases}$
We have
(LHL at $x=0$ )
$\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}\left(\frac{1-\cos 2(-h)}{2(-h)^{2}}\right)=\lim _{h \rightarrow 0}\left(\frac{1-\cos 2 h}{2 h^{2}}\right)=\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{2 \sin ^{2} h}{h^{2}}\right)$
$\left[\because 1-\cos 2 x=2 \sin ^{2} x\right]$
$=\frac{2}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{2} h}{h^{2}}\right)=\lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)^{2}=1$ $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
(RHL at $x=0$ )
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}(1)=1$
Also
$f(0)=k$
If $f\left ( x \right )$ is continuous at $x=0$ .
$\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \\\\ &1=1=k \\\\ &k=1 \end{aligned}$
Hence, the required value of $k$ is $1$ .

Continuity exercise 8.1 question 46

Answer:
$3a-3b=2$
Hint:
$f(x)$ must be defined. The limit of the $f(x)$ approaches the value $x$ must exist.
Given:
$f(x)=\left\{\begin{array}{l} a x+1, \text { if } x \leq 3 \\ b x+3, \text { if } x>3 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{l} a x+1, \text { if } x \leq 3 \\ b x+3, \text { if } x>3 \end{array}\right.$
We have
(LHL at $x=3$ )
$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}[a(3-h)+1]=3 a+1$
(RHL at $x=3$ )
$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0}[b(3+h)+3]=3 b+3$
Also $f(2)=k$
If $f\left ( x \right )$ is continuous at $x=3$ .
$\begin{aligned} &\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{-}} f(x) \\ &3 a+1=3 b+3 \\ &3 a-3 b=2 \end{aligned}$
Hence the required relationship between $a \; and \; i \; is \; 3 a-3 b=2$

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