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

# 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.

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

## Continuity Excercise: 8.1

### Continuity Excercise 8.1 Question 1

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

$x=3$
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

$x=3$
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

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

$x=0$ (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

$x=0$ (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

$x=0$ (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

$x=0$ (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
$x=a$ (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
$f(x)=\left(\begin{array}{ll} 1, & x>a \\ -1, & x
$f(x)=\left(\begin{array}{l} 1, x \geq a \\ -1, x
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)

Continuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{c} |x| \cos \left(\frac{1}{x}\right), \text { if } x \neq 0 \\\\ 0 \quad, \text { if } x=0 \end{array}\right)$
We observe,
\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}|x| \cos \left(\frac{1}{x}\right) \\\\ &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}|x| \lim _{x \rightarrow 0} \cos \left(\frac{1}{x}\right) \end{aligned}
\begin{aligned} &\lim _{x \rightarrow 0} f(x)=0 \times \lim _{x \rightarrow 0} \cos \left(\frac{1}{x}\right)=0 \\\\ &\lim _{x \rightarrow 0} f(x)=0 \end{aligned}
The limit of function at $x$ tends to $0$ is equal to the value of function at that point hence it is continuous.
Hence, $f\left ( x \right )$ is continuous at $x=0$.

Continuity Exercise point 1 Question 10 (ii)

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

Continuity Exercise 8.1 (iii) Question 10

Continuous
Hint:
For a function to be continuous at a point, its LHL RHL and value at that point should be equal.
Solution:
Given,
$f(x)=\left(\begin{array}{c} (x-a) \sin \left(\frac{1}{x-a}\right), \text { if } x \neq a \\\\ 0, \text { if } x=a \end{array}\right)$
Putting $x-a=y$ , we get
\begin{aligned} &\lim _{x \rightarrow a}(x-a) \sin \left(\frac{1}{x-a}\right)=\lim _{y \rightarrow 0} y \sin \left(\frac{1}{y}\right)=\lim _{y \rightarrow 0} y \lim _{y \rightarrow 0} \sin \left(\frac{1}{y}\right)=0 \times \lim _{y \rightarrow 0} \sin \left(\frac{1}{y}\right)=0 \\\\ &\lim _{x \rightarrow a} f(x)=f(a)=0 \end{aligned}
Hence, $f\left ( x \right )$ is continuous at $x=a$.

Continuity Exercise 8.1 Question 10 (iv)
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 $x=0$.

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)

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)

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)

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
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

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

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

$a=\frac{1}{2}$
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

$f\left ( x \right )$ 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

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

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 or left hand limit at that point.
Solution:
Given,
$f(x)=\left(\begin{array}{l} x, i\! f \: 0 \leq x<\frac{1}{2} \\\\ \frac{1}{2}, \text { if } x=\frac{1}{2} \\\\ 1-x, \text { if } \frac{1}{2}
We observe
[LHL at $x=\frac{1}{2}$ ]
$\lim _{x \rightarrow \frac{1^{-}}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{1}{2}-h\right)=\lim _{h \rightarrow 0}\left(\frac{1}{2}-h\right)=\frac{1}{2}$
[RHL at $x=\frac{1}{2}$ ]
$\lim _{x \rightarrow \frac{1^{+}}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{1}{2}+h\right)=\lim _{h \rightarrow 0}\left(1-\left(\frac{1}{2}+h\right)\right)=\frac{1}{2}$
Also $f\left ( \frac{1}{2} \right )=\frac{1}{2}$
$\lim _{x \rightarrow \frac{1^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{1^{-}}{2}}f(x)=f\left(\frac{1}{2}\right)$
Thus, $f\left ( x \right )$ is continuous at $x=\frac{1}{2}$ .

Continuity Exercise 8.1 Question 17

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

$k=-1$
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

$k=\frac{5}{3}$
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

$k=\frac{3}{4}$
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

$k=\frac{2}{5}$
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

$a=-2$
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

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

$k=6$
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

$a=\frac{-3}{2}, b \neq 0, c=\frac{1}{2} ; b \in R-\{0\}$
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

$k=\pm 1$
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

$a=1, b=-1$
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

$k=2$
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

$\frac{a+b}{a b}=f(0)$
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

$k=\frac{1}{2}$
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}

$k=-4$
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

$f(\pi)=\frac{49}{10}$
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}

$f\left ( 0 \right )=1$
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}

$k=1$
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)

$k=\pm 2$
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)

$k=\frac{-2}{\pi}$
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)

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)

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

Given:
$f(x)=\left\{\begin{array}{l} k x+1, \text { if } x \leq \pi \\\\ \cos x, \text { if } x>\pi \end{array}\right.$

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

Continuity Excercise 8.1 Question 36 (v)
$k=\frac{9}{5}$
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)

$k=10$
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)

$k=4$
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)

$k=\frac{1}{2}$
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)

$k=7$
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

$a=3, b=-8$
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
Solution:
$f(x)= \begin{cases}1 & , \text { if } x \leq 3 \\ a x+b, & \text { if } 3
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)

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

$f\left ( x \right )$ 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

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

$k$ 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

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

$k=2$
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

$a=\frac{1}{2}, b=4$
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

$k=1$
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

$3a-3b=2$
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 $a \; and \; i \; is \; 3 a-3 b=2$

There are a couple of exercises, ex 8.1 and ex 8.2 in this chapter. The concepts that the first exercise, ex 8.1 covers are continuous function, absolute continuous function, continuous probability distribution, absolute continuity of a measure concerning another measure and many more. This first exercise consists of 62 questions to be answered by the students. Looking at the number of questions they need not worry if they have the RD Sharma Class 12th Exercise 8.1 material with them.

Most of the students who utilised the RD Sharma Class 8 solution of Continuity Ex 8.1 book to the best have attained excellent scores in their exams. Even the teachers do not miss to refer to the RD Sharma Class 12th Exercise 8.1 solution book before taking the lessons.

Following are a few advantages that the students attain when they use the RD Sharma Class 12 Solutions Chapter 8 exercise 8.1 solution book:

All the sums in the RD Sharma class 12th Exercise 8.1 reference book are solved in various methods that can be easily grasped by the students.

The fact is that, many teachers pick the homework sums from the RD Sharma Class 12th Exercise 8.1 book. It becomes even easier to recheck the answers or clarify the doubts regarding the homework sums with the help of this book.

The RD Sharma Class 12 Chapter 8 exercise 8.1 and the other RD Sharma books are available at the Career 360 website. Students can also download a copy if they want to use it offline.

Studying with outdated books is useless. The RD Sharma Class 12th Exercise 8.1 are updated according to the latest NCERT textbooks. As the PDF format of these books are available for free of cost at the Career 360 website, the students can make the most of it.

## RD Sharma Chapter wise Solutions

1. Who can refer to the RD Sharma Class 12 Chapter 8 ex 8.1 solutions material?

Most of the class 12 students, teachers and tutors who overcome the concept of Continuity use the RD Sharma Class 12 Chapter 8 ex 8.1 solutions.

2. Do the RD Sharma books cost much?

The RD Sharma Class 12 Chapter 8 ex 8.1 book along with the other RD Sharma books are present in the Career 360 website. Anyone can use this resource by downloading it for free.

3. Does the RD Sharma Class 12 Chapter 8 ex 8.1 book contain solution for all the questions asked on the textbook?

Apart from providing worked out solutions for the questions asked in the book, the RD Sharma Class 12 Chapter 8 ex 8.1 contains additional practice questions and answers too.

4. Where can the students find the updated reference materials for class 12 chapter 8?

The updated solution book, RD Sharma Class 12 Chapter 8 ex 8.1 is available at the Career 360 website.

5. What is the most used solution book to crack the public examinations with high scores?

The RD Sharma Class 12 Chapter 8 ex 8.1 is the most used reference material by the previous batch students to crack the public examinations with high scores.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:22 July,2024 - 31 August,2024

Exam Date:19 September,2024 - 19 September,2024