CBSE Class 12 Maths Chapter 5 Continuity and Differentiability – NCERT Solutions 2025–26

Exercise 5.1

1. Prove that \[\text{f}\left( \text{x} \right)\text{=5x-3}\] is a continuous function at \[\text{x=0}\], \[\text{x=-3}\] and \[\text{x=5}\].

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=5x-3}\].

At \[\text{x=0}\], \[\text{f}\left( \text{0} \right)\text{=5 }\!\!\times\!\!\text{ 0-3=-3}\].

Taking limit as $x\to 0$ both sides of the function give

\[\underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 0}{\mathop{\lim }}\,\left( \text{5x-3} \right)\text{=5 }\!\!\times\!\!\text{ 0-3=-3}\]

$\therefore \underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{0} \right)$.

Thus, $\text{f}$ satisfies continuity at \[\text{x=0}\].

Again, at \[\text{x=-3,f}\left( \text{-3} \right)\text{=5 }\!\!\times\!\!\text{ }\left( \text{-3} \right)\text{-3=-18}\].

Now, taking limit as $x\to 3$ both sides of the function give

\[\underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{5x-3} \right)\text{=5 }\!\!\times\!\!\text{ }\left( \text{-3} \right)\text{-3=-18}\]

\[\therefore \underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{-3} \right)\].

Therefore, $\text{f}$ satisfies continuity at \[\text{x=-3}\].

Also, at \[\text{x=5,f}\left( \text{x} \right)\text{=f}\left( \text{5} \right)\text{=5 }\!\!\times\!\!\text{ 5-3=25-3=22}\].

Taking limit as $x\to 5$ both sides of the function give

\[\underset{x\to 5}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 5}{\mathop{\lim }}\,\left( \text{5x-3} \right)\text{=5 }\!\!\times\!\!\text{ 5-3=22}\]

\[\therefore \underset{x\to 5}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{5} \right)\].

Hence, $\text{f}$ satisfies continuity at \[\text{x=5}\].

2. Examine the continuity of the function $\text{f}\left( \text{x} \right)\text{=2}{{\text{x}}^{\text{2}}}\text{-1}$ at \[\text{x=3}\].

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=2}{{\text{x}}^{\text{2}}}\text{-1}$.

Now, at $\text{x=3, f}\left( \text{3} \right)\text{=2 }\!\!\times\!\!\text{ }{{\text{3}}^{\text{2}}}\text{-1=17}$.

Taking limit as $x\to 3$ both sides of the function give

$\underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 3}{\mathop{\lim }}\,\left( \text{2}{{\text{x}}^{\text{2}}}\text{-1} \right)\text{=2 }\!\!\times\!\!\text{ }{{\text{3}}^{\text{2}}}\text{-1=17}$

\[\therefore \underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f=}\left( \text{3} \right)\].

Hence, $\text{f}$ satisfies continuity at \[\text{x=3}\].

3. Examine the following functions for continuity.

(a) $\text{f}\left( \text{x} \right)\text{=x-5}$

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=x-5}$.

It is assured that for every real number $\text{k}$, $\text{f}$ is defined and its value at $k$ is $\text{k-5}$. Also, it can be noted that 

$\underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x-5} \right)\text{=k=k-5=f}\left( \text{k} \right)$.

$\therefore \underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{k} \right)$

Hence, $\text{f}$ satisfies continuity at every real number and so, it is a continuous function.

(b) $\text{f}\left( \text{x} \right)=\frac{\text{1}}{\text{x-5}}\text{,x}\ne \text{5}$

Ans: The given function is

$\text{f}\left( \text{x} \right)\text{=}\frac{\text{1}}{\text{x-5}}$.

Let $\text{k}\ne \text{5}$ is any real number, then taking limit as $\text{x}\to \text{k}$ both sides of the function give

$\underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to k}{\mathop{\lim }}\,\frac{\text{1}}{\text{x-5}}\text{=}\frac{\text{1}}{\text{k-5}}$

Also, $\text{f(k)=}\frac{\text{1}}{\text{k-5}}$ , since $\text{k}\ne \text{5}$

$\therefore \underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{k} \right)$

Therefore, $f$ satisfies continuity at every point in the domain of $f$ and so, it is a continuous function.

(c) $\text{f}\left( \text{x} \right)\text{=}\frac{{{\text{x}}^{\text{2}}}\text{-25}}{\text{x+5}}\text{,x}\ne -\text{5}$

Ans: The given function is

\[\text{f}\left( \text{x} \right)\text{=}\frac{{{\text{x}}^{\text{2}}}\text{-25}}{\text{x+5}}\text{, x}\ne \text{5}\]

Now let $\text{c}\ne \text{-5}$ be any real number, then taking limit as $\text{x}\to \text{c}$ on both sides of the function give

\[\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\frac{{{\text{x}}^{\text{2}}}\text{-25}}{\text{x+5}}\text{=}\underset{x\to c}{\mathop{\lim }}\,\frac{\left( \text{x+5} \right)\left( \text{x-5} \right)}{\text{x+5}}\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=}\left( \text{c-5} \right)\]

Again, \[\text{f}\left( \text{c} \right)\text{=}\frac{\left( \text{c+5} \right)\left( \text{c-5} \right)}{\text{c+5}}\text{=}\left( \text{c-5} \right)\], since $\text{c}\ne \text{5}$.

Hence, \[\text{f}\] satisfies continuity at every point in the domain of $\text{f}$ and so it is a continuous function.

(d) \[\mathbf{f}\left( \mathbf{x} \right)=\left| \text{x-5} \right|\]

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left| \text{x-5} \right|=\left\{ \begin{align} & \text{5-x, if x}<\text{5} \\ & \text{x-5, if x}\,\ge \text{5} \\ \end{align} \right.$

Note that, $\text{f}$ is defined at all points in the real line. So, let assume $\text{c}$ be a point on a real line. 

Then, we have \[\text{c}<\text{5}\] or \[\text{c}=\text{5}\] or \[\text{c}>\text{5}\].

Now, let's discuss these three cases one by one.

Case (i): $\text{c}<\text{5}$

Then, the function becomes $\text{f}\left( \text{c} \right)\text{=5-c}$.

Now, $\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{5-x} \right)\text{=5-c}$.

$\therefore \underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Therefore, $\text{f}$ is continuous at all real numbers which are less than $5$. 

Case (ii): $\text{c=5}$

Then, \[\text{f}\left( \text{c} \right)\text{=f}\left( \text{5} \right)\text{=}\left( \text{5-5} \right)\text{=0}\].

Now,

$\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{x\to 5}{\mathop{\lim }}\,\left( \text{5-x} \right)\text{=}\left( \text{5-5} \right)\text{=0}$ and

$\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 5}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=0}$.

Therefore, we have

$\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Thus, $\text{f}$ satisfies continuity at \[\text{x=5}\], and so $\text{f}$ is continuous at $\text{x=5}$.

Case (iii): $\text{c}>\text{5}$

Then we have, $\text{f}\left( \text{c} \right)\text{=f}\left( \text{5} \right)\text{=c-5}$.

Now,

$\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=c-5}$.

Therefore,

$\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

So, $\text{f}$ is continuous at all real numbers that are greater than $5$.

Thus, $\text{f}$ satisfies continuity at every real number and hence, it is a continuous function.

4. Prove that $\text{f}\left( \text{x} \right)={{\text{x}}^{\text{n}}}$  is continuous at $\text{x=n}$, where $\mathbf{n}$ is a positive integer.

Ans: The given function is  $\text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{n}}}$.

We noticed that the function $\text{f}$ is defined at all positive integers $\text{n}$ and also its value at $\text{x=n}$ is ${{\text{n}}^{\text{n}}}$.

Therefore, $\underset{x\to n}{\mathop{\lim }}\,\text{f}\left( \text{n} \right)=\underset{x\to n}{\mathop{\lim }}\,\text{f}\left( {{\text{x}}^{\text{n}}} \right)\text{=}{{\text{n}}^{\text{n}}}$.

So, \[\underset{\text{x}\to \text{n}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{n} \right)\].

Thus, the function $\text{f}\left( \text{x} \right)\mathbf{=}{{\text{x}}^{\text{n}}}$  is continuous at $\text{x=n}$ , where $\text{n}$ is a positive integer.

5. Is the function f defined by 

\[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{x,}\,\,\text{if}\,\text{x}\,\le \text{1} \\ & \text{5,}\,\,\text{if}\,\text{x 1} \\ \end{align} \right.\]

continuous at x=0, x=1 ? At x=2?

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{x,}\,\text{ if x}\le \text{1} \\ & \text{5, }\,\text{if x}\,>\text{1} \\ \end{align} \right.\]

It is obvious that the function $\text{f}$ is defined at $\text{x=}0$ and its value at $\text{x=}0$ is $0$.

Now, $\underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 0}{\mathop{\lim }}\,\text{x=0}$.

So, $\underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{0} \right)$.

Hence, the function $\text{f}$ satisfies continuity at \[\text{x=0}\].

It can be observed that $\text{f}$ is defined at $\text{x=1}$ and its value at this point is $\text{1}$.

Now, the left-hand limit of the function $\text{f}$ at \[\text{x=1}\] is

$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\text{x=1}$.

Also, the right-hand limit of the function $\text{f}$ at \[\text{x=1}\] is

$\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{5} \right)$

Therefore, $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Thus, $\text{f}$ is not continuous at \[\text{x=1}\]

It can be found that $\text{f}$ is defined at $\text{x=2}$ and its value at this point is $5$.

That is, \[\underset{x\to 2}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 2}{\mathop{\lim }}\,\text{f}\left( \text{5} \right)\text{=5}\].

Therefore, \[\underset{x\to 2}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{2} \right)\]

Hence, $\text{f}$ satisfies continuity at \[\text{x=2}\].

6. Find all points of discontinuity of the function $\mathbf{f}$, where $\mathbf{f}$ is given by.

\[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{2x+3,}\,\,\text{if}\,\text{x}\,\le 2 \\ & \text{2x-3,}\,\,\text{if}\,\text{x 2} \\ \end{align} \right.\]

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{2x+3, if x}\le \text{2} \\ & \text{2x-3, if x}\,>\,\text{2} \\ \end{align} \right.\]

It can be observed that the function $\text{f}$ is defined at all the points in the real line.

Let consider $\text{c}$ be a point on the real line. Then, three cases may arise.

I.      $\text{c}\,<\,\text{2}$

II.     $\text{c}\,>\,\text{2}$

III.    $\text{c}=\text{2}$

Case (i): When $\text{c}\,<\text{2}$

Then, we have $\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to \infty }{\mathop{\lim }}\,\left( \text{2x+3} \right)\text{=2c+3}$.

Therefore,

$\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ attains continuity at all points $\text{x}$, where \[\text{x}<\text{2}\].

Case (ii): When $\text{c}>\text{2}$

Then, we have  $\text{f}\left( \text{c} \right)\text{=2c-3}$.

So,

$\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \infty }{\mathop{\lim }}\,\left( \text{2x-3} \right)\text{=2c-3}$.

Therefore, $\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ satisfies continuity at all points $\text{x}$ , where \[\text{x}>\text{2}\].

Case(iii): When $\text{c=2}$

Then, the left-hand limit of the function $\text{f}$ at \[\text{x=2}\] is

$\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\left( \text{2x+3} \right)\text{=2 }\!\!\times\!\!\text{ 2+3=7}$ and 

the right-hand limit of the function $\text{f}$ at \[\text{x=2}\] is,

$\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to 2+}{\mathop{\lim }}\,\left( \text{2x+3} \right)\text{=2 }\!\!\times\!\!\text{ 2-3=1}$.

Thus, at \[\text{x=2}\], $\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

So, the function  $\text{f}$ does not satisfy continuity at \[\text{x=2}\].

Hence, \[\text{x=2}\] is the only point of discontinuity of the function $\text{f}\left( \text{x} \right)$.

7. Find all points of discontinuity of the function $\mathbf{f}$, where $\mathbf{f}$ is given by

\[\mathbf{f}\left( \mathbf{x} \right)\mathbf{=}\left\{ \begin{align} & \left| \mathbf{x} \right|\mathbf{+3, if x}\le \mathbf{-3} \\ & \mathbf{-2x, if -3}\,<\,\mathbf{x}\,<\,\mathbf{3} \\ & \mathbf{6x+2, if x}\ge \mathbf{3} \\ \end{align} \right.\]

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \left| \text{x} \right|\text{+3, if x}\le \text{-3} \\ & \text{-2x, if -3}<\text{x}<\text{3} \\ & \text{6x+2, if x}\ge \text{3} \\ \end{align} \right.\]

Observe that, $\text{f}$ is defined at all the points in the real line.

Now, let assume $\text{c}$ as a point on the real line.

Then five cases may arise. Either $\text{c}<\text{-3}$, or $\text{c}=\text{-3}$ or $\text{-3}<\text{c}<\text{3}$, or $\text{c}=\text{3}$, or $\text{c}>\text{3}$.

Let's discuss the five cases one by one.

Case I: When \[\text{c}<\text{-3}\]

Then, $\text{f}\left( \text{c} \right)\text{=-c+3}$ and

$\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{-x+3} \right)\text{=-c+3}$.

Therefore, $\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ satisfies continuity at all points $\text{x}$, where \[\text{x}<\text{-3}\].

Case II: When \[\text{c=-3}\]

Then, $\text{f}\left( \text{-3} \right)\text{=-}\left( \text{-3} \right)\text{+3=6}$.

Also, the left-hand limit

$\underset{\text{x}\to -{{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to -{{3}^{-}}}{\mathop{\lim }}\,\left( \text{-x+3} \right)\text{=-}\left( \text{-3} \right)\text{+3=6}$.

and the right-hand limit

$\underset{\text{x}\to -{{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to -{{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( -2\text{x} \right)\text{=-2}\times \left( \text{-3} \right)\text{=6}$.

Therefore, $\underset{\text{x}\to -3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{-3} \right)$.

Hence, $\text{f}$ satisfies continuity at $\text{x=-3}$.

Case III: When  $\text{-3}<\text{c}<\text{3}$.

Then, $f\left( c \right)=-2c$ and also

 $\underset{x\to c}{\mathop{\lim }}\,\text{f}\left( x \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( -2x \right)=-2c$.

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Hence, $\text{f}$ satisfies continuity at $\text{x}$, where $\text{-3}<\text{x}<\text{3}$.

Case IV: When \[\text{c=3}\]

Then, the left-hand limit of the function $\text{f}$ at \[\text{x=3}\] is

$\underset{\text{x}\to {{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{-2x} \right)\text{=-2 }\!\!\times\!\!\text{ 3=-6}$ and

the right-hand limit of the function $\text{f}$ at \[\text{x=3}\] is

$\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{6x+2} \right)\text{=6 }\!\!\times\!\!\text{ 3+2=20}$.

Thus, at $\text{x=3}$, $\underset{\text{x}\to {{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, $\text{f}$ does not satisfy continuity at \[\text{x=3}\].

Case V: When $\text{c}>\text{3}$.

 Then $\text{f}\left( \text{c} \right)\text{=6c+2}$ and also

 $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{6x+2} \right)\text{=6c+2}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

So, $\text{f}$ satisfies continuity at all points $\text{x}$, when $\text{x}>\text{3}$.

Thus, \[\text{x=3}\] is the only point of discontinuity of the function $\text{f}$.

8. Find all points of discontinuity of the function $\mathbf{f}$, where $\mathbf{f}$ is given by \[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{\left| \text{x} \right|}{\text{x}}\text{, if x}\ne \text{0} \\ & \text{0, if x=0} \\ \end{align} \right\}\]

Ans: The given function is \[\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{\left| \text{x} \right|}{\text{x}}\text{, if x}\ne \text{0} \\ & \text{0, if x=0} \\ \end{align} \right\}\]

Now, $\text{f}\left( \text{x} \right)$ can be rewritten as $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{\left| \text{x} \right|}{\text{x}}\text{=}\frac{\text{-x}}{\text{x}}\text{=-1}\,\,\text{if}\,\,\text{x<0} \\ & \text{0, if x=0} \\ & \frac{\left| \text{x} \right|}{\text{x}}\text{=}\frac{\text{x}}{\text{x}}\text{=1 }\,\,\text{if}\,\,\text{ x>0} \\ \end{align} \right\}$

It can be noted that the function $\text{f}$ is defined at all points of the real line.

Now, let assume $\text{c}$ as a point on the real line.

Then three cases may arise, either $\text{c}<\text{0}$, or $\text{c}=\text{0}$, or $\text{c}>\text{0}$.

Let's discuss three cases one by one.

Case I: When $\text{c}<\text{0}$.

Then, $\text{f}\left( \text{c} \right)\text{=-1}$ and

$\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ satisfies continuity at all the points $\text{x}$ where \[\text{x}<\text{0}\].

Case II: When \[\text{c=0}\].

Then, the left-hand limit of the function $\text{f}$ at \[\text{x=0}\] is

$\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$ and

the right-hand limit of the function $\text{f}$ at \[\text{x=0}\] is

$\underset{\text{x}\to 0+}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{1} \right)\text{=1}$.

At $\text{x=0}$, $\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, the function $\text{f}$ does not satisfy continuity at \[\text{x=0}\].

Case III: When $\text{c}>\text{0}$.

Then \[\text{f}\left( \text{c} \right)\text{=1}\] and also

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{1} \right)\text{=1}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

So, the function $\text{f}$ is continuous at all the points $\text{x}$, for $\text{x}>\text{0}$.

Thus, \[\text{x=0}\] is the only point of discontinuity for the function $\text{f}$.

9. Find all points of discontinuity of the function $\mathbf{f}$, where $\mathbf{f}$ is given by $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{\text{x}}{\left| \text{x} \right|}\text{, if x}<\text{0} \\ & \text{-1, if x}\ge \text{0} \\ \end{align} \right.$

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{\text{x}}{\left| \text{x} \right|}\text{, if x}\,<\,\text{0} \\ & \text{-1, if x}\ge \text{0} \\ \end{align} \right.$

Now, we know that, if \[\text{x}<\text{0}\], then  \[\left| \text{x} \right|\text{=-x}\].

Therefore, the $\text{f}\left( \text{x} \right)$ can be written as

$\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \frac{x}{\left| \text{x} \right|}=\frac{\text{x}}{\text{-x}}=\text{-1}\,\,\text{if}\text{x}<\text{0} \\ & \text{-1,}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ if x}=\text{0} \\ & \text{-1, }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{if}\,\text{x}>\text{0} \\ \end{align} \right\}$

$\Rightarrow \text{f}\left( \text{x} \right)\text{=-1}$ for all positive real numbers.

Now, let's assume $c$ as any real number.

Then three cases may arise, either $\text{c}<\text{0}$, or $\text{c}=\text{0}$, or $\text{c}>\text{0}$.

Let's discuss three cases one by one.

Case I: When \[\text{c}<\text{0}\].

Then, $\text{f}\left( \text{c} \right)\text{=-1}$ and

$\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, $\text{f}$ satisfies continuity at all the points $\text{x}$ where \[\text{x}<\text{0}\].

Case II: When \[\text{c=0}\].

Then, the left-hand limit of the function $\text{f}$ at \[\text{x=0}\] is

$\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$ and

the right-hand limit of the function $\text{f}$ at \[\text{x=0}\] is

$\underset{\text{x}\to 0+}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}$.

At $\text{x=0}$, $\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, the function $\text{f}$ satisfies continuity at \[\text{x=0}\].

Case III: When $\text{c}>\text{0}$.

Then \[\text{f}\left( \text{c} \right)\text{=-1}\] and also

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

So, the function $\text{f}$ is continuous at all the points $\text{x}$, for \[\text{x}>\text{0}\].

Then, we have $\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f(x)}=\underset{\text{x}\to c}{\mathop{\lim }}\,(-1)=-1$  and \[\text{f}\left( \text{c} \right)\text{=-1}=\underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\].

Therefore, the function $\text{f}\left( \text{x} \right)$ is a continuous function.

Thus, there does not exist any point of discontinuity.

10. Find all points of discontinuity of the function $\mathbf{f}$, where $\mathbf{f}$ is given by $\text{f(x)=}\left\{ \begin{align} & \text{x+1, if x}\ge \text{1} \\ & {{\text{x}}^{\text{2}}}\text{+1, if x}\,<\,\text{1} \\ \end{align} \right.$Ans: The given function is$\text{f(x)=}\left\{ \begin{align} & \text{x+1, if x}\ge \text{1} \\ & {{\text{x}}^{\text{2}}}\text{+1, if x}\,<\,\text{1} \\ \end{align} \right.$

Note that, $\text{f}\left( \text{x} \right)$ is defined at all the points of the real line.

Now, let's assume $c$ as a point on the real line.

Then three cases may arise, either either $\text{c}<\text{1}$, or $\text{c}=\text{1}$, or $\text{c}>\text{1}$.

Let's discuss the three cases one by one.

Case I: When $\text{c}<\text{1}$.

 Then, \[\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{2}}}\text{+1}\] and also

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{=}{{\text{c}}^{\text{2}}}\text{+1}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Hence, \[\text{f}\] satisfies continuity at all the points $\text{x}$, where $\text{x}<\text{1}$.

Case II: When \[\text{c=1}\].

Then, we have \[\text{f}\left( \text{c} \right)\text{=f}\left( \text{1} \right)\text{=1+1=2}\].

Now, the left-hand limit of $\text{f}$ at \[\text{x=1}\] is

\[\underset{\text{x}\to {{\text{1}}^{-}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{-}}}{\mathop{\lim }}\,\text{(}{{\text{x}}^{\text{2}}}\text{+1)=}{{\text{1}}^{\text{2}}}\text{+1=2}\] and the right-hand limit of $\text{f}$ at $\text{x=1}$ is, \[\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\left( \text{x+1} \right)\text{=1+1=2}\].

Therefore, \[\underset{\text{x}\to 1}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Hence, $\text{f}$ satisfies continuity at \[\text{x=1}\].

Case III: When $\text{c}>\text{1}$.

Then, we have  $\text{f}\left( \text{c} \right)\text{=c+1}$ and 

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{x+1} \right)\text{=c+1}\].

Therefore,

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

So, $\text{f}$ satisfies continuity at all the points $\text{x}$, where $\text{x}>\text{1}$.

Hence, there does not exist any discontinuity points.

11. Find all points of discontinuity of the function $\mathbf{f}$, where        \[\mathbf{f}\left( \mathbf{x} \right)=\left\{ \begin{align} & {{\mathbf{x}}^{\mathbf{3}}}-\mathbf{3},\mathbf{if}\,\,\mathbf{x}\le \mathbf{2} \\ & {{\mathbf{x}}^{\mathbf{2}}}+\mathbf{1},\mathbf{if}\,\,\mathbf{x}>\mathbf{2} \\ \end{align} \right.\]

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & {{\text{x}}^{\text{3}}}\text{-3, if x}\le \text{2} \\ & {{\text{x}}^{\text{2}}}\text{+1, if x}>\text{2} \\ \end{align} \right.$

Observe that, the function $\text{f}$ is defined at all points in the real line.

Now, let assume $\text{c}$ as a point on the real line.

Case I: When $\text{c}<\text{2}$.

Then, we have  $\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{3}}}\text{-3}$ and also \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{3}}}\text{-3} \right)\text{=}{{\text{c}}^{\text{3}}}\text{-3}\].

Therefore, the function $\text{f}$ attains continuity at all the points $\text{x}$, where $\text{x}<\text{2}$.

Case II: When \[\text{c=2}\].

Then, we have $\text{f}\left( \text{c} \right)\text{=f}\left( \text{2} \right)\text{=}{{\text{2}}^{\text{3}}}\text{-3=5}$.

Now the left-hand limit of the function is

$\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\left( {{\text{x}}^{3}}-3 \right)\text{=}{{\text{2}}^{3}}\text{-3=5}$ and the right-hand limit is

$\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{=}{{\text{2}}^{\text{2}}}\text{+1=5}$.

Therefore, $\underset{\text{x}\to 2}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{2} \right)$.

Hence, the function $\text{f}$ is continuous at \[\text{x=2}\].

Case III: When $\text{c}>\text{2}$.

Then, $\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{2}}+1$ and

$\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{=}{{\text{c}}^{\text{2}}}\text{+1}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

So, $\text{f}$ attains continuity at all the points $\text{x}$, where $\text{x}>\text{2}$.

Thus, the function $\text{f}$ is continuous at all the points on the real line.

Hence, $\text{f}$ does not have any point of discontinuity.

12. Find all points of discontinuity of the function $\mathbf{f}$, where $\mathbf{f}$ is given by \[\text{f}\left( \text{x} \right)=\left\{ \begin{align} & {{\text{x}}^{\text{10}}}\text{-1, if x}\le \text{1} \\ & {{\text{x}}^{\text{2}}}\text{, if x}\,>\,\text{1} \\ \end{align} \right.\]

Ans: The given function is $\text{f}\left( \text{x} \right)=\left\{ \begin{align} & {{\text{x}}^{\text{10}}}\text{-1, if x}\le \text{1} \\ & {{\text{x}}^{\text{2}}}\text{, if x}\,>\text{1} \\ \end{align} \right.$

Observe that, the function $\text{f}$ is defined at every point of the real line.

Now, let assume $\text{c}$ as a point on the real number line.

Case I: When $\text{c}<\text{1}$.

Then $\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{10}}}\text{-1}$.

Also,  $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{10}}}\text{-1} \right)\text{=}{{\text{c}}^{\text{10}}}\text{-1}$

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Hence, the function $\text{f}$ attains continuity at every point $\text{x}$, for $\text{x}<\text{1}$.

Case II: When \[\text{c=1}\].

Then the left-hand limit of the function $\text{f}\left( \text{x} \right)$ at \[\text{x=1}\] is

\[\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\left( {{\text{x}}^{10}}-1 \right)={{1}^{10}}-1=1-1=0\] and

the right-hand limit of the function $\text{f}$ at $\text{x=1}$ is

$\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}} \right)\text{=}{{\text{1}}^{\text{2}}}\text{=1}$.

So, we can notice that, $\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{\text{1}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, the function $\text{f}$ does not satisfy continuity at \[\text{x=1}\].

Case III: When $\text{c}>\text{1}$.

Then, $\text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{2}}}$.

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}} \right)\text{=}{{\text{c}}^{\text{2}}}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)\].

Thus, the function $\text{f}$ attains continuity at every point $\text{x}$, for $\text{x}>\text{1}$.

Hence, we can conclude that \[\text{x=1}\] is the only point of discontinuity for the function $\text{f}$.

13. Is the function defined by $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{x+5, if x}\le \text{1} \\ & \text{x-5, if x}\,>\,\text{1} \\ \end{align} \right.$ a continuous function? 

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{x+5, if x}\le \text{1} \\ & \text{x-5, if x}\,>\text{1} \\ \end{align} \right.$

It can be noted that the function $\text{f}$ is defined at every point on the real line. 

Now, let assume $\text{c}$ as a point on the real line.

Case I: When $\text{c}<\text{1}$.

Then, $\text{f}\left( \text{c} \right)\text{=c+5}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{x+5} \right)\text{=c+5}$.

Hence, $\text{f}$ satisfies continuity at every point $\text{x}$, for $\text{x}<\text{1}$.

Case II: When \[\text{c=1}\].

Then, $\text{f}\left( \text{1} \right)\text{=1+5=6}$.

Now, the left-hand limit of the function $\text{f}$ at \[\text{x=1}\] is

$\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\left( \text{x+5} \right)\text{=1+5=6}$ and the right-hand limit of the function at \[\text{x=1}\] is $\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{(x-5)=1-5=4}$.

Thus, it is seen that, $\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, $\text{f}$ does not attain continuity at \[\text{x=1}\].

Case III: When $\text{c}>\text{1}$.

Then $\text{f}\left( \text{c} \right)\text{=c-5}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=c-5}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Thus, the function $\text{f}$ is continuous at every point \[\text{x}\], for $\text{x}>\text{1}$.

Hence, we can conclude that \[\text{x=1}\] is the only point of discontinuity for the function $\text{f}$.

14. Discuss the continuity of the function $\mathbf{f}$, where $\mathbf{f}$ is defined by $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{3, if 0}\le \text{x}\le \text{1} \\ & \text{4, if 1}\,<\,\text{x}\,<\,\text{3} \\ & \text{5, if 3}\le \text{x}\le \text{10} \\ \end{align} \right.$

Ans: The given function is $\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{3, if 0}\le \text{x}\le \text{1} \\ & \text{4, if 1}\,<\,\text{x}\,<\,\text{3} \\ & \text{5, if 3}\le \text{x}\le \text{10} \\ \end{align} \right.$

Therefore, $\text{f}$ is defined in the interval $\text{ }\!\![\!\!\text{ 0,10 }\!\!]\!\!\text{ }$.

Now let's assume $\text{c}$ as a point in the interval $\text{ }\!\![\!\!\text{ 0,10 }\!\!]\!\!\text{ }$.

Then there may arise five cases.

Case I: When \[\text{0}\le \text{c}<\text{1}\].

Then $\text{f}\left( \text{c} \right)\text{=3}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{3} \right)\text{=3}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, the function $\text{f}$ attains continuity at the interval $\text{ }\!\![\!\!\text{ 0,1 }\!\!]\!\!\text{ }$.

Case II: When \[\text{c=1}\].

Then $\text{f}\left( \text{3} \right)\text{=3}$.

Also, the left-hand-limit of the function at \[\text{x=1}\] is

$\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\left( \text{3} \right)\text{=3}$ and the right-hand-limit of the function at \[\text{x=1}\] is

$\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\left( \text{4} \right)\text{=4}$.

Thus, it is noticed that $\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{\text{1}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$.

Hence, the function $\text{f}$ does not satisfy continuity at \[\text{x=1}\].

Case III: When \[\text{1}<\text{c}<\text{3}\].

Then $\text{f}\left( \text{c} \right)\text{=4}$.

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{4} \right)\text{=4}\].

Thus, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, the function $\text{f}$ attains continuity at every point in the interval $\text{ }\!\![\!\!\text{ 1,3 }\!\!]\!\!\text{ }$.

Case IV: When \[\text{c=3}\].

Then $\text{f}\left( \text{c} \right)\text{=5}$.

Now, the left-hand-limit of the function $\text{f}$ at \[\text{x=3}\] is

$\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\left( \text{4} \right)\text{=4}$ and the right-hand-limit of the function $\text{f}$ at \[\text{x=3}\] is

\[\underset{\text{x}\to {{\text{3}}^{\text{+}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\left( \text{5} \right)\text{=5}\].

Therefore, it is noted that $\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{\text{3}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)$ .

Hence, the function $\text{f}$ is not continuous at \[\text{x=3}\].

Case V: When $\text{3}<\text{c}\le \text{10}$.

Then $\text{f}\left( \text{c} \right)\text{=5}$.

Also,  $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{5} \right)\text{=5}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

So, the function $\text{f}$ attains continuity at every point in the interval $\text{ }\!\![\!\!\text{ 3,10 }\!\!]\!\!\text{ }$.

Hence, the function $\text{f}$ is not continuous at \[\text{x=1}\] and \[\text{x=3}\].

15. Discuss the continuity of the function $\mathbf{f}$, where $\mathbf{f}$ such that$\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{2x, if x}\,<\,\text{0} \\ & \text{0, }\,\,\,\text{if 0}\le \text{x}\le \text{1} \\ & \text{4x, if x}\,>\,\text{1} \\ \end{align} \right.$

Ans: The given function is 

$\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{2x, if x}<\text{0} \\ & \text{0, if 0}\le \text{x}\le \text{1} \\ & \text{4x, if x}>\text{1} \\ \end{align} \right.$

Now, let consider $\text{c}$ be a point on the real number line. 

Then, five cases may arrive.

Case I: When \[\text{c}<\text{0}\].

Then, $\text{f}\left( \text{c} \right)\text{=2c}$.

Also,  $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{2x} \right)\text{=2c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right)$.

Hence, the function $\text{f}$ attains continuity at every point $\text{x}$ whenever \[\text{x}<\text{0}\].

Case II: When \[\text{c = 0}\].

Then, \[\text{f}\left( \text{c} \right)\text{=f}\left( \text{0} \right)\text{=0}\].

Now, the left-hand-limit of the function \[\text{f}\]at \[\text{x = 0}\] is

$\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\left( \text{2x} \right)=0$ and the right-hand limit of the function \[\text{f}\]at \[\text{x = 0}\] is, 

$\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{x} \right)\text{=}\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{0} \right)\text{=0}$.

Therefore, $\underset{\text{x}\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{0} \right)$.

Thus, the function \[\text{f}\] attains continuity at \[\text{x = 0}\].

Case III: When $\text{0}<\text{c}<\text{1}$ 

Then, $\text{f(x)=0}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(0)=0}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, $\text{f}$ attains continuity at every point in the interval $\text{(0,1)}$.

Case IV: When \[\text{c =1}\].

Then, $\text{f(c)=f(1)=0}$.

Now, the left-hand-limit at \[\text{x = 1}\] is

\[\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{(0)=0}\] and the right-hand-limit at \[\text{x = 1}\] is

\[\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{(4x)=4 }\!\!\times\!\!\text{ 1=4}\].

Thus, it is noticed that, \[\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)}\ne \underset{\text{x}\to {{\text{1}}^{+}}}{\mathop{\lim }}\,\text{f(x)}\].

Hence, the function $\text{f}$ is not continuous at \[\text{x = 1}\].

Case V: When \[\text{c}>\text{1}\].

Then, \[\text{f(c)=f(1)=0}\].

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(4x)=4c}\]

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

So, the function $\text{f}$ attains continuity at every point $\text{x}$, for \[\text{c}>\text{1}\].

Hence, the function $\text{f}$ is discontinuous only at \[\text{x = 1}\].

16. Discuss the continuity of the function $\mathbf{f}$, where $\mathbf{f}$ is defined by $\text{f(x)=}\left\{ \begin{align} & \text{-2, if x}\le \text{-1} \\ & \text{2x, if -1}\,<\,\text{x}\le \text{1} \\ & \text{2, }\,\,\,\,\text{if x}\,>\,\text{1} \\ \end{align} \right.$

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{-2, if x}\le \text{-1} \\ & \text{2x, if -1}<\text{x}\le \text{1} \\ & \text{2, }\,\,\,\,\text{if x}>\text{1} \\ \end{align} \right.$

Note that, $\text{f}$ is defined at every point in the interval $\left[ -1,\infty  \right)$.

Now, let assume $\text{c}$ is a point on the real number line.

Case I: When \[\text{c}<-\text{1}\].

Then, $\text{f(c)=-2}$.

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(-2)=-2}\].

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, the function $\text{f}$ attains continuity at every point $\text{x}$ , for \[\text{x}<-\text{1}\].

Case II: When \[\text{c=-1}\].

Then, $\text{f(c)=f(-1)=-2}$.

Now, the left-hand-limit of the function at \[\text{x=-1}\] is

$\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{(-2)=-2}$ and the right-hand-limit at \[\text{x=-1}\] is

$\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{(x)=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{=2 }\!\!\times\!\!\text{ (-1)=-2}$.

Therefore, $\underset{\text{x}\to -1}{\mathop{\lim }}\,\text{f(x)=f(-1)}$.

Hence, the function \[\text{f}\] satisfies continuity at \[\text{x=-1}\].

Case III: When $\text{-1}<\text{c}<\text{1}$.

Then, $\text{f(c)=2c}$ and \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(2x)=2c}\].

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, the function $\text{f}$ attains continuity at every point in the interval $\text{(-1,1)}$.

Case IV: When \[\text{c=1}\].

Then, $\text{f(c)=f(1)=2 }\!\!\times\!\!\text{ 1=2}$

Now, the left-hand-limit of the function at \[\text{x = 1}\] is 

$\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{(2x)=2 }\!\!\times\!\!\text{ 1=2}$ and the right-hand-limit at \[\text{x = 1}\] is

$\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{2=2}$.

Therefore, $\underset{\text{x}\to 1}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Thus, the function $\text{f}$ attains continuity at \[\text{x=2}\].

Case V: When \[\text{c}>\text{1}\].

Then \[\text{f(c)=2}\].

Also, $\underset{\text{x}\to 2}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to 2}{\mathop{\lim }}\,\text{(2)=2}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, the function $\text{f}$ is continuous at every point $\text{x}$, for \[\text{x}>\text{1}\].

Thus, it can be concluded that the function $\text{f}$ is continuous for all the points.

17. Find the relationship between $\mathbf{a}$ and $\mathbf{b}$ so that the function $\mathbf{f}$ defined by $\text{f(x)=}\left\{ \begin{align} & \text{ax+1, if x}\le \text{3} \\ & \text{bx+3, if x}\,>\,\text{3} \\ \end{align} \right.$ is continuous at $\text{x=3}$

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{ax+1, if x}\le \text{3} \\ & \text{bx+3, if x}\,>\text{3} \\ \end{align} \right.$

The function $\text{f}$ will be continuous at \[\text{x = 3}\] if

$\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=f(3)}$,                               ..….. (1)

$\underset{x\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(ax+1)=3a+1}$,

$\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f(bx+1)=3b+3}$,                    …… (2)

 and

$\text{f(3)=3a+1}$.                                                     …… (3)

Therefore, from the equation (1), (2), and (3) gives

$\Rightarrow \text{3a+1=3b+3}$

$\Rightarrow \text{3a=3b+2}$

$\Rightarrow \text{a=b+}\frac{\text{2}}{\text{3}}$

Hence, the required relationship between $\text{a}$ and $\text{b}$ is given by $\text{a=b+}\frac{\text{2}}{\text{3}}$.

18. For what value of $\mathbf{\lambda }$ is the function defined by $\mathbf{f(x)=}\left\{ \begin{align} & \mathbf{\lambda (}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-2x), }\,\,\,\mathbf{if x}\le \mathbf{0} \\ & \mathbf{ 4x+1, if x}>\mathbf{0} \\ \end{align} \right.$ is continuous at $\mathbf{x=0}$. Also discuss the continuity of $\mathbf{f}$ at \[\mathbf{x = 1}\]?

Ans: The given function is\[\text{f(x)=}\left\{ \begin{align} & \text{ }\!\!\lambda\!\!\text{ (}{{\text{x}}^{\text{2}}}\text{-2x), if x}\le \text{0} \\ & \text{ 4x+1, if x}\,>\,\text{0} \\ \end{align} \right.\]

Now the function will be continuous at \[\text{x = 0}\] if

$\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(0)}$.

Also, the R.H.L and L.H.L are given by,

$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( 4x+1 \right)=4\left( 0 \right)+1=1$,

$\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{ }\!\!\lambda\!\!\text{ (}{{\text{x}}^{\text{2}}}\text{-2x)}\,\text{=}\,\text{ }\!\!\lambda\!\!\text{ (}{{\text{0}}^{\text{2}}}\text{-2 }\!\!\times\!\!\text{ 0)=0}$.

So, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)$.

Thus, there does not exist any value of $\text{ }\!\!\lambda\!\!\text{ }$ for which $\text{f}$ is continuous at \[\text{x = 0}\].

Now, at \[\text{x = 1}\],

$\text{f(1)=4x+1=4 }\!\!\times\!\!\text{ 1+1=5}$ and

$\underset{\text{x}\to 1}{\mathop{\lim }}\,\text{(4x+1)=4 }\!\!\times\!\!\text{ 1+1=5}$.

Therefore, $\underset{\text{x}\to 1}{\mathop{\lim }}\,\text{f(x)=f(1)}$.

Hence, the function $\text{f}$ is continuous at \[\text{x = 1}\], for all values of $\text{ }\!\!\lambda\!\!\text{ }$.

19. Show that the function defined by $\text{g(x)=x- }\!\![\!\!\text{ x }\!\!]\!\!\text{ }$ is discontinuous at all integral point, here $\mathbf{[x]}$ denotes the greatest integer value of $\mathbf{x}$ that are less than or equal to $\mathbf{x}$. 

Ans: The given function is $\text{g(x)=x-}\left[ \text{x} \right]$.

Note that the function is defined at every integral point.

Now, let assume that $\text{n}$ is an integer.

Then, $\text{g(n)=n-}\left[ \text{n} \right]\text{=n-n=0}$.

Now taking left-hand-limit as $\text{x}\to \text{n}$ to the function $\text{g}$ gives 

$\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\left[ \text{x-}\left[ \text{x} \right] \right]\text{=}\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\left( \text{x} \right)\text{-}\underset{\text{x}\to \text{n-}}{\mathop{\lim }}\,\left[ \text{x} \right]\text{=n-}\left( \text{n-1} \right)\text{=1}$.

Again, the right-hand-limit on the function at $\text{x=n}$ is

\[\underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\left[ \text{x-}\left[ \text{x} \right] \right]\text{=}\underset{\text{x}\to {{\text{n}}^{\text{+}}}}{\mathop{\lim }}\,\left( \text{x} \right)\text{-}\underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\left[ \text{x} \right]\text{=n-n=0}\].

Note that, $\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\text{g(x)}\ne \underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\text{g(x)}$.

Thus, the function $\text{f}$ is cannot be continuous at $\text{x=n,}$ 

Hence, the function $\text{g}$ is not continuous at any integral point. 

20. Is the function defined by $\text{f(x)=}{{\text{x}}^{\text{2}}}\text{-sinx+5}$ is continuous at \[\text{x= }\!\!\pi\!\!\text{ }\]

Ans: The given function is $\text{f(x)=}{{\text{x}}^{\text{2}}}\text{-sinx+5}$.

Now, at \[\text{x= }\!\!\pi\!\!\text{ }\], 

\[\text{f(x)}\,\text{=}\,\text{f( }\!\!\pi\!\!\text{ )}\,\text{=}\,{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{-sin }\!\!\pi\!\!\text{ }\,\text{+}\,\text{5}\,\text{=}\,{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{-0+5}\,\text{=}\,{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{+5}\]

Taking limit as $\text{x}\to \text{ }\!\!\pi\!\!\text{ }$ on the function $\text{f}\left( \text{x} \right)$ gives

\[\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\text{(}{{\text{x}}^{\text{2}}}\text{-sinx+5)}\].

Now substitute $\text{x= }\!\!\pi\!\!\text{ +h}$ into the function $\text{f}\left( \text{x} \right)$.

When $\text{x}\to \text{ }\!\!\pi\!\!\text{ }$, then $\text{h}\to 0$.

Therefore, 

\[\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{-sinx} \right)\text{+5}\].

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \left( \text{ }\!\!\pi\!\!\text{ +}{{\text{h}}^{\text{2}}} \right)\text{-sin}\left( \text{ }\!\!\pi\!\!\text{ +h} \right)\text{+5} \right]\]

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,{{\left( \text{ }\!\!\pi\!\!\text{ +h} \right)}^{\text{2}}}-\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{sin}\left( \text{ }\!\!\pi\!\!\text{ +h} \right)\text{+}\underset{\text{h}\to 0}{\mathop{\lim }}\,5\]

\[\text{=}{{\left( \text{ }\!\!\pi\!\!\text{ +0} \right)}^{\text{2}}}-\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{ }\!\![\!\!\text{ sin }\!\!\pi\!\!\text{ }\cdot \text{cosh+cos }\!\!\pi\!\!\text{ }\cdot \text{sinh }\!\!]\!\!\text{ +5}\]

\[\text{=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}-\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sin }\!\!\pi\!\!\text{ }\cdot \text{cosh} \right)\text{-}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cos }\!\!\pi\!\!\text{ }\cdot \text{sinh} \right)\text{+5}\]

\[\text{=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}-\text{sin }\!\!\pi\!\!\text{ }\cdot \text{cos0}-\text{cos }\!\!\pi\!\!\text{ }\cdot \text{sin0+5}\]

\[\text{=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}-\text{0 }\!\!\times\!\!\text{ 1}-\text{(}-\text{1) }\!\!\times\!\!\text{ 0+5=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{+5}\].

So, \[\underset{\text{x}\to \text{x}}{\mathop{\lim }}\,\text{f(x)=f( }\!\!\pi\!\!\text{ )}\].

Hence, it is concluded that the function $\text{f}$ is continuous at $\text{x=n}$.

21. Discuss the continuity of the following functions:

(a) \[\mathbf{f(x)}=\mathbf{sinx+cosx}\]  (b) \[\mathbf{f(x)=sinx-cosx}\]  (c) \[\mathbf{f(x)=sinx\times cosx}\].

Ans: It is known that if two functions $\text{g}$ and $\text{h}$ are continuous, then $\text{g+h, g-h}$and $\text{g,h}$ are also continuous.

So, let us assume that, $\text{g(x)=sinx}$ and $\text{h(x)=cosx}$ are two continuous functions.

Now, as $\text{g(x)=sinx}$ is defined for every real number, so let $\text{c}$ be a real number. Substitute $\text{x=c+h}$ into the function $\text{g}$.

When $\text{x}\to \text{c}$, then $\text{h}\to 0$.

So, \[\text{g(c)=sinc}\].

Also,

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{sinx}\]

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{sin}\left( \text{c+h} \right)\]

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{sinc}\cdot \text{cosh+cosc}\cdot \text{sinh} \right]\]

\[=\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{cosh} \right)\text{+}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{sinh} \right)\]

\[\text{=sinc}\cdot \text{cos0+cosc}\cdot \text{sin0}\]

\[\text{=sinc+0}\]

 \[\text{=sinc}\]

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)}\].

Hence, the function $\text{g}$ is continuous.

Again, let us assume that $\text{h(x)=cosx}$.

Note that, the function $\text{h(x)=cosx}$ is defined for every real number.

Now, let $\text{c}$ be a real number. 

Substitute $\text{x=c+h}$ into the function.

When $\text{x}\to \text{c}$, then \[\text{h}\to 0\].

So, $\text{h(c)=cosc}$ and

$\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{cosx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{cos(c+h)} \\ & \text{=}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{cosh} \right)\text{-}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{sinh} \right) \\ & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1}-\text{sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align}$

Therefore, $\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{h(x)=h(c)}$.

Thus, the function $\text{h}$ is continuous.

Hence, we conclude that all the following functions are continuous.

(a) \[\text{f(x)=g(x)+h(x)=sinx+cosx}\].

(b) \[\text{f(x)=g(x)-h(x)=sinx-cosx}\].

(c) \[\text{f(x)=g(x) }\!\!\times\!\!\text{ h(x)=sinx }\!\!\times\!\!\text{ cosx}\].

22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Ans: We know that if two functions say $\text{g}$ and $\text{h}$ are continuous, then

        i.  $\frac{\text{h(x)}}{\text{g(x)}}\text{,g(x)}\ne \text{0}$ is continuous.

      ii.  \[\frac{\text{1}}{\text{g(x)}}\text{,g(x)}\ne \text{0}\] is continuous.

    iii.  \[\frac{\text{1}}{\text{h(x)}}\text{,h(x)}\ne \text{0}\] is continuous.

It can be observed that the function $\text{g(x)=sinx}$ is defined for all real numbers.

Now, let's consider $\text{c}$ to be a real number and substitute $\text{x=c+h}$ into the function $\text{g}$.

When, $\text{x}\to \text{c}$, then \[\text{h}\to 0\].

So, \[\text{g(c)=sinc}\] and

\[\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{sinx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{sin}\left( \text{c+h} \right) \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{sinc}\cdot \text{cosh+cosc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{cosh} \right)\text{+}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{sinh} \right) \\ & \text{=sinc}\cdot \text{cos0+cosc}\cdot \text{sin0} \\ & \text{=sinc+0} \\ & \text{=sinc} \\ \end{align}\]

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)}\].

Thus, the function $\text{g}\left( \text{x} \right)\text{=sinx}$ is continuous.

Again, let $\text{h(x)=cosx}$.

It can be noted that $\text{h(x)=cosx}$ is defined for all real numbers.

Now, let's consider $\text{c}$ to be a real number and substitute $\text{x=c+h}$ into the function $\text{h}$.

Then, \[\text{h(c)=cosc}\] and

\[\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{cosx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{cos}\left( \text{c+h} \right) \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{cosh} \right)\text{-}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{sinh} \right) \\ & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1}-\text{sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align}\]

Therefore, \[\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{h(x)=h(c)}\].

Thus, the function $\text{h(x)=cosx}$ is continuous.

Now note that,

$\text{cosec x=}\frac{\text{1}}{\text{sinx}}\text{,}$ and $\text{sinx}\ne \text{0}$ is a continuous function.

Thus, the cosecant function is continuous except at $\text{x=n }\!\!\pi\!\!\text{ ,}\,\,\text{n}\in \mathbb{Z}$.

Again, $\text{secx=}\frac{\text{1}}{\text{cosx}}\text{,}\,\,\text{cosx}\ne \text{0}$ is continuous.

$\Rightarrow \text{secx,}\,\,\text{x}\ne \text{(2n+1)}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{, }\,\text{n}\in \mathbb{Z}$  is a continuous function.

Thus, secant function is also continuous except at $\text{x=(2n+1)}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{, }\,\text{n}\in \mathbb{Z}$.

And the cotangent function is

$\text{cotx=}\frac{\text{cosx}}{\text{sinx}}$ and where $\text{sinx}\ne 0$ is a continuous function.

$\Rightarrow \text{cotx,}\,\,\text{x}\ne \text{n }\!\!\pi\!\!\text{ ,}\,\,\text{n}\in \mathbb{Z}$ is a continuous function.

Hence, the cotangent function is continuous except at $\text{x=n }\!\!\pi\!\!\text{ ,}\,\,\text{n}\in \mathbb{Z}$.

23. Find all points of discontinuity of function $\mathbf{f}$ defined by $\text{f(x)=}\left\{ \begin{align} & \frac{\text{sinx}}{\text{x}}\text{, if x}\,<\,\text{0} \\ & \text{x+1, }\,\,\text{if x}\ge \text{0} \\ \end{align} \right.$

Ans: The given function is  $\text{f(x)=}\left\{ \begin{align} & \frac{\text{sinx}}{\text{x}}\text{, if x}\,<\,\text{0} \\ & \text{x+1, }\,\text{if x}\ge \text{0} \\ \end{align} \right.$

Note that, the function $\text{f}$ is defined at every point on the real number line.

Now, let's consider $\text{c}$ be a real number.

Then there may arise three cases, either $\text{c}<\text{0}$, or $\text{c}>\text{0}$, or $\text{c}=\text{0}$.

Let us discuss one after another.

Case I: When $\text{c}<\text{0}$.

Then, $\text{f(c)=}\frac{\text{sinc}}{\text{c}}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)}\left( \frac{\text{sinx}}{\text{x}} \right)\text{=}\frac{\text{sinc}}{\text{c}}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, the function $\text{f}$ is continuous at every point $\text{x}$, for \[\text{x0}\].

Case II: When $\text{c}>\text{0}$.

Then $\text{f(c)=c+1}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(x+1)=c+1}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}$.

Hence, the function $\text{f}$ is continuous at every point, where \[\text{x0}\].

Case III: When \[\text{c = 0}\].

Then $\text{f(c)=f(0)=0+1=1}$.

Now, the left-hand-limit of the function $\text{f}$ at $\text{x=0}$ is

$\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\frac{\text{sinx}}{\text{x}}\text{=1}$ and the right-hand-limit is

\[\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{(x+1)=1}\]

Therefore, \[\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=f(0)}\].

So, the function $\text{f}$ is continuous at \[\text{x = 0}\].

Thus, the function $\text{f}$ is continuous at every real point.

Hence, the function $\text{f}$ does not have any point of discontinuity.

24. Determine if function $\mathbf{f}$ defined by $\text{f(x)=}\left\{ \begin{align} & {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}}\text{, if x}\ne \text{0} \\ & \text{0, if x=0} \\ \end{align} \right\}$

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}}\text{, if x}\ne \text{0} \\ & \text{0, if x=0} \\ \end{align} \right\}$

We can observe that the function $\text{f}$ is defined at every point on the real number line.

Now, let's consider $\text{c}$ to be a real number.

Then, there may arise two cases, either $\text{c}\ne \text{0}$ or $\text{c=0}$.

Let us discuss the cases one after another.

Case I: When \[\text{c}\ne 0\].

Then $\text{f(c)=}{{\text{c}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{c}}$.

Also,

\[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)=\left( \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,{{\text{x}}^{\text{2}}} \right)\left( \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{sin}\frac{\text{1}}{\text{x}} \right)={{\text{c}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{c}}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}\].

Hence, the function $\text{f}$ is continuous at every point \[\text{x}\ne 0\].

Case II: When \[\text{c = 0}\].

Then $\text{f(0)=0}$ and also

$\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)\text{=}\underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)$

Now, we know that,

 $\text{-1}\le \text{sin}\frac{\text{1}}{\text{x}}\le 1,\text{ x}\ne 0$.

$\begin{align} & \Rightarrow \text{-}{{\text{x}}^{\text{2}}}\le \text{sin}\frac{\text{1}}{\text{x}}\le {{\text{x}}^{\text{2}}} \\ & \Rightarrow \underset{\text{x}\to 0}{\mathop{\lim }}\,\text{(-}{{\text{x}}^{\text{2}}}\text{)}\le \underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)\le 0 \\ & \Rightarrow 0\le \underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)\le 0 \\ & \Rightarrow \underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)=0 \\ \end{align}$

Therefore, $\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=0}$.

Similarly, we have, 

$\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)=\underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\frac{\text{1}}{\text{x}} \right)\text{=0}$

Therefore, $\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=f(0)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)}$.

Thus, the function $\text{f}$ is continuous at the point \[\text{x = 0}\].

So, the function $\text{f}$ is continuous at all real points.

Hence, the function $\text{f}$ is continuous.

25. Determine whether the following function $\mathbf{f}$ is continuous.

$\mathbf{f}$ such that $\text{f(x)=}\left\{ \begin{align} & \text{sinx-cosx, if x}\ne \text{0} \\ & \text{ 1 if x=0} \\ \end{align} \right.$

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{sinx-cosx, if x}\ne \text{0} \\ & \text{ 1 if x=0} \\ \end{align} \right.$

It can be observed that the function $\text{f}$ is defined at every point on the real number line.

Now, let's consider $\text{c}$ to be a real number.

Then, there may arise two cases, either $\text{c}\ne \text{0}$ or $\text{c=0}$.

Let us discuss the cases one after another.

Case I: When $\text{c}\ne 0$.

Then, $\text{f(c)=sinc-cosc}$.

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(sinx-cosx)=sinc-cosc}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)}\].

Hence, the function $\text{f}$ is continuous at every point $\text{x}$ for $\text{x}\ne 0$.

Case II: When \[\text{c =0}\].

Then, $\text{f(0)=-1}$.

Now the left-hand-limit of the function $\text{f}$ at $\text{x=0}$ is

\[\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)}=\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{(sinx-cosx)=sin0-cos0=0-1=-1}\] and the right-hand-limit is 

\[\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{(sinx-cosx)=sin0-cos0=0-1=-1}\].

Therefore, \[\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(0)}\].

So, the function $\text{f}$ is continuous at \[\text{x = 0}\].

Thus, the function $\text{f}$ is continuous at all real points.

Hence, the function $\text{f}$ is continuous.

26. Calculate the values of \[\mathbf{k}\] for which the function $\mathbf{f}$ attains continuity at the given points.

$\text{f(x)}\mathbf{=}\left\{ \begin{align} & \frac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}\text{, if x}\ne \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ & \text{ 3, if x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ \end{align} \right.$

at $\mathbf{x}=\frac{\mathbf{\pi }}{\mathbf{2}}$.

Ans: The given function is $\text{f(x)}\mathbf{=}\left\{ \begin{align} & \frac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}\text{, if x}\ne \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ & \text{ 3, if x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ \end{align} \right.$

Observe that, $\text{f}$ is defined and continuous at $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ , since the value of the $\text{f}$ at $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ is equal with the limiting value of $\text{f}$ at $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.

Since, $\text{f}$ is defined at $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ and $\text{f}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)\text{=3}$, so

$\underset{\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\text{f(x)}=\underset{\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\frac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}$.

Substitute $\text{x=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+h}$ into the function $\text{f}\left( \text{x} \right)$.

So, we have, \[\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\Rightarrow \text{h}\to 0\].

Then,

$\underset{\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\frac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}=\underset{\text{h}\to 0}{\mathop{\lim }}\,\frac{\text{kcos}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+h} \right)}{\text{ }\!\!\pi\!\!\text{ -2}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+h} \right)}$.

 $\Rightarrow \text{k}\underset{\text{h}\to 0}{\mathop{\lim }}\,\frac{\text{-sinh}}{\text{-2h}}\text{=}\frac{\text{k}}{\text{2}}\underset{\text{h}\to 0}{\mathop{\lim }}\,\frac{\text{sinh}}{\text{h}}\text{=}\frac{\text{k}}{\text{2}}\text{.1}=\frac{\text{k}}{\text{2}}$

Therefore, $\underset{\text{x}\to \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\text{f(x)=f}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)$

 $\begin{align} & \Rightarrow \frac{\text{k}}{\text{2}}\text{=3} \\ & \Rightarrow \text{k=6} \\ \end{align}$

Hence, the value of $\text{k}$ is $\text{6}$ for which the function $\text{f}$ is continuous.

27. Find the values of $\mathbf{k}$ so that the function $\mathbf{f}$ satisfies continuity at the given points. 

$\text{f(x)=}\left\{ \begin{align} & \text{k}{{\text{x}}^{\text{2}}}\text{, if x}\le \text{2} \\ & \text{3, if x}\,>\,\text{2} \\ \end{align} \right.$ at $\mathbf{x=2}$

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{k}{{\text{x}}^{\text{2}}}\text{, if x}\le \text{2} \\ & \text{3, if x}>\text{2} \\ \end{align} \right.$

It is known that, $\text{f}$ is continuous at \[\text{x = 2}\] only if $\text{f}$ is defined at $\text{x=2}$ and if the value of $\text{f}$ at \[\text{x = 2}\] is equal with the limiting value of $\text{f}$ at \[\text{x = 2}\].

So, at $x=2$,

Now, the left-hand-limit and right-hand-limit of the function $\text{f}\left( \text{x} \right)$ at $\text{x}=\text{2}$respectively are,

$\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\left( \text{k}{{\text{x}}^{\text{2}}} \right)=\text{k}{{\left( \text{2} \right)}^{\text{2}}}=\text{4k}$, 

and $\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\left( 3 \right)=3$.

Since, the function is continuous at $\text{x}=\text{2}$, so

\[\begin{align} & \underset{\text{x}\to {{2}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(2)} \\ & \Rightarrow \text{4k=3} \\ & \Rightarrow \text{k=}\frac{\text{3}}{\text{4}} \\ \end{align}\]

Hence, the value of $\text{k}$ is $\frac{\text{3}}{\text{4}}$ for which the function $\text{f}$ is continuous.

28. Find the values of $\mathbf{k}$ so that the function $\mathbf{f}$ attains continuity at the given point.

$\text{f(x)=}\left\{ \begin{align} & \text{kx+1, if x}\le \mathbf{\pi } \\ & \mathbf{cosx}\text{, if x}\,>\,\mathbf{\pi } \\ \end{align} \right.$ at $\mathbf{x=\pi }$

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{kx+1, if x}\le \text{ }\!\!\pi\!\!\text{ } \\ & \text{cosx, if x}\,>\,\text{ }\!\!\pi\!\!\text{ } \\ \end{align} \right.$

It is known that, $\text{f}$ is continuous at $\text{x= }\!\!\pi\!\!\text{ }$ only if the value of $\text{f}$ at $\text{x= }\!\!\pi\!\!\text{ }$ is equal with the limiting value of $\text{f}$ at $\text{x= }\!\!\pi\!\!\text{ }$.

It is provided that the function $\text{f}$ is defined at $\text{x= }\!\!\pi\!\!\text{ }$.

Also, $\text{f( }\!\!\pi\!\!\text{ )=k }\!\!\pi\!\!\text{ +1}$.

Now, the left-hand-limit,

\[\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{-}}}{\mathop{\lim }}\,\text{(kx+1)=k }\!\!\pi\!\!\text{ +1}\].

Also, the right-hand-limit,

\[\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{\text{+}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{\text{+}}}}{\mathop{\lim }}\,\text{cosx=cos }\!\!\pi\!\!\text{ =-1}\].

Since, the function $\text{f}$ is continuous, so

\[\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{\text{+}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\text{f}\left( \text{ }\!\!\pi\!\!\text{ } \right)\]

\[\begin{align} & \Rightarrow \text{k }\!\!\pi\!\!\text{ +1=-1} \\ & \Rightarrow \text{k }\!\!\pi\!\!\text{ =-2} \\ & \Rightarrow \text{k=-}\frac{\text{2}}{\text{ }\!\!\pi\!\!\text{ }} \\ \end{align}\]

Hence, the value of $\text{k}$ is $\text{-}\frac{\text{2}}{\text{ }\!\!\pi\!\!\text{ }}$ for which the function $\text{f}$ is continuous at $\text{x= }\!\!\pi\!\!\text{ }$. 

29. Find the values of $\mathbf{k}$ so that the function $\mathbf{f}$ attains continuity at the given point.

$\text{f(x)=}\left\{ \begin{align} & \text{kx+1, if x}\le \text{5} \\ & \text{3x-5, }\,\text{if x}\,>\,\text{5} \\ \end{align} \right.$ at \[\mathbf{x = 5}\]

Ans: The given function is  $\text{f(x)=}\left\{ \begin{align} & \text{kx+1, if x}\le \text{5} \\ & \text{3x-5, }\,\,\text{if x}>\text{5} \\ \end{align} \right.$

Recall that, the function $\text{f}$ is continuous at \[\text{x = 5}\] only if the value of $\text{f}$ at \[\text{x = 5}\] is equal to the limiting value of $\text{f}$ at \[\text{x = 5}\].

Note that, the function $\text{f}$ is defined at \[\text{x = 5}\].

Also, $\text{f(5)=kx+1=5k+1}$.

Then, the left-hand-limit of the function,

$\underset{\text{x}\to {{\text{5}}^{-}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{5}}^{-}}}{\mathop{\lim }}\,\left( \text{kx+1} \right)\text{=5k+1}$.

The right-hand-limit of the function,

$\underset{\text{x}\to {{\text{5}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{\text{5}}^{+}}}{\mathop{\lim }}\,\left( \text{3x-5} \right)\text{=3}\left( 5 \right)-5=15-5=10$.

Since, the function $\text{f}$ is continuous, so

$\begin{align} & \underset{\text{x}\to {{\text{5}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{5}^{+}}}{\mathop{\lim }}\,\text{(3x-5)=5k+1} \\ & \Rightarrow \text{5k+1=10} \\ & \Rightarrow \text{5k=9} \\ & \Rightarrow \text{k=}\frac{\text{9}}{\text{5}} \\ \end{align}$

Hence, the value of $\text{k}$ is $\frac{9}{5}$ for which the function $\text{f}$ is continuous at $\text{x=5}$.

30. Find the values of constants $\mathbf{a}$ and $\mathbf{b}$ such that the function $\mathbf{f}$ defined by is continuous. 

$\mathbf{f}$ such that $\text{f(x)=}\left\{ \begin{align} & \text{5, if x}\le \text{2} \\ & \text{ax+b, if 2}\,<\,\text{x}\,<\,\text{10} \\ & \text{ 21, if x}\ge \text{10} \\ \end{align} \right.$ is a continuous function.

Ans: The given function is $\text{f(x)=}\left\{ \begin{align} & \text{5, }\,\,\,\text{if x}\le \text{2} \\ & \text{ax+b,}\,\text{ if }\,\,\text{2}<\text{x}<\text{10} \\ & \text{ 21, if x}\ge \text{10} \\ \end{align} \right.$

Note that, $\text{f}$ is defined at every point on the real number line.

Now, realise that if the function $\text{f}$ is continuous then $\text{f}$ is continuous at every real number.

So, let $\text{f}$ satisfies continuity at \[\text{x=2}\] and \[\text{x=10}\].

Then, since $\text{f}$ is continuous at $\text{x=2}$, so

\[\begin{align} & \underset{\text{x}\to {{\text{2}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(2)} \\ & \Rightarrow \underset{\text{x}\to {{\text{2}}^{\text{-}}}}{\mathop{\lim }}\,\text{(5)=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{(ax+b)=5} \\ & \Rightarrow \text{5=2a+b} \\ \end{align}\]

$\Rightarrow \text{2a+b=5}$                                                   …… (1)

Again, since $\text{f}$ attains continuity at \[\text{x=10}\], so

\[\begin{align} & \underset{\text{x}\to 1{{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{10}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(10)} \\ & \Rightarrow \underset{\text{x}\to 1{{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{(ax+b)=}\underset{\text{x}\to {{10}^{+}}}{\mathop{\lim }}\,\text{(21)=21} \\ \end{align}\]

\[\Rightarrow 1\text{0a+b=21}\]                                   …… (2)

Subtracting the equation (1) from the equation (2), gives

$\text{8a=16}\Rightarrow \text{a=2}$

Substituting \[\text{a=2}\] in the equation (1), gives

$\begin{align} & \text{2 }\!\!\times\!\!\text{ 2+b=5} \\ & \Rightarrow \text{4+b=5}\Rightarrow \text{b=1} \\ \end{align}$

Hence, the values of $\text{a}$ and $\text{b}$ are $2$ and $1$ respectively for which $\text{f}$ is a continuous function.

31. Show that the function defined by $\text{f(x)=cos(}{{\text{x}}^{\text{2}}}\text{)}$ is a continuous function.

Ans: The given function is $\text{f(x)=cos(}{{\text{x}}^{\text{2}}}\text{)}$.

Note that, $\text{f}$ is defined for all real numbers and so $\text{f}$ can be expressed as the composition of two functions as, $\text{f=g}\circ \text{h}$, where $\text{g(x)=cosx}$ and $\text{h(x)=}{{\text{x}}^{\text{2}}}$.

$\text{ }\!\![\!\!\text{ }\therefore \text{(goh)(x)=g(h(x))=g(}{{\text{x}}^{\text{2}}}\text{)=cos(}{{\text{x}}^{\text{2}}}\text{)=f(x) }\!\!]\!\!\text{ }$

Now, it is to be Proven that, the functions $\text{g(x)=cosx}$ and $\text{h(x)=}{{\text{x}}^{\text{2}}}$ are continuous.

Since the function $\text{g}$ is defined for all the real numbers, let 's consider $\text{c}$ be a real number.

Then, $\text{g(c)=cosc}$.

Substitute $\text{x=c+h}$ into the function $\text{g}$.

When, $\text{x}\to \text{c}$, then \[\text{h}\to 0\].

Then we have,

$\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{cosx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{cos}\left( \text{c+h} \right) \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{cosh} \right)-\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{sinh} \right) \\ & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1}-\text{sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align}$

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g(x)=cosx}$ is continuous.

Again, $\text{h(x)=}{{\text{x}}^{\text{2}}}$ is defined for every real point.

So, let consider $\text{k}$ be a real number, then $\text{h(k)=}{{\text{k}}^{\text{2}}}$ and

$\underset{\text{x}\to \text{k}}{\mathop{\lim }}\,\text{h(x)=}\underset{\text{x}\to \text{k}}{\mathop{\lim }}\,{{\text{x}}^{\text{2}}}\text{=}{{\text{k}}^{\text{2}}}$.

Therefore, $\underset{\text{x}\to \text{k}}{\mathop{\lim }}\,\text{h(x)=h(k)}$.

Hence, the function $\text{h}$ is continuous.

Now, remember that for real valued functions $\text{g}$ and $\text{h}$ , such that $\text{(g }\circ \text{ h)}$ is defined at $\text{c}$ , if $\text{g}$ is continuous at $\text{c}$ and $\text{f}$ is continuous at $\text{g(c)}$, then $\text{(f }\circ \text{ h)}$is continuous at $\text{c}$.

Hence, the function $\text{f(x)=(g }\circ \text{ h)(x)=cos(}{{\text{x}}^{2}}\text{)}$ is continuous.

32. Show that the function defined by $\text{f(x)=}\left| \text{cosx} \right|$ is a continuous function.

Ans: The given function is $\text{f(x)=}\left| \text{cosx} \right|$.

Note that, the function $\text{f}$ is defined for all real numbers. So, the function $\text{f}$ can be expressed as the composition of two functions as, $\text{f=g}\circ \text{h}$, where $\text{g(x)=}\left| \text{x} \right|$ and $\text{h(x)=cosx}$.

$[\because (\text{goh)(x)=g(h(x))=g(cosx)=}\left| \text{cosx} \right|\text{=f(x) }\!\!]\!\!\text{ }$

Now, it is to be proved that the functions $\text{g(x)=}\left| \text{x} \right|$ and $\text{h(x)=cosx}$ are continuous.

Remember that, $\text{g(x)=}\left| \text{x} \right|$, can be written as

$\text{g(x)=}\left\{ \begin{align} & \text{-x, if x}\,<\,\text{0} \\ & \text{ x, if x}\ge \text{0} \\ \end{align} \right.$

Now, since the function $\text{g}$ is defined for every real number, let consider $\text{c}$ be a real number.

Then there may arise three cases, either $\text{c}<\text{0}$, or $\text{c}>\text{0}$, or $\text{c=0}$.

Let's discuss the cases one after another.

Case I: When $\text{c}<\text{0}$.

Then, $\text{g(c)=-c}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(-x)=-c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous at every point $\text{x}$, for $\text{x}<\text{0}$.

Case II: When $\text{c}>\text{0}$.

Then, $\text{g(c)=c}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{x=c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous at every point $\text{x}$ for $\text{x}>\text{0}$.

Case III: When \[\text{c=0}\].

Then, $\text{g(c)=g(0)=0}$.

Now, the left-hand-limit of the function $\text{g}$ at $\text{x=0}$ is

$\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{(-x)=0}$ and the right-hand-limit is

$\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{(x)=0}$.

Therefore, $\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{g(x)=g(0)}$.

Hence, the function $\text{g}$ is continuous at \[\text{x=0}\].

By observing the above three discussions, we can conclude that the function $\text{g}$ is continuous at every real point.

Now, since the function $\text{h(x)=cosx}$ is defined for all real numbers, let 's consider $c$ be a real number. Then, substitute $\text{x=c+h}$ into the function $\text{h}$.

So, when $\text{x}\to \text{c}$, then $\text{h}\to 0$.

Then, we have

\[\text{h}\left( \text{c} \right)=\text{cosc}\] and

\[\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{cosx}\]

\[\text{=}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\text{cos}\left( \text{c+h} \right)\]

\[\text{=}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right]\]

\[\text{=}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{cosc}\cdot \text{cosh} \right)-\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{sinc}\cdot \text{sinh} \right)\]

\[\begin{align} & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1-sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align}\]

Therefore, \[\underset{x\to \text{c}}{\mathop{\lim }}\,\text{h}\left( \text{x} \right)=\text{h}\left( \text{c} \right)\].

Hence, the function \[\text{h(x)=cosx}\] is continuous.

Now remember that, for real valued functions $\text{g}$ and \[\text{h}\], such that $\text{(g}\circ \text{h)}$ is defined at $\text{x=c}$ only if $\text{g}$ is continuous at $\text{c}$ and $\text{f}$ is continuous at $\text{g(c)}$, then the composition functions $\text{(f }\circ \text{ g)}$ is continuous at $\text{x=c}$.

Thus, the function $\text{f(x)=(goh)(x)=g(h(x))=g(cosx)=}\left| \text{cosx} \right|$ is continuous.

33. Examine that $\mathbf{sin}\left| \mathbf{x} \right|$ is continuous. 

Ans: First suppose that, $f\left( \text{x} \right)=\sin \left| \text{x} \right|$.

Now, note that the function $\text{f}$ is defined for all real numbers and so \[\text{f}\] can be expressed as the composition of  functions as, $\text{f=g}\circ \text{h,}$ where $\text{g(x)=sinx}$ and $\text{h(x)=}\left| \text{x} \right|$.

$\left[ \text{(g}\circ \text{h)(x)=g(h(x))=g(}\left| \text{x} \right|\text{)=sin}\left| \text{x} \right|\text{=f(x)} \right]$

So, it is to be proved that the functions $\text{g(x)=sinx}$ and \[\text{h(x)=}\left| \text{x} \right|\] are continuous.

Now, remember that, the function $\text{h(x)=}\left| \text{x} \right|$ can be written as 

$\text{h(x)=}\left\{ \begin{align} & \text{-x, if x}<\text{0} \\ & \text{x, if x}\ge \text{0} \\ \end{align} \right.$

Note that, the function $\text{h}$ is defined for every real number, and so let consider $\text{c}$ be a real number. 

Then, there may arise three cases, either $\text{c}<\text{0}$, or $\text{c}>\text{0}$, or $\text{c=0}$.

Let us discuss the cases one after another.

Case I: When $\text{c}<\text{0}$.

Then \[\text{h(c)=-c}\].

Also, \[\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(-x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{x=-c}\].

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)}\].

Hence, the function $\text{h}$ is continuous at every point $\text{x}$  for $\text{x}<\text{0}$.

Case II: When $\text{c}>\text{0}$.

Then, \[\text{h(c)=c}\]

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(-x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{x=c}$.

Therefore, \[\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)}\].

Thus, the function $\text{h}$ is continuous at every point $\text{x}$  for $\text{x}>\text{0}$.

Case III: When \[\text{c = 0}\].

 Then, $\text{h(c)=h(0)=0}$.

Also, the left-hand-limit of the function $\text{h}$ at $\text{x=0}$ is

$\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{(-x)=0}$ and the right-hand -limit is 

$\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=0}$.

Therefore, $\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=h(0)}$.

Thus, the function $\text{h}$ is continuous at \[\text{x = 0}\].

By observing the above three discussions, we can conclude that the function $\text{h}$ is continuous at every point. 

Again, since the function $\text{g(x)=sinx}$ is defined for all real numbers, so let consider $\text{c}$ be a real number and substitute $\text{x=c+k}$ into the function.

Now, when $\text{x}\to \text{c}$ then \[\text{k }\to \text{ 0}\].

Then, we have

$\text{g(c)=sinc}$.

Also,

$\begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{sinx} \\ & \text{=}\underset{\text{k}\to \text{0}}{\mathop{\text{lim}}}\,\text{sin}\left( \text{c+k} \right) \\ & \text{=}\underset{\text{k}\to \text{0}}{\mathop{\text{lim}}}\,\left[ \text{sinc}\cdot \text{cosk+cosc}\cdot \text{sink} \right] \\ & \text{=}\underset{\text{k}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{sinc}\cdot \text{cosk} \right)\text{+}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{cosc}\cdot \text{sink} \right) \\ & \text{=sinc}\cdot \text{cos0+cosc}\cdot \text{sin0} \\ & \text{=sinc+0} \\ & \text{=sinc} \\ \end{align}$

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous.

Now, remember that, for any two real valued functions $\text{g}$ and $\text{h}$, such that the composition of functions $\text{g}\circ \text{h}$ is defined at $\text{c}$, if $\text{g}$ is continuous at $\text{c}$ and $\text{f}$ is continuous at $\text{g(c)}$, then the composition function $\text{g}\circ \text{h}$ is continuous at $\text{c}$.

Thus, the function $\text{(g}\circ \text{h)(x)=g(h(x))=g(}\left| \text{x} \right|\text{)=sin}\left| \text{x} \right|\text{=f(x)}$ is continuous.

34. Find all the points of discontinuity of function $\mathbf{f}$ defined by $\mathbf{f(x)=}\left| \mathbf{x} \right|\mathbf{-}\left| \mathbf{x}+\mathbf{1} \right|$. 

Ans: The given function is $\text{f(x)=}\left| \text{x} \right|\text{-}\left| \text{x+1} \right|$.

Let consider two functions

$\text{g(x)=}\left| \text{x} \right|$ and $\text{h(x)=}\left| \text{x+1} \right|$.

Then we get,  $\text{f=g-h}$.

Now, the function \[\text{g(x)=}\left| \text{x} \right|\] can be written as

$\text{g(x)=}\left\{ \begin{align} & \text{-x, if x}<\text{0} \\ & \text{x, if x}\ge \text{0} \\ \end{align} \right.$

Note that, the function $\text{g}$ is defined for every real number and so let consider $\text{c}$ be a real number.

Then there may arise three cases, either $\text{c}<\text{0}$, or $\text{c}>\text{0}$, or $\text{c=0}$.

Let us discuss the cases one after another.

Case I: When $\text{c}<\text{0}$.

Then, $\text{g(c)=g(0)=-c}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(-x)=-c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous at every point $\text{x}$ for \[\text{x}<\text{0}\].

Case II: When \[\text{c}>\text{0}\].

 Then $\text{g(c)=c}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{x=c}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=g(c)}$.

Hence, the function $\text{g}$ is continuous at every point $\text{x}$, where \[\text{x}>\text{0}\].

Case III: When \[\text{c = 0}\].

Then $\text{g(c)=g(0)=0}$.

Also, the left-hand-limit of the function $\text{g}$ at $\text{x=0}$ is

$\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{(-x)=0}$ and the right-hand-limit is

$\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=0}$.

Therefore, $\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=g(0)}$.

Hence, the function $\text{g}$ is continuous at \[\text{x = 0}\].

Thus, we can conclude by observing the above three discussions that $\text{g}$ is continuous at every real point.

Now, remember that, the function $\text{h(x)=}\left| \text{x+1} \right|$ can be written as

$\text{h(x)=}\left\{ \begin{align} & \text{-x(x+1), if, x}<\text{-1} \\ & \text{x+1, }\,\,\,\,\,\,\,\,\text{if, x}\ge \text{-1} \\ \end{align} \right.$

Note that, the function $\text{h}$ is defined for all real numbers, and so let consider $\text{c}$ be a real number.

Case I: When \[\text{c}<\text{-1}\].

Then $\text{h(c)=-(c+1)}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\left[ \text{-(x+1)} \right]\text{=-(c+1)}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)}$.

Hence, the function $\text{h}$ attains continuity at every real point $\text{x}$, where \[\text{x}<\text{-1}\].

Case II: When \[\text{c}>\text{-1}\].

Then, $\text{h(c)=c+1}$.

Also, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(x+1)=(c+1)}$.

Therefore, $\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)}$.

Hence, the function $\text{h}$ satisfies continuity at every real point $\text{x}$ for \[\text{x}>\text{-1}\].

Case III: When \[\text{c =-1}\].

Then, $\text{h(c)=h(-1)=-1+1=0}$.

Also, the left-hand-limit of the function $\text{h}$ at $\text{x=1}$ is

$\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\text{lim}}}\,\left[ \text{-(x+1)} \right]\text{=-(-1+1)=0}$ and the right-hand-limit is

$\underset{\text{x}\to {{\text{1}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{1}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x+1)=(-1+1)=0}$.

Therefore, $\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h=}\underset{\text{x}\to {{\text{1}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{h(x)=h(-1)}$.

Thus, the function $\text{h}$ satisfies continuity at \[\text{x=-1}\].

Hence, by observing the above three discussions, we can conclude that the function $\text{h}$ is continuous for every real point.

Now, since the functions $\text{g}$ and $\text{h}$ are both continuous, so the function $\text{f=g-h}$ is also continuous.

Hence, the function $\text{f}$ does not have any discontinuity points.

Conclusion

In conclusion, mastering NCERT Solutions for Class 12 Maths Chapter 5 - Continuity and Differentiability, Class 12 Ex 5.1 is crucial. Focus on understanding the core concepts of continuity and differentiability, particularly at specific points and over intervals. Pay special attention to polynomial, rational, trigonometric, exponential, and logarithmic functions. Prioritize these fundamentals as they are essential for advanced calculus and further mathematical studies.

Class 12 Maths Chapter 5: Exercises Breakdown

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NCERT Solutions for Class 12 Maths | Chapter-wise List

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