# NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.1

## Chapter 5 Ex.5.1 Question 1

Prove that the function $$f\left( x \right) = 5x - 3$$ is continuous at $$x = 0$$, $$x = - 3$$ and at $$x = 5$$.

### Solution

The given function is $$f\left( x \right) = 5x - 3$$

\begin{align}&{\text{At }}x = 0,f\left( 0 \right) = 5\left( 0 \right) - 3 = - 3\\&\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( {5x - 3} \right) = 5\left( 0 \right) - 3 = - 3\\&\therefore \mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)\end{align}

Therefore, $$f$$ is continous at $$x = 0$$.

\begin{align}&{\text{At }}x = - 3,f\left( { - 3} \right) = 5\left( { - 3} \right) - 3 = - 18\\& \mathop {\lim }\limits_{x \to - 3} f\left( x \right) = \mathop {\lim }\limits_{x \to - 3} \left( {5x - 3} \right) = 5\left( { - 3} \right) - 3 = - 18\\&\therefore \mathop {\lim }\limits_{x \to - 3} f\left( x \right) = f\left( { - 3} \right)\end{align}

Therefore, $$f$$ is continous at $$x = - 3$$.

\begin{align}&{\text{At }}x = 5,f\left( 5 \right) = 5\left( 5 \right) - 3 = 22\\&\mathop {\lim }\limits_{x \to 5} f\left( x \right) = \mathop {\lim }\limits_{x \to 5} \left( {5x - 3} \right) = 5\left( 5 \right) - 3 = 22\\&\therefore \mathop {\lim }\limits_{x \to 5} f\left( x \right) = f\left( 5 \right)\end{align}

Therefore, $$f$$ is continous at $$x = 5$$.

## Chapter 5 Ex.5.1 Question 2

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

### Solution

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

\begin{align}&{\text{At }}x = 3,f\left( 3 \right) = 2{\left( 3 \right)^2} - 1 = 17\\&\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \left( {2{x^2} - 1} \right) = 2\left( {{3^2}} \right) - 1 = 17\\&\therefore \mathop {\lim }\limits_{x \to 3} f\left( x \right) = f\left( 3 \right)\end{align}

Therefore, $$f$$is continous at $$x = 3$$.

## Chapter 5 Ex.5.1 Question 3

Examine the following functions for continuity.

(i) $$f\left( x \right) = x - 5$$

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

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

(iv) $$f\left( x \right) = \left| {x - 5} \right|,x \ne 5$$

### Solution

(i) The given function is $$f\left( x \right) = x - 5$$

It is evident that $$f$$ is defined at every real number $$k$$ and its value at $$k$$ is $$k - 5$$.

It is also observed that

\begin{align}\mathop {\lim }\limits_{x \to k} f\left( x \right) &= \mathop {\lim }\limits_{x \to k} \left( {x - 5} \right) = k - 5 = f\left( k \right)\\\therefore \mathop {\lim }\limits_{x \to k} f\left( x \right) &= f\left( k \right)\end{align}

Hence, $$f$$ is continuous at every real number and therefore, it is a continuous function.

(ii) The given function is $$f\left( x \right) = \frac{1}{{x - 5}},x \ne 5$$

For any real number $$k \ne 5$$, we obtain

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

Also,

\begin{align}&f\left( k \right) = \frac{1}{{k - 5}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{\text{As }}k \ne 5} \right)\\&\therefore \mathop {\lim }\limits_{x \to k} f\left( x \right) = f\left( k \right)\end{align}

Hence,$$f$$ is continuous at every point in the domain of $$f$$ and therefore, it is a continuous function.

(iii) The given function is $$f\left( x \right) = \frac{{{x^2} - 25}}{{x + 5}},x \ne - 5$$

For any real number $$c \ne - 5$$, we obtain

$$\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \frac{{{x^2} - 25}}{{x + 5}} = \mathop {\lim }\limits_{x \to c} \frac{{\left( {x + 5} \right)\left( {x - 5} \right)}}{{x + 5}} = \mathop {\lim }\limits_{x \to c} \left( {x - 5} \right) = \left( {c - 5} \right)$$

Also,

\begin{align}&{\text{ }}f\left( c \right) = \frac{{\left( {c + 5} \right)\left( {c - 5} \right)}}{{c + 5}} = \left( {c - 5} \right)\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Hence, $$f$$ is continuous at every point in the domain of $$f$$ and therefore, it is a continuous function.

(iv) The given function is $$f\left( x \right) = \left| {x - 5} \right| = \left\{ \begin{array}{l}5 - x,{\text{ if }}x < 5\\x - 5,{\text{ if }}x \ge 5\end{array} \right\}$$

This function $$f$$ is defined at all points of the real line. Let c be a point on a real line. Then, $$c < 5$$, $$c = 5$$ or $$c > 5$$

Case I: $$c < 5$$

Then, $$f\left( c \right) = 5 - c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {5 - x} \right) = 5 - c\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore,$$f$$ is continuous at all real numbers less than 5.

Case II: $$c = 5$$

Then, $$f\left( c \right) = f\left( 5 \right) = \left( {5 - 5} \right) = 0$$

\begin{align}&\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to 5} \left( {5 - x} \right) = \left( {5 - 5} \right) = 0\\&\mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 5} \left( {x - 5} \right) = 0\\&\therefore \mathop {\lim }\limits_{x \to {c^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {c^ + }} f\left( x \right) = f\left( c \right)\end{align}

Therefore,$$f$$is continuous at $$x = 5$$

Case III: $$c > 5$$

Then, $$f\left( c \right) = f\left( 5 \right) = c - 5$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {x - 5} \right) = c - 5\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore,$$f$$ is continuous at all real numbers greater than 5.

Hence,$$f$$ is continuous at every real number and therefore, it is a continuous function.

## Chapter 5 Ex.5.1 Question 4

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

### Solution

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

It is observed that$$f$$is defined at all positive integers, n, and its value at n is $${n^n}$$.

Then,

\begin{align}&\mathop {\lim }\limits_{x \to n} f\left( n \right) = \mathop {\lim }\limits_{x \to n} \left( {{x^n}} \right) = {x^n}\\&\therefore \mathop {\lim }\limits_{x \to n} f\left( x \right) = f\left( n \right)\end{align}

Therefore,$$f$$ is continuous at n, where n is a positive integer.

## Chapter 5 Ex.5.1 Question 5

Is the function$$f$$defined by $$f\left( x \right) = \left\{ \begin{array}{l}x,\,{\text{if}}\,x \le 1\\5,\,{\text{if}}\,x > 1\end{array} \right.$$ continuous at $$x = 0$$? At $$x = 1$$? At $$x = 2$$?

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}x,\,{\text{if}}\,x \le 1\\5,\,{\text{if}}\,x > 1\end{array} \right.$$

At $$x = 0$$,

It is evident that$$f$$is defined at $$0$$and its value at $$0$$ is $$0$$.

Then,

\begin{align}&\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( x \right) = 0\\&\therefore \mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)\end{align}

Therefore,$$f$$ is continuous at$$x = 0$$.

At $$x = 1$$,

It is evident that$$f$$is defined at $$1$$and its value at $$1$$ is $$1$$.

The left hand limit of $$f$$ at $$x = 1$$ is,

$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( x \right) = 1$$

The right hand limit of$$f$$at $$x = 1$$ is,

\begin{align}&\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( 5 \right) = 5\\&\therefore \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)\end{align}

Therefore, $$f$$ is not continuous at$$x = 1$$.

At $$x = 2$$,

It is evident that $$f$$ is defined at $$2$$ and its value at $$2$$ is $$5$$.

\begin{align}&\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left( 5 \right) = 5\\&\therefore \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 2 \right)\end{align}

Therefore, $$f$$ is continuous at$$x = 2$$.

## Chapter 5 Ex.5.1 Question 6

Find all points of discontinuity of$$f,$$where f is defined by $$f\left( x \right) = \left\{ \begin{array}{l}2x + 3,{\text{ if}}\,x \le 2\\2x - 3,{\text{ if}}\,x > 2\end{array} \right.$$.

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}2x + 3,{\text{ if}}\,x \le 2\\2x - 3,{\text{ if}}\,x > 2\end{array} \right.$$

It is evident that the given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line. Then, three cases arise.

$$c < 2$$

$$c > 2$$

$$c = 2$$

Case I: $$c < 2$$

$$f\left( c \right) = 2c + 3$$

Then,

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {2x + 3} \right) = 2c + 3\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points x, such that $$x < 2$$.

Case II: $$c > 2$$

Then,

\begin{align}&f\left( c \right) = 2c - 3\\&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {2x - 3} \right) = 2c - 3\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points x, such that $$x > 2$$

Case III: $$c = 2$$

Then, the left hand limit of $$f$$ at$$x = 2$$ is,

$$\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {2x + 3} \right) = 2\left( 2 \right) + 3 = 7$$

The right hand limit of $$f$$ at $$x = 2$$is,

$$\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {2x - 3} \right) = 2\left( 2 \right) - 3 = 1$$

It is observed that the left and right hand limit of $$f$$ at $$x = 2$$ do not coincide.

Therefore, $$f$$ is not continuous at$$x = 2$$.

Hence, $$x = 2$$is the only point of discontinuity of $$f$$ .

## Chapter 5 Ex.5.1 Question 7

Find all points of discontinuity of $$f$$ , where $$f$$ is defined by $$f\left( x \right) = \left\{ \begin{array}{l}\left| x \right| + 3,{\text{ if }}x \le - 3\\ - 2x,{\text{ if }} - 3 < x < 3\\6x + 2,{\text{ if }}x \ge 3\end{array} \right.$$

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}\left| x \right| + 3,{\text{ if }}x \le - 3\\ - 2x,{\text{ if }} - 3 < x < 3\\6x + 2,{\text{ if }}x \ge 3\end{array} \right.$$

The given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c < - 3$$, then $$f\left( c \right) = - c + 3$$

\begin{align}\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( { - x + 3} \right) = - c + 3\\\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points x, such that $$x < - 3$$.

Case II:

If $$c = - 3$$, then $$f\left( { - 3} \right) = - \left( { - 3} \right) + 3 = 6$$

\begin{align} &\mathop {\lim }\limits_{x \to - {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {3^ - }} \left( { - x + 3} \right) = - \left( { - 3} \right) + 3 = 6\\&\mathop {\lim }\limits_{x \to - {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {3^ + }} \left( { - 2x} \right) = - 2\left( { - 3} \right) = 6\\& \therefore \mathop {\lim }\limits_{x \to - 3} f\left( x \right) = f\left( { - 3} \right)\end{align}

Therefore, $$f$$ is continuous at $$x = - 3$$.

Case III:

If $$- 3 < c < 3$$, then $$f\left( c \right) = - 2c$$

\begin{align}\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( { - 2x} \right) = - 2c\\\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous in $$\left( { - 3,3} \right)$$.

Case IV:

If $$c = 3$$, then the left hand limit of $$f$$ at$$x = 3$$ is,

$$\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} \left( { - 2x} \right) = - 2\left( 3 \right) = - 6$$

The right hand limit of $$f$$ at $$x = 3$$ is,

$$\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} \left( {6x + 2} \right) = 6\left( 3 \right) + 2 = 20$$

It is observed that the left and right hand limit of $$f$$ at $$x = 3$$ do not coincide.

Therefore, $$f$$ is not continuous at$$x = 3$$.

Case V:

If $$c > 3$$, then $$f\left( c \right) = 6c + 2$$

\begin{align}\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {6x + 2} \right) = 6c + 2\\\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x > 3$$.

Hence, $$x = 3$$ is the only point of discontinuity of $$f$$ .

## Chapter 5 Ex.5.1 Question 8

Find all points of discontinuity of $$f,$$ where f is defined by $$f\left( x \right) = \left\{ \begin{array}{l}\frac{{\left| x \right|}}{x},{\text{if }}x \ne 0\\0,{\text{ if }}x = 0\end{array} \right.$$

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}\frac{{\left| x \right|}}{x},{\text{if }}x \ne 0\\0,{\text{ if }}x = 0\end{array} \right.$$

It is known that, $$x < 0 \Rightarrow \left| x \right| = - x$$and $$x > 0 \Rightarrow \left| x \right| = x$$

Therefore, the given function can be rewritten as

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

The given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c < 0$$, then $$f\left( c \right) = - 1$$

\begin{align}\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( { - 1} \right) = - 1\\\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x < 0$$.

Case II:

If $$c = 0$$, then the left hand limit of $$f$$ at$$x = 0$$ is,

$$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - 1} \right) = - 1$$

The right hand limit of $$f$$ at $$x = 0$$is,

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

It is observed that the left and right hand limit of $$f$$ at $$x = 0$$do not coincide.

Therefore, $$f$$ is not continuous at$$x = 0$$.

Case III:

If $$c > 0$$, then $$f\left( c \right) = 1$$

\begin{align}\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( 1 \right) = 1\\\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x > 0$$.

Hence, $$x = 0$$is the only point of discontinuity of $$f$$ .

## Chapter 5 Ex.5.1 Question 9

Find all points of discontinuity of $$f,$$ where f is defined by$$f\left( x \right) = \left\{ \begin{array}{l}\frac{x}{{\left| x \right|}},{\text{ if}}\,x < 0\\ - 1,{\text{ if}}\,x \ge 0\end{array} \right.$$.

### Solution

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

It is known that $$x < 0 \Rightarrow \left| x \right| = - x$$

Therefore, the given function can be rewritten as

\begin{align}f\left( x \right)& = \left\{ \begin{array}{l}\frac{x}{{\left| x \right|}} = \frac{x}{{ - x}} = - 1,{\text{ if}}\,x < 0\\ &- 1,{\text{ if}}\,x \ge 0\end{array} \right.\\ &\Rightarrow f\left( x \right) = - 1\,\forall \,x \in R\end{align}

Let c be any real number.

\begin{align}{\text{Then}},{\text{ }}\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( { - 1} \right) = - 1\\{\text{Also, }}f\left( c \right) = - 1 = \mathop {\lim }\limits_{x \to c} f\left( x \right)\end{align}

Therefore, the given function is a continuous function.

Hence, the given function has no point of discontinuity.

## Chapter 5 Ex.5.1 Question 10

Find all points of discontinuity of $$f,$$where f is defined by $$f\left( x \right) = \left\{ \begin{array}{l}x + 1,{\text{ if}}\,x \ge 1\\{x^2} + 1,{\text{ if}}\,x < 1\end{array} \right.$$

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}x + 1,{\text{ if}}\,x \ge 1\\{x^2} + 1,{\text{ if}}\,x < 1\end{array} \right.$$

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c < 1$$, then $$f\left( c \right) = {c^2} + 1$$

\begin{align}\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {{x^2} + 1} \right) = {c^2} + 1\\\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points x, such that $$x < 1$$.

Case II:

If $$c = 1$$, then $$f\left( c \right) = f\left( 1 \right) = 1 + 1 = 2$$

The left hand limit of $$f$$ at$$x = 1$$ is,

$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {{x^2} + 1} \right) = {1^2} + 1 = 2$$

The right hand limit of $$f$$ at $$x = 1$$ is,

\begin{align}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {x + 1} \right) = 1 + 1 = 2\\\therefore \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\end{align}

Therefore, $$f$$ is continuous at $$x = 1$$.

Case III:

If $$c > 1$$, then $$f\left( c \right) = c + 1$$

\begin{align}\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {x + 1} \right) = c + 1\\\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x > 1$$.

Hence, the given function $$f$$ has no point of discontinuity.

## Chapter 5 Ex.5.1 Question 11

Find all points of discontinuity of $$f$$ , where f is defined by $$f\left( x \right) = \left\{ \begin{array}{l}{x^3} - 3,{\text{ if}}\,x \le 2\\{x^2} + 1,{\text{ if}}\,x > 2\end{array} \right.$$

### Solution

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

The given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c < 2$$, then $$f\left( c \right) = {c^3} - 3$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {{x^3} - 3} \right) = {c^3} - 3\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points x, such that $$x < 2$$.

Case II:

If $$c = 2$$, then $$f\left( c \right) = f\left( 2 \right) = {2^3} - 3 = 5$$

\begin{align}&\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^3} - 3} \right) = {2^3} - 3 = 5\\&\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {{x^2} + 1} \right) = {2^2} + 1 = 5\\&\therefore \mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 2 \right)\end{align}

Therefore, $$f$$ is continuous at$$x = 2$$.

Case III:

If $$c > 2$$, then $$f\left( c \right) = {c^2} + 1$$

\begin{align}\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {{x^2} + 1} \right) = {c^2} + 1\\\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x > 2$$.

Thus, the given function $$f$$ is continuous at every point on the real line.

Hence, $$f$$ has no point of discontinuity.

## Chapter 5 Ex.5.1 Question 12

Find all points of discontinuity of $$f$$ , where f is defined by $$f\left( x \right) = \left\{ \begin{array}{l}{x^{10}} - 1,{\text{ if}}\,x \le 1\\{x^2},{\text{ if}}\,x > 1\end{array} \right.$$.

### Solution

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

The given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c < 1$$, then $$f\left( c \right) = {c^{10}} - 1$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {{x^{10}} - 1} \right) = {c^{10}} - 1\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points x, such that $$x < 1$$.

Case II:

If $$c = 1$$, then the left hand limit of $$f$$ at$$x = 1$$ is,

$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {{x^{10}} - 1} \right) = {1^{10}} - 1 = 1 - 1 = 0$$

The right hand limit of $$f$$ at $$x = 1$$ is,

$$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2}} \right) = {1^2} = 1$$

It is observed that the left and right hand limit of $$f$$ at $$x = 1$$ do not coincide.

Therefore, $$f$$ is not continuous at$$x = 1$$.

Case III:

If $$c > 1$$, then $$f\left( c \right) = {c^2}$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {{x^2}} \right) = {c^2}\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x > 1$$.

Thus from the above observation, it can be concluded that $$x = 1$$ is the only point of discontinuity of $$f$$ .

## Chapter 5 Ex.5.1 Question 13

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

### Solution

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

The given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c < 1$$, then $$f\left( c \right) = c + 5$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {x + 5} \right) = c + 5\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points x, such that $$x < 1$$.

Case II:

If $$c = 1$$, then $$f\left( 1 \right) = 1 + 5 = 6$$

The left hand limit of $$f$$ at$$x = 1$$ is,

$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {x + 5} \right) = 1 + 5 = 6$$

The right hand limit of $$f$$ at $$x = 1$$is,

$$\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {x - 5} \right) = 1 - 5 = - 4$$

It is observed that the left and right hand limit of $$f$$ at $$x = 1$$ do not coincide.

Therefore, $$f$$ is not continuous at$$x = 1$$.

Case III:

If $$c > 1$$, then $$f\left( c \right) = c - 5$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {x - 5} \right) = c - 5\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x > 1$$.

From the above observation it can be concluded that, $$x = 1$$ is the only point of discontinuity of $$f$$ .

## Chapter 5 Ex.5.1 Question 14

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

### Solution

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

The given function $$f$$ is defined at all the points of the interval $$\left[ {0,10} \right]$$.

Let c be a point in the interval $$\left[ {0,10} \right]$$.

Case I:

If $$0 \le c < 1$$, then $$f\left( c \right) = 3$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( 3 \right) = 3\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous in the interval $$\left[ {0,1} \right)$$.

Case II:

If $$c = 1$$, then $$f\left( 3 \right) = 3$$

The left hand limit of $$f$$ at$$x = 1$$ is,

$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( 3 \right) = 3$$

The right hand limit of $$f$$ at $$x = 1$$ is,

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

It is observed that the left and right hand limit of $$f$$ at $$x = 1$$ do not coincide.

Therefore, $$f$$ is not continuous at$$x = 1$$.

Case III:

If $$1 < c < 3$$, then $$f\left( c \right) = 4$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( 4 \right) = 4\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at in the interval $$\left( {1,3} \right)$$.

Case IV:

If $$c = 3$$, then $$f\left( c \right) = 5$$

The left hand limit of $$f$$ at$$x = 3$$ is,

$$\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} \left( 4 \right) = 4$$

The right hand limit of $$f$$ at $$x = 3$$ is,

$$\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} \left( 5 \right) = 5$$

It is observed that the left and right hand limit of $$f$$ at $$x = 3$$ do not coincide.

Therefore, $$f$$ is discontinuous at$$x = 3$$.

Case V:

If $$3 < c \le 10$$, then $$f\left( c \right) = 5$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( 5 \right) = 5\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points of the interval $$\left( {3,10} \right]$$.

Hence, $$f$$ is discontinuous at $$x = 1$$ and $$x = 3$$ .

## Chapter 5 Ex.5.1 Question 15

Discuss the continuity of the function $$f$$ , where $$f$$ is defined by$$f\left( x \right) = \left\{ \begin{array}{l}2x,{\text{ if}}\,x < 0\\0,{\text{ if}}\,0 \le x \le 1\\4x,{\text{ if}}\,x > 1\end{array} \right.$$

### Solution

The given function is$$f\left( x \right) = \left\{ \begin{array}{l}2x,{\text{ if}}\,x < 0\\0,{\text{ if}}\,0 \le x \le 1\\4x,{\text{ if}}\,x > 1\end{array} \right.$$

The given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c < 0$$, then $$f\left( c \right) = 2c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {2x} \right) = 2c\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x < 0$$.

Case II:

If $$c = 0$$, then $$f\left( c \right) = f\left( 0 \right) = 0$$

The left hand limit of $$f$$ at$$x = 0$$ is,

$$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {2x} \right) = 2\left( 0 \right) = 0$$

The right hand limit of $$f$$ at $$x = 0$$ is,

\begin{align}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( 0 \right) = 0\\\therefore \mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)\end{align}

Therefore, $$f$$ is continuous at $$x = 0$$

Case III:

If $$0 < c < 1$$, then $$f\left( x \right) = 0$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( 0 \right) = 0\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous in the interval $$\left( {0,1} \right)$$.

Case IV:

If $$c = 1$$, then $$f\left( c \right) = f\left( 1 \right) = 0$$

The left hand limit of $$f$$ at$$x = 1$$ is,

$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( 0 \right) = 0$$

The right hand limit of $$f$$ at $$x = 1$$ is,

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

It is observed that the left and right hand limit of $$f$$ at $$x = 1$$ do not coincide.

Therefore, $$f$$ is not continuous at$$x = 1$$.

Case V:

If $$c < 1$$, then $$f\left( c \right) = 4c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {4x} \right) = 4c\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that$$x > 1$$.

Hence,$$f$$ is not continuous only at $$x = 1$$ .

## Chapter 5 Ex.5.1 Question 16

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

### Solution

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

The given function $$f$$ is defined at all the points.

Let c be a point on the real line.

Case I:

If $$c < - 1$$, then $$f\left( c \right) = - 2$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( { - 2} \right) = - 2\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x < - 1$$.

Case II:

If $$c = - 1$$, then $$f\left( c \right) = f\left( { - 1} \right) = - 2$$

The left hand limit of $$f$$ at$$x = - 1$$ is,

$$\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} \left( { - 2} \right) = - 2$$

The right hand limit of $$f$$ at $$x = - 1$$ is,

\begin{align}&\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} \left( {2x} \right) = 2\left( { - 1} \right) = - 2\\&\therefore \mathop {\lim }\limits_{x \to - 1} f\left( x \right) = f\left( { - 1} \right) \end{align}

Therefore, $$f$$ is continuous at $$x = - 1$$

Case III:

If $$- 1 < c < 1$$, then $$f\left( c \right) = 2c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {2x} \right) = 2c\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous in the interval $$\left( { - 1,1} \right)$$.

Case IV:

If $$c = 1$$, then $$f\left( c \right) = f\left( 1 \right) = 2\left( 1 \right) = 2$$

The left hand limit of $$f$$ at$$x = 1$$ is,

$$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {2x} \right) = 2\left( 1 \right) = 2$$

The right hand limit of $$f$$ at $$x = 1$$ is,

\begin{align}&\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( 2 \right) = 2\\&\therefore \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at$$x = 2$$.

Case V:

If $$c > 1$$, then $$f\left( c \right) = 2$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( 2 \right) = 2\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that$$x > 1$$.

Thus, from the above observations, it can be concluded that $$f$$ is continuous at all points of the real line.

## Chapter 5 Ex.5.1 Question 17

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

### Solution

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

For $$f$$ to be continuous at $$x = 3$$, then

$$\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = f\left( 3 \right)\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

Also,

\begin{align}&\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} \left( {ax + 1} \right) = 3a + 1\\&\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} \left( {bx + 3} \right) = 3b + 3\\&f\left( 3 \right) = 3a + 1\end{align}

Therefore, from ($$1$$), we obtain

\begin{align}&3a + 1 = 3b + 3 = 3a + 1\\& \Rightarrow 3a + 1 = 3b + 3\\& \Rightarrow 3a = 3b + 2\\& \Rightarrow a = b + \frac{2}{3}\end{align}

Therefore, the required relationship is given by, $$a = b + \frac{2}{3}$$.

## Chapter 5 Ex.5.1 Question 18

For what value of $$\lambda$$ is the function defined by $$f\left( x \right) = \left\{ \begin{array}{l}\lambda \left( {{x^2} - 2x} \right),{\text{ if }}x \le 0\\4x + 1,{\text{ if }}x > 0\end{array} \right.$$is continous at $$x = 0$$? What about continuity at $$x = 1$$?

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}\lambda \left( {{x^2} - 2x} \right),{\text{ if }}x \le 0\\4x + 1,{\text{ if }}x > 0\end{array} \right.$$

If $$f$$ is continuous at $$x = 0$$, then

\begin{align}&\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = f\left( 0 \right)\\& \Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} \lambda \left( {{x^2} - 2x} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {4x + 1} \right) = \lambda \left( {{0^2} - 2 \times 0} \right)\\ &\Rightarrow \lambda \left( {{0^2} - 2 \times 0} \right) = 4\left( 0 \right) + 1 = 0\\ &\Rightarrow 0 = 1 = 0 \qquad \left[ {{\text{which is not possible}}} \right]\end{align}

Therefore, there is no value of $$\lambda$$ for which $$f$$ is continuous at $$x = 0$$.

At $$x = 1$$

\begin{align}&f\left( 1 \right) = 4x + 1 = 4\left( 1 \right) + 1 = 5\\&\mathop {\lim }\limits_{x \to 1} \left( {4x + 1} \right) = 4\left( 1 \right) + 1 = 5\\&\therefore \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\end{align}

Therefore, for any values of $$\lambda$$, $$f$$ is continuous at $$x = 1$$.

## Chapter 5 Ex.5.1 Question 19

Show that the function defined by $$g\left( x \right) = x - \left[ x \right]$$ is discontinuous at all integral point. Here $$\left[ x \right]$$ denotes the greatest integer less than or equal to$$x$$.

### Solution

The given function is $$g\left( x \right) = x - \left[ x \right]$$

It is evident that $$g$$ is defined at all integral points.

Let $$n$$ be an integer.

Then,

$$g\left( n \right) = n - \left[ n \right] = n - n = 0$$

The left hand limit of $$g$$at$$x = n$$ is,

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

The right hand limit of $$g$$ at $$x = n$$ is,

$$\mathop {\lim }\limits_{x \to {n^ + }} g\left( x \right) = \mathop {\lim }\limits_{x \to {n^ + }} \left( {x - \left[ x \right]} \right) = \mathop {\lim }\limits_{x \to {n^ + }} \left( x \right) - \mathop {\lim }\limits_{x \to {n^ + }} \left[ x \right] = n - n = 0$$

It is observed that the left and right hand limit of $$g$$ at $$x = n$$ do not coincide.

Therefore, $$g$$ is not continuous at$$x = n$$.

Hence, $$g$$ is discontinuous at all integral points.

## Chapter 5 Ex.5.1 Question 20

Is the function defined by $$f\left( x \right) = {x^2} - \sin x + 5$$ continuous at $$x = \pi$$?

### Solution

The given function is $$f\left( x \right) = {x^2} - \sin x + 5$$

It is evident that $$f$$ is defined at $$x = \pi$$.

At $$x = \pi$$, $$f\left( x \right) = f\left( \pi \right) = {\pi ^2} - \sin \pi + 5 = {\pi ^2} - 0 + 5 = {\pi ^2} + 5$$

Consider $$\mathop {\lim }\limits_{x \to \pi } f\left( x \right) = \mathop {\lim }\limits_{x \to \pi } \left( {{x^2} - \sin x + 5} \right)$$

Put $$x = \pi + h$$, it is evident that if $$x \to \pi ,$$ then $$h \to 0$$

\begin{align}&\mathop {\therefore \lim }\limits_{{\text{ }}x \to \pi } f\left( x \right) = \mathop {\lim }\limits_{x \to \pi } \left( {{x^2} - \sin x} \right) + 5\\ &= \mathop {\lim }\limits_{h \to 0} \left[ {{{\left( {\pi + h} \right)}^2} - \sin \left( {\pi + h} \right) + 5} \right]\\ &= \mathop {\lim }\limits_{h \to 0} {\left( {\pi + h} \right)^2} - \mathop {\lim }\limits_{h \to 0} \sin \left( {\pi + h} \right) + \mathop {\lim }\limits_{h \to 0} 5\\& = {\left( {\pi + 0} \right)^2} - \mathop {\lim }\limits_{h \to 0} \left[ {\sin \pi \cos \,h + \cos \pi \sin \,h} \right] + 5\\& = {\pi ^2} - \mathop {\lim }\limits_{h \to 0} \sin \pi \cos \,h - \mathop {\lim }\limits_{h \to 0} \cos \pi \sin \,h + 5\\ &= {\pi ^2} - \sin \pi \cos 0 - \cos \pi \sin 0 + 5\\ &= {\pi ^2} - 0\left( 1 \right) - \left( { - 1} \right)0 + 5\\ &= {\pi ^2} + 5\\ &= f\left( \pi \right)\end{align}

Therefore, the given function $$f$$ is continuous at $$x = \pi$$.

## Chapter 5 Ex.5.1 Question 21

Discuss the continuity of the following functions.

(i) $$f\left( x \right) = \sin x + \cos x$$

(ii) $$f\left( x \right) = \sin x - \cos x$$

(iii) $$f\left( x \right) = \sin x \times \cos x$$

### Solution

It is known that if $$g$$and $$h$$are two continuous functions, then $$g + h,g - h$$and $$g,h$$ are also continuous.

Let $$g\left( x \right) = \sin x$$ and $$h\left( x \right) = \cos x$$are continuous functions.

It is evident that $$g\left( x \right) = \sin x$$is defined for every real number.

Let c be a real number. Put $$x = c + h$$

If $$x \to c$$, then $$h \to 0$$

$$g\left( c \right) = \sin c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} g\left( x \right) = \mathop {\lim }\limits_{x \to c} \sin x\\ &= \mathop {\lim }\limits_{h \to 0} \sin \left( {c + h} \right)\\& = \mathop {\lim }\limits_{h \to 0} \left[ {\sin c\cos \,h + \cos c\sin \,h} \right]\\ &= \mathop {\lim }\limits_{h \to 0} \left( {\sin c\cos \,h} \right) + \mathop {\lim }\limits_{h \to 0} \left( {\cos c\sin \,h} \right)\\ &= \sin c\cos 0 + \cos c\sin 0\\& = \sin c\left( 1 \right) + \cos c\left( 0 \right)\\ &= \sin c\\&\therefore \mathop {\lim }\limits_{x \to c} g\left( x \right) = g\left( c \right)\end{align}

Therefore, $$g\left( x \right) = \sin x$$ is a continuous function.

Let $$h\left( x \right) = \cos x$$

It is evident that $$h\left( x \right) = \cos x$$is defined for every real number.

Let c be a real number. Put $$x = c + h$$

If $$x \to c$$, then $$h \to 0$$

$$h\left( c \right) = \cos c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} h\left( x \right) = \mathop {\lim }\limits_{x \to c} \cos x\\& = \mathop {\lim }\limits_{h \to 0} \cos \left( {c + h} \right)\\ &= \mathop {\lim }\limits_{h \to 0} \left[ {\cos c\cos \,h - \sin c\sin \,h} \right]\\ &= \mathop {\lim }\limits_{h \to 0} \left( {\cos c\cos \,h} \right) - \mathop {\lim }\limits_{h \to 0} \left( {\sin c\sin \,h} \right)\\& = \cos c\cos 0 - \sin c\sin 0\\ &= \cos c\left( 1 \right) - \sin c\left( 0 \right)\\& = \cos c\end{align}

$$\therefore \mathop {\lim }\limits_{x \to c} h\left( x \right) = h\left( c \right)$$

Therefore, $$h\left( x \right) = \cos x$$ is a continuous function.

Therefore, it can be concluded that,

(i) $$f\left( x \right) = g\left( x \right) + h\left( x \right) = \sin x + \cos x$$is a continuous function.

(ii) $$f\left( x \right) = g\left( x \right) - h\left( x \right) = \sin x - \cos x$$is a continuous function.

(iii) $$f\left( x \right) = g\left( x \right) \times h\left( x \right) = \sin x \times \cos x$$is a continuous function.

## Chapter 5 Ex.5.1 Question 22

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

### Solution

It is known that if $$g$$and $$h$$are two continuous functions, then

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

(ii) $$\frac{1}{{g\left( x \right)}},g\left( x \right) \ne 0$$is continuous.

(iii) $$\frac{1}{{h\left( x \right)}},h\left( x \right) \ne 0$$is continuous.

Let $$g\left( x \right) = \sin x$$ and $$h\left( x \right) = \cos x$$are continuous functions.

It is evident that $$g\left( x \right) = \sin x$$is defined for every real number.

Let c be a real number. Put $$x = c + h$$

If $$x \to c$$, then $$h \to 0$$

\begin{align}g\left( c \right) &= \sin c\\\mathop {\lim }\limits_{x \to c} g\left( x \right) &= \mathop {\lim }\limits_{x \to c} \sin x\\ &= \mathop {\lim }\limits_{h \to 0} \sin \left( {c + h} \right)\\ &= \mathop {\lim }\limits_{h \to 0} \left[ {\sin c\cos \,h + \cos c\sin \,h} \right]\\& = \mathop {\lim }\limits_{h \to 0} \left( {\sin c\cos \,h} \right) + \mathop {\lim }\limits_{h \to 0} \left( {\cos c\sin \,h} \right)\\& = \sin c\cos 0 + \cos c\sin 0\\ &= \sin c\left( 1 \right) + \cos c\left( 0 \right)\\& = \sin c\end{align}

$$\therefore \mathop {\lim }\limits_{x \to c} g\left( x \right) = g\left( c \right)$$

Therefore, $$g\left( x \right) = \sin x$$ is a continuous function.

Let $$h\left( x \right) = \cos x$$

It is evident that $$h\left( x \right) = \cos x$$is defined for every real number.

Let c be a real number. Put $$x = c + h$$

If $$x \to c$$, then $$h \to 0$$

\begin{align}h\left( c \right) &= \cos c\\\mathop {\lim }\limits_{x \to c} h\left( x \right) &= \mathop {\lim }\limits_{x \to c} \cos x\\ &= \mathop {\lim }\limits_{h \to 0} \cos \left( {c + h} \right)\\ &= \mathop {\lim }\limits_{h \to 0} \left[ {\cos c\cos \,h - \sin c\sin \,h} \right]\\ &= \mathop {\lim }\limits_{h \to 0} \left( {\cos c\cos \,h} \right) - \mathop {\lim }\limits_{h \to 0} \left( {\sin c\sin \,h} \right)\\ &= \cos c\cos 0 - \sin c\sin 0\\& = \cos c\left( 1 \right) - \sin c\left( 0 \right)\\& = \cos c\end{align}

$$\therefore \mathop {\lim }\limits_{x \to c} h\left( x \right) = h\left( c \right)$$

Therefore, $$h\left( x \right) = \cos x$$ is a continuous function.

Therefore, it can be concluded that,

$${\text{cosec}}\,x = \frac{1}{{\sin x}},\sin x \ne 0$$ is continuous.

$$\Rightarrow \cos {\text{ec}}\,x,x \ne n\pi \left( {n \in Z} \right)$$ is continuous.

Therefore, cosecant is continuous except at $$x = n\pi \left( {n \in Z} \right)$$

$$s{\text{ec}}\,x = \frac{1}{{\cos x}},\cos x \ne 0$$ is continuous.

$$\Rightarrow s{\text{ec}}\,x,x \ne \left( {2n + 1} \right)\frac{\pi }{2}\left( {n \in Z} \right)$$ is continuous.

Therefore, secant is continuous except at $$x = \left( {2n + 1} \right)\frac{\pi }{2}\left( {n \in Z} \right)$$

$$\cot \,x = \frac{{\cos x}}{{\sin x}},\sin x \ne 0$$ is continuous.

$$\Rightarrow \cot \,x,x \ne n\pi \left( {n \in Z} \right)$$ is continuous.

Therefore, cotangent is continuous except at $$x = n\pi \left( {n \in Z} \right)$$.

## Chapter 5 Ex.5.1 Question 23

Find the points of discontinuity of $$f$$ , where $$f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sin x}}{x},{\text{if }}x < 0\\x + 1,{\text{ if }}x \ge 0\end{array} \right.$$

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sin x}}{x},{\text{if }}x < 0\\x + 1,{\text{ if }}x \ge 0\end{array} \right.$$

The given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c < 0$$, then $$f\left( c \right) = \frac{{\sin c}}{c}$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {\frac{{\sin x}}{x}} \right) = \frac{{\sin c}}{c}\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x < 0$$.

Case II:

If $$c > 0$$, then $$f\left( c \right) = c + 1$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {x + 1} \right) = c + 1\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x > 0$$.

Case III:

If $$c = 0$$, then $$f\left( c \right) = f\left( 0 \right) = 0 + 1 = 1$$

The left hand limit of $$f$$ at$$x = 0$$ is,

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

The right hand limit of $$f$$ at $$x = 0$$ is,

\begin{align}&\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {x + 1} \right) = 1\\&\therefore \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = f\left( 0 \right)\end{align}

Therefore, $$f$$ is continuous at $$x = 0$$

From the above observations, it can be concluded that $$f$$ is continuous at all points of the real line.

Thus, $$f$$ has no point of discontinuity.

## Chapter 5 Ex.5.1 Question 24

Determine if $$f$$ defined by $$f\left( x \right) = \left\{ \begin{array}{l}{x^2}\sin \frac{1}{x},{\text{ if }}x \ne 0\\0,{\text{ if }}x = 0\end{array} \right.$$is a continuous function?

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}{x^2}\sin \frac{1}{x},{\text{ if }}x \ne 0\\0,{\text{ if }}x = 0\end{array} \right.$$

The given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c \ne 0$$, then $$f\left( c \right) = {c^2}\sin \frac{1}{c}$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {{x^2}\sin \frac{1}{x}} \right) = \left( {\mathop {\lim }\limits_{x \to c} {x^2}} \right)\left( {\mathop {\lim }\limits_{x \to c} \sin \frac{1}{x}} \right) = {c^2}\sin \frac{1}{c}\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x \ne 0$$.

Case II:

If $$c = 0$$, then $$f\left( 0 \right) = 0$$

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

It is known that, $$- 1 \le \sin \frac{1}{x} \le 1,x \ne 0$$

\begin{align} &\Rightarrow - {x^2} \le {x^2}\sin \frac{1}{x} \le {x^2}\\& \Rightarrow \mathop {\lim }\limits_{x \to 0} \left( { - {x^2}} \right) \le \mathop {\lim }\limits_{x \to 0} \left( {{x^2}\sin \frac{1}{x}} \right) \le \mathop {\lim {x^2}}\limits_{x \to 0} \\& \Rightarrow 0 \le \mathop {\lim }\limits_{x \to 0} \left( {{x^2}\sin \frac{1}{x}} \right) \le 0\\& \Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {{x^2}\sin \frac{1}{x}} \right) = 0\\&\therefore \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = 0\end{align}

Similarly,

\begin{align}&\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{x^2}\sin \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {{x^2}\sin \frac{1}{x}} \right) = 0\\&\therefore \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = f\left( 0 \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\end{align}

Therefore, $$f$$ is continuous at $$x = 0$$.

From the above observations, it can be concluded that $$f$$ is continuous at every point of the real line.

Thus, $$f$$ is a continuous function.

## Chapter 5 Ex.5.1 Question 25

Examine the continuity of $$f$$ , where $$f$$ is defined by $$f\left( x \right) = \left\{ \begin{array}{l}\sin x - \cos x,{\text{ if }}x \ne 0\\ - 1,{\text{ if }}x = 0\end{array} \right.$$

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}\sin x - \cos x,{\text{ if }}x \ne 0\\ - 1,{\text{ if }}x = 0\end{array} \right.$$

The given function $$f$$ is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

If $$c \ne 0$$, then $$f\left( c \right) = \sin c - \cos c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} f\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {\sin x - \cos x} \right) = \sin c - \cos c\\&\therefore \mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\end{align}

Therefore, $$f$$ is continuous at all points $$x$$, such that $$x \ne 0$$.

Case II:

If $$c = 0$$, then $$f\left( 0 \right) = - 1$$

\begin{align}&\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( {\sin x - \cos x} \right) = \sin 0 - \cos 0 = 0 - 1 = - 1\\&\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left( {\sin x - \cos x} \right) = \sin 0 - \cos 0 = 0 - 1 = - 1\\&\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = f\left( 0 \right)\end{align}

Therefore, $$f$$ is continuous at $$x = 0$$.

From the above observations, it can be concluded that $$f$$ is continuous at every point of the real line.

Thus, $$f$$ is a continuous function.

## Chapter 5 Ex.5.1 Question 26

Find the values of $$k$$so that the function $$f$$ is continuous at the indicated point $$f\left( x \right) = \left\{ \begin{array}{l}\frac{{k\cos x}}{{\pi - 2x}},{\text{ if }}x \ne \frac{\pi }{2}\\3,{\text{ if }}x = \frac{\pi }{2}\end{array} \right.$$ at $$x = \frac{\pi }{2}$$

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}\frac{{k\cos x}}{{\pi - 2x}},{\text{ if }}x \ne \frac{\pi }{2}\\3,{\text{ if }}x = \frac{\pi }{2}\end{array} \right.$$

The given function $$f$$ is continuous at $$x = \frac{\pi }{2}$$, if $$f$$ is defined at $$x = \frac{\pi }{2}$$and if the value of the $$f$$ at $$x = \frac{\pi }{2}$$ equals the limit of $$f$$ at $$x = \frac{\pi }{2}$$.

It is evident that $$f$$ is defined at$$x = \frac{\pi }{2}$$ and$$f\left( {\frac{\pi }{2}} \right) = 3$$

$$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} f\left( x \right) = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{k\cos x}}{{\pi - 2x}}$$

Put $$x = \frac{\pi }{2} + h$$

Then $$x \to \frac{\pi }{2} \Rightarrow h \to 0$$

\begin{align}&\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{2}} f\left( x \right) = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{k\cos x}}{{\pi - 2x}} = \mathop {\lim }\limits_{h \to 0} \frac{{k\cos \left( {\frac{\pi }{2} + h} \right)}}{{\pi - 2\left( {\frac{\pi }{2} + h} \right)}}\\&= k\mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \,h}}{{ - 2h}} = \frac{k}{2}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h}}{h} = \frac{k}{2}.1 = \frac{k}{2}\\&\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{2}} f\left( x \right) = f\left( {\frac{\pi }{2}} \right)\\ &\Rightarrow \frac{k}{2} = 3\\& \Rightarrow k = 6\end{align}

Therefore, the value of $$k = 6$$.

## Chapter 5 Ex.5.1 Question 27

Find the values of $$k$$ so that the function $$f$$ is continuous at the indicated point. $$f\left( x \right) = \left\{ \begin{array}{l}k{x^2},{\text{ if }}x \le 2\\3,{\text{ if }}x > 2\end{array} \right.$$ at $$x = 2$$

### Solution

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

The given function $$f$$ is continuous at $$x = 2$$, if $$f$$ is defined at $$x = 2$$and if the value of the $$f$$ at $$x = 2$$ equals the limit of $$f$$ at$$x = 2$$.

It is evident that $$f$$ is defined at $$x = 2$$ and $$f\left( 2 \right) = k{\left( 2 \right)^2} = 4k$$

\begin{align}&\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = f\left( 2 \right)\\& \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \left( {k{x^2}} \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( 3 \right) = 4k\\& \Rightarrow k \times {2^2} = 3 = 4k\\ &\Rightarrow 4k = 3\\ &\Rightarrow k = \frac{3}{4}\end{align}

Therefore, the value of $$k = \frac{3}{4}$$.

## Chapter 5 Ex.5.1 Question 28

Find the values of $$k$$ so that the function $$f$$ is continuous at the indicated point $$f\left( x \right) = \left\{ \begin{array}{l}kx + 1,{\text{ if }}x \le \pi \\\cos x,{\text{ if }}x > \pi\end{array} \right.$$ at $$x = \pi$$

### Solution

The given function is $$f\left( x \right) = \left\{ \begin{array}{l}kx + 1,{\text{ if }}x \le \pi \\\cos x,{\text{ if }}x > \pi\end{array} \right.$$

The given function $$f$$ is continuous at $$x = \pi$$, if $$f$$ is defined at $$x = \pi$$ and if the value of the $$f$$ at $$x = \pi$$ equals the limit of $$f$$ at $$x = \pi$$.

It is evident that $$f$$ is defined at $$x = \pi$$ and $$f\left( \pi \right) = k\pi + 1$$

\begin{align}&\mathop {\lim }\limits_{x \to {\pi ^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {\pi ^ + }} f\left( x \right) = f\left( \pi \right)\\& \Rightarrow \mathop {\lim }\limits_{x \to {\pi ^ - }} \left( {kx + 1} \right) = \mathop {\lim }\limits_{x \to {\pi ^ + }} \left( {\cos x} \right) = k\pi + 1\\ &\Rightarrow k\pi + 1 = \cos \pi = k\pi + 1\\ &\Rightarrow k\pi + 1 = - 1 = k\pi + 1\\ &\Rightarrow k = - \frac{2}{\pi }\end{align}

Therefore, the value of $$k = - \frac{2}{\pi }$$.

## Chapter 5 Ex.5.1 Question 29

Find the values of $$k$$so that the function $$f$$ is continuous at the indicated point $$f\left( x \right) = \left\{ \begin{array}{l}kx + 1,{\text{ if }}x \le 5\\3x - 5,{\text{ if }}x > 5\end{array} \right.$$ at $$x=5$$.

### Solution

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

The given function $$f$$ is continuous at $$x = 5$$, if $$f$$ is defined at $$x = 5$$ and if the value of the $$f$$ at $$x = 5$$ equals the limit of $$f$$ at $$x = 5$$.

It is evident that $$f$$ is defined at $$x = 5$$ and $$f\left( 5 \right) = kx + 1 = 5k + 1$$

\begin{align}&\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) = f\left( 5 \right)\\ &\Rightarrow \mathop {\lim }\limits_{x \to {5^ - }} \left( {kx + 1} \right) = \mathop {\lim }\limits_{x \to {5^ + }} \left( {3x - 5} \right) = 5k + 1\\& \Rightarrow 5k + 1 = 3\left( 5 \right) - 5 = 5k + 1\\& \Rightarrow 5k + 1 = 15 - 5 = 5k + 1\\& \Rightarrow 5k + 1 = 10 = 5k + 1\\ &\Rightarrow 5k + 1 = 10\\&\Rightarrow 5k = 9\\ &\Rightarrow k = \frac{9}{5}\end{align}

Therefore, the value of $$k = \frac{9}{5}$$.

## Chapter 5 Ex.5.1 Question 30

Find the values of $$a{\text{ & }}b$$such that the function defined by $$f\left( x \right) = \left\{ \begin{array}{l}5,{\text{ if }}x \le 2\\ax + b,{\text{ if }}2 < x < 10\\21,{\text{ if }}x \ge 10\end{array} \right.$$, is a continuous function.

### Solution

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

It is evident that $$f$$ is defined at all points of the real line.

If $$f$$ is a continuous function, then $$f$$ is continuous at all real numbers.

In particular, $$f$$ is continuous at $$x = 2$$ and $$x = 10$$

Since $$f$$ is continuous at $$x = 2$$, we obtain

\begin{align}&\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = f\left( 2 \right)\\& \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \left( 5 \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {ax + b} \right) = 5\\ &\Rightarrow 5 = 2a + b = 5\\ &\Rightarrow 2a + b = 5\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Since $$f$$ is continuous at $$x = 10$$, we obtain

\begin{align}&\mathop {\lim }\limits_{x \to {{10}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{10}^ + }} f\left( x \right) = f\left( {10} \right)\\ &\Rightarrow \mathop {\lim }\limits_{x \to {{10}^ - }} \left( {ax + b} \right) = \mathop {\lim }\limits_{x \to {{10}^ + }} \left( {21} \right) = 21\\&\Rightarrow 10a + b = 21 = 21\\ &\Rightarrow 10a + b = 21\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

On subtracting equation (1) from equation (2), we obtain

\begin{align}&8a = 16\\ &\Rightarrow a = 2\end{align}

By putting $$a = 2$$ in equation (1), we obtain

\begin{align}&2\left( 2 \right) + b = 5\\ &\Rightarrow 4 + b = 5\\& \Rightarrow b = 1\end{align}

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

## Chapter 5 Ex.5.1 Question 31

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

### Solution

The given function is $$f\left( x \right) = \cos \left( {{x^2}} \right)$$.

This function $$f$$ is defined for every real number and $$f$$ can be written as the composition of two functions as,

$$f = goh$$, where $$g\left( x \right) = \cos x$$ and $$h\left( x \right) = {x^2}$$

$$\left[ {\because \left( {goh} \right)\left( x \right) = g\left( {h\left( x \right)} \right) = g\left( {{x^2}} \right) = \cos \left( {{x^2}} \right) = f\left( x \right)} \right]$$

It has to be proved first that $$g\left( x \right) = \cos x$$ and $$h\left( x \right) = {x^2}$$ are continuous functions.

It is evident that $$g$$ is defined for every real number.

Let $$c$$be a real number.

Let $$g\left( c \right) = \cos c$$. Put $$x = c + h$$

If $$x \to c$$, then $$h \to 0$$

\begin{align}&\mathop {\lim }\limits_{x \to c} g\left( x \right) = \mathop {\lim }\limits_{x \to c} \cos x\\ &= \mathop {\lim }\limits_{h \to 0} \cos \left( {c + h} \right)\\ &= \mathop {\lim }\limits_{h \to 0} \left[ {\cos c\cos \,h - \sin c\sin \,h} \right]\\ &= \mathop {\lim }\limits_{h \to 0} \left( {\cos c\cos \,h} \right) - \mathop {\lim }\limits_{h \to 0} \left( {\sin c\sin \,h} \right)\\& = \cos c\cos 0 - \sin c\sin 0\\ &= \cos c\left( 1 \right) - \sin c\left( 0 \right)\\ &= \cos c\end{align}

$$\therefore \mathop {\lim }\limits_{x \to c} g\left( x \right) = g\left( c \right)$$

Therefore, $$g\left( x \right) = \cos x$$ is a continuous function.

Let $$h\left( x \right) = {x^2}$$

It is evident that $$h$$is defined for every real number.

Let $$k$$be a real number, then $$h\left( k \right) = {k^2}$$

\begin{align}&\mathop {\lim }\limits_{x \to k} h\left( x \right) = \mathop {\lim }\limits_{x \to k} {x^2} = {k^2}\\&\therefore \mathop {\lim }\limits_{x \to k} h\left( x \right) = h\left( k \right)\end{align}

Therefore, $$h$$ is a continuous function.

It is known that for real valued functions $$g$$and $$h$$, such that $$\left( {goh} \right)$$ is defined at $$c$$, if $$g$$is continuous at $$c$$and if $$f$$ is continuous at $$g~\left( c \right)$$, then $$\left( {fog} \right)$$is continuous at $$c$$.

Therefore, $$f\left( x \right) = \left( {goh} \right)\left( x \right) = \cos \left( {{x^2}} \right)$$is a continuous function.

## Chapter 5 Ex.5.1 Question 32

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

### Solution

The given function is $$f\left( x \right) = \left| {\cos x} \right|$$.

This function $$f$$ is defined for every real number and $$f$$ can be written as the composition of two functions as,

$$f = goh$$, where $$g\left( x \right) = \left| x \right|$$ and $$h\left( x \right) = \cos x$$

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

It has to be proved first that $$g\left( x \right) = \left| x \right|$$ and $$h\left( x \right) = \cos x$$ are continuous functions.

$$g\left( x \right) = \left| x \right|$$can be written as $$g\left( x \right) = \left\{ \begin{array}{l} - x,{\text{ if }}x < 0\\x,{\text{ if }}x \ge 0\end{array} \right.$$

It is evident that $$g$$ is defined for every real number.

Let $$c$$be a real number.

Case I:

If $$c < 0$$, then $$g\left( c \right) = - c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} g\left( x \right) = &\mathop {\lim }\limits_{x \to c} \left( { - x} \right) = - c\\&\therefore \mathop {\lim }\limits_{x \to c} g\left( x \right) = g\left( c \right)\end{align}

Therefore, $$g$$ is continuous at all points $$x$$, such that$$x < 0$$.

Case II:

If $$c > 0$$, then $$g\left( c \right) = c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} g\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( x \right) = c\\&\therefore \mathop {\lim }\limits_{x \to c} g\left( x \right) = g\left( c \right)\end{align}

Therefore, $$g$$ is continuous at all points $$x$$, such that$$x > 0$$.

Case III:

If $$c = 0$$, then $$g\left( c \right) = g\left( 0 \right) = 0$$

\begin{align}&\mathop {\lim }\limits_{x \to {0^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0\\&\mathop {\lim }\limits_{x \to {0^ + }} g\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( x \right) = 0\\&\therefore \mathop {\lim }\limits_{x \to {0^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( x \right) = g\left( 0 \right)\end{align}

Therefore, $$g$$ is continuous at all $$x = 0$$.

From the above three observations, it can be concluded that $$g$$ is continuous at all points.

Let $$h\left( x \right) = \cos x$$

It is evident that $$h\left( x \right) = \cos x$$ is defined for every real number.

Let $$c$$be a real number. Put $$x = c + h$$

If $$x \to c$$, then $$h \to 0$$

\begin{align}h\left( c \right) &= \cos c\\\mathop {\lim }\limits_{x \to c} h\left( x \right) &= \mathop {\lim }\limits_{x \to c} \cos x\\ &= \mathop {\lim }\limits_{h \to 0} \cos \left( {c + h} \right)\\& = \mathop {\lim }\limits_{h \to 0} \left[ {\cos c\cos \,h - \sin c\sin \,h} \right]\\& = \mathop {\lim }\limits_{h \to 0} \left( {\cos c\cos \,h} \right) - \mathop {\lim }\limits_{h \to 0} \left( {\sin c\sin \,h} \right)\\ &= \cos c\cos 0 - \sin c\sin 0\\ &= \cos c\left( 1 \right) - \sin c\left( 0 \right)\\ &= \cos c\end{align}

$$\therefore \mathop {\lim }\limits_{x \to c} h\left( x \right) = h\left( c \right)$$

Therefore, $$h\left( x \right) = \cos x$$ is a continuous function.

It is known that for real valued functions $$g$$and $$h$$, such that $$\left( {goh} \right)$$ is defined at $$c$$, if $$g$$is continuous at $$c$$and if $$f$$ is continuous at $$g~\left( c \right)$$, then $$\left( {fog} \right)$$is continuous at $$c$$.

Therefore, $$f\left( x \right) = \left( {goh} \right)\left( x \right) = g\left( {h\left( x \right)} \right) = g\left( {\cos x} \right) = \left| {\cos x} \right|$$is a continuous function.

## Chapter 5 Ex.5.1 Question 33

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

### Solution

The given function is $$f\left( x \right) = \left| {\sin x} \right|$$.

This function $$f$$ is defined for every real number and $$f$$ can be written as the composition of two functions as,

$$f = goh$$, where $$g\left( x \right) = \left| x \right|$$ and $$h\left( x \right) = \sin x$$

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

It has to be proved first that $$g\left( x \right) = \left| x \right|$$ and $$h\left( x \right) = \sin x$$are continuous functions.

$$g\left( x \right) = \left| x \right|$$can be written as $$g\left( x \right) = \left\{ \begin{array}{l} - x,{\text{ if }}x < 0\\x,{\text{ if }}x \ge 0\end{array} \right.$$

It is evident that $$g$$ is defined for every real number.

Let $$c$$ be a real number.

Case I:

If $$c < 0$$, then $$g\left( c \right) = - c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} g\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( { - x} \right) = - c\\&\therefore \mathop {\lim }\limits_{x \to c} g\left( x \right) = g\left( c \right)\end{align}

Therefore, $$g$$ is continuous at all points $$x$$, such that$$x < 0$$.

Case II:

If $$c > 0$$, then $$g\left( c \right) = c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} g\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( x \right) = c\\&\therefore \mathop {\lim }\limits_{x \to c} g\left( x \right) = g\left( c \right)\end{align}

Therefore, $$g$$ is continuous at all points $$x$$, such that$$x > 0$$.

Case III:

If $$c = 0$$, then $$g\left( c \right) = g\left( 0 \right) = 0$$

\begin{align}&\mathop {\lim }\limits_{x \to {0^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0\\&\mathop {\lim }\limits_{x \to {0^ + }} g\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( x \right) = 0\\&\therefore \mathop {\lim }\limits_{x \to {0^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( x \right) = g\left( 0 \right)\end{align}

Therefore, $$g$$ is continuous at all $$x = 0$$.

From the above three observations, it can be concluded that $$g$$ is continuous at all points.

Let $$h\left( x \right) = \sin x$$

It is evident that $$h\left( x \right) = \sin x$$ is defined for every real number.

Let $$c$$be a real number. Put $$x = c + k$$

If $$x \to c$$, then $$k \to 0$$

\begin{align}h\left( c \right) &= \sin c\\\mathop {\lim }\limits_{x \to c} h\left( x \right) &= \mathop {\lim }\limits_{x \to c} \sin x\\ &= \mathop {\lim }\limits_{k \to 0} \sin \left( {c + k} \right)\\ &= \mathop {\lim }\limits_{k \to 0} \left[ {\sin c\cos \,k + \cos c\sin \,k} \right]\\ &= \mathop {\lim }\limits_{k \to 0} \left( {\sin c\cos \,k} \right) + \mathop {\lim }\limits_{k \to 0} \left( {\cos c\sin \,k} \right)\\& = \sin c\cos 0 + \cos c\sin 0\\& = \sin c\left( 1 \right) + \cos c\left( 0 \right)\\ &= \sin c\end{align}

$$\therefore \mathop {\lim }\limits_{x \to c} h\left( x \right) = h\left( c \right)$$

Therefore, $$h\left( x \right) = \sin x$$ is a continuous function.

It is known that for real valued functions $$g$$and $$h$$, such that $$\left( {goh} \right)$$ is defined at $$c$$, if $$g$$is continuous at $$c$$and if $$f$$ is continuous at $$g~\left( c \right)$$, then $$\left( {fog} \right)$$is continuous at $$c$$.

Therefore, $$f\left( x \right) = \left( {goh} \right)\left( x \right) = g\left( {h\left( x \right)} \right) = g\left( {\sin x} \right) = \left| {\sin x} \right|$$is a continuous function.

## Chapter 5 Ex.5.1 Question 34

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

### Solution

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

The two functions, $$g$$ and $$h$$ are defined as $$g\left( x \right) = \left| x \right|$$ and $$h\left( x \right) = \left| {x + 1} \right|$$.

Then, $$f = g - h$$

The continuity of $$g$$ and $$h$$ are examined first.

$$g\left( x \right) = \left| x \right|$$can be written as $$g\left( x \right) = \left\{ \begin{array}{l} - x,{\text{ if}}\,x < 0\\x,{\text{ if}}\,x \ge 0\end{array} \right.$$

It is evident that $$g$$ is defined for every real number.

Let $$c$$be a real number.

Case I:

If $$c < 0$$, then $$g\left( c \right) = - c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} g\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( { - x} \right) = - c\\&\therefore \mathop {\lim }\limits_{x \to c} g\left( x \right) = g\left( c \right)\end{align}

Therefore, $$g$$ is continuous at all points $$x$$, such that$$x < 0$$.

Case II:

If $$c > 0$$, then $$g\left( c \right) = c$$

\begin{align}&\mathop {\lim }\limits_{x \to c} g\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( x \right) = c\\&\therefore \mathop {\lim }\limits_{x \to c} g\left( x \right) = g\left( c \right)\end{align}

Therefore, $$g$$ is continuous at all points $$x$$, such that$$x > 0$$.

Case III:

If $$c = 0$$, then $$g\left( c \right) = g\left( 0 \right) = 0$$

\begin{align}&\mathop {\lim }\limits_{x \to {0^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0\\&\mathop {\lim }\limits_{x \to {0^ + }} g\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( x \right) = 0\\&\therefore \mathop {\lim }\limits_{x \to {0^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( x \right) = g\left( 0 \right)\end{align}

Therefore, $$g$$ is continuous at all $$x = 0$$.

From the above three observations, it can be concluded that $$g$$ is continuous at all points.

$$h\left( x \right) = \left| {x + 1} \right|$$ can be written as $$h\left( x \right) = \left\{ \begin{array}{l} - \left( {x + 1} \right),{\text{ if }}x < - 1\\x + 1,{\text{ if }}x \ge - 1\end{array} \right.$$

It is evident that $$h$$ is defined for every real number.

Let $$c$$be a real number.

Case I:

If $$c < - 1$$, then $$h\left( c \right) = - \left( {c + 1} \right)$$

\begin{align}&\mathop {\lim }\limits_{x \to c} h\left( x \right) = \mathop {\lim }\limits_{x \to c} \left[ { - \left( {x + 1} \right)} \right] = - \left( {c + 1} \right)\\&\therefore \mathop {\lim }\limits_{x \to c} h\left( x \right) = h\left( c \right)\end{align}

Therefore, $$h$$ is continuous at all points $$x$$, such that$$x < - 1$$.

Case II:

If $$c > - 1$$, then $$h\left( c \right) = c + 1$$

\begin{align}&\mathop {\lim }\limits_{x \to c} h\left( x \right) = \mathop {\lim }\limits_{x \to c} \left( {x + 1} \right) = c + 1\\&\therefore \mathop {\lim }\limits_{x \to c} h\left( x \right) = h\left( c \right)\end{align}

Therefore, $$h$$ is continuous at all points $$x$$, such that$$x > - 1$$.

Case III:

If $$c = - 1$$, then $$h\left( c \right) = h\left( { - 1} \right) = - 1 + 1 = 0$$

\begin{align}&\mathop {\lim }\limits_{x \to - {1^ - }} h\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} \left[ { - \left( {x + 1} \right)} \right] = - \left( { - 1 + 1} \right) = 0\\&\mathop {\lim }\limits_{x \to - {1^ + }} h\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} \left( {x + 1} \right) = \left( { - 1 + 1} \right) = 0\\&\therefore \mathop {\lim }\limits_{x \to - {1^ - }} h\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} h\left( x \right) = h\left( { - 1} \right)\end{align}

Therefore, $$h$$ is continuous at $$x = - 1$$.

From the above three observations, it can be concluded that $$h$$ is continuous at all points.

It concludes that$$g$$ and $$h$$are continuous functions. Therefore, $$f = g - h$$ is also a continuous function.

Therefore, $$f$$ has no point of discontinuity.

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