NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.1

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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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