# NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.2

## Chapter 1 Ex.1.2 Question 1

Show that the function \(f:{R_ \bullet } \to {R_ \bullet }\)defined by \(\left( x \right) = \frac{1}{x}\)is one –one and onto, where \({R_ \bullet }\)is the set of all non –zero real numbers. Is the result true, if the domain\({R_ \bullet }\)is replaced by N with co-domain being same as\({R_ \bullet }\)?

**Solution**

\(f:{R_ \bullet } \to {R_ \bullet }\)is by \(f\left( x \right) = \frac{1}{x}\)

For one-one:

\(x,y \in {R_ \bullet }\) such that \(f\left( x \right) = f\left( y \right)\)

\[\begin{align}\Rightarrow \frac{1}{x} = \frac{1}{y}\\\Rightarrow x = y\end{align}\]

\(\therefore f\)is one-one.

For onto:

For \(y \in R,\)there exists \(x = \frac{1}{y} \in {R_ \bullet }\left[ {{\rm{as }}y \notin 0} \right]\)such that

\(f\left( x \right) = \frac{1}{{\left( {\frac{1}{y}} \right)}} = y\)

\(\therefore f\)is onto.

Given function \(f\)is one-one and onto.

Consider function \(g:N \to {R_ \bullet }\)defined by \(g\left( x \right) = \frac{1}{x}\)

We have, \(g\left( {{x_1}} \right) = g\left( {{x_2}} \right) \Rightarrow \frac{1}{{{x_1}}} = \frac{1}{{{x_2}}} \Rightarrow {x_1} = {x_2}\)

\(\therefore g\)is one-one.

\(g\)is not onto as for \(1.2 \in {R_ \bullet }\)there exist any \(x\)in \(N\)such that \(g\left( x \right) = \frac{1}{{1.2}}\)

Function \(g\)is one-one but not onto.

## Chapter 1 Ex.1.2 Question 2

Check the injectivity and surjectivity of the following functions:

(i) \(f:N \to N\)given by \(f\left( x \right) = {x^2}\)

(ii) \(f:Z \to Z\)given by \(f\left( x \right) = {x^2}\)

(iii) \(f:R \to R\)given by \(f\left( x \right) = {x^2}\)

(iv) \(f:N \to N\)given by \(f\left( x \right) = {x^3}\)

(v) \(f:Z \to Z\)given by \(f\left( x \right) = {x^3}\)

**Solution**

(i) For \(f:N \to N\)given by \(f\left( x \right) = {x^2}\)

\[\begin{align}x,y \in N\\f\left( x \right) = f\left( y \right) \Rightarrow {x^2} = {y^2} \Rightarrow x = y\end{align}\]

\(\therefore f\)is injective.

\(2 \in N.\)But, there does not exist any \(x\)in \(N\)such that \(f\left( x \right) = {x^2} = 2\)

\(\therefore f\)is not surjective

Function \(f\)is injective but not surjective.

(ii) \(f:Z \to Z\)given by \(f\left( x \right) = {x^2}\)

\(f\left( { - 1} \right) = f\left( 1 \right) = 1{\rm{ but }} - 1 \ne 1\)

\(\therefore f\)is not injective.

\( - 2 \in Z\) But, there does not exist any \(x \in Z\)such that \(f\left( x \right) = - 2 \Rightarrow {x^2} = - 2\)

\(\therefore f\)is not surjective.

Function \(f\)is neither injective nor surjective.

(iii) \(f:R \to R\)given by \(f\left( x \right) = {x^2}\)

\(f\left( { - 1} \right) = f\left( 1 \right) = 1{\rm{ but }} - 1 \ne 1\)

\(\therefore f\)is not injective.

\( - 2 \in Z\) But, there does not exist any \(x \in Z\)such that \(f\left( x \right) = - 2 \Rightarrow {x^2} = - 2\)

\(\therefore f\)is not surjective.

Function \(f\)is neither injective nor surjective.

(iv) \(f:N \to N\)given by \(f\left( x \right) = {x^3}\)

\[\begin{align}x,y \in N\\f\left( x \right) = f\left( y \right) \Rightarrow {x^3} = {y^3} \Rightarrow x = y\end{align}\]

\(\therefore f\)is injective.

\(2 \in N.\)But, there does not exist any \(x\)in \(N\)such that \(f\left( x \right) = {x^3} = 2\)

\(\therefore f\)is not surjective

Function \(f\)is injective but not surjective.

(v) \(f:Z \to Z\)given by \(f\left( x \right) = {x^3}\)

\[\begin{align}x,y \in Z\\f\left( x \right) = f\left( y \right) \Rightarrow {x^3} = {y^3} \Rightarrow x = y\end{align}\]

\(\therefore f\)is injective.

\(2 \in Z\). But, there does not exist any \(x\)in \(Z\) such that \(f\left( x \right) = {x^3} = 2\)

\(\therefore f\)is not surjective.

Function \(f\)is injective but not surjective.

## Chapter 1 Ex.1.2 Question 3

Prove that the greatest integer function \(f:R \to R\)given by \(f\left( x \right) = \left[ x \right]\)is neither one-one nor onto, where \(\left[ x \right]\)denotes the greatest integer less than or equal to\(x\).

**Solution**

\(f:R \to R\)given by \(f\left( x \right) = \left[ x \right]\)

\[\begin{align}f\left( {1.2} \right) = \left[ {1.2} \right] = 1,f\left( {1.9} \right) = \left[ {1.9} \right] = 1\\\therefore f\left( {1.2} \right) = f\left( {1.9} \right),{\rm{ but 1}}{\rm{.2}} \ne {\rm{1}}{\rm{.9}}\end{align}\]

\(\therefore f\)is not one-one.

Consider \(0.7 \in R\)

\(f\left( x \right) = \left[ x \right]\)is an integer. There does not exist any element \(x \in R\)such that \(f\left( x \right) = 0.7\)

\(\therefore f\)is not onto.

The greatest integer function is neither one-one nor onto.

## Chapter 1 Ex.1.2 Question 4

Show that the modulus function \(f:R \to R\)given by \(f\left( x \right) = \left| x \right|\)is neither one-one nor onto, where \(\left| x \right|\)is \(x\), if \(x\)is positive or \(0\)and \(\left| x \right|\)is \( - x\), if \(x\)is negative.

**Solution**

\(f:R \to R\)is\[f\left( x \right) = \left| x \right| = \left\{ \begin{array}{l}{\rm{x, if x}} \ge 0\\- x,{\rm{if x}} < 0\end{array} \right\}\]

\[\begin{array}{l}f\left( { - 1} \right) = \left| { - 1} \right| = 1{\rm{ and }}f\left( 1 \right) = \left| 1 \right| = 1\\\therefore f\left( { - 1} \right) = f\left( 1 \right){\rm{ but }} - 1 \ne 1\end{array}\]

\(\therefore f\)is not one-one.

Consider \( - 1 \in R\)

\(f\left( x \right) = \left| x \right|\)is non negative. There exist any element \(x\)in domain \(R\)such that \(f\left( x \right) = \left| x \right| = - 1\)

\(\therefore f\)is not onto.

The modulus function is neither one-one nor onto.

## Chapter 1 Ex.1.2 Question 5

Show that the signum function\(f:R \to R\)given by \(f\left( x \right) = \left\{ \begin{array}{l}1,{\rm{ if x}} > 0\\{\rm{0, if x}} = 0\\{\rm{ - 1, if x}} < 0\end{array} \right\}\)is neither one-one nor onto.

**Solution**

\(f:R \to R\)is\(f\left( x \right) = \left\{ \begin{array}{l}1,{\rm{if x}} > 0\\{\rm{0, if x}} = 0\\{\rm{ - 1, if x}} < 0\end{array} \right\}\)

\(f\left( 1 \right) = f\left( 2 \right) = 1,{\rm{ but 1}} \ne {\rm{2}}\)

\(\therefore f\)is not one-one.

\(f\left( x \right)\)takes only 3 values\(\left( {1,0, - 1} \right)\)for the element \( - 2\)in co-domain

R, there does not exist any \(x\)in domain R such that \(f\left( x \right) = - 2\).

\(\therefore f\)is not onto.

The signum function is neither one-one nor onto.

## Chapter 1 Ex.1.2 Question 6

Let \(A = \left\{ {1,2,3} \right\},\) \(B = \left\{ {4,5,6,7} \right\}\)and let \(f = \left\{ {\left( {1,4} \right),\left( {2,5} \right),\left( {3,6} \right)} \right\}\)be a function from \(A{\rm{ to B}}\). Show that f is one-one.

**Solution**

\(A = \left\{ {1,2,3} \right\}\),\(B = \left\{ {4,5,6,7} \right\}\)

\(f:A \to B\)is defined as \(f = \left\{ {\left( {1,4} \right),\left( {2,5} \right),\left( {3,6} \right)} \right\}\)

\(\therefore f\left( 1 \right) = 4,f\left( 2 \right) = 5,f\left( 3 \right) = 6\)

It is seen that the images of distinct elements of \(A\)under \(f\)are distinct.

\(\therefore f\) is one-one.

## Chapter 1 Ex.1.2 Question 7

In each of the following cases, state whether the function is one-one, onto or bijective.

Justify your answer.

(i) \(f:R \to R\)defined by \(f\left( x \right) = 3 - 4x\)

(ii) \(f:R \to R\)defined by \(f\left( x \right) = 1 + {x^2}\)

**Solution**

(i) \(f:R \to R\)defined by \(f\left( x \right) = 3 - 4x\)

\({x_1},{x_2} \in R\)such that \(f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\)

\[\begin{align}&\Rightarrow 3 - 4{x_1} = 3 - 4{x_2}\\&\Rightarrow - 4{x_{}} = - 4{x_2}\\&\Rightarrow {x_1} = {x_2}\end{align}\]

\(\therefore f\)is one-one.

For any real number \(\left( y \right)\)in \(R\), there exists \(\frac{{3 - y}}{4}\)in \(R\) such that \(f\left( {\frac{{3 - y}}{4}} \right) = 3 - 4\left( {\frac{{3 - y}}{4}} \right) = y\)

\(\therefore f\)is onto.

Hence, \(f\)is bijective.

(ii) \(f:R \to R\)defined by \(f\left( x \right) = 1 + {x^2}\)

\({x_1},{x_2} \in R\)such that \(f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\)

\[\begin{align}&\Rightarrow 1 + {x_1}^2 = 1 + {x_2}^2\\&\Rightarrow {x_1}^2 = {x_2}^2\\&\Rightarrow {x_1} = \pm {x_2}\end{align}\]

\(\therefore f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\) does not imply that \({x_1} = {x_2}\)

Consider \(f\left( 1 \right) = f\left( { - 1} \right) = 2\)

\(\therefore f\)is not one-one.

Consider an element \( - 2\)in co domain \(R\).

It is seen that \(f\left( x \right) = 1 + {x^2}\)is positive for all \(x \in R\).

\(\therefore f\)is not onto.

Hence, \(f\)is neither one-one nor onto.

## Chapter 1 Ex.1.2 Question 8

Let \(A\)and \(B\) be sets. Show that \(f:A \times B \to B \times A\) such that \(\left( {a,b} \right) = \left( {b,a} \right)\) is a bijective function.

**Solution**

\(f:A \times B \to B \times A\)is defined as \(\left( {a,b} \right) = \left( {b,a} \right)\).

\(\left( {{a_1},{b_1}} \right),\left( {{a_2},{b_2}} \right) \in A \times B\)such that \(f\left( {{a_1},{b_1}} \right) = f\left( {{a_2},{b_2}} \right)\)

\[\begin{align}&\Rightarrow \left( {{b_1},{a_1}} \right) = \left( {{b_2},{a_2}} \right)\\&\Rightarrow {b_1} = {b_2}{\rm{ and }}{a_1} = {a_2}\\&\Rightarrow \left( {{a_1},{b_1}} \right) = \left( {{a_2},{b_2}} \right)\end{align}\]

\(\therefore f\)is one-one.

\(\left( {b,a} \right) \in B \times A\) there exist \(\left( {a,b} \right) \in A \times B\)such that \(f\left( {a,b} \right) = \left( {b,a} \right)\)

\(\therefore f\) is onto.

\(f\) is bijective.

## Chapter 1 Ex.1.2 Question 9

Let \(f:N \to N\)be defined as \(\begin{array}{l}f\left( n \right) = \left\{ \begin{array}{l}\frac{{n + 1}}{2},{\rm{ if }}n{\rm{ is odd}}\\\frac{n}{2},{\rm{if }}n{\rm{ is even}}\end{array} \right\}\\\end{array}\)for all \(n \in N\). State whether the function \(f\)is bijective. Justify your answer.

**Solution**

\(f:N \to N\)be defined as \(\begin{array}{l}f\left( n \right) = \left\{ \begin{array}{l}\frac{{n + 1}}{2},{\rm{ if }}n{\rm{ is odd}}\\\frac{n}{2},{\rm{if }}n{\rm{ is even}}\end{array} \right\}\\\end{array}\)for all \(n \in N\).

\[\begin{align}f\left( 1 \right) &= \frac{{1 + 1}}{2} = 1{\rm{ and }}f\left( 2 \right) = \frac{2}{2} = 1\\f\left( 1 \right)& = f\left( 2 \right),{\rm{ where 1}} \ne {\rm{2}}\end{align}\]

\(\therefore f\)is not one-one.

Consider a natural number \(n\)in co domain\(N\).

Case I: \(n\)is odd

\(\therefore n = 2r + 1\)for some \(r \in N\)there exists \(4r + 1 \in N\)such that

\[f\left( {4r + 1} \right) = \frac{{4r + 1 + 1}}{2} = 2r + 1\]

Case II: \(n\)is even

\(\therefore n = 2r\)for some \(r \in N\)there exists \(4r \in N\)such that

\[f\left( {4r} \right) = \frac{{4r}}{2} = 2r\]

\(\therefore f\)is onto.

\(f\)is not a bijective function.

## Chapter 1 Ex.1.2 Question 10

Let \(A = R - \left\{ 3 \right\},B = R - \left\{ 1 \right\}\)and \(f:A \to B\)defined by \(f\left( x \right) = \left( {\frac{{x - 2}}{{x - 3}}} \right)\). Is \(f\)one-one and onto? Justify your answer.

**Solution**

\(A = R - \left\{ 3 \right\},B = R - \left\{ 1 \right\}\)and \(f:A \to B\)defined by \(f\left( x \right) = \left( {\frac{{x - 2}}{{x - 3}}} \right)\)

\(x,y \in A\)such that \(f\left( x \right) = f\left( y \right)\)

\[\begin{align}&\Rightarrow \frac{{x - 2}}{{x - 3}} = \frac{{y - 2}}{{y - 3}}\\&\Rightarrow \left( {x - 2} \right)\left( {y - 3} \right) = \left( {y - 2} \right)\left( {x - 3} \right)\\&\Rightarrow xy - 3x - 2y + 6 = xy - 3y - 2x + 6\\&\Rightarrow - 3x - 2y = - 3y - 2x\\&\Rightarrow 3x - 2x = 3y - 2y\\&\Rightarrow x = y\end{align}\]

\(\therefore f\)is one-one.

Let \(y \in B = R - \left\{ 1 \right\}\), then \(y \ne 1\)

The function \(f\)is onto if there exists \(x \in A\)such that \(f\left( x \right) = y\).

Now,

\[\begin{align}&f\left( x \right) = y\\&\Rightarrow \frac{{x - 2}}{{x - 3}} = y\\&\Rightarrow x - 2 = xy - 3y\\&\Rightarrow x\left( {1 - y} \right) = - 3y + 2\\&\Rightarrow x = \frac{{2 - 3y}}{{1 - y}} \in A \qquad \qquad \qquad \left[ {y \ne 1} \right]\end{align}\]

Thus, for any \(y \in B\), there exists \(\frac{{2 - 3y}}{{1 - y}} \in A\)such that

\[f\left( {\frac{{2 - 3y}}{{1 - y}}} \right) = \frac{{\left( {\frac{{2 - 3y}}{{1 - y}}} \right) - 2}}{{\left( {\frac{{2 - 3y}}{{1 - - y}}} \right) - 3}} = \frac{{2 - 3y - 2 + 2y}}{{2 - 3y - 3 + 3y}} = \frac{{ - y}}{{ - 1}} = y\]

\(\therefore f\)is onto.

Hence, the function is one-one and onto.

## Chapter 1 Ex.1.2 Question 11

Let \(f:R \to R\)defined as \(f\left( x \right) = {x^4}\).Choose the correct answer.

A) \(f\)is one-one onto

B) \(f\)is many-one onto

C) \(f\)is one-one but not onto

D) \(f\)is neither one-one nor onto

**Solution**

\(f:R \to R\)defined as \(f\left( x \right) = {x^4}\)

\(x,y \in R\)such that \(f\left( x \right) = f\left( y \right)\)

\[\begin{align}&\Rightarrow {x^4} = {y^4}\\&\Rightarrow x = \pm y\end{align}\]

\(\therefore f\left( x \right) = f\left( y \right)\)does not imply that \(x = y\).

For example \(f\left( 1 \right) = f\left( { - 1} \right) = 1\)

\(\therefore f\)is not one-one.

Consider an element \(2\) in co domain\(R\)there does not exist any \(x\)in domain \(R\)such that \(f\left( x \right) = 2\).

\(\therefore f\)is not onto.

Function \(f\)is neither one-one nor onto.

The correct answer is D.

## Chapter 1 Ex.1.2 Question 12

Let \(f:R \to R\)defined as \(f\left( x \right) = 3x\).Choose the correct answer.

A) \(f\)is one-one onto

B) \(f\)is many-one onto

C) \(f\)is one-one but not onto

D) \(f\)is neither one-one nor onto

**Solution**

\(f:R \to R\)defined as \(f\left( x \right) = 3x\)

\(x,y \in R\)such that \(f\left( x \right) = f\left( y \right)\)

\[\begin{align}&\Rightarrow 3x = 3y\\&\Rightarrow x = y\end{align}\]

\(\therefore f\)is one-one.

For any real number \(y\)in co domain R, there exist \(\frac{y}{3}\)in R such that \(f\left( {\frac{y}{3}} \right) = 3\left( {\frac{y}{3}} \right) = y\)

\(\therefore f\)is onto.

Hence, function \(f\)is one-one and onto.

The correct answer is A.