# 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.

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