Relations and Functions 1.2 NCERT Solutions

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

Solution

We are given the function

f:\mathbb{R}_*\to\mathbb{R}_*

defined by

f(x)=\frac{1}{x}

where \mathbb{R}_* denotes the set of all non-zero real numbers.

We will check whether the function is one-one and onto.

One-One:

Let

f(x_1)=f(x_2)

Then

\frac{1}{x_1}=\frac{1}{x_2}

Cross multiplying, we get

x_1=x_2

Therefore, the function is one-one (injective).

Onto:

Let y\in\mathbb{R}_* .

We need to show that there exists x\in\mathbb{R}_* such that

f(x)=y

Now,

y=\frac{1}{x}

gives

x=\frac{1}{y}

Since y\neq0 , we have \frac{1}{y}\in\mathbb{R}_* .

Thus, for every y\in\mathbb{R}_* , there exists x=\frac{1}{y} such that

f(x)=y

Therefore, the function is onto (surjective).

Hence, the function is both one-one and onto, i.e., it is bijective.

Now, consider the function

f:\mathbb{N}\to\mathbb{R}_*

defined by

f(x)=\frac{1}{x}

One-One:

Suppose

f(x_1)=f(x_2)

Then

\frac{1}{x_1}=\frac{1}{x_2}

which gives

x_1=x_2

Therefore, the function is still one-one.

Onto:

The range of the function is

\left\{1,\frac12,\frac13,\frac14,\dots\right\}

which is not equal to \mathbb{R}_* .

For example, there is no natural number x such that

\frac{1}{x}=2

Therefore, the function is not onto.

Hence, when the domain is changed to \mathbb{N} , the function remains one-one but is not onto.

2. Check the injectivity and surjectivity of the following functions:

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

Solution

We are given

f(x)=x^2

where

f:\mathbb{N}\to\mathbb{N}

Injective (One-One):

Suppose

f(x_1)=f(x_2)

Then

x_1^2=x_2^2

Taking square roots, we get

x_1=x_2

since x_1,x_2\in\mathbb{N} .

Therefore, the function is one-one.

Surjective (Onto):

The range of the function is

\{1,4,9,16,\dots\}

which does not contain all natural numbers.

For example, 2\in\mathbb{N} , but there is no x\in\mathbb{N} such that

x^2=2

Therefore, the function is not onto.

Hence, the function is one-one but not onto.

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

Solution

We are given

f(x)=x^2

where

f:\mathbb{Z}\to\mathbb{Z}

Injective (One-One):

Consider

f(2)=2^2=4

and

f(-2)=(-2)^2=4

Since

f(2)=f(-2)

but

2\neq-2

therefore, the function is not one-one.

Surjective (Onto):

The range of the function contains only non-negative perfect squares.

Negative integers like -1,-2,-3 are not images of any integer.

Also, integers like 2 or 3 are not perfect squares.

Therefore, the function is not onto.

Hence, the function is neither one-one nor onto.

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

Solution

We are given

f(x)=x^2

where

f:\mathbb{R}\to\mathbb{R}

Injective (One-One):

Consider

f(2)=4

and

f(-2)=4

Since different inputs have the same image, the function is not one-one.

Surjective (Onto):

For every real number x , we have

x^2\geq0

Thus, negative real numbers are not images of any real number.

For example, there is no x\in\mathbb{R} such that

x^2=-1

Therefore, the function is not onto.

Hence, the function is neither one-one nor onto.

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

Solution

We are given

f(x)=x^3

where

f:\mathbb{N}\to\mathbb{N}

Injective (One-One):

Suppose

f(x_1)=f(x_2)

Then

x_1^3=x_2^3

Taking cube roots, we get

x_1=x_2

Therefore, the function is one-one.

Surjective (Onto):

The range of the function is

\{1,8,27,64,\dots\}

which does not contain all natural numbers.

For example, 2 is not the cube of any natural number.

Therefore, the function is not onto.

Hence, the function is one-one but not onto.

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

Solution

We are given

f(x)=x^3

where

f:\mathbb{Z}\to\mathbb{Z}

Injective (One-One):

Suppose

f(x_1)=f(x_2)

Then

x_1^3=x_2^3

Taking cube roots, we get

x_1=x_2

Therefore, the function is one-one.

Surjective (Onto):

The range of the function contains only perfect cubes.

Integers like 2 , 3 and 5 are not perfect cubes.

Therefore, the function is not onto.

Hence, the function is one-one but not onto.

3. Prove that the Greatest Integer Function f:\mathbb{R}\to\mathbb{R} , given by f(x)=[x] , is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x .

Solution

We are given the greatest integer function

f:\mathbb{R}\to\mathbb{R}

defined by

f(x)=[x]

where [x] denotes the greatest integer less than or equal to x .

We will check whether the function is one-one and onto.

One-One (Injective):

A function is one-one if different elements of the domain have different images.

Consider the real numbers

1.2 \text{ and } 1.8

Now,

f(1.2)=[1.2]=1

and

f(1.8)=[1.8]=1

Thus,

f(1.2)=f(1.8)

but

1.2\neq1.8

Therefore, the function is not one-one.

Onto (Surjective):

The codomain of the function is \mathbb{R} .

But the greatest integer function gives only integer values.

Thus, the range of the function is

\mathbb{Z}

which is not equal to the codomain \mathbb{R} .

For example, there is no real number x such that

[x]=\frac12

Therefore, the function is not onto.

Hence, the greatest integer function is neither one-one nor onto.

4. Show that the Modulus Function f:\mathbb{R}\to\mathbb{R} , given by f(x)=|x| , is neither one-one nor onto, where

|x|=\begin{cases} x, & x\geq0 \\ \\ -x, & x<0 \end{cases}

Solution

We are given the modulus function

f:\mathbb{R}\to\mathbb{R}

defined by

f(x)=|x|

where

|x|=\begin{cases} x, & x\geq0 \\ \\ -x, & x<0 \end{cases}

We will check whether the function is one-one and onto.

One-One (Injective):

A function is one-one if different elements of the domain have different images.

Consider the real numbers

2 \text{ and } -2

Now,

f(2)=|2|=2

and

f(-2)=|-2|=2

Thus,

f(2)=f(-2)

but

2\neq-2

Therefore, the function is not one-one.

Onto (Surjective):

For every real number x , we have

|x|\geq0

Thus, the range of the function contains only non-negative real numbers.

Negative real numbers are not images of any real number.

For example, there is no x\in\mathbb{R} such that

|x|=-3

Therefore, the function is not onto.

Hence, the modulus function is neither one-one nor onto.

5. Show that the Signum Function f:\mathbb{R}\to\mathbb{R} , given by

f(x)=\begin{cases} 1, & x>0 \\ \\ 0, & x=0 \\ \\ -1, & x<0 \end{cases}

is neither one-one nor onto.

Solution

We are given the signum function

f:\mathbb{R}\to\mathbb{R}

defined by

f(x)=\begin{cases} 1, & x>0 \\ \\ 0, & x=0 \\ \\ -1, & x<0 \end{cases}

We will check whether the function is one-one and onto.

One-One (Injective):

A function is one-one if different elements of the domain have different images.

Consider the real numbers

2 \text{ and } 5

Now,

f(2)=1

and

f(5)=1

Thus,

f(2)=f(5)

but

2\neq5

Therefore, the function is not one-one.

Onto (Surjective):

The range of the signum function is

\{-1,0,1\}

which is not equal to the codomain \mathbb{R} .

For example, there is no real number x such that

f(x)=2

Therefore, the function is not onto.

Hence, the signum function is neither one-one nor onto.

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

Solution

We are given the function

f:A\to B

where

A=\{1,2,3\} \text{ and } B=\{4,5,6,7\}

The function is defined as

f=\{(1,4),(2,5),(3,6)\}

This means:

• f(1)=4
• f(2)=5
• f(3)=6

We observe that different elements of the domain have different images.

That is,

f(1)\neq f(2),\quad f(2)\neq f(3),\quad f(1)\neq f(3)

since

4\neq5,\quad 5\neq6,\quad 4\neq6

Thus, no two distinct elements of A have the same image in B .

Therefore, the function f is one-one (injective).

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

(i) f:\mathbb{R}\to\mathbb{R} defined by f(x)=3-4x

Solution

We are given

f(x)=3-4x

where

f:\mathbb{R}\to\mathbb{R}

We will check whether the function is one-one and onto.

One-One (Injective):

Suppose

f(x_1)=f(x_2)

Then

3-4x_1=3-4x_2

Subtracting 3 from both sides, we get

-4x_1=-4x_2

Dividing by -4 , we get

x_1=x_2

Therefore, the function is one-one.

Onto (Surjective):

Let y\in\mathbb{R} .

We need to find x\in\mathbb{R} such that

f(x)=y

Now,

y=3-4x

gives

4x=3-y

Therefore,

x=\frac{3-y}{4}

Since \frac{3-y}{4}\in\mathbb{R} for every real number y, therefore. the function is onto.

Hence, the function is both one-one and onto.

Therefore, the function is bijective.

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

Solution

We are given

f(x)=1+x^2

where

f:\mathbb{R}\to\mathbb{R}

We will check whether the function is one-one and onto.

One-One (Injective):

Consider

f(2)=1+2^2=5

and

f(-2)=1+(-2)^2=5

Thus,

f(2)=f(-2)

but

2\neq-2

Therefore, the function is not one-one.

Onto (Surjective):

Since

x^2\geq0

for every x\in\mathbb{R} , we have

1+x^2\geq1

Thus, the range of the function is

[1,\infty)

which is not equal to the codomain \mathbb{R} .

For example, there is no real number x such that

1+x^2=0

Therefore, the function is not onto.

Hence, the function is neither one-one nor onto.

Important Note:

To prove that a function is not one-one, students often use a counterexample such as finding two different inputs having the same output.

However, the standard algebraic method is to start with

f(x_1)=f(x_2)

and simplify it.

If the result gives a unique solution

x_1=x_2

then the function is one-one.

But if we get more than one possibility, such as

x_1=x_2 \text{ or } x_1=-x_2

then the function is many-one because equal outputs are possible for different inputs.

Example: For f(x)=1+x^2 ,

f(x_1)=f(x_2)

gives

1+x_1^2=1+x_2^2

x_1^2=x_2^2

x_1=\pm x_2

Since this does not give only x_1=x_2 , the function is many-one.

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

Solution

We are given the function

f:A\times B\to B\times A

defined by

f(a,b)=(b,a)

We will show that the function is both one-one and onto.

One-One (Injective):

Suppose

f(a_1,b_1)=f(a_2,b_2)

Then

(b_1,a_1)=(b_2,a_2)

Equality of ordered pairs gives

b_1=b_2 \text{ and } a_1=a_2

Therefore,

(a_1,b_1)=(a_2,b_2)

Hence, the function is one-one.

Onto (Surjective):

Let (b,a)\in B\times A .

Consider the element (a,b)\in A\times B .

Then

f(a,b)=(b,a)

Thus, every element of B\times A has a pre-image in A\times B .

Therefore, the function is onto.

Since the function is both one-one and onto, it is bijective.

9. Let f:\mathbb{N}\to\mathbb{N} be defined by

f(n)=\begin{cases} \dfrac{n+1}{2}, & \text{if } n \text{ is odd} \\ \\ \dfrac{n}{2}, & \text{if } n \text{ is even} \end{cases}

for all n\in\mathbb{N} . State whether the function f is bijective. Justify your answer.

Solution

We are given the function

f:\mathbb{N}\to\mathbb{N}

defined by

f(n)=\begin{cases} \dfrac{n+1}{2}, & \text{if } n \text{ is odd} \\ \\ \dfrac{n}{2}, & \text{if } n \text{ is even} \end{cases}

We will check whether the function is one-one and onto.

One-One (Injective):

Consider

n=1

Since 1 is odd,

f(1)=\frac{1+1}{2}=1

Now consider

n=2

Since 2 is even,

f(2)=\frac{2}{2}=1

Thus,

f(1)=f(2)

but

1\neq2

Therefore, the function is not one-one.

Onto (Surjective):

Let y\in\mathbb{N} .

We show that there exists n\in\mathbb{N} such that

f(n)=y

Choose n=2y .

Since 2y is even,

f(2y)=\frac{2y}{2}=y

Thus, every natural number has a pre-image in \mathbb{N} .

Therefore, the function is onto.

Hence, the function is onto but not one-one.

Therefore, the function is not bijective.

10. Let A=\mathbb{R}-\{3\} and B=\mathbb{R}-\{1\} . Consider the function f:A\to B defined by

f(x)=\frac{x-2}{x-3}

Is f one-one and onto? Justify your answer.

Solution

We are given

f(x)=\frac{x-2}{x-3}

where

A=\mathbb{R}-\{3\}, \quad B=\mathbb{R}-\{1\}

We will check whether the function is one-one and onto.

One-One (Injective):

Suppose

f(x_1)=f(x_2)

Then

\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}

Cross-multiplying, we get

(x_1-2)(x_2-3)=(x_2-2)(x_1-3)

Expanding both sides,

x_1x_2-3x_1-2x_2+6=x_1x_2-3x_2-2x_1+6

Cancelling common terms, we get

-3x_1-2x_2=-3x_2-2x_1

Therefore,

x_2-x_1=0

or

x_1=x_2

Hence, the function is one-one.

Onto (Surjective):

Let y\in B . Then

y\neq1

We need to find x\in A such that

y=\frac{x-2}{x-3}

Cross-multiplying, we get

y(x-3)=x-2

yx-3y=x-2

or yx-x=3y-2

i.e. x(y-1)=3y-2

Since y\neq1 , we can divide by y-1 .

Thus,

x=\frac{3y-2}{y-1}

Now,

\frac{3y-2}{y-1}\neq3

because

\frac{3y-2}{y-1}=3

would give

3y-2=3y-3

which is impossible.

Therefore, x\in A .

Hence, for every y\in B , there exists x\in A such that

f(x)=y

Therefore, the function is onto.

Moreover,

for every y\neq1 , we obtained

x=\dfrac{3y-2}{y-1}

Now, substituting this value of x in the given function,

f\left(\dfrac{3y-2}{y-1}\right)=\dfrac{\left(\dfrac{3y-2}{y-1}\right)-2}{\left(\dfrac{3y-2}{y-1}\right)-3}

On simplifying, we get

f\left(\dfrac{3y-2}{y-1}\right)=y

Hence, for every y\in\mathbb{R}-\{1\} , there exists an x\in\mathbb{R}-\{3\} such that f(x)=y .

Therefore, the function is onto.

Since the function is both one-one and onto, it is bijective.

11. Let f:\mathbb{R}\to\mathbb{R} be defined as f(x)=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

We are given

f(x)=x^4

where

f:\mathbb{R}\to\mathbb{R}

We will check whether the function is one-one and onto.

One-One (Injective):

Consider

f(2)=2^4=16

and

f(-2)=(-2)^4=16

Thus,

f(2)=f(-2)

but

2\neq-2

Therefore, the function is not one-one.

Onto (Surjective):

For every real number x , we have

x^4\geq0

Thus, the range of the function contains only non-negative real numbers.

Negative real numbers are not images of any real number.

For example, there is no real number x such that

x^4=-1

Therefore, the function is not onto.

Hence, the function is neither one-one nor onto.

✅️ Therefore, the correct answer is (D).

12. Let f:\mathbb{R}\to\mathbb{R} be defined as f(x)=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

We are given

f(x)=3x

where

f:\mathbb{R}\to\mathbb{R}

We will check whether the function is one-one and onto.

One-One (Injective):

Suppose

f(x_1)=f(x_2)

Then

3x_1=3x_2

Dividing both sides by 3 , we get

x_1=x_2

Therefore, the function is one-one.

Onto (Surjective):

Let y\in\mathbb{R} .

We need to find x\in\mathbb{R} such that

f(x)=y

Now,

y=3x

gives

x=\frac{y}{3}

Since \frac{y}{3}\in\mathbb{R} for every real number y , the function is onto.

Hence, the function is both one-one and onto.

✅️ Therefore, the correct answer is (A).