This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 2 Solution. Solutions for other problems are available at Exercise 1.2 Solutions
Exercise 1.2 Problem 2 Solution
2. Check the injectivity and surjectivity of the following functions:
i.
f : N → N given by {f(x) = x^2}
ii.
f : Z → Z given by {f(x) = x^2}
iii.
f : R → R given by {f(x) = x²}
iv.
f : N → N given by {f(x) = x^3}
v.
f : Z → Z given by {f(x) = x^3}
2.i Check the injectivity and surjectivity of f : N → N given by {f(x) = x²}
To check whether f is injective(one-one):
Consider two elements x_1, x_2 ∈ N such that
{f(x_1) = f(x_2)}
⇒ {x_1^2 = x_2^2}
⇒ {x_1 = \pm x_2}
As both x_1, x_2 ∈ N, we consider only {x_1 = x_2} and ignore {x_1 = -x_2} (as the set of natural numbers N does not contain any negative numbers)
⇒ {x_1 = x_2}
∴ f : N → N is injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^2}
For the function f to be surjective or onto, for every element y ∈ N in the co-domain, there should exist an element x ∈ N in the domain such that
{x = \pm\sqrt{y}}
As we know, every natural number y ∈ N is not a perfect square.
So, there exists some elements i.e. natural numbers which are not perfect squares in the co-domain which are not images of any elements in the domain.
For example, \sqrt{2} = 1.414 but 1.414 ∉ N.
⇒ Element 2 in the co-domain N is not an image of any element in the domain N.
∴ f is not surjective(onto).
∴ f is injective but not surjective
2.ii Check the injectivity and surjectivity of f : Z → Z given by {f(x) = x^2}
To check whether f is injective(one-one):
Consider two elements x_1, x_2 ∈ Z such that
{f(x_1) = f(x_2)}
⇒ {x_1^2 = x_2^2}
⇒ {x_1 = \pm x_2}
As both x_1, x_2 ∈ Z, we need to consider both {x_1 = x_2} and also {x_1 = -x_2} (as the set of integers Z contains both positive and negative numbers)
⇒ The image of x_1 is not unique/distinctive (In otherwords, two distinct elements x_1, x_2 have the same image)
∴ f : Z → Z is not injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^2}
For the function f to be surjective or onto, for every element y ∈ Z in the co-domain, there should exist an element x ∈ Z in the domain such that
{x = ±\sqrt{y}}
As we know, every positive integer y ∈ Z is not a perfect square. Also, all the negative integers are not squares of any integers.
⇒ There exists some elements in the co-domain i.e. positive integers which are not perfect squares as well as all the negative integers, which are not images of any element in the domain.
∴ f is not surjective(onto).
∴ f is neither injective nor surgective.
2.iii Check the injectivity and surjectivity of f : R → R given by {f(x) = x^2}
To check whether f is injective(one-one):
Consider two elements x_1, x_2 ∈ R such that
{f(x_1) = f(x_2)}
⇒ {x_1^2 = x_2^2}
⇒ {x_1 = \pm x_2}
As both x_1, x_2 ∈ R, we need to consider both {x_1 = x_2} and also {x_1 = -x_2} (as the set of real numbers R contains both positive and negative numbers)
⇒ The image of x_1 is not unique/distinctive (In otherwords, two distinct elements x_1 and x_2 in the domain have can have the same image in the co-domain)
∴ f : R → R is not injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^2}
For the function f to be surjective or onto, for every element y ∈ R in the co-domain, there should exist an element x ∈ R in the domain such that
{x = \pm\sqrt{y}}
As we know, all the negative real numbers are not squares of any other real numbers.
⇒ There exists some elements i.e. negative real numbers in the co-domain which are not images of any element in the domain.
∴ f is not sujective(onto).
∴ f is neither injective nor surjective.
2.iv Check the injectivity and surjectivity of f : N → N given by {f(x) = x^3}
To check whether f is injective(one-one):
Consider two elements x_1, x_2 ∈ N such that
{f(x_1) = f(x_2)}
⇒ {x_1^3 = x_2^3}
⇒ {x_1 = x_2}
∴ f : N → N is injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^3}
For the function f to be onto, for every element y ∈ N in the co-domain, there should exist an element x ∈ N in the domain such that
{x = \sqrt[3]{y}}
As we know, every natural number y ∈ N is not a perfect cube.
So, there exists some elements i.e. natural numbers which are not perfect cubes in the co-domain which are not images of any natural number in the domain.
For example, \sqrt[3]{2} ≅ 1.26 but 1.26 ∉ N.
⇒ Element 2 in the co-domain N is not an image of any element in the domain N.
∴ f is not surjective(onto).
∴ f is injective but not surjective.
2.v Check the injectivity and surjectivity of f : Z → Z given by {f(x) = x^3}
To check whether f is injective(one-one):
Consider two elements x_1, x_2 ∈ Z such that
{f(x_1) = f(x_2)}
⇒ {x_1^3 = x_2^3}
⇒ {x_1 = x_2}
∴ f : Z → Z is injective or one-one.
To check whether f is surjective(onto):
The function f is defined as {f(x) = x^3}
For the function f to be surjective or onto, for every element y ∈ Z in the co-domain, there should exist an element x ∈ Z in the domain such that
{x = \sqrt[3]{y}}
As we know, every integer y ∈ Z is not a perfect cube.
So, there exists some elements i.e. natural numbers which are not perfect cubes in the co-domain which are not images of any natural number in the domain.
For example, \sqrt[3]{2} ≅ 1.26 but 1.26 ∉ Z.
⇒ Element 2 in the co-domain N is not an image of any element in the domain N.
∴ f is not surjective(onto).
∴ f is injective but not surjective.
