Rajasthan Board RBSE Class 11 Maths Chapter 2 Relations and Functions Ex 2.4

Question 1.
Classify the following functions is the form of one-one, many one, into and onto, also give reason to support your answer.
(i) f : Q → Q, f(x) = 3x + 7
(ii) f : C → R, f(x + iy) = x
(iii) f : if R → [-1, 1], f(x) = sin x
(iv) f : N → Z, f(x) = |x|
Solution:
(i) Given f : Q → Q and f(x) = 3x + 7
where Q is set of rational numbers.
Let x₁, x₂ ∈ Q are in this way f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ 3x₁ + 7 = 3x₂ + 7
⇒ 3x₁ = 3x₂
⇒ x₁ = x₂
⇒ f(x₁) = f(x₂)
⇒ x₁ = x₂
x₁, x₂ ∈ Q
Hence, f is one-one function.
Now, Let y ∈ Q co-domain
If possible, then let pre-image of y is x in domain Q then
f(x) = y
⇒ 3x + 7 = y
⇒ x = ∈ Q
So, every element in the co-domain of Q has pre-image in Q
Hence, f is onto function.
Thus, f is a one-one, onto function.

(ii) f : C → R : f(x + iy) = x
Given : function
f : C → R and f(x + iy) = x
Here C = set of complex numbers
R = set of real numbers
Let x + iy and x – iy (y ≠ 0) are different elements in domain C.
f(x + iy) = x and f (x – iy) = x
⇒ f(x + iy) = f(x – iy)
So, two different elements of domain R have the same image.
So, f is a many-one function.
Here Range of f = (x : x + iy ∈ C} – R Co-domain
[Range of x + iy in x ∈ R, y ∈ R and i = √-1 ]
f is a onto function.
Thus, f is a many-one, onto function.

(iii) f : R → [-1, 1], f(x) = sin x
Given: f : R → [-1, 1] and f(x) = sinx
where R is set of real numbers.
Let x₁, x₂ ∈ R
If f(x₁) = f(x₂) ⇒ sin x₁ = sin x₂
⇒ x₁ = nπ + (-1)ⁿx₂, n ∈ I (General value of sin θ)
⇒ x₁ ≠ x₂
f is not one-one function so, it is many-one function
Again, Let y ∈ Y
If possible then let pre-image of y under f is x then
f(x) = y
⇒ sin x = y
⇒ x = sin^-1y
So, -1 ≤ y ≤ 1
Many value of sin^-1y present in R, then
x ∈ R ∀ y ∈ [-1, 1]
f is onto function.
Thus, f is many-one, onto function.

(iv) f : N → Z, f(x) = |x|
Given f : N → Z and f(x) = |x|
where N = Set of natural number = {1, 2, 3, 4,…}
and Z = set of natural numbers = {0, ± 1, ± 2, ± 3, …}
Let x₁, x₂ ∈ N
If f(x₁) = f(x₂) ⇒ |x₁| = |x₂| [∵ x₁ > 0, x₂ > 0, x₁, x₂ ∈ N]
⇒ x₁ = x₂
f is one-one function.
Again ∵ Range of f = {|x| : x ∈ N], Z ≠ Z (co-domain)
So, f is not onto function.
Hence, f is into function.
Thus, f is one-one into function.

Question 2.
If A = {x | -1 ≤ x ≤ 1} = B, then find out which function is one-one, into or one-one onto defined from A to B
(i) f(x) =
(ii) g(x) = |x|
(iii) h(x) = x²
(iv) k(x) = sinπx
Solution:
Given equation
A = {x : -1 ≤ x ≤ 1)
and B ={x : -1 ≤ x ≤ 1}
(i) Function is defined from A to B.

f is not onto function.
Hence, it is proved that f is one-one, into function,

(ii) g : A → B, g (x) = |x|
Let x₁, x₂ ∈ A
If f(x₁) = f(x₂)
⇒ g(x₁) = g(x₂)
⇒ |x₁| = |x₂|
⇒ x₁ = ±x₂
g is not one-one function so, g is many one function.
Again, range of g = { x : -1 ≤ x ≤ 1} ≠ B (co-domain)
i.e., pre-image of negative number is not exist in codomain B.
Thus, it is proved that f is many-one, into function.
g is not one-one function.

(iii) h : A → B, h(x) = x²
Let x₁, x₂ ∈ A.
Thus, h(x₁) = h(x₂)
h (x₁) = h (x₂)
⇒ x₁² = x₂²
⇒ x₁ = ± x₂
h is not one-one function.
For example 1 ≠ -1 but h(1) = h(-1) = 1
So, h is a many-one function.
Let y ∈ B if possible pre-image of y is x exist in B, then
h(x) = y
⇒ x² = y
⇒ x = ±√y
when y is positive then, x does not exist
i.e., pre-image of a negative number does not exist.
So, Range of h ={x : 0 ≤ x ≤ 1} ≠ B (co-domain)
So, h is a into function
Hence, it is proved that h is many-one, into function,

(iv) k : A → B, k(x) = sin πx
-1, 1 ∈ A are such numbes
-1 ≠ 1
But k(-1) = sin (-π) = 0
⇒ k(1) = sin π = 0
So, -1 ≠ 1
⇒ k(-1) = k(1)
i.e., different elements have same image.
k is many-one function.
Again, range of k = {sin πx : x ∈ A} = {x : -1 ≤ x ≤ 1} = B (co-domain)
k is onto function
Hence, it is proved that k is many-one, onto function.

Question 3.
If f : C → C, f(x + iy) = (x – iy), then prove that f is an one-one onto function.
Solution:
Given : f : C → C and f(x + iy) = x – iy
where C is set of complex numbers.
Let x₁ + iy₁ and x₂ + iy₂ ∈ C thus.
f(x₁ + iy₁) = f (x₂ + iy₂)
⇒ x₁ – iy₁ = x₂ – iy₂


So, f is one-one function.
Range of f = {x – iy : x + iy ∈ C} = C (co-domain)
f is onto function.
Hence, f is one-one, onto function.
Hence Proved.

Question 4.
Give one example of each of the following function:
(i) One-one into
(ii) Many-one onto
(iii) Onto but not one-one
(iv) One-one but not onto
(v) Neither one-one nor onto
(vi) One-one onto
Solution:
(i) f : N → N, f(x) = 2x
(ii) f : R₀ → R+, f(x) = x²
(iii) f : z₀ → N, f(x) = |x|
(iv) f : Z → Z, f(x) = 2x
(v) f : R → R₁, f(x) = x²
(vi) f : Z → f(z) = -x

Question 5.
Prove that f : R → R, f(x) = cos x is a many- one into function. Change the domain and co-domain of f such that I become:
(i) One-one into
(ii) Many-one onto
(iii) One-one onto
Solution:
Given, f : R → R, f(x) = cos x
Let x₁, x₂ ∈ R
Thus, f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ cos x₁ = cos x₂
⇒ x₂ = 2nπ ± x₂, x₁ ∈ I
f is not one-one function.
Again, let, y ∈ R (co-domain), if possible let pre-image of y is x is domain R, then
f(x) = y
⇒ cos x = y
⇒ x = cos^-1y
x only exist when,
-1 ≤ y ≤ 1
when y ∈ R – [-1, 1]
Pre-image of y is not present in domain R.
So, f is not onto function
Hence, f is many-one, into function.
(i) One-one, into function
f : [0, π] → R, f(x) = cos x
(ii) Many-one, onto function
f : R → [-1, 1], f(x) = cos x
(iii) One-one, onto function.
f : [0, π] → [-1, 1], f(x) = cos x

Question 6.
If N= {1, 2, 3, 4, …), O = (1, 3, 5, 7, …}, E = (2, 4, 6, 8,…..) and f₁, f₂ are function defined as
f1 : N → O, f₁(x) = 2x – 1; f₂ : N → E, f₂(x) = 2x
Then prove that f₁ and f₂ are one-one onto.
Solution:
(i) Given N = {1, 2, 3, …}
O = {1, 3, 5, …}
Function f(x) = 2x – 1
Let x₁, x₂ ∈ N
Thus f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ 2x₁ – 1 = 2x₂ – 1
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂ ∀ x₁, x₂ ∈ N
So, f₁ is one-one function.
Again, let y ∈ O (co-domain), if possible let pre-image of y is x exist in domain N, then f(x) = y.
f(x) = y
⇒ 2x – 1 = y
⇒ x = ∈ N ∀ y ∈ O
So, pre-image of every element of O (co-domain) exist in domain N
So, f₁ is onto function.
Hence, f is one-one onto function

(ii) Given set are
N = {1, 2, 3, …}
E = {2, 4, 6, …}
Function f₂ : N → E, f₂(x) = 2x ∀ x ∈ N
Let x₁, x₂ ∈ N
Thus f₂(x₁) = f₂(x₂)
f₂(x₁) = f₂(x₂)
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂ ∀ a, b ∈ N
f₂ is one-one function.
Again, let y ∈ E (co-domain), if possible, let pre-image of y is x exist in domain N then f₂(x) = y
f₂(x) = y
⇒ 2x = y
⇒ x = ∈ N ∀ y ∈ E
So, pre-image of every element of E (co-domain) exist in domain N.
Hence, f₂ is one-one, onto function.

Question 7.
If function f is defined from set of real numbers R to R in the following way then classify them in the form of one-one, many-one, into or onto.
(i) f(x) = x²
(ii) f(x) = x³
(iii) f(x) = x³ + 3
(iv) f(x) = x³ – x
Solution:
(i) Given, f : R → R, f(x) = x²
where, R is set of real number
Let x₁, x₂ ∈ R
Thus f (x₁) = f (x₂)
f(x₁) = f(x₂)
⇒ x₁² = x₂²
⇒ x₁ = ±x₂
f is not one-one function. So, f is many-one function.
Let y ∈ R (co-domain), if possible, let pre-image of y is x in domain R, then f(x) = y
f (x) = y
⇒ x² = y
⇒ x = ±√y
If is clear, x is not exist for negative values of y.
So, f is into function.
Hence, f is many-one into function.

(ii) Given: f : R → R and f(x) = x³
Let x₁, x₂ ∈ R
Thus f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ x₁³ = x₂³
⇒ x₁ = x₂ ∀ x₁, x₂ ∈ R
f is one-one function
Again, let y ∈ R (co-domain), if possible, let pre-image ofy is x in domain R, then
f (x) = y
f(x) = y
⇒ x³ = y
⇒ x = y^1/3 ∈ R
So, f is onto function.
Hence, f is one-one onto function.

(iii) Given: f : R → R and f(x) = x³ + 3
Where R is set of real number
Let x₁, x₂ ∈ R
Thus f (x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ x₁³ + 3 = x₂² + 3
⇒ x₁³ = x₂³ ∀ x₁, x₂ ∈ R
So, f is one-one function.
Again let y ∈ R (co-domain)
If possible, let pre-image of y is x in f then
f (x) = y
⇒ x³ + 3 = y
⇒ x = (y – 3)^1/3 ∈ R ∀ y ∈ R
So, pre-image of every element R (co-domain) is exist in domain R.
Hence, f is one-one, onto function.

(iv) Given: f : R → R, f(x) = x³ – x
Where R is set of real numbers.
Let 0, 1 ∈ R are such numbers
0 ≠ 1
0 ∈ R ⇒ f(0) = 0 – 0 = 0
1 ∈ R ⇒ f(1) = 1³ – 1 = 0
0 ≠ 1 and f(0) = f(1) = 1
So, f is many-one function.
Again, range of f = f(R) = {x³ – x : x ∈ R) = R (domain)
So, f is onto function
Hence, f is many-one, onto function.