## 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 x_{1}, x_{2} ∈ Q are in this way f(x_{1}) = f(x_{2})

f(x_{1}) = f(x_{2})

⇒ 3x_{1} + 7 = 3x_{2} + 7

⇒ 3x_{1} = 3x_{2}

⇒ x_{1} = x_{2}

⇒ f(x_{1}) = f(x_{2})

⇒ x_{1} = x_{2}

x_{1}, x_{2} ∈ 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 = [latex]\frac { y-7 }{ 3 }[/latex] ∈ 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.

Use our Domain and Range Calculator tool to get the domain and range for your function. Also, get steps to check the domain and range for any type of 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 x_{1}, x_{2} ∈ R

If f(x_{1}) = f(x_{2}) ⇒ sin x_{1} = sin x_{2}

⇒ x_{1} = nπ + (-1)^{n}x_{2}, n ∈ I (General value of sin θ)

⇒ x_{1} ≠ x_{2}

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^{-1}y

So, -1 ≤ y ≤ 1

Many value of sin^{-1}y 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 x_{1}, x_{2} ∈ N

If f(x_{1}) = f(x_{2}) ⇒ |x_{1}| = |x_{2}| [∵ x_{1} > 0, x_{2} > 0, x_{1}, x_{2} ∈ N]

⇒ x_{1} = x_{2}

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) = [latex]\frac { x }{ 2 }[/latex]

(ii) g(x) = |x|

(iii) h(x) = x^{2}

(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 x_{1}, x_{2} ∈ A

If f(x_{1}) = f(x_{2})

⇒ g(x_{1}) = g(x_{2})

⇒ |x_{1}| = |x_{2}|

⇒ x_{1} = ±x_{2}

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) = x^{2}

Let x_{1}, x_{2} ∈ A.

Thus, h(x_{1}) = h(x_{2})

h (x_{1}) = h (x_{2})

⇒ x_{1}^{2} = x_{2}^{2}

⇒ x_{1} = ± x_{2}

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

⇒ x^{2} = 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 x_{1} + iy_{1} and x_{2} + iy_{2} ∈ C thus.

f(x_{1} + iy_{1}) = f (x_{2} + iy_{2})

⇒ x_{1} – iy_{1} = x_{2} – iy_{2}

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 : R_{0} → R+, f(x) = x^{2}

(iii) f : z_{0} → N, f(x) = |x|

(iv) f : Z → Z, f(x) = 2x

(v) f : R → R_{1}, f(x) = x^{2}

(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 x_{1}, x_{2} ∈ R

Thus, f(x_{1}) = f(x_{2})

f(x_{1}) = f(x_{2})

⇒ cos x_{1} = cos x_{2}

⇒ x_{2} = 2nπ ± x_{2}, x_{1} ∈ 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^{-1}y

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 f_{1}, f_{2} are function defined as

f1 : N → O, f_{1}(x) = 2x – 1; f_{2} : N → E, f_{2}(x) = 2x

Then prove that f_{1} and f_{2} are one-one onto.

Solution:

(i) Given N = {1, 2, 3, …}

O = {1, 3, 5, …}

Function f(x) = 2x – 1

Let x_{1}, x_{2} ∈ N

Thus f(x_{1}) = f(x_{2})

f(x_{1}) = f(x_{2})

⇒ 2x_{1} – 1 = 2x_{2} – 1

⇒ 2x_{1} = 2x_{2}

⇒ x_{1} = x_{2} ∀ x_{1}, x_{2} ∈ N

So, f_{1} 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 = [latex]\frac { y+1 }{ 2 }[/latex] ∈ N ∀ y ∈ O

So, pre-image of every element of O (co-domain) exist in domain N

So, f_{1} is onto function.

Hence, f is one-one onto function

(ii) Given set are

N = {1, 2, 3, …}

E = {2, 4, 6, …}

Function f_{2} : N → E, f_{2}(x) = 2x ∀ x ∈ N

Let x_{1}, x_{2} ∈ N

Thus f_{2}(x_{1}) = f_{2}(x_{2})

f_{2}(x_{1}) = f_{2}(x_{2})

⇒ 2x_{1} = 2x_{2}

⇒ x_{1} = x_{2} ∀ a, b ∈ N

f_{2} is one-one function.

Again, let y ∈ E (co-domain), if possible, let pre-image of y is x exist in domain N then f_{2}(x) = y

f_{2}(x) = y

⇒ 2x = y

⇒ x = [latex]\frac { y }{ 2 }[/latex] ∈ N ∀ y ∈ E

So, pre-image of every element of E (co-domain) exist in domain N.

Hence, f_{2} 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) = x^{2}

(ii) f(x) = x^{3}

(iii) f(x) = x^{3} + 3

(iv) f(x) = x^{3} – x

Solution:

(i) Given, f : R → R, f(x) = x^{2}

where, R is set of real number

Let x_{1}, x_{2} ∈ R

Thus f (x_{1}) = f (x_{2})

f(x_{1}) = f(x_{2})

⇒ x_{1}^{2} = x_{2}^{2}

⇒ x_{1} = ±x_{2}

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

⇒ x^{2} = 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) = x^{3}

Let x_{1}, x_{2} ∈ R

Thus f(x_{1}) = f(x_{2})

f(x_{1}) = f(x_{2})

⇒ x_{1}^{3} = x_{2}^{3}

⇒ x_{1} = x_{2} ∀ x_{1}, x_{2} ∈ 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

⇒ x^{3} = 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) = x^{3} + 3

Where R is set of real number

Let x_{1}, x_{2} ∈ R

Thus f (x_{1}) = f(x_{2})

f(x_{1}) = f(x_{2})

⇒ x_{1}^{3} + 3 = x_{2}^{2} + 3

⇒ x_{1}^{3} = x_{2}^{3} ∀ x_{1}, x_{2} ∈ 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

⇒ x^{3} + 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) = x^{3} – 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) = 1^{3} – 1 = 0

0 ≠ 1 and f(0) = f(1) = 1

So, f is many-one function.

Again, range of f = f(R) = {x^{3} – x : x ∈ R) = R (domain)

So, f is onto function

Hence, f is many-one, onto function.

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