(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 = (y -7)/3 ∈ 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.