Function f : N → N, f(x) = 2x + 3 is(A) One-one onto(B) One-one into

1.	Function f : N → N, f(x) = 2x + 3 is(A) One-one onto(B) One-one into(C) Many one-onto(D) Many-one into
Answer» Answer is (B) Given: f : N → N and f(x) = 2x + 3 where N = set of natural numbers Let x₁, x₂ ∈ N is such that f(x₁) = f(x₂) f(x₁) = f(x₂) ⇒ 2x₁ + 3 = 2x₂ + 3 ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂ f(x₁) = f(x₂) ⇒ x₁ = x₂ ∀ x₁, x₂ ∈ N f is one-one function. Again, let y ∈ N (co-domain) if possible than let pre image of x is in domain N then f(x) = y f(x) = y ⇒ 2x + 3 = y ⇒ x = (y - 3)/2 ∈ N At y = 1, then x = (1 - 3)/2 = -1 ∉ N If this way, y has many values for which x is not exist in domain A. So, f is into function.

Function f : N → N, f(x) = 2x + 3 is(A) One-one onto(B) One-one into(C) Many one-onto(D) Many-one into

Answer»

Answer is (B)

Given: f : N → N and f(x) = 2x + 3

where N = set of natural numbers

Let x₁, x₂ ∈ N is such that f(x₁) = f(x₂)

f(x₁) = f(x₂)

⇒ 2x₁ + 3 = 2x₂ + 3

⇒ 2x₁ = 2x₂

⇒ x₁ = x₂

f(x₁) = f(x₂) ⇒ x₁ = x₂ ∀ x₁, x₂ ∈ N

f is one-one function.

Again, let y ∈ N (co-domain) if possible than let pre image of x is in domain N then f(x) = y

f(x) = y

⇒ 2x + 3 = y

⇒ x = (y - 3)/2 ∈ N

At y = 1, then x = (1 - 3)/2 = -1 ∉ N

If this way, y has many values for which x is not exist in domain A.

So, f is into function.