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.