Question

The function $f:N \to N$ (N is set of natural numbers defined by $f(n) = 2n + 3$ is
(A) Surjective only
(B) Injective only
(C) Bijective
(D) Neither one - one nor onto

Answer 1

Hint : To check one – one function we first let f(x) = f(y) and then simplify it if it ends with x = y then called one-one other not one-one. To check onto, we take a given function equal to y and then solve n, if there exist any value of y for which we can’t find any value of ‘n’ on $N \to N$ . Then we can say that function is not onto.

Complete step-by-step answer :
Given function is $f(n) = 2n + 3$ . It is also given that the function is defined on $N \to N$ , where N stands for Natural number.
For one – one function we let
$f(x) = f(y) \\\ \Rightarrow 2x + 3 = 2y + 3 \\\ \Rightarrow 2x = 2y \\\ \Rightarrow x = y \\\$
From above we see that for f(x) = f(y) we have x = y
Therefore, we can say that a given function is one – one or se say Injective.
Now, we will discuss whether given function is onto or not
For this we let f (n) = y
$\Rightarrow y = 2n + 3 \\\ \Rightarrow y - 3 = 2n \\\ \Rightarrow n = \dfrac{{y - 3}}{2} \\\$
From above we see that for y =1 there does not exist any value of $n \in N$ .
Therefore, from above we can say that given function $f(n) = 2n + 3$ is not onto or we can say that function is not Surjective.
Hence, from above we see that the given function $f(n) = 2n + 3$ defined on $N \to N$ only one-one but not onto.
Therefore from given four options, option (B) is the correct option.

So, the correct answer is “Option B”.

Note : While doing this type of problem we must be careful regarding which given function is defined on which system like N (natural number), W (whole numbers), I (integers) etc. As, the same problem leads to different answers due to the given system of function.