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.