Let N be the set of natural numbers and let \(f : N \to N\) be defined by
\( \displaystyle f(n) = \begin{cases} n + 3, & \text{if } n \text{ is odd} \\ n - 1, & \text{if } n \text{ is even} \end{cases} \)
Then f is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → one-one but not onto
Listing the images by type of input:
odd n = 1, 3, 5, … give \(n + 3\) = 4, 6, 8, … (even numbers from 4 on)
even n = 2, 4, 6, … give \(n - 1\) = 1, 3, 5, … (all odd numbers)
One-one: an odd input always gives an even image and an even input always gives an odd image, so the two groups never share an image. Inside each group,
\(p + 3 = q + 3\) gives \(p = q\)
\(p - 1 = q - 1\) gives \(p = q\)
Not onto: 2 is not an image. It is even, so it could only come from an odd n, but then
\(n + 3 = 2\) gives \(n = -1\)
which is not a natural number. So f is one-one but not onto.