Let a function f : N → N be defined by
f(n) = [ 2n,, n = 2,4,6,8,......; n - 1,, n = 3,7,11,15,......; (n + 1)/2,, n = 1,5,9,13,......
then, f is

Solution

Maths sir · 4 steps

See which outputs each rule produces: odd numbers, 2 more than a multiple of 4, and multiples of 4.

1. Split the inputs. Every natural number falls in exactly one of the three cases: even (n = 2k), n = 4k - 1, or n = 4k - 3, with k = 1, 2, 3, …

   Inputs	Rule	Output in terms of k	Outputs you get
   n = 2k: 2, 4, 6, …	2n	4k	4, 8, 12, … (all multiples of 4)
   n = 4k - 1: 3, 7, 11, …	n - 1	4k - 2	2, 6, 10, … (all even numbers that are not multiples of 4)
   n = 4k - 3: 1, 5, 9, …	(n + 1)/2	2k - 1	1, 3, 5, … (all odd numbers)

2. One-one. Inside each case, different k give different outputs. The three output sets are odd numbers, even non-multiples of 4, and multiples of 4, so no two cases share an output. Hence no two inputs map to the same value.
3. Onto. Together the three output sets cover every natural number: every number is odd, or even but not a multiple of 4, or a multiple of 4. So every element of ℕ is hit.
4. So. f is both one-one and onto, which is option (d).

The idea to keep. For a piecewise function, check two things: each piece is one-one on its own, and the output sets of the pieces do not overlap. If those sets also fill the codomain, the function is a bijection.

Answer from the NTA final answer key. Solution by PrepWiser.