Let f, g: ℕ-{1} → ℕ be functions defined by f(a)=α, where α is the maximum of the powers of those primes p such that p^α divides a, and g(a)=a+1, for all a ∈ ℕ-{1}. Then, the function f+g is

Solution

Maths sir · 4 steps

Try the first few values of a: a repeat kills one-one, and a missing number kills onto.

1. Read f properly. Write a as a product of prime powers. Then f(a) is the largest exponent in that factorisation, so f(12) = f(2² × 3) = 2. Every a ≥ 2 has at least one prime factor, so f(a) ≥ 1.

   a	Factorisation	f(a)	g(a) = a + 1	(f+g)(a)
   2	2¹	1	3	4
   3	3¹	1	4	5
   4	2²	2	5	7
   5	5¹	1	6	7
   6	2¹ × 3¹	1	7	8

2. Not one-one. (f+g)(4) = 7 and (f+g)(5) = 7. Two different inputs give the same output, so f+g is not one-one.
3. Not onto. Since a ≥ 2, we have g(a) ≥ 3 and f(a) ≥ 1, so (f+g)(a) ≥ 4. The values 1, 2 and 3 in the codomain ℕ are never reached, so f+g is not onto.
4. So. f+g is neither one-one nor onto, which is option (d).

The idea to keep. For a strange-looking function, first tabulate the first few inputs. One repeated output disproves one-one, and one codomain value that is never reached disproves onto.

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