The correct answer is (D):
f, g : N – {1} → N defined as
f(a) = α, where α is the maximum power of those primes p such that pα divides a.
g(a) = a + 1,
Now, f(2) = 1, g(2) = 3 ⇒ (f + g) (2) = 4
f(3) = 1, g(3) = 4 ⇒ (f + g) (3) = 5
f(4) = 2, g(4) = 5 ⇒ (f + g) (4) = 7
f(5) = 1, g(5) = 6 ⇒ (f + g) (5) = 7
∵ (f + g) (5) = (f + g) (4)
∴ f + g is not one-one
Now, ∵ fmin = 1, gmin = 3
So, there does not exist any x ∈ N – {1} such that
(f + g)(x) = 1, 2, 3
∴ f + g is not onto
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