given : f : N → N such that f (n) = (n+5)2 , n ∊ N Now, For one - one function Let f(n1) = f(n2),n1,n2 ∊ N ⇒ (n1+5)2 = (n2+5)2 ⇒ n12 + 25 + 10n1 = n22 + 25 + 10n2 ⇒ (n1−n2)(n1+n2+10) = 0 ⇒ n1 = n2 ∴ f is one - one function if we put n = 1 , 2 , 3 , 4 , ... ∞ , we get f (1) = (1+5)2 = 36 f (2) = (5+2)2 = 49 f (3) = (3+5)2 = 64 ... etc ⇒ Range of f = (36 , 49 , 64 , ... ∞) ⊂ N ∴ f is not an onto function. Hence, f is one-one but not onto. Therefore, option 'D' is correct.