Question

If f(x + y) = f(x) + f(y) − xy − 1 ∀x, y ∈ R and f(1) = 1, then the number of solutions of the equation f(n) = n, n ∈ N is

(A)  0 

(B)  1 

(C)  2 

(D)  n

Answer 1

Correct option  (B) 1 

Explanation :

(x + y) = f(x) + f(y) − xy − 1 ∀x, y ∈ R

 Put x = y = 1, f(2) =2 f(1) – 2 = 0

Put x = 1, y = 2, f(3) = f(1) + f(2) – 2 – 1 = 3 – 3 = 0

Put x = 1, y = 3, f(4) = f(1) + f(3) – 3 – 1 = –3

Hence, only one solution for f(n) = n.