If `f(x+y)=f(x)+f(y)-x y-1`, `AA x , `y in R `and f(1)=1,` then the number of solution of f(n)=n , `n in N ,` is

Options

A. 0
B. 1
C. 2
D. more than 2

Answer

B

Explanation

Given `f(x+y)=f(x)+f(y)=xy=1 AA x, y in R`
`f(1)=1,f(2)=f(1+1)=f(1)+f(1)-1-1=0`
`f(3)=f(2+1)=f(2)+f(1)-2xx1 -1= -2`
`f(n+1) = f(n)+f(1)-n-1=f(n)-n lt f(n)`
Thus, `f(1) gt f(2) gt f(3) gt …, and f(1) =1.`
Therefore, `f(1)=1 and f(n) gt 1, " for " n gt 1.`
Hence, ` f(n)=n, n in N,` has only one solution `n=1.`