If `f(x)` is a real valued functions satisfying `f(x+y) = f(x) +f(y) -yx -1` for all `x, y in R` such that `f(1)=1` then the number of solutions of `f(n) = n,n in N`, is

Options

A. 1
B. 2
C. 3
D. 4

Answer

A

Explanation

We have,
`f(x+y) = f(x) + f(y) -xy -1` for all `x,y in R`
`rArr f(x+1) = f(x) +f(x) + f(1)-x-1` for all `x in R` [Putting y=1]
`rArr f (x+1)= f(x) -x ` for all `x in R`
`rArr f(x+1) = f(n) -n` for all `n in N`
`rArr f (n+1) le f(n)` for all `nin N`
`rArr f(n+1) le f(n) le f(n-1) le .........le f(2) lt f(1) =1`
`rArr f (n+1) le 1 " "`for all `n in N`
But it is given that `f(1)=1`
Thus, we have
`f(1) = 1` and `f(n) lt 1` for n = 2,3.......
Hence`,f(n)=n ` has only one solution, namely n =1