A function `f (x)` is defined for all `x in R` and satisfies, `f(x + y) = f (x) + 2y^2 + kxy AA x, y in R`, where `k` is a given constant. If `f(1) = 2 and f(2) = 8`, find `f(x)` and show that `f (x+y).f(1/(x+y))=k,x+y != 0`.

Options

A. f(0) = 0
B. f(0) cannot be determined
C. k = 2
D. k cannot be determined

Answer

A, C

Explanation

`f(x+y)-kxy=f(x)+2y^(2)" (1)"`
Put `y=-x`
`rArr" "f(0)=kx^(2)=f(x)+2x^(2)" (2)"`
`rArr" "f(x)=(k-2)x^(2)+f(0)`
Put x = 1
`therefore" "f(1)=(k-2)+f(0)`
`therefore" "k-2+f(0)=2`
`therefore" "k+f(0)=4" (3)"`
Put x = 1
`therefore" "f(2)=(k-2).4+f(0)`
`therefore" "4k+f(0)=16`
Solving we get 3k = 12 or k = 4 and f(0) = 0
`therefore" "f(x)=2x^(2)`