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

A function `f (x)` is defined for all `x in R` and satisfies, `f(x + y

1.	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`.A. f(0) = 0B. f(0) cannot be determinedC. k = 2D. k cannot be determined
Answer» Correct Answer - A::C `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)`