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Question Number 7781 by 314159 last updated on 15/Sep/16

Commented by prakash jain last updated on 15/Sep/16

$$f\left(1\right)=2$$$$f\left(2\right)=8$$$$f(x+y)-kxy=f\left(x\right)+2y^2$$$$x=0$$$$f\left(y\right)=f\left(0\right)+2y^2$$$$f\left(1\right)=2\Rightarrow 2=f\left(0\right)+2\Rightarrow f\left(0\right)=0$$$$f\left(y\right)=2y^2$$$$f\left(x\right)=2x^2$$$$f(x+y)=2{(x+y)}^2$$$$f(x+y)=2x^2+2y^2+4xy$$$$f(x+y)-4xy=2x^2+2y^2$$$$f(x+y)-4xy=f\left(x\right)+2y^2$$$$k=4$$$$f(x+y)f\left(\frac{1}{x+y}\right)=2{(x+y)}^2\times 2\frac{1}{{(x+y)}^2}=4=k$$

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