A function f- R - R is defined by fx-y - kxy - fx - 2y2 - x-y - R and f1 - 2- f2 - 8 where k is some constant- then fx-y-f1-x-y - x-y - 0

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Convexity

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Question

A function $f : R \to R$ is defined by $f (x + y) - k x y = f (x) + 2 y^{2} \forall x, y \in R$ and $f (1) = 2; f (2) = 8$ where k is some constant, then $f (x + y) . f (\frac{1}{x + y}) = (x + y \neq 0)$

1

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4

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k^{2}

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k^{2} + 4

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Solution

The correct option is B $4$
$f (x + y) - k x y = f (x) + 2 y^{2}$
$f (x + y) = f (x) + 2 y^{2} + k x y$
Put $x = 1, y = 1 ⟹ f (1 + 1) = f (1) + 2 + k ⟹ k = 4$
Put $x = 1, f (1 + y) = f (1) + 2 y^{2} + 4 (1) y = 2 (y + 1)^{2} ⟹ f (y) = 2 y^{2}$
$f (x + y) . f (\frac{1}{x + y}) = 2. (x + y)^{2} .2 (\frac{1}{x + y})^{2} = 4$

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A function f:R→R is defined by f(x+y)−kxy=f(x)+2y2∀x,y∈R and f(1)=2;f(2)=8 where k is some constant, then f(x+y).f(1x+y)=(x+y≠0)

The correct option is B 4f(x+y)−kxy=f(x)+2y2f(x+y)=f(x)+2y2+kxyPut x=1,y=1⟹f(1+1)=f(1)+2+k⟹k=4Put x=1,f(1+y)=f(1)+2y2+4(1)y=2(y+1)2⟹f(y)=2y2f(x+y).f(1x+y)=2.(x+y)2.2(1x+y)2=4

A function $f : R \to R$ is defined by $f (x + y) - k x y = f (x) + 2 y^{2} \forall x, y \in R$ and $f (1) = 2; f (2) = 8$ where k is some constant, then $f (x + y) . f (\frac{1}{x + y}) = (x + y \neq 0)$