Let f:R+→R be a function which satisfy fxfy=fxy+21x+1y+1 for all x, y>0, then the value of f12 is

x=1, y=1, f(1)=−2, 3f(x)=2(2x+1)x(f(1)−1)⇒f(1)=−2, f(12)=−83

Let f:R+→R be a function which satisfy fxfy=fxy+21x+1y+1 for all x, y>0, then the value of f12 is

x=1, y=1, f(1)=−2, 3f(x)=2(2x+1)x(f(1)−1)⇒f(1)=−2, f(12)=−83

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Let $f : R^{+} \to R$ be a function which satisfy $f (x) f (y) = f (x y) + 2 (\frac{1}{x} + \frac{1}{y} + 1)$ for all $x, y > 0,$ then the value of $f (\frac{1}{2})$ is

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–4

$- \frac{8}{3}$

–5

answer is C.

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Detailed Solution

$x = 1, y = 1, f (1) = - 2, 3$

$f (x) = \frac{2 (2 x + 1)}{x (f (1) - 1)} \Rightarrow f (1) = - 2, f (\frac{1}{2}) = - \frac{8}{3}$

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Let f:R+→R be a function which satisfy fxfy=fxy+21x+1y+1 for all x, y>0, then the value of f12 is

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Let $f : R^{+} \to R$ be a function which satisfy $f (x) f (y) = f (x y) + 2 (\frac{1}{x} + \frac{1}{y} + 1)$ for all $x, y > 0,$ then the value of $f (\frac{1}{2})$ is