If log42fx1-fx=x then f2010+f-2009 is equal to

log42fx1-fx=x⇒2fx1-fx=4x 1Replacing x by 1-x, we get 2f1-x1-f1-x=41-x 2Multiplying (1) and (2), we get2fx1-fx2f1-x1-f1-x=41-x⇒fxf1-x=1-fx-f1-x+fxf1-x⇒fx+f1-x=1Putting x=2010, we getf2010+f-2009=1

If log42fx1-fx=x then f2010+f-2009 is equal to

log42fx1-fx=x⇒2fx1-fx=4x 1Replacing x by 1-x, we get 2f1-x1-f1-x=41-x 2Multiplying (1) and (2), we get2fx1-fx2f1-x1-f1-x=41-x⇒fxf1-x=1-fx-f1-x+fxf1-x⇒fx+f1-x=1Putting x=2010, we getf2010+f-2009=1

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If $\log_{4} (\frac{2 f (x)}{1 - f (x)}) = x$ then $(f (2010) + f (- 2009))$ is equal to

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Detailed Solution

$\begin{array}{l} \log_{4} (\frac{2 f (x)}{1 - f (x)}) = x \\ \Rightarrow \frac{2 f (x)}{1 - f (x)} = 4^{x} (1) \\ Replacing x by 1 - x, we get \\ \frac{2 f (1 - x)}{1 - f (1 - x)} = 4^{1 - x} (2) \end{array}$

Multiplying (1) and (2), we get

$\begin{array}{l} \frac{2 f (x)}{1 - f (x)} (\frac{2 f (1 - x)}{1 - f (1 - x)}) = 4^{1 - x} \\ \Rightarrow f (x) f (1 - x) = 1 - f (x) - f (1 - x) + f (x) f (1 - x) \\ \Rightarrow f (x) + f (1 - x) = 1 \\ Putting x = 2010, we get \\ f (2010) + f (- 2009) = 1 \end{array}$