If F(X) = Cos (Log X), Then the Value of F(X) F(Y) − 1 2 { F ( X Y ) + F ( X Y ) } Is(A) −1 (B) 1/2 (C) −2 (D) None of These - Concept of Functions

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If f(x) = cos (log x), then the value of f(x) f(y) &minus;$$\frac{1}{2}\left\{ f\left( \frac{x}{y} \right) + f\left( xy \right) \right\}$$ is

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(d) None of these Given: $$f\left( x \right) = \cos\left( \log x \right)$$ &there4; $$f\left( y \right) = \cos\left( \log y \right)$$ Now, $$f\left( \frac{x}{y} \right) = \cos\left( \log\left( \frac{x}{y} \right) \right) = \cos\left( \log x - \log y \right)$$ and $$f\left( xy \right) = \cos\left( \log xy \right) = \cos\left( \log x + \log y \right)$$ $$\Rightarrow f\left( \frac{x}{y} \right) + f\left( xy \right) = \cos\left( \log x - \log y \right) + \cos\left( \log x + \log y \right)$$$$ \Rightarrow f\left( \frac{x}{y} \right) + f\left( xy \right) = 2\cos\left( \log x \right)\cos\left( \log y \right)$$$$ \Rightarrow \frac{1}{2}\left[ f\left( \frac{x}{y} \right) + f\left( xy \right) \right] = \cos\left( \log x \right)\cos\left( \log y \right)$$ $$\Rightarrow f\left( x \right)f\left( y \right) - \frac{1}{2}\left\{ f\left( xy \right) + f\left( \frac{x}{y} \right) \right\} = \cos\left( \log x \right)\cos\left( \log y \right) - \cos\left( \log x \right)\cos\left( \log y \right) = 0$$ $$\Rightarrow f\left( x \right)f\left( y \right) - \frac{1}{2}\left\{ f\left( xy \right) + f\left( \frac{x}{y} \right) \right\} = \cos\left( \log x \right)\cos\left( \log y \right) - \cos\left( \log x \right)\cos\left( \log y \right) = 0$$

If F(X) = Cos (Log X), Then the Value of F(X) F(Y) − 1 2 { F ( X Y ) + F ( X Y ) } Is(A) −1 (B) 1/2 (C) −2 (D) None of These - Mathematics

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