Question

If f(x) = cos (log x), then the value of f(x) f(y) −\[\frac{1}{2}\left\{ f\left( \frac{x}{y} \right) + f\left( xy \right) \right\}\] is

 

विकल्प

• (a) −1

• (b) 1/2

• (c) −2

• (d) None of these

   

Answer 1

(d) None of these

Given:

\[f\left( x \right) = \cos\left( \log x \right)\]

∴ \[f\left( y \right) = \cos\left( \log y \right)\]

 

Now,

\[f\left( \frac{x}{y} \right) = \cos\left( \log\left( \frac{x}{y} \right) \right) = \cos\left( \log x - \log y \right)\] and

\[f\left( xy \right) = \cos\left( \log xy \right) = \cos\left( \log x + \log y \right)\]  \[\Rightarrow f\left( \frac{x}{y} \right) + f\left( xy \right) = \cos\left( \log x - \log y \right) + \cos\left( \log x + \log y \right)\]
\[ \Rightarrow f\left( \frac{x}{y} \right) + f\left( xy \right) = 2\cos\left( \log x \right)\cos\left( \log y \right)\]
\[ \Rightarrow \frac{1}{2}\left[ f\left( \frac{x}{y} \right) + f\left( xy \right) \right] = \cos\left( \log x \right)\cos\left( \log y \right)\] \[\Rightarrow f\left( x \right)f\left( y \right) - \frac{1}{2}\left\{ f\left( xy \right) + f\left( \frac{x}{y} \right) \right\} = \cos\left( \log x \right)\cos\left( \log y \right) - \cos\left( \log x \right)\cos\left( \log y \right) = 0\]

\[\Rightarrow f\left( x \right)f\left( y \right) - \frac{1}{2}\left\{ f\left( xy \right) + f\left( \frac{x}{y} \right) \right\} = \cos\left( \log x \right)\cos\left( \log y \right) - \cos\left( \log x \right)\cos\left( \log y \right) = 0\]