Question

If f(x) = cos (log x), then the value of f(x²) f(y²) −

\[\frac{1}{2}\left\{ f\left( \frac{x^2}{y^2} \right) + f\left( x^2 y^2 \right) \right\}\] is

 

विकल्प

• (a) −2

• (b) −1

• (c) 1/2

• (d) None of these

   

Answer 1

(d) None of these

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

\[\Rightarrow f\left( x^2 \right) = \cos\left( \log\left( x^2 \right) \right)\]
\[ \Rightarrow f\left( x^2 \right) = \cos\left( 2\log\left( x \right) \right)\]

Similarly,

\[f\left( y^2 \right) = \cos\left( 2\log\left( y \right) \right)\]

Now,

 

\[f\left( \frac{x^2}{y^2} \right) = \cos\left( \log\left( \frac{x^2}{y^2} \right) \right) = \cos\left( \log x^2 - \log y^2 \right)\]and

\[f\left( x^2 y^2 \right) = \cos\left( \log x^2 y^2 \right) = \cos\left( \log x^2 + \log y^2 \right)\]

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

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