# 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

MCQ

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

#### Options

• (a) −1

• (b) 1/2

• (c) −2

• (d) None of these

#### Solution

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Q 5 | Page 43