# If F(X) = Cos (Loge X), Then F ( 1 X ) F ( 1 Y ) − 1 2 { F ( X Y ) + F ( X Y ) } is Equal To(A) Cos (X − Y) (B) Log (Cos (X − Y)) (C) 1 (D) Cos (X + Y) - Mathematics

MCQ

If f(x) = cos (loge x), then $f\left( \frac{1}{x} \right)f\left( \frac{1}{y} \right) - \frac{1}{2}\left\{ f\left( xy \right) + f\left( \frac{x}{y} \right) \right\}$ is equal to

#### Options

• (a) cos (x − y)

• (b) log (cos (x − y))

• (c) 1

• (d) cos (x + y)

• (e) 0

#### Solution

Given:

$f\left( x \right) = \cos\left( \log_e x \right)$
$\Rightarrow f\left( \frac{1}{x} \right) = \cos\left( \log_e \left( \frac{1}{x} \right) \right)$
$\Rightarrow f\left( \frac{1}{x} \right) = \cos\left( - \log_e \left( x \right) \right)$
$\Rightarrow f\left( \frac{1}{x} \right) = \cos\left( \log_e \left( x \right) \right)$
Similarly,
$f\left( \frac{1}{y} \right) = \cos\left( \log_e y \right)$
Now,
$f\left( xy \right) = \cos\left( \log_e xy \right) = \cos\left( \log_e x + \log_e y \right)$
and

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

#### Notes

Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Q 18 | Page 43