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