Question

$\bold{if \: f(x) = \cos( log(x) ) \: then \: f( \frac{1}{x})f( \frac{1}{y} ) - \frac{1}{2} [f(xy) + f( \frac{x}{y}) ]}$
Please answer this question.. from the chapter functions.. no irrelevant answers please..​

Answer 1

Hello mate..

Solution:

GIVEN:

$\bold{if \: f(x) = \cos( log(x) ) \: then \: f( \frac{1}{x})f( \frac{1}{y} ) - \frac{1}{2} [f(xy) + f( \frac{x}{y}) ]}$

TO FIND:

$\bold{ f( \frac{1}{x})f( \frac{1}{y} ) - \frac{1}{2} [f(xy) + f( \frac{x}{y}) ]}$

SOLUTION:

$\bold{ \implies cos(log \frac{1}{x}) \times cos(log \frac{1}{y}) - \frac{1}{2}[cos(log \: xy) + cos(log \frac{x}{y} ) ] } \\ \bold{ \implies cos[log \: 1 - log \: x] \times cos[log \: 1 - log \: y] - \frac{1}{2}[cos(log \: x + log \: y) + cos(log \: x - log \: y)] } \\ \bold{ \implies cos( - log \: x) \times cos(log \: y) - \frac{1}{2}[2cos(log \: x) \times (log \: y) } \\ \bold{ \implies cos (log \: x )\times cos ( log \: y) - cos(log \: x) \times cos(log \: y)} \\ \bold{ = 0}$

$\huge\underline\mathcal \pink{ formulas \: used : }$

$\implies\bold \green{ log_{a}(1) = 0} \\ \implies\bold \green{cos(a + b) + \cos(a - b) = 2 \cos(a) \cos(b) } \\ \implies\bold \green{log \: xy = log \: x + log \: y} \\ \implies\bold \green{log \frac{x}{y} = log \: x - log \: y } \\ \implies\bold \green{ log_{ {a}^{m} }(b) = m log_{b}(a) } \\ \implies\bold \green{ \cos( - \alpha ) = \cos( \alpha ) }$

Note:

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Thanks!

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