If f(x) = cos(log x),then the value of f(x) c | vedclass

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(C) Given $f(x) = \cos(\log x)$.We need to evaluate the expression $E = f(x) \cdot f(y) - \frac{1}{2} \left( f\left(\frac{x}{y}\right) + f(xy) \right)

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(C) Given $f(x) = \cos(\log x)$.We need to evaluate the expression $E = f(x) \cdot f(y) - \frac{1}{2} \left( f\left(\frac{x}{y}\right) + f(xy) \right)$.Substitute the function definition:$E = \cos(\log x) \cdot \cos(\log y) - \frac{1}{2} \left( \cos(\log(x/y)) + \cos(\log(xy)) \right)$.Using logarithmic properties $\log(x/y) = \log x - \log y$ and $\log(xy) = \log x + \log y$:$E = \cos(\log x) \cdot \cos(\log y) - \frac{1}{2} \left( \cos(\log x - \log y) + \cos(\log x + \log y) \right)$.Apply the trigonometric identity $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$ where $A = \log x$ and $B = \log y$:$E = \cos(\log x) \cdot \cos(\log y) - \frac{1}{2} \left( 2 \cos(\log x) \cos(\log y) \right)$.$E = \cos(\log x) \cdot \cos(\log y) - \cos(\log x) \cdot \cos(\log y) = 0$.

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

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