Question

If a function is given by $f\left( x \right)=\cos \left( \log x \right)$ , then $f\left( x \right)f\left( y \right)-\dfrac{1}{2}\left( f\left( \dfrac{x}{y} \right)+f\left( xy \right) \right)$ is equal to:
(a) 1
(b) 2
(c) -2
(d) 0

Answer 1

Hint : Start by using the definition of the function and putting the terms in the given expression. Then use the formula ${{\log }_{a}}x-{{\log }_{a}}y={{\log }_{a}}\dfrac{x}{y}$ and ${{\log }_{a}}x+{{\log }_{a}}y={{\log }_{a}}xy$ to simplify the expression. Finally, use the formula $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and eliminate the removable terms to get the answer.

Complete step-by-step answer :
Let us start the solution to the above question by looking at some of the identities related to logarithmic functions.
${{\log }_{a}}x+{{\log }_{a}}y={{\log }_{a}}xy$
${{\log }_{a}}x-{{\log }_{a}}y={{\log }_{a}}\dfrac{x}{y}$
${{\log }_{{{a}^{b}}}}x=\dfrac{1}{b}{{\log }_{a}}x$
${{\log }_{a}}{{x}^{b}}=b{{\log }_{a}}x$
Now let us start the simplification of the expression given in the question.
$f\left( x \right)f\left( y \right)-\dfrac{1}{2}\left( f\left( \dfrac{x}{y} \right)+f\left( xy \right) \right)$
If we use the definition $f\left( x \right)=\cos \left( \log x \right)$ , we get
$\cos \left( \log x \right).\cos \left( \log y \right)-\dfrac{1}{2}\left( \cos \left( \log \left( \dfrac{x}{y} \right) \right)+\cos \left( \log \left( xy \right) \right) \right)$
Now we know that ${{\log }_{a}}x-{{\log }_{a}}y={{\log }_{a}}\dfrac{x}{y}$ and ${{\log }_{a}}x+{{\log }_{a}}y={{\log }_{a}}xy$ . So, if we use this in our expression, we get
$\cos \left( \log x \right).\cos \left( \log y \right)-\dfrac{1}{2}\left( \cos \left( \log x-\log y \right)+\cos \left( \operatorname{logx}+logy \right) \right)$
Now, we will use the formula $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ , such that $A=\log x-\log y$ and $B=\log x-\log y$ . On doing so, we get
$\cos \left( \log x \right).\cos \left( \log y \right)-\dfrac{1}{2}\left( 2\cos \left( \dfrac{\log x-\log y+\log x+\log y}{2} \right)\cos \left( \dfrac{\log x-\log y-\log x-\log y}{2} \right) \right)$ $=\cos \left( \log x \right).\cos \left( \log y \right)-\dfrac{1}{2}\left( 2\cos \left( \log x \right)\cos \left( -\log y \right) \right)$
Now, we know that cos(-x)=cosx. So, using this in our expression, we get
$\cos \left( \log x \right).\cos \left( \log y \right)-\cos \left( \log x \right)\cos \left( \log y \right)$
Now, both the terms have the same magnitude but opposite signs, so they are cancelled. So, the final answer comes out to be:
$\cos \left( \log x \right).\cos \left( \log y \right)-\cos \left( \log x \right)\cos \left( \log y \right)=0$

So, the correct answer is “Option d”.

Note : The key to the above question is applying the identities related to logarithmic function, if you apply the formulas correctly with correct signs then it is a sure thing that you will reach your answer. Also, don’t miss the half in the expression, because it is a general mistake that students miss the constant factors in the questions involving large numbers of unknown terms.