Question

If f(x) = cos(log x) and f(y) = cos(logy) then find the value of f(x).f(y)-1/2[f(x/y)+f(xy)]​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:f\:(\:x\:)\:.\:f\:(\:y\:)\:-\:\dfrac{1}{2}\:\left[\:f\:\left(\:\dfrac{x}{y}\:\right)\:+\:f\:(\:xy\:)\:\right]\:=\:0\:}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:f\:(\:x\:)\:=\:\cos\:(\:\log\:x\:)}$

$\displaystyle{\sf\:f\:(\:y\:)\:=\:\cos\:(\:\log\:y\:)}$

We have to find the value of

$\displaystyle{\sf\:f\:(\:x\:)\:.\:f\:(\:y\:)\:-\:\dfrac{1}{2}\:\left[\:f\:\left(\:\dfrac{x}{y}\:\right)\:+\:f\:(\:xy\:)\:\right]}$

$\displaystyle{\implies\sf\:\cos\:(\:\log\:x\:)\:\cos\:(\:\log\:y\:)\:-\:\dfrac{1}{2}\:\left[\:\cos\:\left(\:\log\:\left(\:\dfrac{x}{y}\:\right)\:\right)\:+\:\cos\:(\:\log\:(\:xy\:))\:\right]}$

$\displaystyle{\implies\sf\:\cos\:(\:\log\:x\:)\:\cos\:(\:\log\:y\:)\:-\:\dfrac{1}{2}\:\left[\:\cos\:(\:\log\:x\:-\:\log\:y\:)\:+\:\cos\:(\:\log\:x\:+\:\log\:y\:)\:\right]}$

We know that,

$\displaystyle{\pink{\sf\:\cos\:(\:a\:-\:b\:)\:+\:\cos\:(\:a\:+\:b\:)\:=\:2\:\cos\:a\:\cos\:b}}$

$\displaystyle{\implies\sf\:\cos\:(\:\log\:x\:)\:\cos\:(\:\log\:y\:)\:-\:\dfrac{1}{\cancel{2}}\:\times\:\cancel{2}\:\cos\:(\:\log\:x\:)\:\cos\:(\:\log\:y\:)}$

$\displaystyle{\implies\sf\:\cos\:(\:\log\:x\:)\:\cos\:(\:\log\:y\:)\:-\:\cos\:(\:\log\:x\:)\:\cos\:(\:\log\:y\:)}$

$\displaystyle{\implies\sf\:0}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:f\:(\:x\:)\:.\:f\:(\:y\:)\:-\:\dfrac{1}{2}\:\left[\:f\:\left(\:\dfrac{x}{y}\:\right)\:+\:f\:(\:xy\:)\:\right]\:=\:0\:}}}}$

Answer 2

Step-by-step explanation:

topic :

• If f(x) = cos(log x) and f(y) = cos(logy) then find the value of f(x).f(y)-1/2[f(x/y)+f(xy)]

to find :

• value of f(x).f(y)-1/2[f(x/y)+f(xy)]

solution :

• Substituting the values we get

• 1/2 [ cos ( log (x/y)) + cos (log (xy)) ]

• = cos (log (x)). cos (log (y)) - 1/2 [cos (log (x) - log (y)) + cos (log (x) + log (y))]

• = cos (log (x)). cos (log (y)) -

• 1/2 [2 cos (log (x)).c . cos (log (y))] = 0