Question

First to give correct answer gets brainliest.​

Answer 1

Solution:-

given:-

$f(x) = \rm\cos( log x )$

To find value of

$f(x)f(4) - \dfrac{1}{2} \bigg\{f \big( \dfrac{x}{4} \big) + f \big(4x) \bigg \}$

Let

$\rm \: f(x)f(4) = x_1$

$\rm \: f \big( \dfrac{x}{4} \big) + f \big(4x) = x_2$

So we have to find

$\rm \: x_1 - \frac{1}{2} x_2$

Now take

$\rm \: f(x)f(4) = x_1$

and put the value

$f(x) = \rm\cos( log x )$

$\rm \: x_1 = \cos( log x ) . \cos( log4 )$

Now take

$\rm x_2 = \: f \big( \dfrac{x}{4} \big) + f \big(4x)$

and put value

$f(x) = \rm\cos( log x )$

we get

$\rm x_2 = \cos( log \frac{x}{4} ) + \cos( log4x )$

Using logarithm property

$\rm log(ab) = log(a) + log(b)$

$\rm \: log( \frac{a}{b} ) = log(a) - log(b)$

we get

$\rm \cos( logx - log4 ) + \cos( log4+ logx )$

Apply trigonometry identity

$\rm \cos(a + b) = \cos(a) \cos(b) - \sin(a) \sin(b)$

$\rm \cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b)$

Now

$\rm \cos( logx) \times \cos( log4) + \sin( logx ) \times \sin( log4 ) + cos( logx) \times \cos( log4) - \sin( logx ) \times \sin( log4 )$

Now we get

$\rm \: x_2 = \rm \: 2 \cos( log4 ) . \cos( logx )$

$\rm \: x_1 = \cos( log x ) . \cos( log4 )$

So we can write

$\rm \: x_2 = 2 x_1$

$\rm\dfrac{ x_2}{2} = x_1$

$\rm \: x_1 - \frac{1}{2} x_2 = 0$

So we get

$\rm \: value \: of \: \rm \: x_1 - \frac{1}{2} x_2 \: is \: 0$

Option c is correct