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Solution:-
given:-
[tex]f(x) = \rm\cos( log x ) [/tex]
To find value of
[tex]f(x)f(4) - \dfrac{1}{2} \bigg\{f \big( \dfrac{x}{4} \big) + f \big(4x) \bigg \}[/tex]
Let
[tex] \rm \: f(x)f(4) = x_1[/tex]
[tex] \rm \: f \big( \dfrac{x}{4} \big) + f \big(4x) = x_2[/tex]
So we have to find
[tex] \rm \: x_1 - \frac{1}{2} x_2 [/tex]
Now take
[tex] \rm \: f(x)f(4) = x_1[/tex]
and put the value
[tex]f(x) = \rm\cos( log x ) [/tex]
[tex] \rm \: x_1 = \cos( log x ) . \cos( log4 ) [/tex]
Now take
[tex]\rm x_2 = \: f \big( \dfrac{x}{4} \big) + f \big(4x) [/tex]
and put value
[tex]f(x) = \rm\cos( log x ) [/tex]
we get
[tex] \rm x_2 = \cos( log \frac{x}{4} ) + \cos( log4x ) [/tex]
Using logarithm property
[tex] \rm log(ab) = log(a) + log(b) [/tex]
[tex] \rm \: log( \frac{a}{b} ) = log(a) - log(b) [/tex]
we get
[tex] \rm \cos( logx - log4 ) + \cos( log4+ logx ) [/tex]
Apply trigonometry identity
[tex] \rm \cos(a + b) = \cos(a) \cos(b) - \sin(a) \sin(b) [/tex]
[tex]\rm \cos(a - b) = \cos(a) \cos(b) + \sin(a) \sin(b) [/tex]
Now
[tex] \rm \cos( logx) \times \cos( log4) + \sin( logx ) \times \sin( log4 ) + cos( logx) \times \cos( log4) - \sin( logx ) \times \sin( log4 ) [/tex]
Now we get
[tex] \rm \: x_2 = \rm \: 2 \cos( log4 ) . \cos( logx ) [/tex]
[tex] \rm \: x_1 = \cos( log x ) . \cos( log4 ) [/tex]
So we can write
[tex] \rm \: x_2 = 2 x_1[/tex]
[tex] \rm\dfrac{ x_2}{2} = x_1[/tex]
[tex] \rm \: x_1 - \frac{1}{2} x_2 = 0[/tex]
So we get
[tex] \rm \: value \: of \: \rm \: x_1 - \frac{1}{2} x_2 \: is \: 0[/tex]
Option c is correct