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1.Relation and Function
medium
જો $f(x) = \cos (\log x)$, તો $f(x).f(4) - \frac{1}{2}\left[ {f\left( {\frac{x}{4}} \right) + f(4x)} \right] =$
A
$1$
B
$-1$
C
$0$
D
$ \pm 1$
Solution
(c) $f(x) = \cos \,(\log x)$
Now let $y = f(x)\,\,.\,\,f(4) – \frac{1}{2}\,\left[ {f\left( {\frac{x}{4}} \right) + f(4x)} \right]$
==> $y = \cos \,(\log x).\cos \,(\log 4) $
$- \frac{1}{2}\,\left[ {\cos \,\log \,\left( {\frac{x}{4}} \right) + \cos \,(\log 4x)} \right]$
==> $y = \cos \,(\log x)\,\cos \,(\log 4)$
$ – \frac{1}{2}\,\left[ {\cos \,(\log x – \log 4) + \cos \,(\log x + \log 4)} \right]$
==> $y = \cos \,(\log x)\,\cos \,(\log 4) – \frac{1}{2}\,\left[ {2\,\cos \,(\log x)\,\cos \,(\log 4)} \right]$
==> $y = 0$.
Standard 12
Mathematics
