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$f(x) + f\left( {\frac{1}{{1 - x}}} \right) = x,\forall x \in \mathbb{R}$

$f(x) = ?$

$x \to 1 - \frac{1}{x}$ yields,

$f\left( {1 - \frac{1}{x}} \right) + f\left[ {\frac{1}{{1 - \left( {1 - \frac{1}{x}} \right)}}} \right] = 1 - \frac{1}{x}$

$ \Rightarrow f\left( {\frac{{x - 1}}{x}} \right) + f(x) = \frac{{x - 1}}{x}........(A)$

$x \to \frac{1}{{1 - x}}$ yields,

$f\left( {\frac{1}{{1 - x}}} \right) + f\left( {\frac{1}{{1 - \frac{1}{{1 - x}}}}} \right) = \frac{1}{{1 - x}}$

$ \Rightarrow f\left( {\frac{1}{{1 - x}}} \right) + f\left( {\frac{{x - 1}}{x}} \right) = \frac{1}{{1 - x}}........(B)$

(A) - (B) yields,

$f(x) - f\left( {\frac{1}{{1 - x}}} \right) = \frac{{x - 1}}{x} - \frac{1}{{1 - x}}........(C)$

(C) + The original equation yields,

$2f(x) = x + \frac{{x - 1}}{x} - \frac{1}{{1 - x}} = \frac{{{x^2}(1 - x) - {{(1 - x)}^2} - x}}{{x(1 - x)}}$

$ \Rightarrow f(x) = \frac{{{x^2} - {x^3} - 1 - {x^2} + 2x - x}}{{2x(1 - x)}} = \frac{{ - {x^3} + x - 1}}{{2x(1 - x)}}$

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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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