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Find $f(x)$ such that,

$f(x) + f\left( {\frac{1}{x}} \right) + 3f( - x) = x$

$x \to \frac{1}{x}$ in the original equation yields,

$f\left( {\frac{1}{x}} \right) + f(x) + 3f\left( { - \frac{1}{x}} \right) = \frac{1}{x}$ ........(A)

Original equation - (A) yields,

$3f( - x) - 3f\left( { - \frac{1}{x}} \right) = x - \frac{1}{x}$

$x \to  - x$ above gives us $3f(x) - 3f\left( {\frac{1}{x}} \right) =  - x + \frac{1}{x}$

$ \Rightarrow f(x) - f\left( {\frac{1}{x}} \right) =  - \frac{x}{3} + \frac{1}{{3x}}$ ........(B)

Original equation + (B) gives us $2f(x) + 3f( - x) = \frac{{2x}}{3} + \frac{1}{{3x}}$ ........(C)

$x \to  - x$ above yields,

$2f( - x) + 3f(x) =  - \frac{{2x}}{3} - \frac{1}{{3x}}$ ........(D)

Now, $(D) \times 3 - (C) \times 2$

$5f(x) =  - 2x - \frac{1}{x} - \frac{{4x}}{3} - \frac{2}{{3x}} =  - \frac{{10x}}{3} - \frac{5}{{3x}}$

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

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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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