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$x^2-x-1=0$

$R(x)=\frac{{{x^{16}} - 1}}{{{x^8} + 2{x^7}}} = ?$

$R(x)=\frac{{{x^{16}} - 1}}{{{x^8} + 2{x^7}}} = \frac{{({x^2} - 1)({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^7}(x + 2)}}$

$\because {x^2} - x - 1 = 0,{x^2} - 1 = x$

$R(x) = \frac{{x({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^7}(x + 2)}} = \frac{{({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^6}(x + 2)}}$

$\because {x^2} - x - 1 = 0,{x^2} = x + 1$

$R(x) = \frac{{(x + 1 + 1)({x^4} + 1)({x^8} + 1)}}{{{x^6}(x + 2)}} = \frac{{({x^4} + 1)({x^8} + 1)}}{{{x^6}}}$

${x^2} = x + 1,{x^4} = {(x + 1)^2} = {x^2} + 2x + 1$

${x^4} + 1 = {x^2} + 2x + 2 = {x^2} + 2(x + 1) = {x^2} + 2{x^2} = 3{x^2}$

${({x^4} + 1)^2} = 9{x^4},{x^8} + 2{x^4} + 1 = 9{x^4},{x^8} + 1 = 7{x^4}$

$R(x) = \frac{{({x^4} + 1)({x^8} + 1)}}{{{x^6}}} = \frac{{3{x^2}\times 7{x^4}}}{{{x^6}}} = 21$

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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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