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$\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 - x}}{{1 + x}} + 1 + \sin x}}{{1 + \cos 2x}}} dx = ?$

The given integral can be split as,

$ \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 - x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}} dx + \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{1 + \cos 2x}}} dx$

$\frac{{6{x^5} + \frac{{{x^3}}}{3} + \ln \frac{{1 - x}}{{1 + x}} + \sin x}}{{1 + \cos 2x}}$ is an odd function. Thus, the given integral

$ = 0 + \int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {\frac{1}{{2{{\cos }^2}x}}} dx = \frac{1}{2}\int\limits_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} {{{\sec }^2}xdx}  = \frac{1}{2}\left. {\tan x} \right|_{ - \frac{\pi }{6}}^{\frac{\pi }{6}} = \frac{1}{{\sqrt 3 }}$

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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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