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$I = \int\limits_{ - \pi }^\pi {\frac{{{{\sin }^2}x}}{{1 + {e^x}}}dx} = ?$

Let $x=-t$, so $dx=-dt$

$I = \int\limits_\pi ^{ - \pi } { - \frac{{{{\sin }^2}t}}{{1 + {e^{ - t}}}}dt}  = \int\limits_{ - \pi }^\pi  {\frac{{{e^t}{{\sin }^2}t}}{{1 + {e^t}}}dt}  = \int\limits_{ - \pi }^\pi  {\frac{{{e^x}{{\sin }^2}x}}{{1 + {e^x}}}dx} $

$I + I = \int\limits_{ - \pi }^\pi  {\frac{{{{\sin }^2}x}}{{1 + {e^x}}}dx}  + \int\limits_{ - \pi }^\pi  {\frac{{{e^x}{{\sin }^2}x}}{{1 + {e^x}}}dx} $

$ \Rightarrow 2I = \int\limits_{ - \pi }^\pi  {\frac{{(1 + {e^x}){{\sin }^2}x}}{{1 + {e^x}}}dx}  = \int\limits_{ - \pi }^\pi  {{{\sin }^2}xdx} $

$ \Rightarrow 2I = \int\limits_0^\pi  {2{{\sin }^2}xdx}  = \int\limits_0^\pi  {(1 - \cos 2x)dx} $

$ \Rightarrow 2I = \left. {\left( {x - \frac{{\sin 2x}}{2}} \right)} \right|_0^\pi  = \pi $

$ \Rightarrow I = \frac{\pi }{2}$

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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)