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Let $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}x}}}$

$\frac{{dI}}{{d\alpha }} = ?$

We have, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}(\pi /2 - x)}}} $

$ \Rightarrow I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\cot }^{\sqrt {\tan \alpha } }}x}}}  = \int\limits_0^{\pi /2} {\frac{{{{\sin }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\sin }^{\sqrt {\tan \alpha } }}x + {{\cos }^{\sqrt {\tan \alpha } }}x}}} $ ........(A)

Also, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}x}} = } \int\limits_0^{\pi /2} {\frac{{{{\cos }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\cos }^{\sqrt {\tan \alpha } }}x + {{\sin }^{\sqrt {\tan \alpha } }}x}}}$ ........(B)

(A) + (B) gives, $2I = \int\limits_0^{\pi /2} {dx}  = \frac{\pi }{2}$

$ \Rightarrow I = \frac{\pi }{4}$

$\therefore \frac{{dI}}{{d\alpha }} = 0$

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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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