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Evaluate,

$\int\limits_{ - 2/3}^0 {\frac{{dx}}{{{{(x + 1)}^{1/2}} + {{(x + 1)}^{3/2}}}}} $

 Let, $\sqrt {x + 1}  = t$

Hence, $\frac{1}{{2\sqrt {x + 1} }}dx = dt$

Thus, $dx = 2\sqrt {x + 1} dt$

The given integral = $\int\limits_{1/\sqrt 3 }^1 {\frac{{2\sqrt {x + 1} dt}}{{{{(x + 1)}^{1/2}} + {{(x + 1)}^{3/2}}}}} $

=$\int\limits_{1/\sqrt 3 }^1 {\frac{{2tdt}}{{t + {t^3}}}} $

=$2\int\limits_{1/\sqrt 3 }^1 {\frac{{dt}}{{1 + {t^2}}}} $

=$2\left[ {{{\tan }^{ - 1}}t|_{1/\sqrt 3 }^1} \right]$

=$2\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) = \frac{\pi }{6}$

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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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