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$I = \int\limits_0^\infty {\frac{{{e^{ - nx}} - {e^{ - x}}}}{x}dx} = ?$

$n \in \mathbb{N}$

Let, ${e^{ - x}} = t$

$ \Rightarrow  - {e^{ - x}}dx = dt$

$ \Rightarrow  - tdx = dt$

$I = \int\limits_1^0 {\frac{{{t^n} - t}}{{ - \ln t}}\frac{{dt}}{{ - t}}}  =  - \int\limits_0^1 {\frac{{{t^{n - 1}} - 1}}{{\ln t}}dt} $

But, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } - 1}}{{\ln x}}dx = \ln (\lambda  + 1)} $ (Proof can be seen in this problem)

So, $I =  - \ln (n - 1 + 1) =  - \ln n$

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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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