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$f(x) = {\left( {\frac{x}{\pi }} \right)^x} + {\left( {\frac{\pi }{x}} \right)^x}$

$\int {f(x)dx + } \int {f(x)\ln xdx} - \int {f(x)\ln \pi dx} = ?$

Combining given three integrals yields $\int {f(x)[1 + \ln x - \ln \pi ]dx} $

$ = \int {f(x)(\ln e + \ln x - \ln \pi )dx} $

$ = \int {\left[ {{{\left( {\frac{x}{\pi }} \right)}^x} + {{\left( {\frac{\pi }{x}} \right)}^x}} \right]\left( {\ln \frac{{ex}}{\pi }} \right)dx}  = I$

Let, ${\left( {\frac{x}{\pi }} \right)^x} = t$

$\therefore x\ln \frac{x}{\pi } = \ln t$

$\therefore x(\ln x - \ln \pi ) = \ln t$

Differentiation yields $x.\frac{1}{x} + \ln x - \ln \pi  = \frac{1}{t}.\frac{{dt}}{{dx}}$

$\therefore (1 + \ln x - \ln \pi )dx = \frac{{dt}}{t}$

$\therefore (\ln e + \ln x - \ln \pi )dx = \frac{{dt}}{t}$

$ \Rightarrow \ln \left( {\frac{{ex}}{\pi }} \right)dx = \frac{{dt}}{t}$

$I = \int {\left( {t + \frac{1}{t}} \right)\frac{{dt}}{t}} $

$ = \int {\left( {1 + \frac{1}{{{t^2}}}} \right)dt} $

$ = t - \frac{1}{t} + C$

$ = {\left( {\frac{x}{\pi }} \right)^x} - {\left( {\frac{\pi }{x}} \right)^x} + C$

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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)