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$f\left( {\frac{{2x - 1}}{{x - 3}}} \right) = {x^2}$,

$\int {f(x)dx = ?} $

 Let, $\frac{{2x - 1}}{{x - 3}} = t$

$\therefore x = \frac{{3t - 1}}{{t - 2}}$

$\therefore f(t) = {\left( {\frac{{3t - 1}}{{t - 2}}} \right)^2}$

Now, $\int {f(t)dt}  = \int {{{\left( {\frac{{3t - 1}}{{t - 2}}} \right)}^2}dt} $

$ = \int {{{\left[ {\frac{{3(t - 2) + 5}}{{t - 2}}} \right]}^2}dt} $

$ = \int {{{\left( {3 + \frac{5}{{t - 2}}} \right)}^2}dt} $

$ = \int {9 + \frac{{30}}{{t - 2}} + \frac{{25}}{{{{(t - 2)}^2}}}dt} $

$ = 9t + 30\ln |t - 2| - \frac{{25}}{{t - 2}} + C$

The integral asked $=9x + 30\ln |x - 2| - \frac{{25}}{{x - 2}} + C$

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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)
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