Skip to main content

Visit this link for 1 : 1 LIVE Classes.

$I = \int\limits_0^\infty {\ln \left( {\frac{{{e^x} + 1}}{{{e^x} - 1}}} \right)dx} $

(A) I > 2
(B) I < 1
(C) I = 2
(D) 1 < I < 2

$I = \int\limits_0^\infty  {\ln \left( {\frac{{1 + {e^{ - x}}}}{{1 - {e^{ - x}}}}} \right)dx} $

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

So, $ - {e^{ - x}}dx = dt$

$\therefore dx =  - \frac{{dt}}{t}$

$I = \int\limits_1^0 {\ln \left( {\frac{{1 + t}}{{1 - t}}} \right)\left( { - \frac{1}{t}} \right)dt}  = \int\limits_0^1 {\frac{1}{t}\ln \left( {\frac{{1 + t}}{{1 - t}}} \right)dt} $

$ = \int\limits_0^1 {\frac{{\ln (1 + t)}}{t}} dt - \int\limits_0^1 {\frac{{\ln (1 - t)}}{t}dt} $

$ = \int\limits_0^1 {\frac{{t - \frac{{{t^2}}}{2} + \frac{{{t^3}}}{3} - \frac{{{t^4}}}{4} + ...........}}{t}} dt - \int\limits_0^1 {\frac{{ - t - \frac{{{t^2}}}{2} - \frac{{{t^3}}}{3} - \frac{{{t^4}}}{4} + ...........}}{t}} dt$

$ = \int\limits_0^1 {\left( {1 - \frac{t}{2} + \frac{{{t^2}}}{3} - \frac{{{t^3}}}{4} + ...........} \right)} dt - \int\limits_0^1 {\left( { - 1 - \frac{t}{2} - \frac{{{t^2}}}{3} - \frac{{{t^3}}}{4} - ...........} \right)} dt$

$ = \left. {\left( {t - \frac{{{t^2}}}{{{2^2}}} + \frac{{{t^3}}}{{{3^2}}} - \frac{{{t^4}}}{{{4^2}}} + .............} \right)} \right|_0^1 + \left. {\left( {t + \frac{{{t^2}}}{{{2^2}}} + \frac{{{t^3}}}{{{3^2}}} + \frac{{{t^4}}}{{{4^2}}} + .............} \right)} \right|_0^1$

$ = \left( {1 - \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} - \frac{1}{{{4^2}}} + ...........} \right) + \left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ............} \right)$

$ = 2\left( {1 + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{7^2}}}...........} \right)$

Hence, Option (A).

Popular posts from this blog

${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

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)
Read more