QJEEM21Mab6de: JEE Main 2021 Mathematics - Class 12 JEEM21Mab6de Solution

$\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2}$

$= \sqrt {{{\left( {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right)}^2}}$

$= \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|$

$= \left| {{{{{(x + 1)}^2} - {{(x - 1)}^2}} \over {{x^2} - 1}}} \right|$

$= \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|$

$= \left| {{{4x} \over {{x^2} - 1}}} \right|$

$\therefore$ $I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx}$

$I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx}$

Let $f(x) = \left| {{{4x} \over {{x^2} - 1}}} \right|$

$\therefore$ $f( - x) = \left| {{{4( - x)} \over {{{( - x)}^2} - 1}}} \right| = \left| {{{ - 4x} \over {{x^2} - 1}}} \right| = \left| {{{4x} \over {{x^2} - 1}}} \right|$

$\Rightarrow f(x) = f( - x)$

$\therefore$ f(x) is a even function.

$\therefore$ $I = 2\int_0^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx}$

Using property, If f(x) is an even function then,

$\int_{ - a}^a {f(x) = 2\int_0^a {f(x)dx} }$

x > 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$ x² > 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$ x² $-$ 1 < 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

\Rightarrow {1 \over {{x^2} - 1}} &lt; 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

\Rightarrow {{4x} \over {{x^2} - 1}} &lt; 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\therefore$ $\left| {{{4x} \over {{x^2} - 1}}} \right| = - {{4x} \over {{x^2} - 1}}$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\therefore$ $I = 2\int_0^{{1 \over {\sqrt 2 }}} { - \left( {{{4x} \over {{x^2} - 1}}} \right)dx}$

$= - 4\int_0^{{1 \over {\sqrt 2 }}} {{{2x} \over {{x^2} - 1}}dx}$

$= - 4\left[ {{{\log }_e}\left| {{x^2} - 1} \right|} \right]_0^{{1 \over {\sqrt 2 }}}$

$= - 4\left[ {\log \left| {{1 \over 2} - 1} \right| - \log \left| { - 1} \right|} \right]$

$= - 4{\log _e}\left| { - {1 \over 2}} \right|$

$= - 4{\log _e}{1 \over 2}$

$= - 4\log _e^{{2^{ - 1}}}$

$= 4\log _e^2$

$= \log _e^{{2^4}}$

$= \log _e^{16}$