QJEEAdv22Maa29c: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Maa29c Solution: Consider the equation Which of the following statements is/are…

Given,

$\int_1^e {{{{{\left( {\log _e^x} \right)}^{1/2}}} \over {x{{\left( {a - {{\left( {\log _e^x} \right)}^{3/2}}} \right)}^2}}} = 1}$

Let $\log _e^x = t$

$\Rightarrow x = {e^t}$

$\Rightarrow dx = {e^t}dt$

When $x = 1$ then $t = \log _e^1 = 0$

and when $x = e$ then $t = \log _e^e = 1$

$\Rightarrow \int_0^1 {{{\sqrt t \,.\,{e^t}dt} \over {{e^t}{{(a - t\sqrt t )}^2}}} = 1}$

$\Rightarrow \int_0^1 {{{\sqrt t \,dt} \over {{{(a - t\sqrt t )}^2}}} = 1}$

Let ${t^{1/2}} = z$

$\Rightarrow {1 \over {2\sqrt t }}dt = dz$

$\Rightarrow dt = 2\sqrt t \,dz \Rightarrow dt = 2z\,dz$

When $t = 0$ , then $z = 0$

When $t = 1$ , then $z = {1^{1/2}} = 1$

$\therefore$ $\int_0^1 {{{z \times 2z\,dz} \over {{{(a - {z^3})}^2}}} = 1}$

$\Rightarrow \int_0^1 {{{2{z^2}\,dz} \over {{{(a - {z^3})}^2}}} = 1}$

Now let $a - {z^3} = p$

$\Rightarrow - 3{z^2}\,dz = dp$

When $z = 0$ then $p = a - 0 = a$

When $z = 1$ then $p = a - {1^3} = a - 1$

$\therefore$ $\int_a^{a - 1} {{{2\,.\,\left( { - {{dp} \over 3}} \right)} \over {{p^2}}} = 1}$

$\Rightarrow - {2 \over 3}\int_a^{a - 1} {{{dp} \over {{p^2}}} = 1}$

$\Rightarrow - {2 \over 3}\left[ { - {1 \over p}} \right]_a^{a - 1} = 1$

$\Rightarrow - {2 \over 3}\left[ { - {1 \over {a - 1}} + {1 \over a}} \right] = 1$

$\Rightarrow - {2 \over 3}\left[ {{{ - a + a - 1} \over {a(a - 1)}}} \right] = 1$

$\Rightarrow {2 \over 3}\left( {{1 \over {a(a - 1)}}} \right) = 1$

$\Rightarrow 3{a^2} - 3a = 2$

$\Rightarrow 3{a^2} - 3a - 2 = 0$

$\Rightarrow a = {{3\, \pm \,\sqrt {9 - 4(3)( - 2)} } \over {2\,.\,3}}$

$\therefore$ $a = {{3\, \pm \,\sqrt {33} } \over 6},\,{{3 - \sqrt {33} } \over 6}$