QJEEM22Ma4b28: JEE Main 2022 Mathematics - Class 12 JEEM22Ma4b28 Solution: The set of all values of k for which , is the interval :

${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3}$

Let $f(t) = {t^3} + {\left( {{\pi \over 2} - t} \right)^3}$

Where $t = {\tan ^{ - 1}}x$ ; $x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$

$= {t^3} + {\left( {{\pi \over 2}} \right)^3} - {{3{\pi ^2}t} \over 4} + {{3\pi } \over 2}{t^2} - {t^3}$

$f(t) = {{3\pi } \over 2}{t^2} - {{3{\pi ^2}} \over 4}\,.\,t + {{{\pi ^3}} \over 8}$

This is a quadratic equation of t.

Here, coefficient of t² term is ${{3\pi } \over 2}$ which is > 0.

$\therefore$ It is a upward parabola.

Now, $f'(t) = 3\pi t - {{3{\pi ^2}} \over 4}$

f''(t) = 3\pi > 0

$\therefore$ $3\pi t - {{3{\pi ^2}} \over 4} = 0$

$\Rightarrow t = {\pi \over 4}$ (minima)

$\therefore$ vertex of graph at ${\pi \over 4}$

$\therefore$ Minimum value at ${\pi \over 4}$ and maximum value at $-$ ${\pi \over 2}$ .

$\therefore$ $f\left( {{\pi \over 4}} \right) = {{{\pi ^3}} \over {64}} + {\left( {{\pi \over 2} - {\pi \over 4}} \right)^3} = {{{\pi ^3}} \over {32}}$

$f\left( { - {\pi \over 2}} \right) = - {{{\pi ^3}} \over 8} + {\pi ^3}$

$= {{7{\pi ^3}} \over 8}$

$\therefore$ $k{\pi ^3} \in \left[ {{{{\pi ^3}} \over {32}},\,{{7{\pi ^3}} \over 8}} \right)$

$\Rightarrow k \in \left[ {{1 \over {32}},\,{7 \over 8}} \right)$