JEE Main 2022 (Online) 24th June Evening ShiftQuestion 1
If $y=\tan^{-1}\left(\sec x^{3}-\tan x^{3}\right)$, $\frac{\pi}{2}<x^{3}<\frac{3\pi}{2}$, then
Differentiation and its Application
Solution
Concept. With $u=x^3$, $\sec u-\tan u=\tan\left(\frac\pi4-\frac u2\right)$; the angle lies in $\left(-\frac\pi2,0\right)$, so $\tan^{-1}$ returns it. Formulas used. (1) $\dfrac{1-\sin u}{\cos u}=\tan\left(\dfrac\pi4-\dfrac u2\right)$Step 1 - Simplify. By (1), with $\frac\pi4-\frac u2\in\left(-\frac\pi2,0\right)$: $$y=\frac\pi4-\frac{x^3}{2}.$$ Step 2 - Derivatives. $y'=-\frac{3x^2}{2}$, $y''=-3x$. Step 3 - Check (B). $$x^2y''-6y=-3x^3-\frac{3\pi}{2}+3x^3=-\frac{3\pi}{2}.$$ So $x^2y''-6y+\frac{3\pi}{2}=0$. Answer: (B)