If $y=\tan ^{-1}\left(\sec x^{3}-\tan x^{3}\right), \frac{\pi}{2}<x^{3}<\frac{3 \pi}{2}$, then

(B) $x^{2} y^{\prime \prime}-6 y+\frac{3 \pi}{2}=0$

(A) $x y^{\prime \prime}+2 y^{\prime}=0$

(C) $x^{2} y^{\prime \prime}-6 y+3 \pi=0$

(D) $x y^{\prime \prime}-4 y^{\prime}=0$

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Differentiation

y = tan^{- 1} (sec x^{3} - tan x^{3}), \frac{π}{2} < x^{3} < \frac{3 π}{2}

, then

Answer: (B)

Simplify:

sec x^{3} - tan x^{3} = \frac{1 - s i n x ^{3}}{c o s x ^{3}} = \frac{1 - c o s ( \frac{π}{2} - x ^{3} )}{s i n ( \frac{π}{2} - x ^{3} )} = tan (\frac{π}{4} - \frac{x ^{3}}{2})

. Since

\frac{π}{4} - \frac{x ^{3}}{2} \in (- \frac{π}{2}, 0)

for the given range,

y = \frac{π}{4} - \frac{x ^{3}}{2}

. Then

y^{'} = - \frac{3 x ^{2}}{2}

and

y^{''} = - 3 x

. From

y = \frac{π}{4} - \frac{x ^{3}}{2}

, we get

4 y = π - 2 x^{3}

. Substituting

x^{3} = - \frac{y ^{''}}{3} x^{2}

4 y = π + \frac{2 x ^{2} y ^{''}}{3}

, which gives

12 y = 3 π + 2 x^{2} y^{''}

, or

x^{2} y^{''} - 6 y + \frac{3 π}{2} = 0

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option