If y=tan−1(secx3−tanx3)y=\tan^{-1}\left(\sec x^3-\tan x^3\right)y=tan−1(secx3−tanx3), π2<x3<3π2\dfrac\pi2<x^3<\dfrac{3\pi}22π<x3<23π, then
Official answer: B
Simplify. With t=x3t=x^3t=x3: sect−tant=1−sintcost=tan(π4−t2)\sec t-\tan t=\dfrac{1-\sin t}{\cos t}=\tan\left(\dfrac\pi4-\dfrac t2\right)sect−tant=cost1−sint=tan(4π−2t), and π4−t2∈(−π2,0)\frac\pi4-\frac t2\in\left(-\frac\pi2,0\right)4π−2t∈(−2π,0) for π2<t<3π2\frac\pi2<t<\frac{3\pi}22π<t<23π.
So y=π4−x32y=\dfrac\pi4-\dfrac{x^3}2y=4π−2x3, y′=−3x22y'=-\dfrac{3x^2}2y′=−23x2, y′′=−3xy''=-3xy′′=−3x.
Check (B). x2y′′−6y+3π2=−3x3−3π2+3x3+3π2=0x^2y''-6y+\frac{3\pi}2=-3x^3-\frac{3\pi}2+3x^3+\frac{3\pi}2=0x2y′′−6y+23π=−3x3−23π+3x3+23π=0.
(B)