JEE Main 202224 Jun, Evening shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002If y=tan−1(secx3−tanx3),π2<x3<3π2y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}y=tan−1(secx3−tanx3),2π<x3<23π, thenAOption A. xy′′+2y′=0xy'' + 2y' = 0xy′′+2y′=0BOption B. x2y′′−6y+3π2=0{x^2}y'' - 6y + {{3\pi } \over 2} = 0x2y′′−6y+23π=0COption C. x2y′′−6y+3π=0{x^2}y'' - 6y + 3\pi = 0x2y′′−6y+3π=0DOption D. xy′′−4y′=0xy'' - 4y' = 0xy′′−4y′=0CheckSkip · show solution
