JEE Main 2022 — Mathematics Calculus

Given,

$$y={\mathrm{tan}}^{-1}(\mathrm{sec}{x}^{3}-\mathrm{tan}{x}^{3})$$y=tan−1(secx3−tanx3)

$$={\mathrm{tan}}^{-1}(\frac{1-\mathrm{sin}{x}^{3}}{\mathrm{cos}{x}^{3}})$$=tan−1(cosx31−sinx3​)

$$={\mathrm{tan}}^{-1}(\frac{1-\mathrm{cos}(\frac{\pi }{2}-{x}^{3})}{\mathrm{sin}(\frac{\pi }{2}-{x}^{3})})$$=tan−1(sin(2π​−x3)1−cos(2π​−x3)​)

$$={\mathrm{tan}}^{-1}(\mathrm{tan}(\frac{\pi }{4}-\frac{{x}^{3}}{2}))$$=tan−1(tan(4π​−2x3​))

Since $\frac{\pi }{4}-\frac{{x}^{3}}{2}\in (-\frac{\pi }{2},0)$4π​−2x3​∈(−2π​,0) as $(\frac{\pi }{2}<{x}^{3}<\frac{3\pi }{2})$(2π​<x3<23π​)

So, $y=(\frac{\pi }{4}-\frac{{x}^{3}}{2})$y=(4π​−2x3​)

Now differentiating we get,

$${y}^{'}=\frac{-3{x}^{2}}{2},{y}^{''}=-3x$$y′=2−3x2​,y′′=−3x

Now putting the value of $x$x in term of ${y}^{''}$y′′ in $4y=\pi -2{x}^{3}$4y=π−2x3

We get,

$$4y=\pi -2{x}^{2}(\frac{-{y}^{''}}{3})$$4y=π−2x2(3−y′′​)

$$12y=3\pi +2{x}^{2}{y}^{''}$$12y=3π+2x2y′′

$${x}^{2}{y}^{''}-6y+\frac{3\pi }{2}=0$$x2y′′−6y+23π​=0