JEE Main 2024 — 4 April, Shift 2 — Question 30

Sol.

$\quad \frac{d y}{d x}=(x+y+2)^{2} \ldots(1), \quad y(0)=-2$

Let $x+y+2=v$

$1+\frac{d y}{d x}=\frac{d v}{d x}$

from (1) $\frac{d v}{d x}=1+v^{2}$

$\int \frac{\mathrm{dv}}{1+\mathrm{v}^{2}}=\int \mathrm{dx}$

$\tan ^{-1}(v)=x+C$

$\tan ^{-1}(x+y+2)=x+C$

at $x=0 \quad y=-2 \Rightarrow C=0$

$\Rightarrow \tan ^{-1}(x+y+2)=x$

$\mathrm{y}=\tan \mathrm{x}-\mathrm{x}-2$

$f(x)=\tan x-x-2, x \in\left[0, \frac{\pi}{3}\right]$

$f^{\prime}(x)=\sec ^{2} x-1>0 \Rightarrow f(x) \uparrow$

$f_{\min }=f(0)=-2=\beta$

$f_{\max }=f\left(\frac{\pi}{3}\right)=\sqrt{3}-\frac{\pi}{3}-2=\alpha$

now $(3 \alpha+\pi)^{2}+\beta^{2}=\gamma+\delta \sqrt{3}$

$\Rightarrow(3 \alpha+\pi)^{2}+\beta^{2}=(3 \sqrt{3}-6)^{2}+4$

$\gamma+\delta \sqrt{3}=67-36 \sqrt{3}$

$\Rightarrow \gamma=67$ and $\delta=-36 \Rightarrow \gamma+\delta=31$