JEE Main 2024 — Mathematics Calculus

$\frac{d y}{d x}=(x+y+2)^2$dxdy​=(x+y+2)2 ...(1) $y(0)=-2$y(0)=−2

Let $\mathrm{x}+\mathrm{y}+2=\mathrm{v}$x+y+2=v $1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}$1+dxdy​=dxdv​ from (1) $\frac{d v}{d x}=1+v^2$dxdv​=1+v2 $\begin{aligned}
& \int \frac{d v}{1+v^2}=\int d x \
& \tan ^{-1}(v)=x+C \
& \tan ^{-1}(x+y+2)=x+C \
& \text { at } x=0 y=-2 \Rightarrow C=0
\end{aligned}$$\begin{aligned}
& \Rightarrow \tan ^{-1}(\mathrm{x}+\mathrm{y}+2)=\mathrm{x} \
& \mathrm{y}=\tan \mathrm{x}-\mathrm{x}-2 \
& \mathrm{f}(\mathrm{x})=\tan \mathrm{x}-\mathrm{x}-2, \mathrm{x} \in\left[0, \frac{\pi}{3}\right]
\end{aligned}$$\begin{aligned}
& 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
\end{aligned}$now$now(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}$$\begin{aligned}
& \Rightarrow(3 \alpha+\pi)^2+\beta^2=(3 \sqrt{3}-6)^2+4 \
& \gamma+\delta \sqrt{3}=67-36 \sqrt{3} \
& \Rightarrow \gamma=67 \text { and } \delta=-36 \Rightarrow \gamma+\delta=31
\end{aligned}$