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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Differential Equations

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation (x+y+2)^2 d...

Let y=y(x)y=y(x) be the solution of the differential equation (x+y+2)2dx=dy,y(0)=−2(x+y+2)^2 d x=d y, y(0)=-2. Let the maximum and minimum values of the function y=y(x)y=y(x) in [0,π3]\left[0, \frac{\pi}{3}\right] be α\alpha and β\beta, respectively. If (3α+π)2+β2=γ+δ3,γ,δ∈Z(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}, \gamma, \delta \in \mathbb{Z}, then γ+δ\gamma+\delta equals ______

Correct answer

0

Step-by-step solution

dydx=(x+y+2)2\frac{d y}{d x}=(x+y+2)^2 ...(1) y(0)=−2y(0)=-2
Let x+y+2=v1+dydx=dvdx\mathrm{x}+\mathrm{y}+2=\mathrm{v}1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}} from (1) dvdx=1+v2\frac{d v}{d x}=1+v^2∫dv1+v2=∫dxtan⁡−1(v)=x+Ctan⁡−1(x+y+2)=x+C at x=0y=−2⇒C=0\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}⇒tan⁡−1(x+y+2)=xy=tan⁡x−x−2f(x)=tan⁡x−x−2,x∈[0,π3]\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}f′(x)=sec⁡2x−1>0⇒f(x)↑fmin⁡=f(0)=−2=βfmax⁡=f(π3)=3−π3−2=α\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 $(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{\text{ and } } \delta=-36 \Rightarrow \gamma+\delta=31 \end{aligned}$$$
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