Let y=y(x) be the solution of the differential equation (x+y+2)² d x=d y, y(0)=-2. Let the maximum and minimum values of the function y=y(x) in [0, π/3] be α and β, respectively. If (3 α+π)²+β²=γ+δ √3, γ, δ ∈ ℤ, then γ+δ equals _________.

Solution

Maths sir · 4 steps

Put u = x + y + 2 and the equation separates at once; then the solution only ever climbs.

1. Substitute. Let u = x + y + 2, so du/dx = 1 + dy/dx = 1 + u². Now the variables separate: du/(1 + u²) = dx.
2. Integrate. This gives tan⁻¹ u = x + C. At x = 0, y = -2, so u = 0 and C = 0. Hence u = tan x, which means y = tan x - x - 2.
3. Find max and min. y' = sec² x - 1 = tan² x ≥ 0, so y is increasing on [0, π/3]. The minimum is at x = 0: β = -2. The maximum is at x = π/3: α = √3 - π/3 - 2.

   (3α + π)² + β² = (3√3 - 6)² + 4 = 27 - 36√3 + 36 + 4 = 67 - 36√3

4. So. γ = 67 and δ = -36, so γ + δ = 31.

The common slip: 103. Taking δ = 36 instead of -36 drops the minus sign from -36√3 and gives 103.

The idea to keep. When dy/dx is a function of ax + by + c, substitute that whole expression as u. For max and min on an interval, check the sign of y' before trying critical points.

Answer from the NTA final answer key. Solution by PrepWiser.