Let a curve y = y(x) be given by the solution of the differential equation cos ( 1/2cos ⁻¹(e⁻ˣ) )dx = √(e²ˣ - 1) dy. If it intersects y-axis at y = -1, and the intersection point of the curve with x-axis is (α, 0), then e α is equal to __________________.

Solution

Maths sir · 4 steps

Call the inverse cosine θ and use the half-angle formula. The ugly cosine becomes a plain square root.

1. Half-angle. Let θ = cos⁻¹(e⁻ˣ), so cos θ = e⁻ˣ and 0 ≤ θ ≤ π/2. Then cos θ/2 = √((1+cos θ)/2) = √((1+e⁻ˣ)/2) = √((eˣ+1)/2eˣ). The positive root is right because θ/2 lies in the first quadrant.
2. Separate variables. Write e²ˣ-1 = (eˣ-1)(eˣ+1). The factor √(eˣ+1) cancels, so dy = dx/(√2 √eˣ √(eˣ-1)) = (e⁻ˣ dx)/(√2 √(1-e⁻ˣ)).
3. Integrate. Put u = e⁻ˣ, so du = -e⁻ˣ dx. Then ∫ (e⁻ˣ dx)/√(1-e⁻ˣ) = ∫ -du/√(1-u) = 2√(1-u). The curve passes through (0, -1), so -1 = 0 + C and C = -1.

   y = √2 √(1-e⁻ˣ) - 1

4. So. At (α, 0) we need √2 √(1-e^(-α)) = 1. So 1 - e^(-α) = 1/2, which gives e^(-α) = 1/2 and e^α = 2.

The common slip: 0.5. Stopping at e^(-α) = 1/2 gives 0.5. The question asks for e^α, which is the reciprocal, 2.

The idea to keep. When you see cos(1/2 cos⁻¹ t), name the angle and use cos θ/2 = √((1+cos θ)/2). Then factor the other side and look for what cancels.

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