Let ∫α^(log ₑ 4) (d x)/√(eˣ-1)=π/6. Then e^α and e^(-α) are the roots of the equation :

Solution

Maths sir · 4 steps

Put t = √(eˣ - 1) and the integral turns into a plain 2 tan⁻¹ t.

1. Substitute. Let t = √(eˣ - 1), so eˣ = 1 + t² and eˣ dx = 2t dt. That gives dx = (2t dt)/(1 + t²), and the integrand becomes (2 dt)/(1 + t²).
2. Change the limits. At x = logₑ 4, t = √(4 - 1) = √3. At x = α, t = √(e^α - 1), which we call t₀.

   2 tan⁻¹√3 - 2 tan⁻¹ t₀ = 2π/3 - 2 tan⁻¹ t₀ = π/6

3. Find e^α. So 2 tan⁻¹ t₀ = π/2, which gives tan⁻¹ t₀ = π/4 and t₀ = 1. Then e^α - 1 = 1, so e^α = 2 and e^(-α) = 1/2.
4. So. The roots are 2 and 1/2: sum 5/2, product 1. The equation is x² - 5/2x + 1 = 0, that is 2x² - 5x + 2 = 0, option (a).

The idea to keep. When you see √(eˣ - 1) under an integral, substitute the whole root as t. And roots k and 1/k always have product 1, so the constant term equals the leading coefficient.

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