The slope of the tangent to a curve C: y=y(x) at any point (x, y) on it is (2 e²ˣ-6 e⁻ˣ+9)/(2+9 e⁻²ˣ). If C passes through the points (0, 1/2+π/(2 √2)) and (α, 1/2 e^(2 α)), then e^α is equal to :

Solution

Maths sir · 5 steps

Multiply through by e²ˣ and the slope splits into e²ˣ plus a neat tan⁻¹ piece.

1. Clear negative powers. Multiply the top and bottom by e²ˣ. The slope becomes (2e⁴ˣ+9e²ˣ-6eˣ)/(2e²ˣ+9). The first two terms are e²ˣ(2e²ˣ+9), so they divide out exactly.

   dy/dx = e²ˣ - 6eˣ/(2e²ˣ+9)

2. Substitute t = e^x. With t = eˣ, dt = eˣ dx, the second part is ∫ (6 dt)/(2t²+9) = 3∫ dt/(t²+9/2) = √2 tan⁻¹(√2 t)/3. So y = 1/2e²ˣ - √2 tan⁻¹(√2 eˣ)/3 + c.
3. Use both points. At x = 0: c = π/2√2 + √2 tan⁻¹√2/3. At x = α the 1/2e^(2α) terms cancel, which leaves √2 tan⁻¹(√2 e^α)/3 = c. Divide by √2: tan⁻¹(√2 e^α)/3 = π/4 + tan⁻¹√2/3.
4. Take tangent. Use tan(A+B) with tan π/4 = 1: (√2 e^α)/3 = (1+√2/3)/(1-√2/3) = (3+√2)/(3-√2).
5. So. e^α = 3/√2((3+√2)/(3-√2)), which is option (b).

The idea to keep. When a slope mixes e²ˣ, eˣ and e⁻ˣ, first multiply through to clear the negative powers. Then look for a factor that divides out before you substitute t = eˣ.

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