Let f: ℝ → ℝ be a function defined by f(x)=4ˣ/(4ˣ+2) and M=∫f(a)ᶠ⁽¹⁻ᵃ⁾ x sin ⁴(x(1-x)) d x, N=∫f(a)ᶠ⁽¹⁻ᵃ⁾ sin ⁴(x(1-x)) d x ; a ≠ 1/2. If α M=β N, α, β ∈ ℕ, then the least value of α²+β² is equal to __________.

Solution

Maths sir · 4 steps

The limits add up to 1, so replace x by 1 − x in M. N then appears and gives N = 2M.

1. Limits add to 1. f(1-a) = 4¹⁻ᵃ/(4¹⁻ᵃ+2) = 4/(4+2· 4ᵃ) = 2/(4ᵃ+2). Adding f(a) = 4ᵃ/(4ᵃ+2) gives f(a)+f(1-a) = 1. So the limits are p and 1-p, where p = f(a).
2. Use the King property. Replace x by p + (1-p) - x = 1-x in M. The factor sin⁴(x(1-x)) does not change, because (1-x)x = x(1-x).

   M = ∫ₚ¹⁻ᵖ (1-x)sin⁴(x(1-x)) dx = N - M ⇒ N = 2M

3. Find α and β. Since a ≠ 1/2, the limits differ. On this range x(1-x) > 0, so N ≠ 0 and also M ≠ 0. Then α M = β N = 2β M gives α = 2β.
4. So. The smallest natural numbers are β = 1 and α = 2. So α² + β² = 4 + 1 = 5.

The idea to keep. When f(a) + f(1-a) = 1, the limits are symmetric about 1/2. With a factor symmetric in x ↔ 1-x, swap x for 1-x: ∫ x g = 1/2∫ g.

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