If f: ℝ → ℝ be a continuous function satisfying ∫0^(π/2) f(sin 2 x) sin x d x+α ∫0^(π/4) f(cos 2 x) cos x d x=0, then the value of α is :

Solution

Maths sir · 4 steps

Reflect both integrals with the King property and they become the same integral.

1. Fold the first integral. Call the first integral I₁. Replace x by π/2-x: sin 2x does not change and sin x becomes cos x. So I₁ = ∫₀^(π/2) f(sin 2x)cos x dx as well. Adding the two forms gives 2I₁ = ∫₀^(π/2) f(sin 2x)(sin x+cos x) dx.
2. Halve the interval. The new integrand does not change under x → π/2-x, so it is symmetric about π/4. That gives I₁ = ∫₀^(π/4) f(sin 2x)(sin x+cos x) dx.
3. Reflect the second. Call the second integral I₂. Replace x by π/4-x: cos 2x becomes sin 2x, and cos x becomes cos(π/4-x) = (cos x+sin x)/√2.

   I₂ = 1/√2∫₀^(π/4) f(sin 2x)(sin x+cos x) dx = I₁/√2

4. So. The condition becomes I₁ + α/√2 I₁ = 0. This must hold for every such f, including ones with I₁ ≠ 0, so 1 + α/√2 = 0 and α = -√2, option (c).

The idea to keep. When two integrals look unrelated, use ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a-x) dx on each. Pick the reflection that makes their integrands match, then compare them directly.

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