Let S={sin ² 2 θ:(sin ⁴ θ+cos ⁴ θ) x²+(sin 2 θ) x+(sin ⁶ θ+cos ⁶ θ)=0 has real roots }. If α and β be the smallest and largest elements of the set S, respectively, then 3((α-2)²+(β-1)²) equals __________.

Solution

Maths sir · 4 steps

Write every coefficient in terms of t = sin² 2θ, then remember t can never go above 1.

1. Rewrite coefficients. Use sin² θ cos² θ = t/4 with t = sin² 2θ. Then sin⁴ θ + cos⁴ θ = 1 - 2 sin² θ cos² θ = 1 - t/2, and sin⁶ θ + cos⁶ θ = 1 - 3 sin² θ cos² θ = 1 - 3t/4.
2. Real roots condition. The leading coefficient 1 - t/2 is at least 1/2, so the equation is a true quadratic. Real roots need D ≥ 0, where b² = sin² 2θ = t. Expanding 4(1 - t/2)(1 - 3t/4) = 4 - 5t + 3t²/2 gives the inequality below.

   t - (4 - 5t + 3t²/2) ≥ 0 ⇒ 3t² - 12t + 8 ≤ 0

3. Solve for t. The roots are t = (12 ± √48)/6 = 2 ± 2/√3, so 2 - 2/√3 ≤ t ≤ 2 + 2/√3. But t = sin² 2θ also satisfies 0 ≤ t ≤ 1. Since 2 - 2/√3 ≈ 0.845, the set is S = [2 - 2/√3, 1].
4. So. α = 2 - 2/√3 and β = 1. That gives (α - 2)² = 4/3 and (β - 1)² = 0, so 3(4/3 + 0) = 4.

The idea to keep. When a trig expression is used as a variable, solve the inequality first, then always cut it down to that expression's own range. Here sin² 2θ lives in [0, 1].

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