JEE Main 2021 · 25 February · evening · Single correct · +4 / −1
Let α and β be the roots of x² - 6x - 2 = 0. If aₙ = αⁿ - βⁿ for n ≥ 1, then the value of (a₁₀ - 2a₈)/3a₉ is :
- (a) 3
- (b) 2
- (c) 4
- (d) 1
Answer and solution
Answer (b) 2
Solution
Maths sir · 4 steps
Do not find the roots: use α² = 6α + 2 and multiply through by α⁸.
- Use the equation. Since α is a root of x² - 6x - 2 = 0, we have α² = 6α + 2. In the same way, β² = 6β + 2.
- Raise the powers. Multiply the first relation by α⁸ to get α¹⁰ = 6α⁹ + 2α⁸. Similarly, β¹⁰ = 6β⁹ + 2β⁸.
- Subtract. Subtract the second from the first and group the terms into aₙ = αⁿ - βⁿ. This gives a recurrence.
a₁₀ = 6a₉ + 2a₈ ⇒ a₁₀ - 2a₈ = 6a₉
- So. (a₁₀ - 2a₈)/3a₉ = 6a₉/3a₉ = 2, which is option (b).
The idea to keep. When aₙ = αⁿ ± βⁿ and the roots satisfy x² = px + q, then aₙ₊₂ = paₙ₊₁ + qaₙ. Use the equation itself and never compute the actual roots.
Answer from the NTA final answer key. Solution by PrepWiser.