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JEE Main 2024 Mathematics question

MathematicsQuadratic Equations and InequalitiesJEE Main 2024

Let α, β; α > β, be the roots of the equation x² − √2x − √3 = 0. Let Pₙ = αⁿ − βⁿ, n ∈ ℕ. Then (11√3 − 10√2)P₁₀ + (11√2 + 10)P₁₁ − 11P₁₂ is equal to :

Answer: (B)

Solution

Since α² = √2α + √3 and β² = √2β + √3, Pₙ₊₂ = √2 Pₙ₊₁ + √3 Pₙ. So P₁₂ = √2P₁₁ + √3P₁₀ and P₁₁ = √2P₁₀ + √3P₉. The expression

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Question ID VK-016463