Mathematics · 2024
JEE Main · 9 April 2024, Shift 2 · Q3
Let α, β; α>β, be the roots of the equation x^2-√ 2 x-√ 3 =0. Let P_n=α^n-β^n, n ∈ N. Then (11 √ 3 -10 √ 2 ) P_10+(11 √ 2 +10) P_11-11 P_12 is equal…
Let $\displaystyle \alpha, \beta ; \alpha>\beta$, be the roots of the equation $\displaystyle x^2-\sqrt{2} x-\sqrt{3}=0$. Let $\displaystyle \mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{~N}$. Then $\displaystyle (11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$ is equal to
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 10 \sqrt{3} \mathrm{P}_9$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.