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 in N . Then (11 √(3)-10 √(2)) P 10 +(11 √(2)+10) P 11 -11 P 12 is equal to :

Mathematics · Quadratic Equations

JEE Main 2024 — 9 April, Shift 2 — Question 19

Let $\alpha, \beta ; \alpha>\beta$ , be the roots of the equation $x^{2}-\sqrt{2} x-\sqrt{3}=0$ .

Let $P_{n}=\alpha^{n}-\beta^{n}, n \in N$ . Then $(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$ is equal to :

Option A:
$10 \sqrt{2} P_{9}$
Option B:
$10 \sqrt{3} \mathrm{P}_{9}$
Correct
Option C:
$11 \sqrt{2} P_{9}$
Option D:
$11 \sqrt{3} P_{9}$

Answer: B

Step-by-step solution

$\quad x^{2}-\sqrt{2 x}-\sqrt{3}=0\left\langle_{\beta}^{\alpha}\right.$

$\alpha^{\mathrm{n}+2}-\sqrt{2} \alpha^{\mathrm{n}+1}-\sqrt{3} \alpha^{\mathrm{n}}=0$ and $\beta^{\mathrm{n}+2}-\sqrt{2} \beta^{\mathrm{n}+1}-\sqrt{3} \beta^{\mathrm{n}}=0$

Subtracting $\left(\alpha^{n+2}-\beta^{n+2}\right)-\sqrt{2}\left(\alpha^{n+1}-\beta^{n+1}\right)-\sqrt{3}\left(\alpha^{n}-\beta^{n}\right)=0$

$\Rightarrow P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_{n}=0$

Put $\mathrm{n}=10$

$P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0$

Put $\mathrm{n}=9$

$P_{11}-\sqrt{2} P_{10}-\sqrt{3} P_{9}=0$

$11\left(\sqrt{3} \cdot \mathrm{P}_{10}+\sqrt{2} \mathrm{P}_{11}-\mathrm{P}_{11}\right)-10\left(\sqrt{2} \mathrm{P}_{10}-\mathrm{P}_{11}\right)$

$=0-10\left(-\sqrt{3} \mathrm{P}_{9}\right)=10 \sqrt{3} \mathrm{P}_{9}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Theory of Quadratic Equations