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JEE Main 2024
09-04-2024 S2
Question
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 $(11 \sqrt{3}- 10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12}$ is equal to :
Select the correct option:
A
$10 \sqrt{2} \mathrm{P}_9$
B
$10 \sqrt{3} \mathrm{P}_9$
C
$11 \sqrt{2} P_9$
D
$11 \sqrt{3} \mathrm{P}_9$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$ \begin{aligned} & 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 \end{aligned} $
and $\beta^{\mathrm{n}+2}-\sqrt{2} \beta^{\mathrm{n}+1}-\sqrt{3} \beta^{\mathrm{n}}=0$
Subtracting
$ \begin{aligned} & \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 \mathrm{P}_{\mathrm{n}+2}-\sqrt{2} \mathrm{P}_{\mathrm{n}+1}-\sqrt{3} \mathrm{P}_{\mathrm{n}}=0 \end{aligned} $
Put $\mathrm{n}=10$
$ \mathrm{P}_{12}-\sqrt{2} \mathrm{P}_{11}-\sqrt{3} \mathrm{P}_{10}=0 $
$\begin{aligned} & n=9 \\ & P_{11}-\sqrt{2} P_{10}-\sqrt{3} P_9=0 \\ & \begin{aligned} 11\left(\sqrt{3} \cdot P_{10}+\sqrt{2} P_{11}-P_{11}\right)-10\left(\sqrt{2} P_{10}\right. & \left.-P_{11}\right) \\ & =0-10\left(-\sqrt{3} P_9\right)=10 \sqrt{3} P_9\end{aligned}\end{aligned}$
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