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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Quadratic Equation

Let α, β ; α>β , be the roots of the equation x^2- 2 x- 3 =0 . Let...

Let α,β;α>β\alpha, \beta ; \alpha>\beta, be the roots of the equation x2−2x−3=0x^2-\sqrt{2} x-\sqrt{3}=0. Let Pn=αn−βn,n∈N\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{N}. Then (113−102)P10+(112+10)P11−11P12(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12} is equal to

Options

  1. A103P910 \sqrt{3} \mathrm{P}_9
  2. B113P911 \sqrt{3} P_9
  3. C102P910 \sqrt{2} \mathrm{P}_9
  4. D112P911 \sqrt{2} \mathrm{P}_9

Correct answer

A. 103P910 \sqrt{3} \mathrm{P}_9

Step-by-step solution

x2−2x−3=0⟨βααn+2−2αn+1−3αn=0 and  βn+2−2βn+1−3βn=0\begin{aligned} & \mathrm{x}^2-\sqrt{2 \mathrm{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 \\ & \text{\text{ and } } \beta^{\mathrm{n}+2}-\sqrt{2} \beta^{\mathrm{n}+1}-\sqrt{3} \beta^{\mathrm{n}}=0 \end{aligned}Subtracting (αn+2−βn+2)−2(αn+1−βn+1)−3(αn−βn)=0⇒Pn+2−2Pn+1−3Pn=0\begin{aligned} & \left(\alpha^{\mathrm{n}+2}-\beta^{\mathrm{n}+2}\right)-\sqrt{2}\left(\alpha^{\mathrm{n}+1}-\beta^{\mathrm{n}+1}\right)-\sqrt{3}\left(\alpha^{\mathrm{n}}-\beta^{\mathrm{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 n=10P12−2P11−3P10=0n=9P11−2P10−3P9=011(3⋅P10+2P11−P11)−10(2P10−P11)=0−10(−3P9)=103P9\begin{aligned} & \text{ Put } \mathrm{n}=10 \\ & \mathrm{P}_{12}-\sqrt{2} \mathrm{P}_{11}-\sqrt{3} \mathrm{P}_{10}=0 \\ & \mathrm{n}=9 \\ & \mathrm{P}_{11}-\sqrt{2} \mathrm{P}_{10}-\sqrt{3} \mathrm{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\end{aligned}$
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