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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Quadratic Equation

Let α, β be the distinct roots of the equation x^2-(t^2-5 t+6)...

Let α,β\alpha, \beta be the distinct roots of the equation x2−(t2−5t+6)x+1=0,t∈Rx^2-\left(t^2-5 t+6\right) x+1=0, t \in \mathbb{R} and an=αn+βna_n=\alpha^n+\beta^n. Then the minimum value of a2023+a2025a2024\frac{a_{2023}+a_{2025}}{a_{2024}} is

Options

  1. A−1/4-1 / 4
  2. B−1/4-1 / 4
  3. C−1/2-1 / 2
  4. D1/41 / 4

Correct answer

B. −1/4-1 / 4

Step-by-step solution

by newton's theorem an+2−(t2−5t+6)an+1+an=0∴a2025+a2023=(t2−5t+6)a2024∴a2025+a2023a2024=t2−5t+6∵t2−5t+6=(t−52)2−14∴ minimum value =−14\begin{aligned} & \mathrm{a}_{\mathrm{n}+2}-\left(\mathrm{t}^2-5 \mathrm{t}+6\right) \mathrm{a}_{\mathrm{n}+1}+\mathrm{a}_{\mathrm{n}}=0 \\ & \therefore \mathrm{a}_{2025}+\mathrm{a}_{2023}=\left(\mathrm{t}^2-5 \mathrm{t}+6\right) \mathrm{a}_{2024} \\ & \therefore \frac{\mathrm{a}_{2025}+\mathrm{a}_{2023}}{\mathrm{a}_{2024}}=\mathrm{t}^2-5 \mathrm{t}+6 \\ & \because \mathrm{t}^2-5 \mathrm{t}+6=\left(\mathrm{t}-\frac{5}{2}\right)^2-\frac{1}{4} \\ & \therefore \text{ minimum value }=-\frac{1}{4} \end{aligned}
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