Medium MCQ +4 / -1 PYQ · JEE Mains 2024

Let $\alpha, \beta$ be the distinct roots of the equation $x^2-\left(t^2-5 t+6\right) x+1=0, t \in \mathbb{R}$ and $a_n=\alpha^n+\beta^n$. Then the minimum value of $\frac{a_{2023}+a_{2025}}{a_{2024}}$ is

  1. A $-1 / 2$
  2. B $-1 / 4$ Correct answer
  3. C $1 / 4$
  4. D $1 / 2$

Solution

<p>$$\begin{aligned} & x^2-\left(t^2-5 t+6\right) x+1=0 \\ & \therefore a_{2025}-\left(t^2-5 t+6\right) a_{2024}+a_{2023}=0 \\ & \Rightarrow \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \\ & =\left(t+\frac{5}{2}\right)^2+\left(\frac{-1}{4}\right) \\ & \text { Minimum value }=\frac{-1}{4} \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane

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