Let $\alpha, \beta$ be the distinct roots of the equation $x^2-\left(t^2-5 t+6\right) x+1=0, t \in…

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. $-1 / 4$
  2. $-1 / 4$
  3. $-1 / 2$
  4. $1 / 4$

Solution

by newton's theorem $\begin{aligned} & \mathrm{a}_{\mathrm{n}+2}-\left(\mathrm{t}^2-5 \mathrm{t}+6\right) \mathrm{a}_{\mathrm{n}+1}+\mathrm{a}_{\mathrm{n}}=0 \\ & \therefore \mathrm{a}_{2025}+\mathrm{a}_{2023}=\left(\mathrm{t}^2-5 \mathrm{t}+6\right) \mathrm{a}_{2024} \\ & \therefore \frac{\mathrm{a}_{2025}+\mathrm{a}_{2023}}{\mathrm{a}_{2024}}=\mathrm{t}^2-5 \mathrm{t}+6 \\ & \because \mathrm{t}^2-5 \mathrm{t}+6=\left(\mathrm{t}-\frac{5}{2}\right)^2-\frac{1}{4} \\ & \therefore \text { minimum value }=-\frac{1}{4} \end{aligned}$

Asked in: JEE Main 2024 (06 Apr Shift 1)

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