JEE MAIN - Mathematics (2024 - 6th April Morning Shift - No. 11)
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
$$-1 / 2$$
$$-1 / 4$$
$$1 / 4$$
$$1 / 2$$
Explanation
$$\begin{aligned}
& x^2-\left(t^2-5 t+6\right) x+1=0 \\
& \therefore a_{2025}-\left(t^2-5 t+6\right) a_{2024}+a_{2023}=0 \\
& \Rightarrow \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \\
& =\left(t+\frac{5}{2}\right)^2+\left(\frac{-1}{4}\right) \\
& \text { Minimum value }=\frac{-1}{4}
\end{aligned}$$
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