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4-2.Quadratic Equations and Inequations
hard
Let, $\alpha, \beta$ be the distinct roots of the equation $\mathrm{x}^2-\left(\mathrm{t}^2-5 \mathrm{t}+6\right) \mathrm{x}+1=0, \mathrm{t} \in \mathrm{R}$ and $\mathrm{a}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$. Then the minimum value of $\frac{\mathrm{a}_{2023}+\mathrm{a}_{2025}}{\mathrm{a}_{2024}}$ is
A
$1 / 4$
B
$-1 / 2$
C
$-1 / 4$
D
$1 / 2$
(JEE MAIN-2024)
Solution
by netwton's theroem
$ a_{n+2}-\left(t^2-5 t+6\right) a_{n+1}+a_n=0 $
$ \therefore a_{2025}+a_{2023}=\left(t^2-5 t+6\right) a_{2024} $
$ \therefore \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 $
$ \because t^2-5 t+6=\left(t-\frac{5}{2}\right)^2-\frac{1}{4} $
$ \therefore \text { minimum value }=-\frac{1}{4}$
Standard 11
Mathematics
