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4-2.Quadratic Equations and Inequations
hard
ધારોકે $\alpha, \beta$ એ સમીકરણ $x^2-\left(t^2-5 t+6\right) x+1=0, t \in \mathbb{R}$ નાં ભિન્ન બીજ છે અને $a_n=\alpha^n+\beta^n$. તો $\frac{a_{2023}+a_{2025}}{a_{2024}}$ નું ન્યૂનતમ મૂલ્ય .............છે.
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
