JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
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 \(1 / 2\)
Answer & Solution
Correct Answer
(C) \(-1 / 4\)
Step-by-step Solution
Detailed explanation
by netwton's theroem \( a_{n+2}-\left(t^2-5 t+6\right) a_{n+1}+a_n=0 \) \( \therefore a_{2025}+a_{2023}=\left(t^2-5 t+6\right) a_{2024} \) \( \therefore \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \) \( \because t^2-5 t+6=\left(t-\frac{5}{2}\right)^2-\frac{1}{4} \)…
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