JEE Main 2024 — Quadratic Equation Question with Solution
JEE Main 2024 (06 Apr Shift 1)
Question
Let be the distinct roots of the equation and . Then the minimum value of is
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
by newton's theorem
$\begin{aligned}
& \mathrm{a}_{\mathrm{n}+2}-\left(\mathrm{t}^2-5 \mathrm{t}+6\right) \mathrm{a}_{\mathrm{n}+1}+\mathrm{a}_{\mathrm{n}}=0 \\
& \therefore \mathrm{a}_{2025}+\mathrm{a}_{2023}=\left(\mathrm{t}^2-5 \mathrm{t}+6\right) \mathrm{a}_{2024} \\
& \therefore \frac{\mathrm{a}_{2025}+\mathrm{a}_{2023}}{\mathrm{a}_{2024}}=\mathrm{t}^2-5 \mathrm{t}+6 \\
& \because \mathrm{t}^2-5 \mathrm{t}+6=\left(\mathrm{t}-\frac{5}{2}\right)^2-\frac{1}{4} \\
& \therefore \text { minimum value }=-\frac{1}{4}
\end{aligned}$
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This is a previous-year question from JEE Main 2024, covering the Quadratic Equation chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.