Quadratic Equation | JEE Main 2025 Solution

JEE MAIN 2025

22-01-2025 SHIFT-2

Question

Let $\alpha_\theta$ and $\beta_\theta$ be the distinct roots of $2 x^2+(\cos \theta) x-1=0, \theta \in(0,2 \pi)$. If m and M are the minimum and the maximum values of $\alpha_\theta^4+\beta_\theta^4$, then $16(M+m)$ equals :

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Solution

\begin{aligned} & \left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2 \\ & {\left[(\alpha+\beta)^2-2 \alpha \beta\right]^2-2(\alpha \beta)^2} \\ & {\left[\frac{\cos ^2 \theta}{4}+1\right]^2-2 \cdot \frac{1}{4}} \\ & \left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{1}{2} \\ & \mathrm{M}=\frac{25}{16}-\frac{1}{2}=\frac{17}{16} \\ & \mathrm{~m}=\frac{1}{2}, 16(\mathrm{M}+\mathrm{m})=25 \end{aligned}

Question Tags

JEE Main

Mathematics

Easy

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