JEE Main 2025 — Quadratic Equation Question with Solution

$\begin{array}{rl}& 2x^2+(\cos \theta )x-1=0\\ & {\alpha }_{\theta }+{\beta }_{\theta }=\frac{-\cos \theta }{2}\\ & {\alpha }_{\theta }\cdot {\beta }_{\theta }=\frac{-1}{2}\\ & {\alpha }_{\theta }^2+{\beta }_{\theta }^2={\left({\alpha }_{\theta }+{\beta }_{\theta }\right)}^2-2{\alpha }_{\theta }{\beta }_{\theta }\frac{{\cos }^2\theta }{4}+1\end{array}$ $\begin{aligned} & \alpha_\theta^4+\beta_\theta^4=\left(a_\theta^2+\beta_\theta^2\right)^2-2 \alpha_\theta^2 \beta_\theta^2=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{2}{4} \\ & =\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{1}{2} \end{aligned}$ Maximum when $\cos \theta =1$ $\begin{aligned} & M=\left(\frac{1}{4}+1\right)^2-\frac{1}{2} \\ & M=\frac{17}{16} \end{aligned}$ Minimum when $\cos \theta =0$ $\begin{aligned} & m=1-\frac{1}{2}=\frac{1}{2} \\ & 16(M+m)=16\left(\frac{17}{16}+\frac{1}{2}\right)=25 \end{aligned}$