JEE Main 2025 — Mathematics Algebra

$\begin{aligned} & 2 x^2+(\cos \theta) x-1=0 \\ & \alpha_\theta+\beta_\theta=\frac{-\cos \theta}{2} \\ & \alpha_\theta \cdot \beta_\theta=\frac{-1}{2} \\ & \alpha_\theta^2+\beta_\theta^2=\left(\alpha_\theta+\beta_\theta\right)^2-2 \alpha_\theta \beta_\theta \frac{\cos ^2 \theta}{4}+1\end{aligned}$​2x2+(cosθ)x−1=0αθ​+βθ​=2−cosθ​αθ​⋅βθ​=2−1​αθ2​+βθ2​=(αθ​+βθ​)2−2αθ​βθ​4cos2θ​+1​ $\begin{aligned}
& \alpha_\theta^4+\beta_\theta^4=\left(a_\theta^2+\beta_\theta^2\right)^2-2 \alpha_\theta^2 \beta_\theta^2=\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{2}{4} \
& =\left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{1}{2}
\end{aligned}$Maximum when$Maximumwhen\cos \theta=1$$\begin{aligned}
& M=\left(\frac{1}{4}+1\right)^2-\frac{1}{2} \
& M=\frac{17}{16}
\end{aligned}$Minimum when$Minimumwhen\cos \theta=0$$\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}$