JEE Main 2025 Jan 22 Shift 2, Mathematics Q12: Quadratic Equations

By Vieta's formulas on $2x^2+(\cos \theta )x-1=0$: $\alpha +\beta =-\frac{\cos \theta }{2}$ and $\alpha \beta =-\frac{1}{2}$.

Express ${\alpha }^4+{\beta }^4$ using power-sum identity:

${\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2-2{\alpha }^2{\beta }^2$ where ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =\frac{{\cos }^2\theta }{4}+1$.

Therefore:

${\alpha }^4+{\beta }^4={\left(\frac{{\cos }^2\theta }{4}+1\right)}^2-2\cdot \frac{1}{4}={\left(\frac{{\cos }^2\theta }{4}+1\right)}^2-\frac{1}{2}$.

Let $u={\cos }^2\theta \in [0,1]$. The expression ${\left(\frac{u}{4}+1\right)}^2-\frac{1}{2}$ is increasing in $u$.

Minimum at $u=0$: $m=1-\frac{1}{2}=\frac{1}{2}$.

Maximum at $u=1$: $M={\left(\frac{5}{4}\right)}^2-\frac{1}{2}=\frac{25}{16}-\frac{8}{16}=\frac{17}{16}$.

$16(M+m)=16\left(\frac{17}{16}+\frac{1}{2}\right)=17+8=25$.