Quadratic Equations
Min/Max of Symmetric Function of Roots
nta_pyq_2025_apr
Grade 11
Question:
Let $\alpha_\theta$ and $\beta_\theta$ be the distinct roots of $2x^2 + (\cos\theta)x - 1 = 0$, $\theta \in [0,2\pi)$. If $m$ and $M$ are respectively the minimum and maximum values of $\alpha_\theta^4 + \beta_\theta^4$, then $16(M+m)$ is equal to
Step-by-Step Solution
Key Concept: Express $\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2$ in terms of $\cos\theta$ using Vieta's formulas, then optimise over $\theta$.
By Vieta's: $\alpha+\beta = -\dfrac{\cos\theta}{2}$, $\alpha\beta = -\dfrac{1}{2}$.
$\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = \dfrac{\cos^2\theta}{4}+1$.
$\alpha^4+\beta^4 = \left(\dfrac{\cos^2\theta}{4}+1\right)^2 - 2\cdot\dfrac{1}{4} = \left(\dfrac{\cos^2\theta}{4}+1\right)^2 - \dfrac{1}{2}$.
Let $u=\cos^2\theta\in[0,1]$. $f(u)=\left(\dfrac{u}{4}+1\right)^2-\dfrac{1}{2}$ is increasing in $u$.
**Maximum** at $u=1$: $M = \left(\dfrac{5}{4}\right)^2-\dfrac{1}{2}=\dfrac{25}{16}-\dfrac{1}{2}=\dfrac{17}{16}$.
**Minimum** at $u=0$: $m = 1-\dfrac{1}{2}=\dfrac{1}{2}$.
$16(M+m)=16\left(\dfrac{17}{16}+\dfrac{1}{2}\right)=17+8=25$.
Correct Answer: 2