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 the minimum and the maximum values of $\alpha_{\theta}^{4}+\beta_{\theta}^{4}$, then $16(M+m)$ equals:
Step-by-Step Solution
Key Concept: Use Newton's identities: with $s=\alpha+\beta$, $p=\alpha\beta$, $\alpha^{2}+\beta^{2}=s^{2}-2p$ and $\alpha^{4}+\beta^{4}=(s^{2}-2p)^{2}-2p^{2}$. Parametrize by $u=\cos^{2}\theta\in[0,1]$ and optimize over $u$.
Vieta: $\alpha+\beta=-\dfrac{\cos\theta}{2}$, $\alpha\beta=-\dfrac{1}{2}$. So
$$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta = \frac{\cos^{2}\theta}{4}+1.$$
And
$$\alpha^{4}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2}-2(\alpha\beta)^{2} = \left(\frac{\cos^{2}\theta}{4}+1\right)^{2}-\frac{1}{2}.$$
Let $u=\cos^{2}\theta\in[0,1]$ (the value $u=1$ is attained at $\theta=\pi$, well inside $(0,2\pi)$).
At $u=0$:\quad $m = 1^{2}-\tfrac{1}{2}=\tfrac{1}{2}$.
At $u=1$:\quad $M = \left(\tfrac{5}{4}\right)^{2}-\tfrac{1}{2}=\tfrac{25}{16}-\tfrac{8}{16}=\tfrac{17}{16}$.
Therefore
$$16(M+m)=16\!\left(\frac{17}{16}+\frac{1}{2}\right)=17+8=25.$$
Correct Answer: 2