Let _, _ be the roots of 2x^2 + ()x - 1 = 0, (0,2).

Question

Let $\alpha_	heta, \beta_	heta$ be the roots of $2x^2 + (\cos	heta)x - 1 = 0$, $	heta \in (0,2\pi)$. If $m$ and $M$ are the minimum and the maximum values of $\alpha_	heta^4+\beta_	heta^4$, then $16(M+m)$ equals:

Accepted Answer

$\alpha+\beta = -\dfrac{\cos\theta}{2}$, $\alpha\beta = -\dfrac12$.

$$\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = \frac{\cos^2\theta}{4}+1, \qquad \alpha^2\beta^2 = \frac14$$

$$\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2\alpha^2\beta^2 = \left(\frac{\cos^2\theta}{4}+1\right)^2 - \frac12$$

Since $\cos^2\theta \in [0,1]$, the quantity $\left(\dfrac{\cos^2\theta}{4}+1\right)$ ranges over $\left[1, \dfrac54\right]$.

- Maximum ($\cos^2\theta=1$): $\left(\dfrac54\right)^2 - \dfrac12 = \dfrac{25}{16}-\dfrac{8}{16}=\dfrac{17}{16} = M$
- Minimum ($\cos^2\theta=0$): $1 - \dfrac12 = \dfrac12 = m$

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

Solutions · 1