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:
- (1) 24
- (2) 25
- (3) 27
- (4) 22
Solution:
Sum of roots $\alpha + \beta = -\frac{\cos\theta}{2}$.
Product of roots $\alpha\beta = -\frac{1}{2}$.
We need $\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 – 2(\alpha\beta)^2$.
First, $\alpha^2+\beta^2 = (\alpha+\beta)^2 – 2\alpha\beta = \frac{\cos^2\theta}{4} – 2(-\frac{1}{2}) = \frac{\cos^2\theta}{4} + 1$.
Now, $\alpha^4 + \beta^4 = \left(\frac{\cos^2\theta}{4} + 1\right)^2 – 2\left(-\frac{1}{2}\right)^2$
$$ = \left(\frac{\cos^2\theta}{4} + 1\right)^2 – \frac{1}{2} $$
To find M and m, we analyze $\cos^2\theta$ which ranges from $[0, 1]$.
Max value (M): Put $\cos^2\theta = 1$.
$$ M = \left(\frac{1}{4} + 1\right)^2 – \frac{1}{2} = \left(\frac{5}{4}\right)^2 – \frac{1}{2} = \frac{25}{16} – \frac{8}{16} = \frac{17}{16} $$
Min value (m): Put $\cos^2\theta = 0$.
$$ m = (0 + 1)^2 – \frac{1}{2} = 1 – \frac{1}{2} = \frac{1}{2} $$
We need $16(M+m)$:
$$ 16\left(\frac{17}{16} + \frac{1}{2}\right) = 17 + 8 = 25 $$
Ans. (2)