The given function is:
\[ f(\theta) = \frac{\sin^4\theta + 3\cos^2\theta}{\sin^4\theta + \cos^2\theta}. \]
Substitute \(\cos^2\theta = x\), with \(x \in [0, 1]\):
\[ f(x) = \frac{\sin^4\theta + 3x}{\sin^4\theta + x}. \]
Simplify:
\[ f(x) = \frac{2x}{x + 1} + 1. \]
The range of \(f(x)\) can be computed as:
\[ f_{\min} = 1, \quad f_{\max} = 3. \]
Thus:
\[ \alpha = 1, \quad \beta = 3. \]
The infinite geometric series is:
\[ S = \frac{\text{first term}}{1 - \text{common ratio}}. \]
Substitute:
\[ S = \frac{64}{1 - \frac{1}{3}} = \frac{64}{\frac{2}{3}} = 64 \cdot \frac{3}{2} = 96. \]
Final Answer: 96.
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: