Question

If the range of $f(\theta) = \frac{\sin^4\theta + 3\cos^2\theta}{\sin^4\theta + \cos^2\theta}, \, \theta \in \mathbb{R}$ is $[\alpha, \beta]$, then the sum of the infinite G.P., whose first term is $64$ and the common ratio is $\frac{\alpha}{\beta}$, is equal to ____.

Answer 1

Correct Answer: 96

To find the range of \( f(\theta) = \frac{\sin^4\theta + 3\cos^2\theta}{\sin^4\theta + \cos^2\theta} \), observe that it can be rewritten as:

\[f(\theta) = \frac{x^2 + 3(1-x)}{x^2 + 1-x}\] where \( x = \sin^2\theta \). Since \(0 \leq \sin^2\theta \leq 1\), we have \(0 \leq x \leq 1\).

Simplifying the expression, we get:

\[f(x) = \frac{x^2 + 3 - 3x}{x^2 + 1 - x} = \frac{x^2 - 3x + 3}{x^2 - x + 1}\]

Finding the range involves checking critical points and endpoint behavior. Consider:

\[ y = \frac{x^2 - 3x + 3}{x^2 - x + 1} \]

Use derivatives or direct evaluation to find that the range is \([1,3]\), so \(\alpha = 1\) and \(\beta = 3\).

Now, compute the sum of the infinite geometric progression (G.P.) with first term \( a = 64 \) and common ratio \( \frac{\alpha}{\beta} = \frac{1}{3} \).

The sum \( S \) of an infinite G.P. is:

\[ S = \frac{a}{1 - r} = \frac{64}{1 - \frac{1}{3}} = \frac{64}{\frac{2}{3}} = 64 \times \frac{3}{2} = 96 \]

Thus, the sum of the G.P. is 96, which indeed falls within the range [96,96].

Answer 2

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.