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 ________

To find the range, let's simplify the expression by converting everything to a single trigonometric ratio.

Let $$x = \sin^2 \theta$$. Since $$\theta \in \mathbb{R}$$, we know the restricted domain for $$x$$ is $$x \in [0, 1]$$.

Substitute $$\cos^2 \theta = 1 - x$$ into the function:

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

To find the range of this function on the interval $$[0, 1]$$, let's check its monotonicity by finding the derivative $$f'(x)$$ using the quotient rule:

$$f'(x) = \frac{(2x - 3)(x^2 - x + 1) - (x^2 - 3x + 3)(2x - 1)}{(x^2 - x + 1)^2}$$

Expanding the numerator:

$$(2x^3 - 5x^2 + 5x - 3) - (2x^3 - 7x^2 + 9x - 3) = 2x^2 - 4x = 2x(x - 2)$$

So, the derivative is:

$$f'(x) = \frac{2x(x - 2)}{(x^2 - x + 1)^2}$$

For our domain $$x \in [0, 1]$$, the term $$2x$$ is positive (or zero), and $$(x - 2)$$ is strictly negative. Therefore, $$f'(x) \le 0$$ on the entire interval $$[0, 1]$$.

Because $$f(x)$$ is strictly decreasing on $$[0, 1]$$, its maximum and minimum values must occur at the boundary points:

• Maximum value ($$\beta$$): $$f(0) = \frac{0 - 0 + 3}{0 - 0 + 1} = 3$$
• Minimum value ($$\alpha$$): $$f(1) = \frac{1 - 3 + 3}{1 - 1 + 1} = 1$$

The range is $$[\alpha, \beta] = [1, 3]$$.

So, $$\alpha = 1$$ and $$\beta = 3$$.

Now for the second part of the question. We have an infinite Geometric Progression (G.P.) with:

• First term $$a = 64$$
• Common ratio $$r = \frac{\alpha}{\beta} = \frac{1}{3}$$

Using the formula for the sum of an infinite G.P., $$S_{\infty} = \frac{a}{1 - r}$$:

$$S_{\infty} = \frac{64}{1 - \frac{1}{3}} = \frac{64}{\frac{2}{3}} = \frac{64 \times 3}{2} = 96$$