Key Concepts and Formulas:

1. Trigonometric Simplification: Utilize identities like $\sin^2 \theta + \cos^2 \theta = 1$ to transform trigonometric expressions into algebraic forms.
2. Domain of Trigonometric Functions: For $\sin^2 \theta$ and $\cos^2 \theta$, their values are restricted to $[0, 1]$ for $\theta \in \mathbb{R}$.
3. Range of a Function: To find the range of a function on a closed interval, evaluate the function at the endpoints and any critical points within the interval. The minimum and maximum of these values constitute the range.
4. Calculus for Extrema: The derivative of a function helps identify critical points where the function might attain maximum or minimum values. The sign of the derivative indicates whether the function is increasing or decreasing.
5. Sum of an Infinite Geometric Progression (G.P.): If the first term is $a$ and the common ratio is $r$, the sum $S$ exists if $|r| < 1$, and is given by $S = \frac{a}{1-r}$.

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Step 1: Simplify the Trigonometric Function

The given function is $f(\theta)=\frac{\sin ^4 \theta+3 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta}$. To find its range, we can simplify it by making a substitution. Let $x = \sin^2 \theta$. Since $\theta \in \mathbb{R}$, the possible values for $\sin^2 \theta$ are between 0 and 1, inclusive. Thus, the domain for $x$ is $[0, 1]$. Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$, we have $\cos^2 \theta = 1 - x$. Also, $\sin^4 \theta = (\sin^2 \theta)^2 = x^2$.

Substituting these into the function $f(\theta)$: $f(x) = \frac{x^2 + 3(1-x)}{x^2 + (1-x)}$ Now, simplify the numerator and the denominator: Numerator: $x^2 + 3 - 3x = x^2 - 3x + 3$ Denominator: $x^2 + 1 - x = x^2 - x + 1$

So, the function in terms of $x$ is: $f(x) = \frac{x^2 - 3x + 3}{x^2 - x + 1}, \quad \text{for } x \in [0, 1]$

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Step 2: Determine the Range of $f(x)$ on $[0, 1]$

To find the range of $f(x)$ on the interval $[0, 1]$, we will analyze its behavior using calculus. First, we evaluate the function at the endpoints of the interval.

At $x = 0$: $f(0) = \frac{0^2 - 3(0) + 3}{0^2 - 0 + 1} = \frac{3}{1} = 3$

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

Next, we find the derivative of $f(x)$ to identify any critical points. Using the quotient rule, $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, with $u = x^2 - 3x + 3$ and $v = x^2 - x + 1$. Then $u' = 2x - 3$ and $v' = 2x - 1$.

$f'(x) = \frac{(2x - 3)(x^2 - x + 1) - (x^2 - 3x + 3)(2x - 1)}{(x^2 - x + 1)^2}$ Expanding the numerator: $(2x^3 - 2x^2 + 2x - 3x^2 + 3x - 3) - (2x^3 - x^2 - 6x^2 + 3x + 6x - 3)$ $(2x^3 - 5x^2 + 5x - 3) - (2x^3 - 7x^2 + 9x - 3)$ $2x^3 - 5x^2 + 5x - 3 - 2x^3 + 7x^2 - 9x + 3 = 2x^2 - 4x$ So, the derivative is: $f'(x) = \frac{2x^2 - 4x}{(x^2 - x + 1)^2} = \frac{2x(x - 2)}{(x^2 - x + 1)^2}$

To find critical points, we set $f'(x) = 0$: $2x(x - 2) = 0 \implies x = 0$ or $x = 2$.

We are interested in critical points within the interval $[0, 1]$. The only critical point from the derivative calculation that falls within or at the boundary of our domain is $x=0$. The value $x=2$ is outside the domain $[0, 1]$.

Now, let's analyze the sign of $f'(x)$ for $x \in (0, 1)$. The denominator $(x^2 - x + 1)^2$ is always positive because the discriminant of $x^2 - x + 1$ is $(-1)^2 - 4(1)(1) = -3 < 0$, and the leading coefficient is positive. For $x \in (0, 1)$:

• $2x$ is positive.
• $(x - 2)$ is negative (e.g., for $x=0.5$, $x-2 = -1.5$). Therefore, $f'(x) = \frac{(\text{positive})(\text{negative})}{(\text{positive})} < 0$ for $x \in (0, 1)$.

Since $f'(x) < 0$ on $(0, 1)$, the function $f(x)$ is strictly decreasing on the interval $[0, 1]$. For a strictly decreasing function on a closed interval, the maximum value occurs at the left endpoint and the minimum value occurs at the right endpoint. Maximum value: $f(0) = 3$. Minimum value: $f(1) = 1$.

Thus, the range of $f(\theta)$ is $[1, 3]$.

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Step 3: Identify $\alpha$ and $\beta$

The problem states that the range of $f(\theta)$ is $[\alpha, \beta]$. From our analysis, the range is $[1, 3]$. Therefore, $\alpha = 1$ and $\beta = 3$.

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Step 4: Calculate the Sum of the Infinite G.P.

We are given an infinite G.P. with: First term, $a = 64$. Common ratio, $r = \frac{\alpha}{\beta}$.

Substituting the values of $\alpha$ and $\beta$: $r = \frac{1}{3}$

We must check if the sum converges. The condition for convergence is $|r| < 1$. Here, $|r| = \left|\frac{1}{3}\right| = \frac{1}{3}$, which is less than 1. So, the sum converges.

The sum of an infinite G.P. is given by the formula $S = \frac{a}{1-r}$. $S = \frac{64}{1 - \frac{1}{3}}$ $S = \frac{64}{\frac{3}{3} - \frac{1}{3}}$ $S = \frac{64}{\frac{2}{3}}$ $S = 64 \times \frac{3}{2}$ $S = 32 \times 3$ $S = 96$

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Common Mistakes & Tips:

• Domain of Substitution: Always determine the correct domain for the substituted variable (e.g., $x = \sin^2 \theta \in [0, 1]$).
• Monotonicity: Ensure the function is monotonic on the interval before concluding that endpoints give the extrema. Use derivatives to confirm.
• G.P. Convergence: Verify that the absolute value of the common ratio is strictly less than 1 before applying the infinite G.P. sum formula.

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Summary

The problem required finding the range of a trigonometric function. By substituting $x = \sin^2 \theta$, the function was transformed into a rational algebraic function $f(x) = \frac{x^2 - 3x + 3}{x^2 - x + 1}$ with the domain $x \in [0, 1]$. Calculus was used to analyze the derivative $f'(x) = \frac{2x(x - 2)}{(x^2 - x + 1)^2}$, which showed that $f(x)$ is strictly decreasing on $[0, 1]$. This established the range as $[f(1), f(0)] = [1, 3]$. Thus, $\alpha = 1$ and $\beta = 3$. Finally, the sum of an infinite G.P. with first term 64 and common ratio $\frac{\alpha}{\beta} = \frac{1}{3}$ was calculated using the formula $S = \frac{a}{1-r}$, yielding a sum of 96.

The final answer is $\boxed{96}$.