https://file.aiquickdraw.com/mathbot/file/1725010188155juaebftf.jpg

options are A.   A E (0,2)        B.     A E (0,1)                C.         A E [0,1]          D. A E [-1,1]

The problem asks us to find the values of \(\alpha\) for which the expression 

\[
\frac{\alpha^2 + 2\alpha + \alpha}{\alpha^2 + 4\alpha + 3\alpha}
\]

can take all real values.

First, let's simplify the given expression.

### Simplification of the Expression:
The expression can be rewritten as:

\[
\frac{\alpha^2 + 3\alpha}{\alpha^2 + 7\alpha + 3}
\]

Now, for this rational function to take all real values, the denominator must never be zero, and the function must be able to achieve any real value. This means the range of the function should be all real numbers.

### Analyzing the Expression:
1. **Denominator Condition:** The denominator \( \alpha^2 + 7\alpha + 3 \) should not be zero. Let's find the roots of this quadratic equation:
   \[
   \alpha^2 + 7\alpha + 3 = 0
   \]
   The discriminant of this quadratic is:
   \[
   \Delta = 7^2 - 4 \times 1 \times 3 = 49 - 12 = 37
   \]
   Since the discriminant is positive, the quadratic equation has two real roots. However, the exact roots do not need to be computed unless required.

2. **Range of the Expression:** To find the range, we analyze the function:
   \[
   y = \frac{\alpha^2 + 3\alpha}{\alpha^2 + 7\alpha + 3}
   \]
   We need to ensure that this expression can take all real values. One approach is to check the behavior as \(\alpha\) approaches specific values (like large positive or negative values) and also check if there's any restriction that limits the range of \(y\).

   The expression can take all real values if the numerator and the denominator behave similarly as \(\alpha\) becomes large, without introducing asymptotes that limit the range. This occurs if the function can be manipulated into a linear form over large ranges of \(\alpha\), ensuring the slope does not limit the range.

   The correct condition to allow the expression to take all real values will depend on ensuring that the denominator does not introduce additional zeroes that would restrict the range.

### Conclusion:
The correct condition for \(\alpha\) can be analyzed through the quadratic discriminant of the resulting function, as well as ensuring that the function does not introduce restrictions based on specific ranges of \(\alpha\).

Given the choices:
- **Option A: \(\alpha \in (0,2)\)** appears correct because it limits \(\alpha\) in a way that does not introduce roots or restrictions.
- The other options introduce ranges that either restrict \(\alpha\) too much or not enough, leading to potential undefined points.

Thus, the answer is **Option A: \(\alpha \in (0,2)\)**.

Would you like any further details or clarifications on this solution? Here are some related questions:

1. How do we solve for the roots of a quadratic equation?
2. What are the properties of a quadratic discriminant?
3. How do asymptotes influence the range of a function?
4. What is the significance of the denominator in a rational function?
5. How do we find the range of a rational function?

**Tip:** Always check the discriminant when dealing with quadratic expressions to understand the nature of the roots, which can affect the range and domain of a function.

Explore how to determine the values of alpha for which a specific rational function takes all real values. This problem involves simplifying the expression, analyzing the denominator conditions, and finding the range of the function. Suitable for advanced high school students.

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