Let f be a strictly decreasing function defined on R such that f(x) `gt` 0 `AA` x `in` R, and `x^2/(f(a^2+5a+3))+y^2/(f(3a+15))` =1 represents an ellipse with major axis along the y-axis , then

Options

A. `a cancelin (-oo, -6)`
B. `a cancelin(2,oo)`
C. `a cancelin(-6,2)`
D. `a gt 0`

Answer

To solve the problem, we need to analyze the given equation of the ellipse and the properties of the function \( f \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f \) is strictly decreasing and positive for all \( x \in \mathbb{R} \). This means that if \( x_1 < x_2 \), then \( f(x_1) > f(x_2) \) and \( f(x) > 0 \) for all \( x \).

**Hint**: Remember that a strictly decreasing function will have larger values for smaller inputs.

2. **Equation of the Ellipse**:
The equation given is:
\[ \frac{x^2}{f(a^2 + 5a + 3)} + \frac{y^2}{f(3a + 15)} = 1 \]
Here, the major axis is along the y-axis. For an ellipse, the standard form is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( b^2 > a^2 \) when the major axis is vertical.

**Hint**: Identify which part of the equation corresponds to \( a^2 \) and \( b^2 \).

3. **Identifying \( a^2 \) and \( b^2 \)**:
From the equation, we can identify:
\[ b^2 = f(3a + 15) \quad \text{and} \quad a^2 = f(a^2 + 5a + 3) \]
Since the major axis is along the y-axis, we have:
\[ f(3a + 15) > f(a^2 + 5a + 3) \]

**Hint**: Use the property of the strictly decreasing function to compare the values of \( f \).

4. **Using the Strictly Decreasing Property**:
Since \( f \) is strictly decreasing, if \( 3a + 15 < a^2 + 5a + 3 \), then:
\[ f(3a + 15) > f(a^2 + 5a + 3) \]
Rearranging gives:
\[ 3a + 15 < a^2 + 5a + 3 \]
Simplifying this inequality:
\[ 0 < a^2 + 5a - 3a + 3 - 15 \]
\[ 0 < a^2 + 2a - 12 \]

**Hint**: Factor the quadratic expression to find the critical points.

5. **Factoring the Quadratic**:
The inequality \( a^2 + 2a - 12 > 0 \) can be factored as:
\[ (a - 2)(a + 6) > 0 \]
The critical points are \( a = 2 \) and \( a = -6 \).

**Hint**: Use a number line to determine the intervals where the product is positive.

6. **Finding the Intervals**:
Testing intervals:
- For \( a < -6 \): both factors are negative, product is positive.
- For \( -6 < a < 2 \): one factor is negative, one is positive, product is negative.
- For \( a > 2 \): both factors are positive, product is positive.

Thus, the solution to the inequality is:
\[ a < -6 \quad \text{or} \quad a > 2 \]

**Hint**: Remember to check the endpoints of the intervals to see if they satisfy the inequality.

### Final Answer:
The values of \( a \) for which the given ellipse condition holds are:
\[ a < -6 \quad \text{or} \quad a > 2 \]