Let \(\frac{x^2}{f(a^2 + 7a + 3)} + \frac{y^2}{f(3a + 15)} = 1\) represent an ellipse with major axis along y-axis, where \(f\) is a strictly decreasing positive function on \(\mathbb{R}\). If the set of all possible values of \(a\) is \(\mathbb{R} - [\alpha, \beta]\), then \(\alpha^2 + \beta^2\) is equal to:
- 28
- 40
- 61
- 24
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
For an ellipse $\frac{x^2}{A} + \frac{y^2}{B} = 1$ to have its major axis along the $y$-axis, we must have $B>A$. Since $f$ is a strictly decreasing function, $f(x_1)>f(x_2)$ implies $x_1 < x_2$.
Step 2: Key Formula or Approach:
1. Condition: $f(3a + 15)>f(a^2 + 7a + 3)$.
2. Decreasing property: $x_1 < x_2$ if $f(x_1)>f(x_2)$.
Step 3: Detailed Explanation:
Given $B>A$: \[ f(3a + 15)>f(a^2 + 7a + 3) \] Since $f$ is strictly decreasing: \[ 3a + 15 < a^2 + 7a + 3 \] Rearranging: \[ a^2 + 4a - 12>0 \] Factorizing the quadratic: \[ (a + 6)(a - 2)>0 \] The solution to this inequality is $a \in (-\infty, -6) \cup (2, \infty)$. This can be written as $\mathbb{R} - [-6, 2]$. Here, $\alpha = -6$ and $\beta = 2$. Calculate $\alpha^2 + \beta^2$: \[ (-6)^2 + (2)^2 = 36 + 4 = 40 \]
Step 4: Final Answer:
The value of \(\alpha^2 + \beta^2\) is 40.
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