Question

Let \( \frac{x^2}{f(a^2 + 2a + 7)} + \frac{y^2}{f(3a + 14)} = 1 \) represent an ellipse. The major axis of the given ellipse is the y-axis and \( f \) is a decreasing function. If the range of \( a \) is \( R - [\alpha, \beta] \), then \( \alpha + \beta \) is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For the major axis to be the y-axis in \(\frac{x^2}{A} + \frac{y^2}{B} = 1\), the condition is $B > A > 0$. Since $f$ is a decreasing function, $f(x₁) > f(x₂)$ implies $x₁ < x₂$.

Step 2: Key Formula or Approach:
Set the denominator under $y²$ greater than the denominator under $x²$: \[ f(3a + 14) > f(a^2 + 2a + 7) \] Using the decreasing property of $f$: \[ 3a + 14 < a^2 + 2a + 7 \]

Step 3: Detailed Explanation:
1. Simplify the inequality: \[ a^2 + 2a - 3a + 7 - 14 > 0 \] \[ a^2 - a - 7 > 0 \] 2. This expression relates to the range where the condition is satisfied. In standard problems like this, the range of $a$ where the ellipse is NOT formed (the "complement" $R - [α, β]$) comes from the roots of the quadratic equation. 3. For \(a^2 - a - 7 = 0\), the sum of roots \(\alpha + \beta = -(-1)/1 = 1\).

Step 4: Final Answer:
The value of \(\alpha + \beta\) is 1.