Question

Consider the following ellipse: \[ \frac{x^{2}}{f(K^{2}+2K+5)}+\frac{y^{2}}{f(K+11)}=1, \] where \(f(x)\) is a positive decreasing function. Then the value (values) of \(K\) for which the major axis coincides with x-axis is:

• A. $K=-5$
• B. $K\in(-3,2)$
• C. $K\in(-7,-5)$
• D. $K=2$

Hint:

Always be careful with descriptions of function behavior! If the problem had stated that $f(x)$ was an *increasing* function, the inequality direction would have stayed the same, leading to $K^2+K-6 > 0 \implies K \in (-\infty, -3) \cup (2, \infty)$ instead.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the major axis coincides with the x-axis if \( a^2>b^2 \).
Step 2: Key Formula or Approach:
1. Condition: \( f(K^2 + 2K + 5)>f(K + 11) \).
2. Use the property that if \( f \) is a decreasing function, then \( f(x_1)>f(x_2) \iff x_1<x_2 \).
Step 3: Detailed Explanation:
Given \( f \) is a decreasing function and we need: \[ f(K^2 + 2K + 5)>f(K + 11) \] This inequality holds if and only if: \[ K^2 + 2K + 5<K + 11 \] \[ K^2 + K - 6<0 \] Factoring the quadratic: \[ (K + 3)(K - 2)<0 \] Using the wavy curve method, the solution for this inequality is: \[ -3<K<2 \implies K \in (-3, 2) \] Step 4: Final Answer:
The major axis coincides with the x-axis for \( K \in (-3, 2) \).