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:
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:
Show 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.
Updated On: May 28, 2026
- $K=-5$
- $K\in(-3,2)$
- $K\in(-7,-5)$
- $K=2$
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For the major axis of a standard ellipse centered at the origin to coincide with the horizontal $x$-axis, the denominator parameter located under the $x^2$ term must be strictly greater than the denominator parameter located under the $y^2$ term:
$$\text{Horizontal Denominator } (a^2) > \text{Vertical Denominator } (b^2)$$
Step 1: Set up the denominator inequality condition.
Using the structural coefficients from the given ellipse equation: $$f(K^2 + 2K + 5) > f(K + 11)$$
Step 2: Apply the decreasing function property to the arguments.
We are given that $f(x)$ is a decreasing function. By definition, a decreasing function reverses the direction of inequality signs when comparing its internal arguments: $$y_1 > y_2 \implies f(y_1) < f(y_2)$$ Applying this rule to our inequality allows us to drop the function layers by flipping the direction pointer: $$K^2 + 2K + 5 < K + 11$$
Step 3: Solve the resulting quadratic inequality.
Collect all terms on the left side of the inequality to form a standard quadratic expression: $$K^2 + 2K - K + 5 - 11 < 0$$ $$K^2 + K - 6 < 0$$ Factor the quadratic polynomial expression: $$(K + 3)(K - 2) < 0$$
Step 4: Identify the solution interval region.
Using the standard sign-chart boundary intervals, the product of the two linear binomial factors is strictly negative when the parameter $K$ lies strictly between the two root zero points: $$K \in (-3, 2)$$ This matches option (B) perfectly.
Using the structural coefficients from the given ellipse equation: $$f(K^2 + 2K + 5) > f(K + 11)$$
Step 2: Apply the decreasing function property to the arguments.
We are given that $f(x)$ is a decreasing function. By definition, a decreasing function reverses the direction of inequality signs when comparing its internal arguments: $$y_1 > y_2 \implies f(y_1) < f(y_2)$$ Applying this rule to our inequality allows us to drop the function layers by flipping the direction pointer: $$K^2 + 2K + 5 < K + 11$$
Step 3: Solve the resulting quadratic inequality.
Collect all terms on the left side of the inequality to form a standard quadratic expression: $$K^2 + 2K - K + 5 - 11 < 0$$ $$K^2 + K - 6 < 0$$ Factor the quadratic polynomial expression: $$(K + 3)(K - 2) < 0$$
Step 4: Identify the solution interval region.
Using the standard sign-chart boundary intervals, the product of the two linear binomial factors is strictly negative when the parameter $K$ lies strictly between the two root zero points: $$K \in (-3, 2)$$ This matches option (B) perfectly.
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