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Question #40781

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all real values of $k$ for which the points $A\left(k, \frac{1}{k-1}\right)$ , $B\left(2k, \frac{3}{k-1}\right)$ , and $C\left(3k, \frac{k+3}{k-1}\right)$ form a valid triangle whose area is strictly less than 1. Then $S$ is equal to:

Options

A.
$(-\sqrt{2}, 2-\sqrt{2}) \cup (\sqrt{2}, 2+\sqrt{2}) \setminus \{0, 2\}$
B.
$(-\sqrt{2}, 2-\sqrt{2}) \cup (\sqrt{2}, 2+\sqrt{2})$
C.
$(-\sqrt{2}, 1) \cup (1, 2+\sqrt{2}) \setminus \{0, 2\}$
D.
$(-\infty, 2-\sqrt{2}) \cup (\sqrt{2}, \infty) \setminus \{0, 1, 2\}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Condition of Collinearity|Slope Sensitivity|Wavy Curve (Method of Intervals)

Why This Question Is Tricky

Common Mistakes

This question tests the 'Wavy Curve' method within the context of collinearity and triangle existence.

The primary trap lies in the definition of a 'valid triangle'.

Students often successfully set up the area inequality $\Delta < 1$ and solve the resulting rational inequalities using the method of intervals.

However, they frequently forget that for a triangle to exist, the points cannot be collinear, meaning the area must be strictly greater than zero ( $\Delta > 0$ ).

This oversight leads them to include $k=0$ and $k=2$ in their final solution set (Option B).

Additionally, the problem rigorously tests the correct handling of sign flips around the asymptote $k=1$ during the intersection of two rational inequalities.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Condition of Collinearity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free