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Question #41253

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all real values of $x$ for which the points $A(1, |x|)$ , $B(2, x)$ , and $C(3, 1)$ form a non-degenerate triangle whose area is at most $1$ . Then $S$ is equal to:

Options

A.
$\left[-\frac{1}{3}, 3\right] \setminus \{1\}$
B.
$\left[-\frac{1}{3}, \infty\right) \setminus \{1\}$
C.
$\left[-\frac{1}{3}, 3\right]$
D.
$\left[-3, \frac{1}{3}\right] \setminus \{1\}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Singular vs. Non-Singular Threshold|Condition of Collinearity|Sign Flipping Rule

Why This Question Is Tricky

Common Mistakes

This question tests the condition of collinearity and the handling of nested absolute value inequalities.

The primary trap lies in the 'Sign Flipping Rule' when dealing with the outer modulus of the area formula.

Students often drop the outer absolute value and only solve $|x| - 2x + 1 \le 2$ , completely missing the negative branch $-2 \le |x| - 2x + 1$ .

This omission leads to the incomplete, unbounded interval $[-1/3, \infty)$ .

Furthermore, failing to account for the non-singular threshold (the points must form a non-degenerate triangle, meaning the determinant cannot be zero) causes students to forget to exclude $x=1$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Singular vs. Non-Singular Threshold
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free