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Question #40874

Difficulty: 8/10 | 🟢 Free

The Question

Let A=(1,−1,0)A = (1, -1, 0), B=(0,1,−1)B = (0, 1, -1), and C=(−1,0,1)C = (-1, 0, 1) be three points in R3\mathbb{R}^3. A fourth point DD is given by the coordinates D=(sgn(x2−2x−3),sgn(−2x2+x+1),xsgn(x))D = \left( \text{sgn}(x^2 - 2x - 3), \text{sgn}(-2x^2 + x + 1), \frac{x}{\text{sgn}(x)} \right) where sgn(⋅)\text{sgn}(\cdot) denotes the signum function. Let SS be the set of all real values of xx for which the points A,B,CA, B, C, and DD are coplanar. Find the product of all elements in SS.

Options

  • A.

    -2

  • B.

    0

  • C.

    2

  • D.

    -4

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Correct Answer

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Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Collinearity and Coplanarity|Singular vs. Non-Singular Threshold|Signum Function Discontinuity

Why This Question Is Tricky

Common Mistakes

The problem tests the student's ability to handle domain restrictions rigorously when dealing with piecewise and signum functions in coordinate geometry.

The coplanarity condition simplifies to an equation involving ∣x∣|x|.

The trap is deliberately set at x=0x=0, where the limit of the coordinate exists (and satisfies the equation), but the function itself evaluates to the indeterminate form 0/00/0.

Students who blindly simplify xsgn(x)\frac{x}{\text{sgn}(x)} to ∣x∣|x| without noting x≠0x \neq 0 will incorrectly include 00 in their solution set, leading to a product of 00.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntroduction to Three-dimensional Geometry - Class 11
ConceptCollinearity and Coplanarity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree