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Question #40497

Difficulty: 8/10 | 🟢 Free

The Question

Let $\alpha, \beta, \gamma$ be the real roots of the equation $x^3 - px^2 + qx - r = 0$ . Consider the points $A(\alpha, \beta, \gamma)$ , $B(\beta, \gamma, \alpha)$ , and $C(\gamma, \alpha, \beta)$ in $\mathbb{R}^3$ . If the matrix formed by the position vectors of $A, B, C$ is singular, and the area of $\triangle ABC$ is $2\sqrt{3}$ , then the maximum possible value of $r$ is:

Options

A.
$\frac{16}{27}$
B.
$\frac{8}{27}$
C.
$\frac{4}{9}$
D.
$\frac{32}{27}$

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

Vieta’s Formulas in Complex Domain|Coplanarity Condition|Singular vs. Non-Singular Threshold|Area of Triangle via Determinants

Why This Question Is Tricky

Common Mistakes

This question creates a 'Lethal Collision' between 3D Geometry, Determinants, and the Theory of Equations.

The first trap is the singular matrix condition: students must recognize that the determinant of the cyclic coordinate matrix simplifies to $-p(p^2-3q)$ .

Equating this to zero, while knowing the triangle area $\frac{\sqrt{3}}{2}(p^2-3q)$ is non-zero, forces $p=0$ .

The second trap is the area calculation: using standard cross-products is a massive time sink unless students identify the triangle as equilateral.

The final, most lethal trap is applying the condition for real roots to the resulting cubic $x^3 - \frac{4}{3}x - r = 0$ .

Students often forget that for a cubic to have real roots, its local extrema must have opposite signs, which bounds the value of $r$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Complex Numbers and Quadratic Equations - Class 11
Concept	Vieta’s Formulas in Complex Domain
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free