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Question #40774

Difficulty: 8/10 | 🟢 Free

The Question

Let $A, B,$ and $C$ be the vertices of a triangle in the Cartesian plane. The coordinates $(x_k, y_k)$ of the vertices for $k = 0, 1, 2$ are given by: $x_k = \sin^{-1}\left( \frac{\sqrt{3}}{2} \cos\frac{k\pi}{2} + \frac{1}{2} \sin\frac{k\pi}{2} \right)$ $y_k = \cos^{-1}\left( \frac{1}{2} \cos\frac{k\pi}{2} - \frac{\sqrt{3}}{2} \sin\frac{k\pi}{2} \right)$ Find the area of the triangle $ABC$ .

Options

A.
0
B.
$\frac{5\pi^2}{36}$
C.
$\frac{\pi^2}{36}$
D.
$\frac{\pi^2}{3}$

Check Your Answer

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

Area of Triangle via Determinants|Compound Angle Expansion|Domain Constraint Intersections

Why This Question Is Tricky

Common Mistakes

The question tests the integration of compound angle expansions with determinant-based area calculations, while embedding a critical trap involving inverse trigonometric domain constraints.

Students must first recognize the arguments as expansions of $\sin(\pi/3 + k\pi/2)$ and $\cos(\pi/3 + k\pi/2)$ .

The primary trap lies in the assumption that $\sin^{-1}(\sin \theta) = \theta$ and $\cos^{-1}(\cos \theta) = \theta$ for all $\theta$ .

If a student blindly applies this false identity, all vertices will evaluate to $(\pi/3 + k\pi/2, \pi/3 + k\pi/2)$ , meaning they all lie on the line $y = x$ , resulting in an area of 0 (Option A).

Correctly navigating the principal value branches for $k=1$ and $k=2$ is required to find the true non-degenerate vertices and compute the correct area.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Area of Triangle via Determinants
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free