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Question #41179

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(x)$ be a $3 \times 3$ matrix with real entries such that $\det(A(x)) = \frac{1}{8} \left( \sin^{-1}(x-1) + \cos^{-1}(x-1) + \sin^{-1}(2x) \right)$ . If $B(x) = 2A(x)$ , find the domain of the real-valued function $f(x) = \sqrt{ \frac{2\pi}{3} - \det(B(x)) }$ .

Options

A.
$\left[ 0, \frac{1}{4} \right]$
B.
$\left[ -\frac{1}{2}, \frac{1}{4} \right]$
C.
$\left[ 0, \frac{1}{2} \right]$
D.
$\left[ 0, \frac{\sqrt{3}}{4} \right]$

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

Domain Constraint Intersections|Singular vs. Non-Singular Threshold|Determinant Scaling Property

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of domain constraints for inverse trigonometric functions embedded within a matrix determinant, combined with the determinant scaling property.

The primary trap is the incorrect application of the scalar multiplier to the determinant (e.g., $\det(2A) = 2\det(A)$ instead of $2^3\det(A)$ ), which drastically alters the inequality threshold.

A secondary, equally lethal trap is forgetting to intersect the final inequality solution with the base domain of the initial inverse trigonometric terms (specifically $\sin^{-1}(x-1)$ ), leading students to incorrectly include negative values in the domain.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Domain Constraint Intersections
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free