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Question #41168

Difficulty: 8/10 | 🟢 Free

The Question

Let a rectangle $OABC$ be constructed such that vertex $O$ is at the origin, vertex $A$ lies on the positive $x$ -axis, vertex $C$ lies on the positive $y$ -axis, and vertex $B$ lies on the curve $y = \frac{1}{2} \cos^{-1}(1 - 2x^2) + \sin^{-1}(2x\sqrt{1-x^2})$ in the first quadrant ( $0 \le x \le 1$ ). If $P(x)$ denotes the perimeter of this rectangle, find the maximum value of $P(x)$ .

Options

A.
$\frac{3\pi}{2} + \sqrt{2}$
B.
$3\pi + 2$
C.
$\pi + 2$
D.
$2\pi + \sqrt{2}$

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

Variable Transformation Subsitution|Geometry Constraint Modeling|The Modulus Trap

Why This Question Is Tricky

Common Mistakes

Students often blindly apply the identity $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ without checking the domain, falling into the 'Modulus Trap' where the principal branch changes.

By ignoring the branch condition $x \le 1/\sqrt{2}$ , they incorrectly assume the perimeter function is strictly increasing on the entire interval $[0, 1]$ .

This leads to a false maximum at the boundary $x=1$ , yielding the trap answer $3\pi + 2$ .

The problem tests the rigorous application of variable transformation and critical point analysis at the boundary of geometric constraints.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Variable Transformation Subsitution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free