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Question #40375

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \sqrt{ \sin^{-1}(2x\sqrt{1-x^2}) - 2\sin^{-1}x } + \sec^{-1}(2x)$ . Find the domain of $f(x)$ .

Options

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

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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

Real-Valued Function Domain|The Sum-Difference Formula Constraints|Closed vs Open Intervals

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to evaluate the domain of a composite function involving inverse trigonometric sum-difference identities.

The primary trap lies in the assumption that $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ universally, or that the expression under the square root is only non-negative where this identity strictly holds ( $x \in [-1/\sqrt{2}, 1/\sqrt{2}]$ ).

Students must correctly apply the principal value branches of $\sin^{-1}(\sin 2\theta)$ to discover that the inequality $\sin^{-1}(\sin 2\theta) \ge 2\theta$ is also satisfied for $\theta \in [-\pi/2, -\pi/4)$ , extending the valid domain to $x \in [-1, 1/\sqrt{2}]$ .

Failing to account for this branch leads to the extraneous exclusion of $[-1, -1/\sqrt{2})$ , a classic JEE Advanced negative-marking trap.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Real-Valued Function Domain
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free