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Question #41394

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \sqrt{ \text{Re} \left( (3x - x^3) \sqrt{x^2 - 1 + 2ix} \right) }$ , where $\sqrt{z}$ denotes the principal square root of the complex number $z$ . Find the set of all real numbers $x$ for which $f(x)$ is strictly concave upward.

Options

A.
$\left( -\infty, -\frac{3}{\sqrt{2}} \right)$
B.
$\emptyset$
C.
$\left( \frac{3}{\sqrt{2}}, \infty \right)$
D.
$\left( -\sqrt{3}, -\frac{3}{\sqrt{2}} \right)$

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

Concavity Analysis|Square Root of Complex Numbers|Real-Valued Function Domain

Why This Question Is Tricky

Common Mistakes

This question exploits a lethal collision between complex branch cuts and real-valued domain constraints.

Students commonly assume the algebraic identity $\sqrt{(x+i)^2} = x+i$ holds for all real $x$ .

If they blindly apply this, they obtain $f(x) = \sqrt{3x^2 - x^4}$ for all $x$ , which restricts the domain to $[-\sqrt{3}, \sqrt{3}]$ .

Under this false domain, the concavity condition $f''(x) > 0$ yields $x < -\frac{3}{\sqrt{2}} \approx -2.12$ , which falls outside the fake domain, leading them to incorrectly choose the 'empty set' trap (Option B).

Recognizing the principal square root branch cut for $x < 0$ flips the inner function, expanding the true domain to $(-\infty, -\sqrt{3}]$ and validating the interval.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	Concavity Analysis
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free