Before You Begin

Question #41555

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{ z \in \mathbb{C} : |z - \text{Re}(z)| \le \text{Re}(z) e^{\text{Re}(z)}, 0 \le \text{Re}(z) \le 2 \}$ . Let $S'$ be the image of $S$ under the transformation $w = z e^{i\pi/2}$ , and let $S'' = \{ w \in S' : \text{Re}(w) \ge 0 \}$ . If $A$ denotes the area of the region $S''$ in the complex plane, and $B = \int_0^{2e^2} f^{-1}(x) dx$ where $f(x) = x e^x$ for $x \ge 0$ , find the value of $3A - B$ .

Options

A.
4
B.
6e^2 - 2
C.
2e^2 + 2
D.
4e^2

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Inverse Function Integration|Euler's Form and Rotations|Geometrical Interpretation of Modulus

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of complex plane transformations and inverse function integration.

The trap lies in the geometrical interpretation of the rotated boundaries.

When $S$ is rotated by $\pi/2$ , the axes swap ( $u = -y, v = x$ ).

Students frequently fail to visualize the new region $S''$ correctly, assuming that since it is bounded by $u = f(v)$ , its area is the integral of $f^{-1}(u)$ along the new real axis from $0$ to $2e^2$ .

This Jacobian-like boundary error leads them to incorrectly calculate $A = 3e^2 - 1$ , confusing the area of $S''$ with the area under the inverse function, yielding the trap answer $6e^2 - 2$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Inverse Function Integration
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free