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Question #41464

Difficulty: 8/10 | 🟢 Free

The Question

Let $C$ be the curve defined parametrically by $x(t) = \sin 2t - 2\sin t + 2\cos t$ $y(t) = \ln(\sin t - \cos t)$ for $t \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$ . Let $A$ be the area of the region bounded by the curve $C$ , the x-axis, and the y-axis. Then $A$ is equal to

Options

A.
$3 - \sqrt{2} - \ln(\sqrt{2}+1)$
B.
$3 \ln 3 - \ln(\sqrt{2}+1) + 1 - \sqrt{2}$
C.
$\frac{3 - \sqrt{2}}{2} - \frac{1}{2} \ln(\sqrt{2}+1)$
D.
$3 + \sqrt{2} + \ln(\sqrt{2}-1)$

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

Logarithmic Integration via Trig Manipulation|Parametric Curve Area|Extraneous Solution Rejection

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to perform logarithmic integration via trigonometric substitution while navigating a severe 'Extraneous Solution Rejection' trap.

When finding the y-axis intersection ( $x=0$ ), students must solve the quadratic $u^2 + 2u - 1 = 0$ , yielding $u = -1 \pm \sqrt{2}$ .

The trap lies in blindly accepting the negative root or attempting to eliminate the parameter to form $y = \ln|-1 - \sqrt{2-x}|$ .

Failing to restrict the domain of the logarithmic argument ( $u > 0$ ) or the natural range of the trigonometric transformation ( $u \in (0, \sqrt{2}]$ ) leads directly to evaluating the integral over the wrong branch, resulting in the distractor option involving $3 \ln 3$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Logarithmic Integration via Trig Manipulation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free