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Question #41291

Difficulty: 8/10 | 🟢 Free

The Question

Let $C$ be a curve defined by the locus of points $(x,y)$ satisfying the equation $\begin{vmatrix} x+1 & y & 1 \\ y & x-1 & -1 \\ 1 & 1 & 1 \end{vmatrix} = 5$ Let $R$ be the region in the first quadrant bounded by the curve $C$ , the $x$ -axis, and the line $x = 3$ . The area of the region $R$ is:

Options

A.
$\frac{3\sqrt{5}}{2} - 2 \ln\left(\frac{3+\sqrt{5}}{2}\right)$
B.
$\frac{3\sqrt{5}}{2} - 2 \ln(3+\sqrt{5})$
C.
$3\sqrt{5} - 4 \ln\left(\frac{3+\sqrt{5}}{2}\right)$
D.
$\frac{3\sqrt{5}}{2} + 2 \ln\left(\frac{3+\sqrt{5}}{2}\right)$

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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

Determinant as Geometric Area/Volume|Rectangular Hyperbola|Axis of Integration Selection

Why This Question Is Tricky

Common Mistakes

The problem requires evaluating a determinant to identify a rectangular hyperbola, then finding the area under it.

The primary trap lies in the integration step: students typically default to evaluating $\int \sqrt{x^2-a^2} dx$ .

In doing so, they frequently assume the lower limit $x=a$ evaluates to zero, forgetting that the logarithmic term $- \frac{a^2}{2} \ln(x + \sqrt{x^2-a^2})$ yields $- \frac{a^2}{2} \ln(a) \neq 0$ .

This omission leads directly to the distractor Option B.

Alternatively, recognizing the geometric property of the hyperbolic sector area bypasses the integration entirely, avoiding the trap.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Determinant as Geometric Area/Volume
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free