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Question #41565

Difficulty: 8/10 | 🟢 Free

The Question

A curve $C$ is defined parametrically by the system of differential equations: $\frac{dx}{dt} = 2(t-2), \quad \frac{dy}{dt} = -2(t-2)y^2$ with initial conditions $x(2) = -1$ and $y(2) = -1$ . Let $A$ be the area of the region bounded by the curve $C$ , the $x$ -axis, and the vertical lines corresponding to $t = 2$ and $t = 4$ . Find the value of $A$ .

Options

A.
$\ln 3$
B.
$2 \ln 3$
C.
0
D.
$\text{Diverges to } \infty$

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

Parametric Curve Area|Variable Separable Constraints|Logarithmic Integration Modulus|Parametric Derivative Endpoint Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Logarithmic Modulus Singularity Trap' in parametric area evaluation.

Students must first solve a variable-separable system of differential equations to find the parametric equations of the curve.

The lethal trap lies in the area integral: naive application of the formula $\int y(t) x'(t) dt$ yields a neat, finite real number ( $\ln 3$ ) because the antiderivative $\ln|t^2 - 4t + 3|$ evaluates cleanly at the endpoints.

However, the parameter $t$ forces the argument of the logarithm through zero at $t=3$ , representing a vertical asymptote ( $x=0$ ).

Failing to check the domain of the integrand leads students to select the trap option, missing that the improper integral strictly diverges.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Parametric Curve Area
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free