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Question #40555

Difficulty: 8/10 | 🟢 Free

The Question

Let $f, g : [0, 2] \to \mathbb{R}$ be continuous functions such that both satisfy the differential equation $(x^3 - 1)\frac{dy}{dx} = 3x^2 y$ for all $x \in (0, 2) \setminus \{1\}$ . Given that $f(0) = 2$ , $f(2) = 14$ , $g(0) = -1$ , and $g(2) = 7$ . Let $R$ be the region in the $xy$ -plane defined by $R = \{(x, y) : 0 \le x \le 2, g(x) \le y \le f(x)\}$ The area of the region $R$ is:

Options

A.
$5$
B.
$\frac{21}{2}$
C.
$\frac{7}{2}$
D.
$6$

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

Graphical Feasible Region|General vs Particular Solutions|Integration Constant Trap|Logarithmic Integration Modulus

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Integration Constant Trap' and the 'Logarithmic Integration Modulus' in differential equations.

When solving $(x^3-1)dy = 3x^2 y dx$ , students often integrate blindly to get $y = C(x^3-1)$ globally, ignoring the absolute value from the logarithmic integration.

Because the differential equation is undefined at $x=1$ , the domain is split into two disjoint intervals, $[0, 1)$ and $(1, 2]$ .

The integration constant can (and does) take different values on each side of the singularity.

If a student assumes a single global constant, the boundary conditions $f(0)=2$ and $f(2)=14$ will appear contradictory, or they will force an incorrect global function like $f(x) = 2-2x^3$ , leading to an incorrect area calculation.

By correctly applying piecewise constants and ensuring continuity at $x=1$ , the true piecewise nature of the boundary curves is revealed.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Inequalities - Class 11
Concept	Graphical Feasible Region
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free