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Question #41789

Difficulty: 8/10 | 🟢 Free

The Question

Let $L$ be the orthogonal trajectory of the family of curves $xy = c$ passing through the point $(-1, 1)$ . Let $S$ be the set of all differentiable functions $f: [-2, -1] \to \mathbb{R}$ that satisfy the differential inequality $x f'(x) - f(x) \le -x^2$ with the initial condition $f(-1) = 1$ . Let $R$ be the region in the $xy$ -plane defined by $R = \{ (x,y) : -2 \le x \le -1, \ 0 \le y \le f(x) \text{ for some } f \in S \}.$ If $A$ is the area of the region $R$ , and $B$ is the area of the region bounded by the curve $L$ , the $x$ -axis, $x = -2$ , and $x = -1$ , find the value of $A + B$ .

Options

A.
$\frac{13}{6}$
B.
$\frac{23}{6}$
C.
$\frac{5}{6}$
D.
$\frac{17}{6}$

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

Inconsistency Detection|Orthogonal Trajectories|Integrating Factor (IF) Sign Sensitivity

Why This Question Is Tricky

Common Mistakes

This question tests the student's rigorous handling of differential inequalities and orthogonal trajectories.

The primary trap lies in the Integrating Factor (IF) sign sensitivity.

When solving $x f'(x) - f(x) \le -x^2$ for $x \in [-2, -1]$ , dividing by $x$ flips the inequality.

Furthermore, the IF is $e^{\int -1/x dx} = 1/|x| = -1/x$ .

Students who blindly use $1/x$ as the IF and ignore the inequality flips will incorrectly deduce $f(x) \le x^2$ , leading to a vastly inflated area (Option B).

The problem elegantly combines this inconsistency detection with the geometric concept of orthogonal trajectories to form a comprehensive JEE Advanced level constraint.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Programming - Class 12
Concept	Inconsistency Detection
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free