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Question #40762

Difficulty: 8/10 | 🟢 Free

The Question

Let $L_1: x - y = 0$ and $L_2: x + 7y - 8 = 0$ be two lines in the $xy$ -plane. Let $B_1(x,y) = 0$ be the acute angle bisector of $L_1$ and $L_2$ , normalized such that the coefficient of $x$ is $1$ . Let $B_2(x,y) = 0$ be the obtuse angle bisector of $L_1$ and $L_2$ , normalized such that the coefficient of $x$ is $3$ . Consider the feasible region $R$ defined by the system of inequalities: $B_1(x,y) \ge 0$ $B_2(x,y) \le 0$ $x \ge 0$ Find the maximum value of the objective function $Z = 4x - y$ on the region $R$ .

Options

A.
3
B.
$\frac{14}{3}$
C.
$-\frac{2}{3}$
D.
\text{Does not exist}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Angle Bisector Logic|Graphical Feasible Region|Unbounded Region Trap

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to integrate coordinate geometry (angle bisectors) with linear programming, specifically targeting the 'Unbounded Region Trap'.

Students often mechanically find the intersection points of the constraints (vertices) and evaluate the objective function there, which yields a false maximum of 3 at the vertex (1,1).

However, they must analyze the direction of the inequality signs to realize the feasible region is unbounded in the fourth quadrant.

Since the objective function Z = 4x - y increases as x approaches infinity and y approaches negative infinity, the maximum value does not exist.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Angle Bisector Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free