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Question #40384

Difficulty: 8/10 | 🟢 Free

The Question

Let sets $A$ and $B$ be defined as follows: $A = \{x \in \mathbb{Z} : (x+4)^2 (x-1)^3 (x-5)^4 (x-7) \le 0\}$ $B = \{y \in \mathbb{Z} : (y+2)^4 (y-3)^5 (y-6)^2 (y-9) \le 0\}$ Find the number of ordered pairs $(x, y) \in A \times B$ such that $x \neq y$ .

Options

A.
59
B.
44
C.
64
D.
179

Check Your Answer

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

Cartesian Product Cardinality|Total Selection (at least one)|Even Multiplicity Roots|Complementary Counting

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to correctly identify the elements of sets defined by polynomial inequalities with even-multiplicity roots.

The primary trap lies in incorrectly assuming that terms like $(x+4)^2$ or $(y+2)^4$ can be strictly ignored because they are non-negative, which leads to missing the isolated integer roots $x=-4$ and $y=-2$ .

This cascades into incorrect cardinalities for sets $A$ and $B$ .

Furthermore, the problem requires complementary counting to exclude trivial selections (where $x=y$ ) from the Cartesian product $A \times B$ .

Failing to account for the isolated roots yields a trap answer of 44, while forgetting to subtract the intersection yields 64.

The prime answer 59 acts as an anchor to prevent heuristic guessing.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Cartesian Product Cardinality
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free