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Question #40523

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the set of all integers xx that satisfy the inequality: lim⁡t→x(1−cos⁡(2t−6))⋅(∣t2−4t−5∣−9)(t−3)2(t2+1)≤0\lim_{t \to x} \frac{(1 - \cos(2t - 6)) \cdot (|t^2 - 4t - 5| - 9)}{(t-3)^2 (t^2 + 1)} \le 0 Find the number of elements in SS.

Options

  • A.

    8

  • B.

    9

  • C.

    10

  • D.

    7

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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

Even Multiplicity Roots|Modulus Inequality Logic|Limit Existence at Discontinuity

Why This Question Is Tricky

Common Mistakes

This question tests the 'Limit Existence at Discontinuity' trap within a modulus inequality.

Students commonly apply the wavy-curve method or sign-chart logic directly to the expression, noting that the denominator (t−3)2(t-3)^2 makes the function undefined at x=3x=3.

Consequently, they incorrectly exclude x=3x=3 from the solution set, yielding 8 integers.

However, the inequality is defined via a limit.

Because the zero at x=3x=3 has an even multiplicity in both the numerator (via the cosine expansion) and the denominator, the limit exists, is negative, and satisfies the inequality, making x=3x=3 a valid solution.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLinear Inequalities - Class 11
ConceptEven Multiplicity Roots
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree