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Question #40686

Difficulty: 8/10 | 🟢 Free

The Question

Let α\alpha and β\beta be the roots of the quadratic equation x2−2kx+(k2+k−2)=0x^2 - 2kx + (k^2 + k - 2) = 0, where k∈Rk \in \mathbb{R}. An Arithmetic-Geometric Progression is defined by its partial sum Sn=∑m=1n(m∣α∣+∣β∣)rm−1S_n = \sum_{m=1}^{n} (m|\alpha| + |\beta|) r^{m-1}, where the common ratio is r=∣α+β∣∣α∣+∣β∣r = \frac{|\alpha + \beta|}{|\alpha| + |\beta|}. Let DD be the set of all values of kk for which lim⁡n→∞Sn\lim_{n \to \infty} S_n exists and is finite. Which of the following represents the set DD?

Options

  • A.

    (−2,1)∪(2,∞)(-2, 1) \cup (2, \infty)

  • B.

    [−2,1]∪[2,∞)[-2, 1] \cup [2, \infty)

  • C.

    (−∞,−2)∪(1,2)(-\infty, -2) \cup (1, 2)

  • D.

    (−2,∞)∖{1,2}(-2, \infty) \setminus \{1, 2\}

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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

Arithmetic-Geometric Progression (AGP)|Quadratic Roots in Sequences|Triangle Inequality Mind Games|Closed vs Open Intervals

Why This Question Is Tricky

Common Mistakes

This question traps students who mechanically apply the infinite AGP sum formula without verifying the open interval of convergence (∣r∣<1|r| < 1).

By using the Triangle Inequality, rr is naturally bounded by 1.

Students often mistakenly solve for r≤1r \le 1, leading them to include the boundaries k=−2,1,2k=-2, 1, 2 where r=1r=1 and the series diverges.

Furthermore, it elegantly combines the condition for complex roots and the opposite-sign condition for real roots, forcing a comprehensive analysis of the quadratic's discriminant and root product.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSequences and Series - Class 11
ConceptArithmetic-Geometric Progression (AGP)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree