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Question #40982

Difficulty: 8/10 | 🟢 Free

The Question

Let r(x)=x2−5x+4x2+x−2r(x) = \frac{x^2 - 5x + 4}{x^2 + x - 2}. Let S(x)S(x) denote the sum of the infinite series 1+3r(x)+5r(x)2+7r(x)3+…1 + 3r(x) + 5r(x)^2 + 7r(x)^3 + \dots If S(x)S(x) is equal to the arithmetic mean of the functions f(x)=4r(x)2+18r(x)(1−r(x))2f(x) = \frac{4r(x)^2 + 18r(x)}{(1-r(x))^2} and g(x)=2r(x)−8(1−r(x))2g(x) = \frac{2r(x) - 8}{(1-r(x))^2}, then the sum of all valid values of xx is:

Options

  • A.

    10

  • B.

    9

  • C.

    11

  • D.

    -1

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

Properties of Arithmetic Mean|Arithmetic-Geometric Progression (AGP)|Wavy Curve (Method of Intervals)

Why This Question Is Tricky

Common Mistakes

The problem tests the intersection of infinite series convergence and algebraic manipulation.

The trap lies in the fact that the algebraic formula for the sum of an infinite AGP, S=1+r(1−r)2S = \frac{1+r}{(1-r)^2}, is only valid when the common ratio satisfies ∣r∣<1|r| < 1.

Students who blindly equate the algebraic expressions will solve the quadratic 2r2+9r−5=02r^2 + 9r - 5 = 0 to find r=1/2r = 1/2 and r=−5r = -5, leading to x=10x = 10 and x=−1x = -1.

By ignoring the Wavy Curve method required to establish the domain of convergence (x>1x > 1), they will incorrectly include x=−1x = -1 in their final sum, arriving at the trap answer of 9.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterStatistics - Class 11
ConceptProperties of Arithmetic Mean
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree