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Question #40630

Difficulty: 8/10 | 🟢 Free

The Question

Let f(x)=∣x−2∣17+∣x+2∣17f(x) = |x - 2|^{17} + |x + 2|^{17}. For x∈(−2,2)x \in (-2, 2), f(x)f(x) can be expressed as a polynomial P(x)=∑r=017CrxrP(x) = \sum_{r=0}^{17} C_r x^r. Find the value of rr for which CrC_r is maximum.

Options

  • A.

    4

  • B.

    5

  • C.

    6

  • D.

    8

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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

Coefficient of x^r in Expansion|Greatest Binomial Coefficient|Modulus Sign Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle modulus sign dependency in binomial expansions.

A common trap is to ignore the domain x∈(−2,2)x \in (-2, 2) and incorrectly expand ∣x−2∣17|x - 2|^{17} as (x−2)17(x - 2)^{17}.

This flawed approach yields the polynomial Q(x)=(x−2)17+(x+2)17Q(x) = (x - 2)^{17} + (x + 2)^{17}, where the surviving terms have odd powers of xx, leading to an incorrect maximum coefficient at r=5r = 5.

By correctly applying the modulus definition ∣x−2∣=2−x|x - 2| = 2 - x, the surviving terms shift to even powers of xx, changing the polynomial entirely and shifting the maximum coefficient to r=6r = 6.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterBinomial Theorem - Class 11
ConceptCoefficient of x^r in Expansion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree