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Question #40573

Difficulty: 8/10 | 🟢 Free

The Question

Let mm be the number of ways to arrange 5 identical black balls and 8 identical white balls in a row such that no two black balls are adjacent. Let nn be the number of such arrangements that are symmetric with respect to the center. Consider the function f:R→Rf:\mathbb{R} \to \mathbb{R} defined by f(x)=(x2−mnx+114)∣x−n∣f(x) = \left(x^2 - \frac{m}{n}x + 114\right) |x - n| Let ScS_c be the sum of all critical points of f(x)f(x) and SsS_s be the sum of all stationary points of f(x)f(x). Find the value of Sc+SsS_c + S_s.

Options

  • A.

    36

  • B.

    42

  • C.

    48

  • D.

    24

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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

Gap Method (Non-adjacency)|Symmetry Utilization|Critical Point vs Stationary Point

Why This Question Is Tricky

Common Mistakes

This question bridges combinatorics (Gap Method) with differential calculus.

The combinatorics phase requires parity analysis to deduce the center element of a symmetric arrangement.

The calculus phase introduces the primary trap: distinguishing between critical points and stationary points.

Students often conflate the two, either missing the non-differentiable point at the gap boundary x=nx=n or incorrectly classifying it as a stationary point.

By asking for a combined sum, the problem strictly penalizes any misunderstanding of these fundamental calculus definitions.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterPermutations and Combinations - Class 11
ConceptGap Method (Non-adjacency)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree