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Question #40596

Difficulty: 8/10 | 🟢 Free

The Question

Let NN be the number of ways to select at least one fruit from a basket containing 2 identical apples, 3 identical mangoes, and 4 identical oranges. Let P(x)=1N∑k=0Nk2(Nk)xk(1−x)N−k,x∈RP(x) = \frac{1}{N} \sum_{k=0}^{N} k^2 \binom{N}{k} x^k (1-x)^{N-k}, \quad x \in \mathbb{R} If the solution set of the inequality [P(x)−26x]≤1\left[ P(x) - 26x \right] \le 1 is x∈(a,b)x \in (a, b), where [⋅][\cdot] denotes the greatest integer function, then the value of b−ab - a is:

Options

  • A.

    3358\frac{33}{58}

  • B.

    2558\frac{25}{58}

  • C.

    85758\frac{\sqrt{857}}{58}

  • D.

    3359\frac{33}{59}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Selection from Identical Items|Properties of Binomial Coefficients|Greatest Integer Function (GIF)|Wavy Curve (Method of Intervals)

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of combinatorics, binomial probability identities, and algebraic inequalities.

The primary trap lies in the Greatest Integer Function property: students frequently misinterpret [y]≤1[y] \le 1 as y≤1y \le 1, leading to an incorrect discriminant and irrational roots (Option C).

A secondary trap is forgetting to subtract 1 for the 'at least one' condition in the identical items selection, yielding N=60N=60 instead of 5959 (Option D).

The problem elegantly reduces a complex binomial summation to a quadratic inequality, rewarding students who recognize the second moment of the Binomial distribution E[K2]=Nx(1−x)+N2x2E[K^2] = Nx(1-x) + N^2x^2.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterPermutations and Combinations - Class 11
ConceptSelection from Identical Items
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree