Start Learning Free

Before You Begin

Question #41068

Difficulty: 8/10 | 🟢 Free

The Question

Let n=lim⁡x→3−⌊x2+52⌋n = \lim_{x \to 3^-} \left\lfloor \frac{x^2 + 5}{2} \right\rfloor and k=lim⁡x→0(⌊3sin⁡xx⌋+⌊2tan⁡xx⌋)k = \lim_{x \to 0} \left( \left\lfloor \frac{3 \sin x}{x} \right\rfloor + \left\lfloor \frac{2 \tan x}{x} \right\rfloor \right), where ⌊⋅⌋\lfloor \cdot \rfloor denotes the greatest integer function. If nn distinct balls are randomly distributed into kk distinct boxes, what is the probability that no box remains empty?

Options

  • A.

    195512\frac{195}{512}

  • B.

    5251024\frac{525}{1024}

  • C.

    72625\frac{72}{625}

  • D.

    6723125\frac{672}{3125}

Check Your Answer

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

Complementary Counting|Division into Groups|Step Function Discontinuities

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly evaluate limits involving step functions before applying combinatorial probability.

The primary trap lies in the discontinuity of the greatest integer function.

Students often incorrectly commute the limit and the floor function, assuming lim⁡⌊f(x)⌋=⌊lim⁡f(x)⌋\lim \lfloor f(x) \rfloor = \lfloor \lim f(x) \rfloor.

For nn, f(x)→7−f(x) \to 7^-, so the floor is 6, not 7.

For kk, sin⁡xx→1−\frac{\sin x}{x} \to 1^- and tan⁡xx→1+\frac{\tan x}{x} \to 1^+, making the floors 2 and 2 respectively, yielding k=4k=4 instead of 3+2=53+2=5.

Failing to recognize these directional limits leads to incorrect sample spaces and partitions, specifically testing the boundary conditions of complementary counting (Inclusion-Exclusion).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 11
ConceptComplementary Counting
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree