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Question #40987

Difficulty: 8/10 | 🟢 Free

The Question

A population $S_n$ is divided into subgroups indexed by triplets of non-negative integers $(r, s, t)$ such that $r + s + t = n$ . The weight of the subgroup $(r, s, t)$ is given by $W_{r,s,t} = \frac{n!}{r!s!t!} \left(\frac{1}{n}\right)^s \left(\frac{\lambda}{n^2}\right)^t$ , where $\lambda > 0$ . The mean of the subgroup $(r, s, t)$ is $X_{r,s,t} = s + 2t$ . Let $M_n$ be the combined mean of the population $S_n$ . If $\lim_{n \to \infty} (M_n)^n = e^7$ , then the value of $\lambda$ is:

Options

A.
$3$
B.
$\frac{7}{2}$
C.
$4$
D.
$\frac{9}{2}$

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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

Combined Mean of Subgroups|Multinomial Theorem|The 1^∞ Form Shortcut

Why This Question Is Tricky

Common Mistakes

This question integrates the combined mean of subgroups with the Multinomial Theorem and limits.

The core trap lies in prematurely applying the $1^\infty$ limit shortcut $e^{\lim n(f(x)-1)}$ to the generating function $g(x) = (1 + \frac{x}{n} + \frac{\lambda x^2}{n^2})^n$ .

If a student approximates the base as $e^{x + \frac{\lambda x^2}{n}}$ before differentiating to find the mean, they miss the $-\frac{x^2}{2n}$ term from the Taylor expansion of $\ln(1+y)$ .

This leads to an incorrect mean $M_n \approx 1 + \frac{2\lambda}{n}$ and a limit of $e^{2\lambda}$ , yielding the trap answer $\lambda = 3.5$ .

Exact algebraic evaluation of $M_n$ is strictly required before taking the limit.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Statistics - Class 11
Concept	Combined Mean of Subgroups
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free