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Question #60793

Difficulty: 8/10 | 🟢 Free

The Question

Let $a_1 < a_2 < \dots < a_{2m+1}$ (where $m \in \mathbb{Z}, m \ge 2$ ) be an arithmetic progression of real numbers with common difference $d > 0$ . Let $\mu$ be the arithmetic mean of these numbers. A function $f: \mathbb{R} \to \mathbb{R}$ is defined as $f(x) = \frac{1}{2m+1} \sum_{i=1}^{2m+1} |x - a_i|$ . For which of the following intervals is Lagrange's Mean Value Theorem strictly applicable to $f(x)$ ?

Options

A.
$\left[\mu - \frac{d}{2}, \mu + \frac{d}{2}\right]$
B.
$\left[a_m, a_{m+2}\right]$
C.
$\left[\mu + \frac{d}{3}, \mu + \frac{2d}{3}\right]$
D.
$\left[a_1, \mu\right]$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Properties of Arithmetic Mean|Mean Value Theorem Applicability|Modulus Function Breakdown

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the applicability conditions of Lagrange's Mean Value Theorem (LMVT) against the piecewise non-differentiability of a sum of modulus functions.

The core trap lies in the 'smoothing illusion' of the arithmetic mean operation; students often incorrectly assume that averaging the absolute deviations creates a globally differentiable function or that the function is well-behaved around its centroid $\mu$ .

By setting $\mu$ exactly at a point of non-differentiability ( $a_{m+1}$ ), the problem forces students to recognize that LMVT strictly requires the open interval to be free of any cusps.

Failing to check differentiability at the mean's centroid effectively masks the function's internal discontinuities, leading to incorrect interval selection.

Quick Facts About This Question

Question Information

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