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Question #70316

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function. Define $F(x) = \lim_{n \to \infty} \sum_{k=1}^n \frac{x^2}{n + k x^2} f\left( \ln\left( 1 + \frac{k x^2}{n} \right) \right)$ If the Lagrange Mean Value Theorem is applied to $F(x)$ on the interval $[a, b]$ (where $0 < a < b$ ), then there exists $c \in (a, b)$ such that $f(\ln(1+c^2))$ is equal to:

Options

A.
$\frac{(1+c^2) \int_{\ln(1+a^2)}^{\ln(1+b^2)} f(t) dt}{2c (b - a)}$
B.
$\frac{\int_{\ln(1+a^2)}^{\ln(1+b^2)} f(t) dt}{b - a}$
C.
$\frac{(1+c^2) \int_{\ln(1+a^2)}^{\ln(1+b^2)} f(t) dt}{b - a}$
D.
$\frac{(1+c^2) \int_{a}^{b} f(t) dt}{2c (b - a)}$

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

Integral as the Limit of a Sum|LMVT Tangent Parallelism|Chain Rule Inner Function Differentiability

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 synthesis of Riemann sums and the Lagrange Mean Value Theorem, specifically targeting the Chain Rule trap in the Leibniz integral rule.

Students must first correctly identify the limit of a sum and use substitution to evaluate the definite integral, yielding $F(x) = \int_0^{\ln(1+x^2)} f(u) du$ .

The primary trap occurs during the application of LMVT: when differentiating $F(x)$ , students frequently forget to apply the chain rule to the upper limit $\ln(1+x^2)$ , or they differentiate it incorrectly (missing the $2x$ factor).

This leads directly to the incorrect options B and C, exposing a superficial understanding of composite function differentiability within integral calculus.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Integral as the Limit of a Sum
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free