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Question #41590

Difficulty: 8/10 | 🟢 Free

The Question

A sequence of functions $\{y_n(x)\}_{n=1}^\infty$ satisfies the differential equation $\frac{dy_n}{dx} - \lambda y_n = 0$ with the initial condition $y_n(0) = C_n(\lambda)$ , where $\lambda > 0$ is a parameter. It is given that $C_0(\lambda) = 0$ and for all $n \ge 1$ , the constants $C_n(\lambda)$ satisfy the relation: $\int_0^1 y_n(x) dx - \int_0^1 y_{n-1}(x) dx = \frac{e^\lambda + \lambda - 1}{\lambda^4 n(n+1)}$ Let $C_\infty(\lambda) = \lim_{n \to \infty} C_n(\lambda)$ . Find the value of $\lim_{\lambda \to 0^+} (e^\lambda - 1) C_\infty(\lambda)$ .

Options

A.
$\frac{1}{6}$
B.
$\frac{1}{2}$
C.
0
D.
$\infty$

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

Arbitrary Constants vs Parameters|Method of Differences|L'Hopital's Pitfall

Why This Question Is Tricky

Common Mistakes

This question tests the resolution of a sequence of arbitrary constants derived from a differential equation using the Method of Differences, followed by a critical limit evaluation.

The primary trap lies in the final limit evaluation of $(e^\lambda + \lambda - 1)/\lambda^3$ as $\lambda \to 0^+$ .

Students often blindly apply L'Hôpital's rule multiple times without verifying the indeterminate form condition.

After one application, the form becomes $2/0$ , which diverges to $\infty$ .

However, applying L'Hôpital's rule two more times erroneously yields $1/6$ , a classic pitfall for students who treat algorithmic differentiation as a substitute for rigorous limit analysis.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Differential Equations - Class 12
Concept	Arbitrary Constants vs Parameters
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free