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Question #40963

Difficulty: 8/10 | 🟢 Free

The Question

The velocity $v(t)$ of a particle moving along the x-axis is given by $v(t) = \frac{\ln(1+t) \cdot (e^t - 1) - t^2}{t^4}$ for $t \neq 0$ . If the acceleration $a(t)$ of the particle is continuous at $t=0$ , what is the value of $a(0)$ ?

Options

A.
$-\frac{1}{8}$
B.
$\frac{1}{12}$
C.
$-\frac{1}{6}$
D.
$0$

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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

Physical Interpretation|Logarithmic Series|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Zero-Factor Cancellation Trap' via premature truncation of Taylor series in a physical kinematics context.

Students often truncate the expansions of $\ln(1+t)$ and $e^t-1$ at $O(t^3)$ , assuming higher-order terms won't affect the linear term of $v(t)$ .

However, because the functions are multiplied, an $O(t^4)$ term in one series multiplied by the $O(t)$ term in the other directly contributes to the $t^5$ coefficient of the product.

Truncating at $t^3$ yields an incorrect acceleration of $1/12$ , while truncating at $t^2$ yields $0$ .

Rigorous tracking of the $O(t^n)$ remainder is required to find the true instantaneous rate of change.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Limits and Derivatives - Class 11
Concept	Physical Interpretation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free