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Question #40962

Difficulty: 8/10 | 🟢 Free

The Question

A particle moves along a straight line such that its position $x(t)$ at time $t$ is given by $x(t) = \ln(t^2 - 8t + 20) - |t-4|$ Let $N$ be the number of points in the time interval $t \in (1, 7)$ where the particle comes to instantaneous rest (i.e., $\frac{dx}{dt} = 0$ ). Let $M$ be the number of points in the same interval where $x(t)$ attains a local extremum. The value of $N + M$ is:

Options

A.
0
B.
1
C.
2
D.
3

Check Your Answer

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

Physical Interpretation|Rolle's Theorem Anchors|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of Rolle's Theorem and the distinction between physical extrema and mathematical stationary points.

Students often observe $x(1) = x(7)$ and incorrectly invoke Rolle's Theorem to conclude that $x'(t) = 0$ must have at least one solution in the interval, leading them to assume $N \ge 1$ .

However, Rolle's Theorem requires differentiability on the open interval.

The presence of the modulus function creates a non-differentiable cusp at $t=4$ .

By evaluating the left and right derivatives, it becomes clear that the velocity is never zero ( $N=0$ ), even though the particle physically reverses direction at a local extremum ( $M=1$ ).

This traps students who rely on formulaic theorems without verifying domain constraints.

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