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Question #41640

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{r}(t) = (2t+1)\hat{i} + (t-2)\hat{j} + (3t+4)\hat{k}$ be the parametric equation of a line. Let $\vec{u} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{v} = \hat{i} - \hat{j} + \hat{k}$ . A function $f: \mathbb{R} \to \mathbb{R}$ is defined as $f(x) = \int_0^x |y| e^{-y^2} dy$ . Let $g(t) = f([\vec{r}(t) \ \vec{u} \ \vec{v}])$ . Let $t_0$ be the value of $t$ where $\vec{r}(t)$ becomes linearly dependent on $\vec{u}$ and $\vec{v}$ . If $L = \lim_{h \to 0^+} \frac{g'(t_0+h) - g'(t_0)}{h}$ and $R = \lim_{h \to 0^-} \frac{g'(t_0+h) - g'(t_0)}{h}$ , find the value of $|L - R|$ .

Options

A.
128
B.
64
C.
0
D.
256

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

Parametric form pitfalls|Linear Independence vs Dependence|Composition Continuity|Second Order Derivative Continuity

Why This Question Is Tricky

Common Mistakes

This question tests the subtle distinction between first-order and second-order differentiability when composing a parametric path with an absolute-value-based integral function.

The primary trap lies at the point of linear dependence ( $t_0 = -1/2$ ), where the scalar triple product crosses zero.

Students often correctly identify that $g(t)$ is continuously differentiable ( $C^1$ ) because $g'(t_0) = 0$ , leading them to falsely assume it is twice differentiable ( $C^2$ ).

The advanced pitfall is failing to recognize that the chain rule for the second derivative, $g''(t) = f''(V(t))(V'(t))^2$ , inherits the jump discontinuity of $f''(x)$ at $x=0$ .

Furthermore, the negative slope of $V(t)$ reverses the left and right limits, testing rigorous application of the chain rule under directional limits.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Parametric form pitfalls
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free