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Question #41928

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{u}(t) = (t^2 - 5t + 6)\hat{i} + (t^2 - 3t + 2)\hat{j}$ and $\vec{v}(t) = \hat{i} - \hat{j}$ . Define a vector function $\vec{r}(t) = \text{sgn}(\vec{u}(t) \cdot \vec{v}(t)) \hat{i} + (|t-4| - 2) \hat{j}$ and let $\vec{w}(t) = (t-5) \hat{i} + \hat{j}$ , where $\text{sgn}(\cdot)$ denotes the signum function. A function $F(x)$ is defined for $x > 0$ as: $F(x) = \int_{0}^{x} \vec{r}(t) \cdot \vec{w}(t) \, dt$ If $x_1$ and $x_2$ are the points of local minimum and local maximum of $F(x)$ respectively, find the value of $4(F(x_2) - F(x_1))$ .

Options

A.
7
B.
9
C.
11
D.
13

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

Dot Product Projection Sign|Newton-Leibniz Formula with variable limits|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the Newton-Leibniz formula on piecewise functions defined by dynamic vector projections and modulus operators.

The primary trap lies in identifying the critical points of $F(x)$ .

Students who blindly set $F'(x) = 0$ will only find $x = 3.5$ and completely miss the local minimum at $x = 2$ , where the derivative is undefined due to the jump discontinuity of the signum function (toggled by the dot product's sign change).

Furthermore, the modulus critical point at $x = 4$ acts as a decoy; the derivative is continuous and non-zero there, meaning it is not an extremum.

This forces a meticulous interval-by-interval breakdown of the derivative's sign rather than relying on naive differentiation.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Dot Product Projection Sign
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free