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Question #41920

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a}$ and $\vec{b}$ be two linearly independent vectors in $\mathbb{R}^3$ such that $|\vec{a}| = 2$ and the scalar projection of $\vec{b}$ on $\vec{a}$ is $-\frac{3}{2}$ . A differentiable vector-valued function $\vec{y}(x)$ satisfies the differential equation $(x + \vec{a} \cdot \vec{b}) \frac{d\vec{y}}{dx} = \vec{a} + (x + \vec{a} \cdot \vec{b}) \vec{b}$ for $x < 3$ , with the initial condition $\vec{y}(1) = \vec{b}$ . Find the value of $\vec{y}(2) \cdot \vec{a}$ .

Options

A.
$-4\ln 2 - 6$
B.
$4\ln 2 - 6$
C.
$-4\ln 2 + 6$
D.
$4\ln 2 + 6$

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

Logarithmic Integration Modulus|Linear Independence vs Dependence|Dot Product Projection Sign

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the logarithmic integration modulus when the domain forces a negative argument.

The scalar projection sets the constant $k = -3$ , creating the integral $\int 1/(x-3) dx$ .

Because the domain is $x < 3$ , students must integrate to $\ln|x-3|$ .

A lethal trap is laid for students who either incorrectly factor out a negative sign to get $-\ln(3-x)$ or drop the modulus to get $\ln(x-3)$ and subsequently 'fudge' the resulting complex logarithm by assuming $\ln(-x) = -\ln x$ .

Both of these common algebraic errors lead to an incorrect constant of integration $\vec{C} = (\ln 2)\vec{a}$ , which perfectly generates the distractor option $4\ln 2 - 6$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Differential Equations - Class 12
Concept	Logarithmic Integration Modulus
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free