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Question #40916

Difficulty: 8/10 | 🟢 Free

The Question

Let $\Pi$ be the plane passing through the points $(1,0,0)$ , $(0,1,0)$ , and $(0,0,1)$ . A point $P(x, x, z(x))$ moves in the region $x < 0$ such that its distance $p(x)$ from the plane $\Pi$ satisfies the differential equation: $|x| \frac{dp}{dx} - x p = \sqrt{3} x^3$ As $x \to 0^-$ , the point $P$ approaches the plane $\Pi$ . Given that $z(x) > 1 - 2x$ for all $x < 0$ , find the value of $z(-1)$ .

Options

A.
$6e - 12$
B.
$6 - 6e^{-1}$
C.
$6e^{-1}$
D.
$6e$

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

Collinearity and Coplanarity|Distance from point to plane|Integrating Factor (IF) Sign Sensitivity

Why This Question Is Tricky

Common Mistakes

This question tests the student's vigilance regarding domain constraints and absolute values in differential equations.

The trap lies in the term $|x|$ within the DE.

For the given domain $x < 0$ , $|x| = -x$ , which changes the sign of the entire equation and results in an Integrating Factor of $e^x$ .

If a student ignores the modulus and treats $|x|$ as $x$ , they incorrectly derive an IF of $e^{-x}$ .

This leads to a completely different locus that yields negative distances.

Students might then artificially force the distance to be positive, leading directly to trap option B, or blindly proceed to trap option C, completely invalidating the geometric coplanarity and distance constraints.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Collinearity and Coplanarity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free