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Question #40868

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(0,0)$ and $B(12\pi, 0)$ be two points. A point $P$ divides the line segment $AB$ internally in the ratio $k:1$ . Let $C$ be the curve defined by $y = 4 + \sin x + \sqrt{3}\cos x$ . If $k$ is chosen such that the vertical distance from $P$ to the curve $C$ is minimized, and the horizontal distance from $P$ to the point $M(6\pi, 0)$ is also minimized among all such possible positions of $P$ , then the value of $k$ is:

Options

A.
$\frac{31}{41}$
B.
$\frac{7}{65}$
C.
$\frac{43}{29}$
D.
1

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

Ratio of Division (k:1)|Geometry Constraint Modeling|Symmetry and Periodicity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle periodic constraints within a restricted geometric domain.

The objective function for the vertical distance relies on a trigonometric expression, yielding infinite periodic minima.

The primary trap lies in rote algebraic derivation: a student who merely solves the derivative for the first principal value will incorrectly find $x_P = 7\pi/6$ , leading to the trap answer $7/65$ .

To succeed, the student must recognize that the internal division ratio $k>0$ restricts the domain to $(0, 12\pi)$ , generating a finite set of valid periodic solutions.

They must then apply the secondary geometric constraint (minimum distance to $M$ ) to select the correct specific period $n=2$ , successfully intersecting the periodic solution set with the geometric boundary.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Ratio of Division (k:1)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free