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Question #40902

Difficulty: 8/10 | 🟢 Free

The Question

Let $L$ be the locus of the center of a variable circle which touches the circle $x^2 + y^2 - 4x = 0$ externally and also touches the y-axis. Let $f(x)$ be a real-valued function defined on its maximal domain in $(0, \infty)$ by $f(x) = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{x^2-4}{x^2-4} \cdot \frac{n}{(r+nx)^2}$ Let $S$ be the set of points $(x,y)$ on $L$ such that $x > 0$ and $y^2 = 16 x (x+1) f(x)$ . Find the number of elements in $S$ .

Options

A.
0
B.
1
C.
2
D.
4

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

Locus via Distance Constraints|Integral as the Limit of a Sum|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

The question integrates coordinate geometry (locus via distance constraints) with integral calculus (Riemann sums).

The locus simplifies to a parabola $y^2 = 8x$ .

The Riemann sum evaluates to $1/(x(x+1))$ , which simplifies the given condition to $y^2 = 16$ .

Intersecting this with the locus yields $x=2$ .

The core trap is the 'Zero-Factor Cancellation Trap': the expression for $f(x)$ contains the term $(x^2-4)/(x^2-4)$ , making $f(x)$ strictly undefined at $x=2$ .

Students who naively cancel this factor before evaluating the domain will incorrectly include $x=2$ as a valid solution and find two points $(2, 4)$ and $(2, -4)$ , leading to the trap answer 2.

Recognizing that the intersection point lies exactly at the removable discontinuity ensures the correct answer of 0.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Locus via Distance Constraints
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free