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Question #40873

Difficulty: 8/10 | 🟢 Free

The Question

Let $C: y = f(x)$ be a curve in the first quadrant with $f'(x) > 0$ for all $x > 0$ . For any point $P(x,y)$ on $C$ , let the tangent at $P$ meet the $y$ -axis at $A$ and the normal at $P$ meet the $x$ -axis at $B$ . Let $T_x$ and $T_y$ be the lengths of the projections of the line segment $PA$ onto the $x$ -axis and $y$ -axis, respectively. Let $L_y$ be the length of the projection of the line segment $PB$ onto the $y$ -axis. If the curve satisfies the relation $T_x T_y = T_x^2 + T_x L_y - L_y^2$ and passes through the point $(1, 1)$ , then the value of $f(2024)$ is

Options

A.
1012
B.
2024
C.
4048
D.
e^{2024}

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

Coordinate Axis Projections|Homogeneous Function Homogeneity|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Zero-Factor Cancellation Trap' in homogeneous differential equations.

When students substitute $y=vx$ and separate variables, they routinely divide by a function of $v$ (here, $1-v^2$ ) without considering the case where it equals zero.

This algebraic oversight eliminates the singular solution $v=1$ ( $y=x$ ).

Because the general solution family cannot satisfy the initial condition $f(1)=1$ (it results in an undefined logarithmic term), students who fall for the trap will incorrectly conclude that no such curve exists or guess randomly, effectively filtering out those who lack rigorous foundational logic.

Quick Facts About This Question

Question Information

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