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Question #40748

Difficulty: 8/10 | 🟢 Free

The Question

Let $C$ be a curve passing through the point $P$ , where $P$ is the point of contact of the tangent to the parabola $y^2 = 4x$ that is perpendicular to the line $x - y = 0$ . If the curve $C$ satisfies the differential equation $\frac{dy}{dx} = y - x + 1$ , and intersects the line $x = 1 + \ln 2$ at the point $(1 + \ln 2, y_0)$ , then the value of $y_0$ is

Options

A.
$\ln 2 - 5$
B.
$\ln 2 + 7$
C.
$5 - \ln 2$
D.
$7 - \ln 2$

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

Point-Line Relative Position|Condition of Tangency|Logarithmic Integration Modulus

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to correctly apply boundary conditions to a logarithmic integral arising from a differential equation.

The substitution $v = y - x$ leads to the integral $\int \frac{dv}{v} = \ln|v|$ .

The critical trap lies in the modulus.

At the boundary point $P(1, -2)$ , the argument $v = -3$ is negative.

Students who blindly integrate to $\ln(v)$ or drop the modulus will either incorrectly assume the problem is invalid or force a positive argument, leading to the erroneous branch $y - x = 3e^{x-1}$ and the trap answer $\ln 2 + 7$ .

Recognizing that the trajectory remains on the $y - x < 0$ branch is essential for the correct solution.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Point-Line Relative Position
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free