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Question #40344

Difficulty: 8/10 | 🟢 Free

The Question

Let $y: (0, \infty) \to \mathbb{R}$ be a differentiable function such that $y(x) \neq x$ for all $x \neq 1$ , $y(1) = 1$ , and $x \frac{dy}{dx} = y + 161 x \left| \frac{y}{x} - 1 \right|^{\frac{297}{298}}$ for all $x > 0$ . If $y(e) > e$ and $y(1/e) < 1/e$ , evaluate the limit: $\lim_{x \to 1} \frac{y(x) - x}{(x-1)^{298}}$

Options

A.
$\left( \frac{161}{298} \right)^{298}$
B.
$-\left( \frac{161}{298} \right)^{298}$
C.
0
D.
\text{Does not exist}

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

Identity Function Behavior|Substitution Logic in Homogeneous DE|The Modulus Trap|L'Hopital's Pitfall

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle absolute values in differential equations and limits, specifically targeting the 'L'Hopital's Pitfall'.

The differential equation is structured so the solution $y(x)$ mimics the identity function $y=x$ near $x=1$ .

The modulus creates a non-differentiable cusp in the derivative of the substituted variable $v(x) = y/x - 1$ at the origin of the shifted coordinates.

Students who blindly apply L'Hopital's rule or drop the modulus will incorrectly assume a symmetric solution $v(x) = (\frac{161}{298} \ln x)^{298}$ for all $x$ , leading them to erroneously choose option A.

However, proper interval analysis reveals that $v(x)$ behaves as an odd-like function around $x=1$ , causing the left and right limits to diverge in sign, meaning the two-sided limit does not exist.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Identity Function Behavior
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free