Before You Begin

Question #41875

Difficulty: 8/10 | 🟢 Free

The Question

Let $y(x)$ be a continuous function on $[0, 20]$ such that $y(0) = 0$ and its derivative is given by $\frac{dy}{dx} = \{x\} \ln \left| \sum_{k=0}^{2025} \binom{2025}{k} (-2\{x\})^k \right|$ for all $x \in (0, 20)$ where the derivative is defined, and $\{x\}$ denotes the fractional part of $x$ . Find the value of $y(20)$ .

Options

A.
-20250
B.
-10125
C.
0
D.
20250

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

Integral Part and Fractional Part|Logarithmic Integration Modulus|Modulus Sign Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the 'Modulus Sign Dependency' trap within logarithmic integration, combined with binomial identities and periodic functions.

Students must first recognize the binomial sum as $(1 - 2\{x\})^{2025}$ .

The critical trap lies in pulling down the odd exponent: writing $\ln(1 - 2\{x\})^{2025} = 2025\ln(1 - 2\{x\})$ without the absolute value makes the integral undefined in the real domain for $x \in (1/2, 1)$ .

By correctly maintaining the modulus $\ln|1 - 2\{x\}|$ , students can use the substitution $u = 1 - 2x$ to shift the singularity to the origin.

This reveals an odd function $u\ln|u|$ over a symmetric interval $[-1, 1]$ , elegantly collapsing a difficult integration by parts (which would otherwise have boundary blow-ups at the singularity $x=1/2$ ) into a simple evaluation.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Integral Part and Fractional Part
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free