Problem Analysis #6

A.X-4.0 (72B) · Run 1

45.41s

7,410 toks

Problem Statement

\subsection*{I.}Find the general solution $y(x)$ of the following differential equation:

\frac{dy}{dx} = y(1-y),

where $0 < y < 1$ .\subsection*{II.}Find the value of the definite integral

I = \int_{-1}^{1} \frac{\arccos\left(\dfrac{x}{2}\right)}{\cos^2\left(\dfrac{\pi}{3}x\right)} \, dx,

where $0 \leq \arccos\left(\dfrac{x}{2}\right) \leq \pi$ .\subsection*{III.}For any positive variable $x$ , define

f(x) = \sum_{m=0}^{\infty} \frac{1}{(2m)!} x^{2m},\qquadg(x) = \frac{d}{dx} f(x).

For any non-negative integer $n$ , define

I_n(x) = \int_0^x \left(\frac{g(X)}{f(X)}\right)^n \, dX.

You may use

\exp(x) = \sum_{m=0}^{\infty} \frac{x^m}{m!}.

[(1)] Calculate $f(x)^2 - g(x)^2$ .
[(2)] Express $I_{n+2}(x)$ using $I_n(x)$ . \subsection*{IV. Final quantity}Let $y_*(x)$ be the solution of the differential equation in {\rm I} that satisfies

y_*(0) = \frac{1}{2},

and define

a := y_*(\ln 3).

In {\rm II}, let $I$ be the value of the definite integral. In {\rm III}, let

x_0 = \ln(2+\sqrt{3}),\qquadF_0 := f(x_0),\quad G_0 := g(x_0).

Define

S := 4a + \frac{I}{G_0} + F_0,\qquad\mathcal{K} := 999\,S^2 + 7.

Compute the integer $\mathcal{K}$ .\

Ground Truth

63943

Execution Trace

Reasoning Process

[THOUGHT] Let's solve each part of the problem step-by-step using Python code verification.

Part I: Differential Equation

The given differential equation is:

\frac{dy}{dx} = y(1-y)

This is a separable differential equation. We can rewrite it as:

\frac{1}{y(1-y)} \, dy = dx

Using partial fractions, we get:

\frac{1}{y(1-y)} = \frac{1}{y} + \frac{1}{1-y}

So the equation becomes:

\left( \frac{1}{y} + \frac{1}{1-y} \right) dy = dx

Integrating both sides:

\int \left( \frac{1}{y} + \frac{1}{1-y} \right) dy = \int dx

\ln |y| - \ln |1-y| = x + C

\ln \left| \frac{y}{1-y} \right| = x + C

Exponentiating both sides:

\left| \frac{y}{1-y} \right| = e^{x+C}

Let $e^C = K$ , then:

\frac{y}{1-y} = K e^x

Solving for $y$ :

y = \frac{K e^x}{1 + K e^x}

Given $0 < y < 1$ , we can drop the absolute value.

Part II: Definite Integral

The given integral is:

I = \int_{-1}^{1} \frac{\arccos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{3}x\right)} \, dx

To evaluate this integral, we will use numerical integration in Python.

Part III: Series Functions

(1) Calculate $f(x)^2 - g(x)^2$

Given:

f(x) = \sum_{m=0}^{\infty} \frac{1}{(2m)!} x^{2m},\qquad g(x) = \frac{d}{dx} f(x)

First, let's find $g(x)$ :

g(x) = \sum_{m=1}^{\infty} \frac{2m}{(2m)!} x^{2m-1} = \sum_{m=1}^{\infty} \frac{1}{(2m-1)!} x^{2m-1}

We recognize that:

f(x) = \cosh(x),\qquad g(x) = \sinh(x)

Using the hyperbolic identity:

\cosh^2(x) - \sinh^2(x) = 1

So:

f(x)^2 - g(x)^2 = 1

(2) Express $I_{n+2}(x)$ using $I_n(x)$

Given:

I_n(x) = \int_0^x \left(\frac{g(X)}{f(X)}\right)^n \, dX

We know:

\frac{g(x)}{f(x)} = \frac{\sinh(x)}{\cosh(x)} = \tanh(x)

So:

I_n(x) = \int_0^x \tanh^n(X) \, dX

Using integration by parts for $I_{n+2}(x)$ : Let $u = \tanh^{n+1}(X)$ and $dv = \tanh(X) \, dX$ . Then $du = (n+1) \tanh^n(X) \text{sech}^2(X) \, dX$ and $v = \ln(\cosh(X))$ .

Applying integration by parts:

I_{n+2}(x) = \int_0^x \tanh^{n+2}(X) \, dX = \left[ \tanh^{n+1}(X) \ln(\cosh(X)) \right]_0^x - \int_0^x (n+1) \tanh^n(X) \text{sech}^2(X) \ln(\cosh(X)) \, dX

Since $\tanh(0) = 0$ and $\ln(\cosh(0)) = \ln(1) = 0$ :

I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) \int_0^x \tanh^n(X) \text{sech}^2(X) \ln(\cosh(X)) \, dX

Notice that:

\text{sech}^2(X) = 1 - \tanh^2(X)

Thus:

I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) \left( \int_0^x \tanh^n(X) \ln(\cosh(X)) \, dX - \int_0^x \tanh^{n+2}(X) \ln(\cosh(X)) \, dX \right)

Let $J_n(x) = \int_0^x \tanh^n(X) \ln(\cosh(X)) \, dX$ :

I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) J_n(x) + (n+1) J_{n+2}(x)

Since $J_{n+2}(x)$ contains $I_{n+2}(x)$ and $J_n(x)$ contains $I_n(x)$ , we focus on simplifying:

I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) J_n(x) + (n+1) \left( \int_0^x \tanh^{n+2}(X) \ln(\cosh(X)) \, dX \right)

I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) \left( \int_0^x \tanh^n(X) \ln(\cosh(X)) \, dX - \int_0^x \tanh^{n+2}(X) \ln(\cosh(X)) \, dX \right)

I_{n+2}(x) = \tanh^{n+1}(x) \ln(\cosh(x)) - (n+1) I_n(x) + n I_{n+2}(x)

I_{n+2}(x) = \frac{1}{n+1} \tanh^{n+1}(x) \ln(\cosh(x)) - I_n(x)

Part IV: Final Quantity

First, find $y_*(x)$ that satisfies $y_*(0) = \frac{1}{2}$ :

y_*(x) = \frac{K e^x}{1 + K e^x}

Using the initial condition:

\frac{1}{2} = \frac{K}{1 + K}

1 + K = 2K

K = 1

Thus:

y_*(x) = \frac{e^x}{1 + e^x}

Now find $a = y_*(\ln 3)$ :

a = y_*(\ln 3) = \frac{e^{\ln 3}}{1 + e^{\ln 3}} = \frac{3}{1 + 3} = \frac{3}{4}

Next, evaluate the integral $I$ using Python:

I = \int_{-1}^{1} \frac{\arccos\left(\frac{x}{2}\right)}{\cos^2\left(\frac{\pi}{3}x\right)} \, dx

Then, compute $F_0$ and $G_0$ where $x_0 = \ln(2+\sqrt{3})$ :

F_0 = f(x_0) = \cosh(x_0),\quad G_0 = g(x_0) = \sinh(x_0)

Finally, compute $S$ and $\mathcal{K}$ .

Let's execute the Python code for these calculations. [/THOUGHT] [PYTHON] import numpy as np from scipy.integrate import quad

Part I: Differential Equation

We already have the solution y_*(x) = e^x / (1 + e^x)

a = y_*(ln 3)

a = 3 / 4

Part II: Definite Integral

def integrand(x): return np.arccos(x / 2) / np.cos(np.pi / 3 * x)**2

I, _ = quad(integrand, -1, 1)

Part III: Series Functions

x_0 = ln(2 + sqrt(3))

x_0 = np.log(2 + np.sqrt(3))

F_0 = f(x_0) = cosh(x_0)

F_0 = np.cosh(x_0)

G_0 = g(x_0) = sinh(x_0)

G_0 = np.sinh(x_0)

Part IV: Final Quantity

S = 4 * a + I / G_0 + F_0 K = 999 * S**2 + 7

print(K)

Tool Output

[PYTHON OUTPUT] 63943.0 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The Python output confirms the final value of $\mathcal{K}$ to be 63943. Therefore, the integer $\mathcal{K}$ is $\boxed{63943}$ .

Final Model Output

63943

Comparative Analysis (P6) (Average across runs)

Avg Time Taken

K-EXAONE-236B-A23B

330.0

GPT-5.2 (high)

111.5

Kanana-2-30B-Thinking-2601

76.9

Solar-Open-100B

67.2

GPT-5.1 (high)

57.7

HCX-007(high)

55.6

Llama-VARCO-8B-Instruct

54.7

Deepseek-V3.2

48.3

Gemini-3-Pro-Preview

47.6

A.X-4.0 (72B)

45.4

Kanana-2-30B-Thinking

42.7

Solar-Pro-2 (31B)(high)

40.2

Grok-4.1-fast

29.0

K-EXAONE-236B-A23B

22.8

Claude-Opus-4.5

20.6

EXAONE-4.0.1-32B (high)

11.1

Avg Token Usage

K-EXAONE-236B-A23B

39,933

GPT-5.2 (high)

26,193

Kanana-2-30B-Thinking-2601

19,787

Solar-Open-100B

12,619

HCX-007(high)

10,581

EXAONE-4.0.1-32B (high)

9,149

Solar-Pro-2 (31B)(high)

9,027

GPT-5.1 (high)

8,547

Grok-4.1-fast

8,351

Kanana-2-30B-Thinking

8,050

Gemini-3-Pro-Preview

7,889

A.X-4.0 (72B)

7,410

K-EXAONE-236B-A23B

6,804

Claude-Opus-4.5

5,887

Deepseek-V3.2

5,395

Llama-VARCO-8B-Instruct

1,716

Problem Analysis #6

Part I: Differential Equation

Part II: Definite Integral

Part III: Series Functions

(1) Calculate f(x)2−g(x)2f(x)^2 - g(x)^2f(x)2−g(x)2

(2) Express In+2(x)I_{n+2}(x)In+2​(x) using In(x)I_n(x)In​(x)

Part IV: Final Quantity

Part I: Differential Equation

We already have the solution y_*(x) = e^x / (1 + e^x)

a = y_*(ln 3)

Part II: Definite Integral

Part III: Series Functions

x_0 = ln(2 + sqrt(3))

F_0 = f(x_0) = cosh(x_0)

G_0 = g(x_0) = sinh(x_0)

Part IV: Final Quantity

Comparative Analysis (P6) (Average across runs)

Avg Time Taken

Avg Token Usage

(1) Calculate $f(x)^2 - g(x)^2$

(2) Express $I_{n+2}(x)$ using $I_n(x)$