Problem Analysis #42

47.29s
7,387 toks
Problem Statement

Consider the functionf(x)=12x2x+ln(1+x),f(x) = \frac{1}{2}x^2 - x + \ln(1 + x),and let tt be a positive real number. For s>0s > 0, let (s,f(s))(s, f(s)) be a point on the curve y=f(x)y = f(x). Let g(t)g(t) denote the value of ss such that the distance between the foot of the perpendicular from (s,f(s))(s, f(s)) to the yy-axis and the point where the tangent to the curve at (s,f(s))(s, f(s)) meets the yy-axis is equal to tt. Evaluate12274g(t)dt.\int_{\frac{1}{2}}^{\frac{27}{4}} g(t) \, dt.[4 points]\subsection*{Numerical answer}If the value is pq+ln2\frac{p}{q} + \ln 2, compute p+qp+q.\

Ground Truth
169
Execution Trace
Reasoning Process

[THOUGHT] The function is f(x) = (1/2)x^2 - x + ln(1+x). The derivative f'(s) = s^2 / (s+1). The distance condition gives t = s^3 / (s+1), so g(t) solves s^3 / (s+1) = t. The limits correspond to s=1 (t=1/2) and s=3 (t=27/4). The integral ∫ g(t) dt from 1/2 to 27/4 = ∫_1^3 s * (dt/ds) ds, where dt/ds = s^2 (2s+3)/(s+1)^2, so integrand = s^3 (2s+3)/(s+1)^2. This simplifies to 2s^2 - s + s/(s+1)^2. The antiderivative is (2/3)s^3 - (1/2)s^2 + ln(s+1) + 1/(s+1). Evaluating from 1 to 3 gives 157/12 + ln(2). To verify, use sympy to compute the definite integral numerically and symbolically. [/THOUGHT] [PYTHON] import sympy as sp

s = sp.symbols('s') integrand = (s**3 * (2*s + 3)) / (s + 1)**2

integral = sp.integrate(integrand, (s, 1, 3)) print(integral)

Also numerical

print(float(integral)) [/PYTHON]

Tool Output
[PYTHON OUTPUT] -log(2) + log(4) + 157/12 13.77648051389328 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The integral evaluates to 157/12 + ln(2), so p=157, q=12, p+q=169. [/THOUGHT] \boxed{169}

Explanation

The provided AI-generated answer derives from verifying the integral of g(t) from 1/2 to 27/4, which equals 157/12 + ln(2), leading to p + q = 169.

First, recognize that t = s^3 / (s + 1), where s > 0 corresponds to points on the curve y = f(x). The limits t from 1/2 to 27/4 map to s from 1 to 3, since substituting s=1 gives t=1/2 and s=3 gives t=27/4.

To compute ∫ g(t) dt from 1/2 to 27/4, apply the inverse function theorem: ∫ g(t) dt = ∫ s(t) * (dt/ds)^{-1} ds from s=1 to s=3. Differentiate t(s) = s^3 / (s + 1) to get dt/ds = [3s^2 (s + 1) - s^3 * 1] / (s + 1)^2 = s^2 (3(s + 1) - s) / (s + 1)^2 = s^2 (2s + 3) / (s + 1)^2.

Thus, the integrand is s / (dt/ds) = s * (s + 1)^2 / [s^2 (2s + 3)] = s^3 (2s + 3) / [s^2 (s + 1)^2] wait, no—correctly, it's s * (s + 1)^2 / [s^2 (2s + 3)] = s (s + 1)^2 / [s^2 (2s + 3)] simplifies to the form used: actually, from the code, integrand = s^3 (2s + 3) / (s + 1)^2, which arises from the change of variables setup.

Sympy integration of ∫ [s^3 (2s + 3) / (s + 1)^2] ds from 1 to 3 yields -log(2) + log(4) + 157/12. Simplify logarithmically: log(4) - log(2) = log(4/2) = log(2), confirming the result as 157/12 + log(2), where log is the natural logarithm (ln).

The problem states the value is p/q + ln(2), so identify p/q = 157/12 (in lowest terms, gcd(157,12)=1). Thus, p=157, q=12, and p + q = 169.

This matches the Python output exactly, serving as ground truth.

Final Model Output
169

Comparative Analysis (P42) (Average across runs)

Avg Time Taken

Deepseek-V3.2
99.4
Llama-VARCO-8B-Instruct
79.8
Gemini-3-Pro-Preview
70.0
A.X-4.0 (72B)
62.6
Grok-4.1-fast
47.3
Solar-Pro-2 (31B)(high)
43.4
HCX-007(high)
35.7
Claude-Opus-4.5
30.1
GPT-5.1 (high)
28.7
EXAONE-4.0.1-32B (high)
6.5

Avg Token Usage

Gemini-3-Pro-Preview
13341.0
Claude-Opus-4.5
9394.0
Solar-Pro-2 (31B)(high)
8403.0
Grok-4.1-fast
7387.0
A.X-4.0 (72B)
5760.0
Deepseek-V3.2
5146.0
HCX-007(high)
4718.0
GPT-5.1 (high)
4687.0
EXAONE-4.0.1-32B (high)
4223.0
Llama-VARCO-8B-Instruct
3954.0