Problem Analysis #17

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Problem Statement

Let $a$ and $b$ be real numbers such that $\lim_{x \to 0} \frac{1}{x^3} \left( \frac{a}{2} \int_0^x \frac{1}{1-t^2}\, dt + bx\cos x \right) = 2.$ Find the value of $100(a + b)^2$ .\

Ground Truth

576

Execution Trace

Reasoning Process

To solve the problem, we are given:

\lim_{x \to 0} \frac{1}{x^3} \left( \frac{a}{2} \int_0^x \frac{1}{1 - t^2}\, dt + b x \cos x \right) = 2

We are to find the value of $100(a + b)^2$ .

Step 1: Expand the Integral

We begin by evaluating the integral:

\int_0^x \frac{1}{1 - t^2} dt = \frac{1}{2} \ln\left(\frac{1 + x}{1 - x}\right)

However, for small $x$ , we can use the Taylor expansion:

\frac{1}{1 - t^2} = 1 + t^2 + t^4 + \cdots \quad \text{(for } |t| < 1\text{)}

Integrating term-by-term from 0 to $x$ :

\int_0^x \frac{1}{1 - t^2} dt = x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots

So,

\frac{a}{2} \int_0^x \frac{1}{1 - t^2} dt = \frac{a}{2}x + \frac{a}{6}x^3 + \cdots

Step 2: Expand the Second Term

Next, expand $x \cos x$ :

x \cos x = x\left(1 - \frac{x^2}{2} + \cdots\right) = x - \frac{x^3}{2} + \cdots

Thus,

b x \cos x = b x - \frac{b}{2}x^3 + \cdots

Step 3: Combine Both Expansions

Now combine the two parts:

N(x) = \left( \frac{a}{2}x + \frac{a}{6}x^3 \right) + \left(b x - \frac{b}{2}x^3 \right) + o(x^3)

Grouping like terms:

Coefficient of $x$ : $\frac{a}{2} + b$
Coefficient of $x^3$ : $\frac{a}{6} - \frac{b}{2}$

So:

N(x) = \left(\frac{a}{2} + b\right)x + \left(\frac{a}{6} - \frac{b}{2}\right)x^3 + o(x^3)

Step 4: Apply the Limit Condition

We're told:

\lim_{x \to 0} \frac{N(x)}{x^3} = 2

For this limit to exist and be finite, the coefficient of $x$ in $N(x)$ must be zero (otherwise, the limit would diverge). So we require:

\frac{a}{2} + b = 0 \tag{1}

Then, the leading term becomes:

\left( \frac{a}{6} - \frac{b}{2} \right)x^3

So the limit is:

\lim_{x \to 0} \frac{N(x)}{x^3} = \frac{a}{6} - \frac{b}{2} = 2 \tag{2}

Step 5: Solve the System of Equations

From (1):

b = -\frac{a}{2}

Substitute into (2):

\frac{a}{6} - \frac{-a/2}{2} = \frac{a}{6} + \frac{a}{4} = \frac{2a + 3a}{12} = \frac{5a}{12} = 2

Solve for $a$ :

\frac{5a}{12} = 2 \Rightarrow a = \frac{24}{5}

Then:

b = -\frac{a}{2} = -\frac{12}{5}

Step 6: Compute Final Answer

a + b = \frac{24}{5} - \frac{12}{5} = \{12}{5}

(a + b)^2 = \left(\frac{12}{5}\right)^2 = \frac{144}{25}

100(a + b)^2 = 100 \cdot \frac{144}{25} = \boxed{576}

✅ Final Answer:

\boxed{576}

Reasoning Process

[THOUGHT] [PYTHON] a = 24/5 b = -12/5 print(int(100 * (a + b)**2)) [/PYTHON]

Tool Output

[PYTHON OUTPUT] 576 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The previous Python code correctly computed the value of $100(a + b)^2$ as 576. Since the calculation was verified through symbolic expansion and confirmed via Python, the final answer is:

\boxed{576}

Final Model Output

576

Comparative Analysis (P17) (Average across runs)

Avg Time Taken

Solar-Open-100B

632.5

K-EXAONE-236B-A23B

317.1

Llama-VARCO-8B-Instruct

57.7

Deepseek-V3.2

51.0

Kanana-2-30B-Thinking

41.5

K-EXAONE-236B-A23B

38.6

HCX-007(high)

38.3

Kanana-2-30B-Thinking-2601

33.1

Gemini-3-Pro-Preview

30.4

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

27.4

GPT-5.2 (high)

24.4

GPT-5.1 (high)

23.9

Grok-4.1-fast

23.5

A.X-4.0 (72B)

21.9

Claude-Opus-4.5

15.2

axk1

12.2

EXAONE-4.0.1-32B (high)

7.2

Avg Token Usage

K-EXAONE-236B-A23B

42879.3

Grok-4.1-fast

8530.0

Solar-Open-100B

7810.0

Kanana-2-30B-Thinking-2601

7014.5

Kanana-2-30B-Thinking

6949.0

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

5828.0

GPT-5.2 (high)

5763.0

Gemini-3-Pro-Preview

5699.0

Deepseek-V3.2

5224.0

GPT-5.1 (high)

4873.0

K-EXAONE-236B-A23B

4597.0

EXAONE-4.0.1-32B (high)

4374.0

HCX-007(high)

4370.0

Claude-Opus-4.5

3675.0

axk1

3283.0

A.X-4.0 (72B)

2081.0

Llama-VARCO-8B-Instruct

1031.0