Problem Analysis #common_09

Motif 3 · Run 1

Python ● execution succeeded

Correct

63.75s

13,433 toks

Infron · Motif 3 · 46문항 × 3회 · Python TAR. 131,072 output tokens · 기본 1,800초 · 6+1 turns. ACC는 전체 138회, Pass@1은 각 문항의 첫 유효 시도, Pass@3는 3회 내 정답으로 계산합니다.

Problem Statement

양수 $a$ 에 대하여 함수 $f(x)$ 를 $f(x)=x^3+3ax^2-9a^2x+4$ 라 하자. 직선 $y=5$ 가 곡선 $y=f(x)$ 에 접할 때, $f(2)$ 의 값은?

① $11$ ② $12$ ③ $13$ ④ $14$ ⑤ $15$

최종 답은 선택지 번호 1~5 중 하나를 $\boxed{}$ 안에 쓰세요.

Ground Truth

Recorded model response / solution

Reasoning Process

Recorded Python call:

import sympy as sp

a, x = sp.symbols('a x', real=True)
f = x**3 + 3*a*x**2 - 9*a**2*x + 4
fp = sp.diff(f, x)

# Critical points where f'(x)=0
crit = sp.solve(fp, x)
crit_vals = {c: sp.simplify(f.subs(x, c)) for c in crit}
crit_vals

Tool Output

Execution succeeded

Reasoning Process

Recorded Python call:

import sympy as sp

a, x = sp.symbols('a x', real=True)
f = x**3 + 3*a*x**2 - 9*a**2*x + 4
fp = sp.diff(f, x)
crit = sp.solve(fp, x)
for c in crit:
    val = sp.simplify(f.subs(x, c))
    print(c, val)
    sol = sp.solve(sp.Eq(val, 5), a)
    print("solutions for y=5:", sol)

Tool Output

Execution succeeded -3*a 27*a**3 + 4 solutions for y=5: [1/3] a 4 - 5*a**3 solutions for y=5: [-5**(2/3)/5]

Reasoning Process

Recorded Python call:

a_val = sp.Rational(1,3)
f2 = sp.simplify(f.subs({a:a_val, x:2}))
f2

Tool Output

Execution failed name 'sp' is not defined

Reasoning Process

Recorded Python call:

import sympy as sp
a, x = sp.symbols('a x')
f = x**3 + 3*a*x**2 - 9*a**2*x + 4
f2 = sp.simplify(f.subs({a: sp.Rational(1,3), x:2}))
f2

Tool Output

Execution succeeded

Reasoning Process

Recorded Python call:

import sympy as sp
a, x = sp.symbols('a x')
f = x**3 + 3*a*x**2 - 9*a**2*x + 4
print(sp.simplify(f.subs({a: sp.Rational(1,3), x:2})))

Tool Output

Execution succeeded 14

Reasoning Process

수평선 $y=5$ 가 곡선 $y=f(x)$ 에 접하려면 접점 $x$ 에서 $f'(x)=0$ 이고 $f(x)=5$ 여야 한다.

f'(x)=3x^2+6ax-9a^2=3(x+3a)(x-a)

따라서 임계점은 $x=-3a$ , $x=a$ 이다.

$x=-3a$ 일 때

f(-3a)=27a^3+4

27a^3+4=5 \Rightarrow a^3=\frac{1}{27} \Rightarrow a=\frac13

$x=a$ 일 때

f(a)=4-5a^3

양수 $a$ 에 대해 $4-5a^3<4$ 이므로 $5$ 가 될 수 없다.

따라서 $a=\frac13$ 이고,

f(2)=2^3+3\cdot \frac13\cdot 2^2-9\cdot \left(\frac13\right)^2\cdot 2+4 =8+4-2+4=14

이는 선택지 $④$ 에 해당한다.

\boxed{4}

Final Model Output

Comparative Analysis (Pcommon_09) (Average across runs)

Averages use reported measurements only. Missing usage is not treated as zero.

Avg Time Taken

MiMo V2.6 Pro

228.7

Solar Pro 4

85.0

Motif 3

63.1

Union Alpha

44.8

Grok 4.7

24.7

Muse Spark 1.3

12.1

DeepSeek V4.1 Flash

8.4

Solar Mini 4 Preview

7.2

Avg Token Usage

MiMo V2.6 Pro

8,958

Motif 3

7,904

Union Alpha

7,684

Solar Pro 4

7,679

Grok 4.7

7,556

Solar Mini 4 Preview

6,093

Muse Spark 1.3

4,587

DeepSeek V4.1 Flash

3,781