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
최고차항의 계수가 양수인 삼차함수 와 실수 t에 대하여 함수 는 실수 전체의 집합에서 연속이고 다음 조건을 만족시킨다. (가) 모든 실수 a에 대하여 의 값이 존재한다. (나) 의 값이 음수가 되도록 하는 자연수 m의 집합은 이다. 의 값을 구하시오. (단, )
최종 정수 답을 안에 쓰세요.
Ground Truth
65
Recorded model response / solution
Reasoning Process
Recorded Python call:
import sympy as sp
# Define symbols
a, s = sp.symbols('a s', positive=True, real=True)
# We have a = 3/14, s = 11/3
# Let's verify the conditions and compute g(-5)
a_val = sp.Rational(3,14)
s_val = sp.Rational(11,3)
t_val = 2
# Define f(x)
x = sp.symbols('x')
f = a_val * x * (x-2) * (x - s_val)
sp.simplify(f)
Tool Output
Execution succeeded
Reasoning Process
Recorded Python call:
a_val = 3/14
s_val = 11/3
def f(x):
return a_val * x * (x-2) * (x - s_val)
def g(x):
if x < 2:
return -f(x)
else:
return f(x)
# Check conditions
print("f(-5) =", f(-5))
print("g(-5) =", g(-5))
print("g(-1) =", g(-1))
print("g(1) =", g(1))
print("-7/2 g(1) =", -7/2 * g(1))
# Check limits
import math
def h(x):
return g(x) / (x * (x-2))
# Check limits at m=1,2,3,4
for m in [1,2,3,4,5]:
# right-hand limit
eps = 1e-6
if m == 0:
pass
else:
val = h(m + eps)
print(f"L_{m} = {val}")
Tool Output
Execution succeeded
f(-5) = -65.0
g(-5) = 65.0
g(-1) = 2.999999999999999
g(1) = -0.5714285714285714
-7/2 g(1) = 2.0
L_1 = 0.5714283571428571
L_2 = -0.3571426428571428
L_3 = -0.1428569285714285
L_4 = 0.07142878571428576
L_5 = 0.2857145
Reasoning Process
Recorded Python call:
# Exact verification using rational arithmetic
from fractions import Fraction
a = Fraction(3,14)
s = Fraction(11,3)
t = 2
def f_exact(x):
return a * x * (x-2) * (x - s)
def g_exact(x):
if x < t:
return -f_exact(x)
else:
return f_exact(x)
print("f(-5) =", f_exact(-5))
print("g(-5) =", g_exact(-5))
print("g(-1) =", g_exact(-1))
print("g(1) =", g_exact(1))
print("-7/2 g(1) =", -Fraction(7,2) * g_exact(1))
# Check limits exactly for m=1,2,3,4
for m in [1,2,3,4]:
# h(m) = g(m)/(m(m-2))
if m == 0 or m == 2:
# limit from right
# For m=2, limit is a*(2-s) since t=2 and x>2 => g(x)=f(x)
val = a * (m - s)
else:
val = g_exact(m) / (m * (m-2))
print(f"L_{m} = {val}")
Tool Output
Execution succeeded
f(-5) = -65
g(-5) = 65
g(-1) = 3
g(1) = -4/7
-7/2 g(1) = 2
L_1 = 4/7
L_2 = -5/14
L_3 = -1/7
L_4 = 1/14
Reasoning Process
The solution is unique and verified. The function with satisfies all conditions, giving .
Final Model Output
65
Comparative Analysis (Pcommon_21) (Average across runs)
Averages use reported measurements only. Missing usage is not treated as zero.
Avg Time Taken
Motif 3
1140.1
MiMo V2.6 Pro
651.3
Solar Pro 4
583.2
Grok 4.7
207.1
DeepSeek V4.1 Flash
84.4
Muse Spark 1.3
81.4
Solar Mini 4 Preview
78.2
Union Alpha
0.5
Avg Token Usage
Motif 3
126,933
MiMo V2.6 Pro
75,681
Solar Pro 4
53,446
Grok 4.7
48,743
DeepSeek V4.1 Flash
45,275
Muse Spark 1.3
30,851
Solar Mini 4 Preview
16,968