Problem Analysis #33

Motif 3 · Run 1

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Manual web chat · medium response level. The saved transcript may include the prompt and web interface text.

Problem Statement

This is a Korean CSAT mathematics multiple-choice problem. Solve it and put only the correct choice symbol, one of ①, ②, ③, ④, ⑤, inside \boxed{}.

Problem 26 Consider the solid whose base in thexy-plane is the region bounded by the curve y= √ x+xlnx, thex-axis, and the vertical linesx= 1 andx= 2. The cross-sections of this solid taken by planes perpendicular to thex-axis are all equilateral triangles. Find the volume of this solid. [3 points] ① √ 3(3 + 8 ln 2) 16 ② √ 3(5 + 12 ln 2) 24 ③ √ 3(1 + 12 ln 2) 16 ④ √ 3(1 + 2 ln 2) 4 ⑤ √ 3(1 + 9 ln 2) 12 14

Ground Truth
Recorded model response / solution
Reasoning Process

대화가 없습니다

로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.

For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.

Problem 26 Consider the solid whose base in thexy-plane is the region bounded by the curve y= √ x+xlnx, thex-axis, and the vertical linesx= 1 andx= 2. The cross-sections of this solid taken by planes perpendicular to thex-axis are all equilateral triangles. Find the volume of this solid. [3 points] ① √ 3(3 + 8 ln 2) 16 ② √ 3(5 + 12 ln 2) 24 ③ √ 3(1 + 12 ln 2) 16 ④ √ 3(1 + 2 ln 2) 4 ⑤ √ 3(1 + 9 ln 2) 12 14 더보기

The base of the solid is the region bounded by 𝑦

𝑥 + 𝑥 ln ⁡ 𝑥 y= x+xlnx ​

, the 𝑥 x-axis, and the lines 𝑥

1 x=1 and 𝑥

2 x=2. Cross-sections perpendicular to the 𝑥 x-axis are equilateral triangles with side length equal to the 𝑦 y-value of the curve.

The area of an equilateral triangle with side length 𝑠 s is 3 4 𝑠 2 4 3 ​

s 2 . Thus the volume is

𝑉

∫ 1 2 3 4 ( 𝑥 + 𝑥 ln ⁡ 𝑥 ) 2   𝑑 𝑥

3 4 ∫ 1 2 ( 𝑥 + 𝑥 ln ⁡ 𝑥 )   𝑑 𝑥 . V=∫ 1 2 ​

4 3 ​

( x+xlnx ​

) 2 dx= 4 3 ​

∫ 1 2 ​

(x+xlnx)dx.

Compute the integral:

∫ 1 2 𝑥   𝑑 𝑥

[ 𝑥 2 2 ] 1 2

2 − 1 2

3 2 . ∫ 1 2 ​

xdx=[ 2 x 2 ​

] 1 2 ​

=2− 2 1 ​

= 2 3 ​

.

For ∫ 𝑥 ln ⁡ 𝑥   𝑑 𝑥 ∫xlnxdx, use integration by parts with 𝑢

ln ⁡ 𝑥 u=lnx, 𝑑 𝑣

𝑥   𝑑 𝑥 dv=xdx:

∫ 𝑥 ln ⁡ 𝑥   𝑑 𝑥

𝑥 2 2 ln ⁡ 𝑥 − ∫ 𝑥 2 2 ⋅ 1 𝑥   𝑑 𝑥

𝑥 2 2 ln ⁡ 𝑥 − 1 2 ∫ 𝑥   𝑑 𝑥

𝑥 2 2 ln ⁡ 𝑥 − 𝑥 2 4 . ∫xlnxdx= 2 x 2 ​

lnx−∫ 2 x 2 ​

⋅ x 1 ​

dx= 2 x 2 ​

lnx− 2 1 ​

∫xdx= 2 x 2 ​

lnx− 4 x 2 ​

.

Evaluating from 1 to 2:

[ 𝑥 2 2 ln ⁡ 𝑥 − 𝑥 2 4 ] 1 2

( 2 ln ⁡ 2 − 1 ) − ( 0 − 1 4 )

2 ln ⁡ 2 − 3 4 . [ 2 x 2 ​

lnx− 4 x 2 ​

] 1 2 ​

=(2ln2−1)−(0− 4 1 ​

)=2ln2− 4 3 ​

.

Adding the two parts:

∫ 1 2 ( 𝑥 + 𝑥 ln ⁡ 𝑥 )   𝑑 𝑥

3 2 + 2 ln ⁡ 2 − 3 4

3 4 + 2 ln ⁡ 2. ∫ 1 2 ​

(x+xlnx)dx= 2 3 ​

+2ln2− 4 3 ​

= 4 3 ​

+2ln2.

Therefore,

𝑉

3 4 ( 3 4 + 2 ln ⁡ 2 )

3 4 ⋅ 3 + 8 ln ⁡ 2 4

3 ( 3 + 8 ln ⁡ 2 ) 16 . V= 4 3 ​

( 4 3 ​

+2ln2)= 4 3 ​

⋅ 4 3+8ln2 ​

= 16 3 ​

(3+8ln2) ​

.

This matches option ①.

\boxed{①}

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Final Model Output

Comparative Analysis (P33) (Average across runs)

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

Avg Time Taken

Solar Pro 4
47.2
Motif 3
45.3
K-EXAONE-2
13.3
Kimi K3
7.7

Avg Token Usage

K-EXAONE-2
4276.7
Solar Pro 4
2521.0
Kimi K3
717.3
    CSAT 2026 · Problem 33 · Motif 3 | EntropyMath