Datasets:

cfpark00
/

KoreanSAT

@@ -15,10 +15,10 @@ tags:
 configs:
 - config_name: default
   data_files:
   - split: 2023_math
     path: data/2023_math-*
-  - split: 2024_math
-    path: data/2024_math-*
 dataset_info:
   features:
   - name: id
@@ -32,16 +32,18 @@ dataset_info:
   - name: score
     dtype: int64
   - name: review
-    dtype: float64
   splits:
-  - name: 2023_math
-    num_bytes: 23272
     num_examples: 46
-  - name: 2024_math
-    num_bytes: 22985
     num_examples: 46
-  download_size: 30248
-  dataset_size: 46257
 ---
 # KoreanSAT Benchmark

 configs:
 - config_name: default
   data_files:
+  - split: 2022_math
+    path: data/2022_math-*
   - split: 2023_math
     path: data/2023_math-*
 dataset_info:
   features:
   - name: id
   - name: score
     dtype: int64
   - name: review
+    dtype: string
+  - name: incomplete
+    dtype: bool
   splits:
+  - name: 2022_math
+    num_bytes: 24028
     num_examples: 46
+  - name: 2023_math
+    num_bytes: 22618
     num_examples: 46
+  download_size: 31188
+  dataset_size: 46646
 ---
 # KoreanSAT Benchmark

data/{2024_math-00000-of-00001-589645d2d672c249.parquet → 2022_math-00000-of-00001-b73cb6283dc02610.parquet} RENAMED Viewed

@@ -1,3 +1,3 @@
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+size 15937

data/2023_math-00000-of-00001-3ef9ab05717b4a30.parquet ADDED Viewed

	@@ -0,0 +1,3 @@

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data/json/2023/math.json CHANGED Viewed

@@ -1,3 +1,4 @@
 {"id": 1, "name": "1", "problem": "1. $\\left( \\frac{4}{2^{\\sqrt{2}}} \\right)^{2 + \\sqrt{2}}$ 의 값은? [2점] \\begin{itemize} \\item[1] $\\frac{1}{4}$ \\item[2] $\\frac{1}{2}$ \\item[3] $1$ \\item[4] $2$ \\item[5] $4$ \\end{itemize}", "answer": 5, "score": 2, "review": null, "incomplete": false}
 {"id": 2, "name": "2", "problem": "2. $\\lim_{x \\to \\infty} \\frac{\\sqrt{x^2 - 2} + 3x}{x + 5}$ 의 값은? [2점] \\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}", "answer": 4, "score": 2, "review": null, "incomplete": false}
 {"id": 3, "name": "3", "problem": "3. 공비가 양수인 등비수열$\\{a_n\\}$이 \\[ a_2 + a_4 = 30, \\quad a_4 + a_6 = \\frac{15}{2} \\] 를 만족시킬 때, $a_1$ 의 값은? [3점] \\begin{itemize} \\item[1] 48 \\item[2] 56 \\item[3] 64 \\item[4] 72 \\item[5] 80 \\end{itemize}", "answer": 1, "score": 3, "review": null, "incomplete": false}
@@ -44,3 +45,51 @@
 {"id": 44, "name": "28_geom", "problem": "28. 두 초점이 $( \\mathrm{F}(c, 0) )$, $( \\mathrm{F'}(-c, 0) \\ (c > 0) )$인 쌍곡선 $( C )$와 $( y )$축 위의 점 $( \\mathrm{A} )$가 있다. 쌍곡선 $( C )$가 선분 $( \\mathrm{AF} )$와 만나는 점을 $( \\mathrm{P} )$, 선분 $( \\mathrm{AF'} )$와 만나는 점을 $( \\mathrm{P'} )$이라 하자. 직선 $( \\mathrm{AF} )$는 쌍곡선 $( C )$의 한 점근선과 평행하고\n\n\\[ \\overline{\\mathrm{AP}}:\\overline{\\mathrm{PP'}} = 5:6, \\quad \\overline{\\mathrm{PF}} = 1 \\]\n\n일 때, 쌍곡선 $( C )$의 주축의 길이는? [4점]\n\n\\begin{itemize} \\item[1] \\frac{13}{6} \\item[2] \\frac{9}{4} \\item[3] \\frac{7}{3} \\item[4] \\frac{29}{12} \\item[5] \\frac{5}{2} \\end{itemize}", "answer": 2, "score": 4, "review": "Removed figure.", "incomplete": false}
 {"id": 45, "name": "29_geom", "problem": "29. 평면 $\\alpha$ 위에 $\\overline{\\mathrm{AB}} = \\overline{\\mathrm{CD}} = \\overline{\\mathrm{AD}} = 2$, $\\angle \\mathrm{ABC} = \\angle \\mathrm{BCD} = \\frac{\\pi}{3}$ 인 사다리꼴 $\\mathrm{ABCD}$가 있다. 다음 조건을 만족시키는 평면 $\\alpha$ 위의 두 점 $\\mathrm{P}$, $\\mathrm{Q}$에 대하여 $\\overrightarrow{\\mathrm{CP}} \\cdot \\overrightarrow{\\mathrm{DQ}}$의 값을 구하시오. [4점]\n\n\\begin{itemize} \\item[(가)] $\\overrightarrow{\\mathrm{AC}} = 2 \\left( \\overrightarrow{\\mathrm{AD}} + \\overrightarrow{\\mathrm{BP}} \\right)$ \\item[(나)] $\\overrightarrow{\\mathrm{AC}} \\cdot \\overrightarrow{\\mathrm{PQ}} = 6$ \\item[(다)] $2 \\times \\angle \\mathrm{BQA} = \\angle \\mathrm{PBQ} < \\frac{\\pi}{2}$ \\end{itemize}", "answer": 12, "score": 4, "review": "Removed figure.", "incomplete": false}
 {"id": 46, "name": "30_geom", "problem": "30. 좌표공간에 정사면체 $\\mathrm{ABCD}$가 있다. 정삼각형 $\\mathrm{BCD}$의 외심을 중심으로 하고 점 $\\mathrm{B}$를 지나는 구를 $S$라 하자.\n\n구 $S$와 선분 $\\mathrm{AB}$가 만나는 점 중 $\\mathrm{B}$가 아닌 점을 $\\mathrm{P}$, 구 $S$와 선분 $\\mathrm{AC}$가 만나는 점 중 $\\mathrm{C}$가 아닌 점을 $\\mathrm{Q}$, 구 $S$와 선분 $\\mathrm{AD}$가 만나는 점 중 $\\mathrm{D}$가 아닌 점을 $\\mathrm{R}$라 하고, 점 $\\mathrm{P}$에서 구 $S$에 접하는 평면을 $\\alpha$라 하자.\n\n구 $S$의 반지름의 길이가 $6$일 때, 삼각형 $\\mathrm{PQR}$의 평면 $\\alpha$ 위로의 정사영의 넓이는 $k$이다. $k^2$의 값을 구하시오. [4점]", "answer": 24, "score": 4, "review": "Removed figure.", "incomplete": false}

+<<<<<<< HEAD
 {"id": 1, "name": "1", "problem": "1. $\\left( \\frac{4}{2^{\\sqrt{2}}} \\right)^{2 + \\sqrt{2}}$ 의 값은? [2점] \\begin{itemize} \\item[1] $\\frac{1}{4}$ \\item[2] $\\frac{1}{2}$ \\item[3] $1$ \\item[4] $2$ \\item[5] $4$ \\end{itemize}", "answer": 5, "score": 2, "review": null, "incomplete": false}
 {"id": 2, "name": "2", "problem": "2. $\\lim_{x \\to \\infty} \\frac{\\sqrt{x^2 - 2} + 3x}{x + 5}$ 의 값은? [2점] \\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}", "answer": 4, "score": 2, "review": null, "incomplete": false}
 {"id": 3, "name": "3", "problem": "3. 공비가 양수인 등비수열$\\{a_n\\}$이 \\[ a_2 + a_4 = 30, \\quad a_4 + a_6 = \\frac{15}{2} \\] 를 만족시킬 때, $a_1$ 의 값은? [3점] \\begin{itemize} \\item[1] 48 \\item[2] 56 \\item[3] 64 \\item[4] 72 \\item[5] 80 \\end{itemize}", "answer": 1, "score": 3, "review": null, "incomplete": false}
 {"id": 44, "name": "28_geom", "problem": "28. 두 초점이 $( \\mathrm{F}(c, 0) )$, $( \\mathrm{F'}(-c, 0) \\ (c > 0) )$인 쌍곡선 $( C )$와 $( y )$축 위의 점 $( \\mathrm{A} )$가 있다. 쌍곡선 $( C )$가 선분 $( \\mathrm{AF} )$와 만나는 점을 $( \\mathrm{P} )$, 선분 $( \\mathrm{AF'} )$와 만나는 점을 $( \\mathrm{P'} )$이라 하자. 직선 $( \\mathrm{AF} )$는 쌍곡선 $( C )$의 한 점근선과 평행하고\n\n\\[ \\overline{\\mathrm{AP}}:\\overline{\\mathrm{PP'}} = 5:6, \\quad \\overline{\\mathrm{PF}} = 1 \\]\n\n일 때, 쌍곡선 $( C )$의 주축의 길이는? [4점]\n\n\\begin{itemize} \\item[1] \\frac{13}{6} \\item[2] \\frac{9}{4} \\item[3] \\frac{7}{3} \\item[4] \\frac{29}{12} \\item[5] \\frac{5}{2} \\end{itemize}", "answer": 2, "score": 4, "review": "Removed figure.", "incomplete": false}
 {"id": 45, "name": "29_geom", "problem": "29. 평면 $\\alpha$ 위에 $\\overline{\\mathrm{AB}} = \\overline{\\mathrm{CD}} = \\overline{\\mathrm{AD}} = 2$, $\\angle \\mathrm{ABC} = \\angle \\mathrm{BCD} = \\frac{\\pi}{3}$ 인 사다리꼴 $\\mathrm{ABCD}$가 있다. 다음 조건을 만족시키는 평면 $\\alpha$ 위의 두 점 $\\mathrm{P}$, $\\mathrm{Q}$에 대하여 $\\overrightarrow{\\mathrm{CP}} \\cdot \\overrightarrow{\\mathrm{DQ}}$의 값을 구하시오. [4점]\n\n\\begin{itemize} \\item[(가)] $\\overrightarrow{\\mathrm{AC}} = 2 \\left( \\overrightarrow{\\mathrm{AD}} + \\overrightarrow{\\mathrm{BP}} \\right)$ \\item[(나)] $\\overrightarrow{\\mathrm{AC}} \\cdot \\overrightarrow{\\mathrm{PQ}} = 6$ \\item[(다)] $2 \\times \\angle \\mathrm{BQA} = \\angle \\mathrm{PBQ} < \\frac{\\pi}{2}$ \\end{itemize}", "answer": 12, "score": 4, "review": "Removed figure.", "incomplete": false}
 {"id": 46, "name": "30_geom", "problem": "30. 좌표공간에 정사면체 $\\mathrm{ABCD}$가 있다. 정삼각형 $\\mathrm{BCD}$의 외심을 중심으로 하고 점 $\\mathrm{B}$를 지나는 구를 $S$라 하자.\n\n구 $S$와 선분 $\\mathrm{AB}$가 만나는 점 중 $\\mathrm{B}$가 아닌 점을 $\\mathrm{P}$, 구 $S$와 선분 $\\mathrm{AC}$가 만나는 점 중 $\\mathrm{C}$가 아닌 점을 $\\mathrm{Q}$, 구 $S$와 선분 $\\mathrm{AD}$가 만나는 점 중 $\\mathrm{D}$가 아닌 점을 $\\mathrm{R}$라 하고, 점 $\\mathrm{P}$에서 구 $S$에 접하는 평면을 $\\alpha$라 하자.\n\n구 $S$의 반지름의 길이가 $6$일 때, 삼각형 $\\mathrm{PQR}$의 평면 $\\alpha$ 위로의 정사영의 넓이는 $k$이다. $k^2$의 값을 구하시오. [4점]", "answer": 24, "score": 4, "review": "Removed figure.", "incomplete": false}
+=======
+{"id":1,"name":"1","problem":"1. \\left( \\frac{4}{2^{\\sqrt{2}}} \\right)^{2 + \\sqrt{2}} \\text{\uc758 \uac12\uc740? [2\uc810]}\n\n\\begin{itemize}\n    \\item[1] $\\frac{1}{4}$\n    \\item[2] $\\frac{1}{2}$\n    \\item[3] $1$\n    \\item[4] $2$\n    \\item[5] $4$\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":2,"name":"2","problem":"2. \\lim_{x \\to \\infty} \\frac{\\sqrt{x^2 - 2 + 3x}}{x + 5} \\text{\uc758 \uac12\uc740? [2\uc810]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":3,"name":"3","problem":"3. \\text{\uacf5\ube44\uac00 \uc591\uc218\uc778 \ub4f1\ube44\uc218\uc5f4 } \\{a_n\\}\\text{\uc774}\n\n\\[ a_2 + a_4 = 30, \\quad a_4 + a_6 = \\frac{15}{2} \\]\n\\text{\ub97c \ub9cc\uc871\uc2dc\ud0ac \ub54c, } a_1 \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] 48\n    \\item[2] 56\n    \\item[3] 64\n    \\item[4] 72\n    \\item[5] 80\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":4,"name":"4","problem":"4. \\text{\ub2e4\ud56d\ud568\uc218 } f(x) \\text{\uc5d0 \ub300\ud558\uc5ec \ud568\uc218 } g(x) \\text{\ub97c}\n\n\\[ g(x) = x^2 f(x) \\]\n\\text{\ub77c \ud558\uc790. } f(2) = 1, \\ f'(2) = 3 \\text{\uc77c \ub54c, } g'(2) \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] 12\n    \\item[2] 14\n    \\item[3] 16\n    \\item[4] 18\n    \\item[5] 20\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":5,"name":"5","problem":"5. \\tan \\theta < 0 \\text{\uc774\uace0} \\cos \\left( \\frac{\\pi}{2} + \\theta \\right) = \\frac{\\sqrt{5}}{5} \\text{\uc77c \ub54c, } \\cos \\theta \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] - \\frac{2 \\sqrt{5}}{5}\n    \\item[2] - \\frac{\\sqrt{5}}{5}\n    \\item[3] 0\n    \\item[4] \\frac{\\sqrt{5}}{5}\n    \\item[5] \\frac{2 \\sqrt{5}}{5}\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":6,"name":"6","problem":"6. \\text{\ud568\uc218 } f(x) = 2x^3 - 9x^2 + ax + 5 \\text{\ub294 } x = 1 \\text{\uc5d0\uc11c \uadf9\ub300\uc774\uace0, } x = b \\text{\uc5d0\uc11c \uadf9\uc18c\uc774\ub2e4. } a + b \\text{\uc758 \uac12\uc740? (\ub2e8, } a, b \\text{\ub294 \uc0c1\uc218\uc774\ub2e4.) [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] 12\n    \\item[2] 14\n    \\item[3] 16\n    \\item[4] 18\n    \\item[5] 20\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":7,"name":"7","problem":"7. \\text{\ubaa8\ub4e0 \ud56d\uc774 \uc591\uc218\uc774\uace0 \uccab\uc9f8\ud56d\uacfc \uacf5\ucc28\uac00 \uac19\uc740 \ub4f1\ucc28\uc218\uc5f4 } \\{a_n\\}\\text{\uc774}\n\n\\[ \\sum_{k=1}^{15} \\frac{1}{\\sqrt{a_k} + \\sqrt{a_{k+1}}} = 2 \\]\n\\text{\ub97c \ub9cc\uc871\uc2dc\ud0ac \ub54c, } a_4 \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] 6\n    \\item[2] 7\n    \\item[3] 8\n    \\item[4] 9\n    \\item[5] 10\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":8,"name":"8","problem":"8. \\text{\uc810 } (0, 4) \\text{\uc5d0\uc11c \uace1\uc120 } y = x^3 - x + 2 \\text{\uc5d0 \uadf8\uc740 \uc811\uc120\uc758 } x \\text{\uc808\ud3b8\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] -\\frac{1}{2}\n    \\item[2] -1\n    \\item[3] -\\frac{3}{2}\n    \\item[4] -2\n    \\item[5] -\\frac{5}{2}\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":9,"name":"9","problem":"9. \\text{\ud568\uc218}\n\n\\[ f(x) = a - \\sqrt{3} \\tan 2x \\]\n\\text{\uac00 \ub2eb\ud78c\uad6c\uac04} \\left[ -\\frac{\\pi}{6}, b \\right] \\text{\uc5d0\uc11c \ucd5c\ub300\uac12 7, \ucd5c\uc19f\uac12 3\uc744 \uac00\uc9c8 \ub54c, } a \\times b \\text{\uc758 \uac12\uc740? (\ub2e8, } a, b \\text{\ub294 \uc0c1\uc218\uc774\ub2e4.) [4\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{\\pi}{2}\n    \\item[2] \\frac{5\\pi}{12}\n    \\item[3] \\frac{\\pi}{3}\n    \\item[4] \\frac{\\pi}{4}\n    \\item[5] \\frac{\\pi}{6}\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":10,"name":"10","problem":"10. \\text{\ub450 \uace1\uc120 } y = x^3 + x^2, \\ y = -x^2 + k \\text{\uc640 } y \\text{\ucd95\uc73c\ub85c \ub458\ub7ec\uc2f8\uc778 \ubd80\ubd84\uc758 \ub113\uc774\ub97c } A, \\text{ \ub450 \uace1\uc120 } y = x^3 + x^2, \\ y = -x^2 + k \\text{\uc640 \uc9c1\uc120 } x = 2 \\text{\ub85c \ub458\ub7ec\uc2f8\uc778 \ubd80\ubd84\uc758 \ub113\uc774\ub97c } B \\text{\ub77c \ud558\uc790.} A = B \\text{\uc77c \ub54c, \uc0c1\uc218 } k \\text{\uc758 \uac12\uc740? (\ub2e8, } 4 < k < 5) [4\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{25}{6}\n    \\item[2] \\frac{13}{3}\n    \\item[3] \\frac{9}{2}\n    \\item[4] \\frac{14}{3}\n    \\item[5] \\frac{29}{6}\n\\end{itemize}\n","answer":-1,"score":-1,"review":1.0}
+{"id":11,"name":"11","problem":"11. \\text{\uadf8\ub9bc\uacfc \uac19\uc774 \uc0ac\uac01\ud615 } ABCD \\text{\uac00 \ud55c \uc6d0\uc5d0 \ub0b4\uc811\ud558\uace0}\n\n\\[ \\overline{AB} = 5, \\quad \\overline{AC} = 3 \\sqrt{5}, \\quad \\overline{AD} = 7, \\quad \\angle BAC = \\angle CAD \\]\n\\text{\uc77c \ub54c, \uc774 \uc6d0\uc758 \ubc18\uc9c0\ub984\uc758 \uae38\uc774\ub294? [4\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{5 \\sqrt{2}}{2}\n    \\item[2] \\frac{8 \\sqrt{5}}{5}\n    \\item[3] \\frac{5 \\sqrt{5}}{3}\n    \\item[4] \\frac{8 \\sqrt{2}}{3}\n    \\item[5] \\frac{9 \\sqrt{3}}{4}\n\\end{itemize}\n","answer":-1,"score":-1,"review":1.0}
+{"id":12,"name":"12","problem":"12. \\text{\uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \uc5f0\uc18d\uc778 \ud568\uc218 } f(x) \\text{\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a8\ub2e4.}\n\n\\[ n - 1 \\leq x < n \\text{\uc77c \ub54c, } |f(x)| = |6(x - n + 1)(x - n)| \\text{\uc774\ub2e4. (\ub2e8, } n \\text{\uc740 \uc790\uc5f0\uc218\uc774\ub2e4.)} \\]\n\n\\text{\uc5f4\ub9b0\uad6c\uac04 } (0, 4) \\text{\uc5d0\uc11c \uc815\uc758\ub41c \ud568\uc218} \n\\[ g(x) = \\int_0^x f(t) dt - \\int_x^4 f(t) dt \\]\n\\text{\uac00 } x = 2 \\text{\uc5d0\uc11c \ucd5c\uc19f\uac12 0\uc744 \uac00\uc9c8 \ub54c, } \\int_{\\frac{1}{2}}^4 f(x) dx \\text{\uc758 \uac12\uc740? [4\uc810]}\n\n\\begin{itemize}\n    \\item[1] -\\frac{3}{2}\n    \\item[2] -\\frac{1}{2}\n    \\item[3] \\frac{1}{2}\n    \\item[4] \\frac{3}{2}\n    \\item[5] \\frac{5}{2}\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":13,"name":"13","problem":"13. \\text{\uc790\uc5f0\uc218 } m(m \\geq 2) \\text{\uc5d0 \ub300\ud558\uc5ec } m^{12} \\text{\uc758 } n \\text{\uc81c\uacf1\uadfc \uc911\uc5d0\uc11c \uc815\uc218\uac00 \uc874\uc7ac\ud558\ub3c4\ub85d \ud558\ub294 2 \uc774\uc0c1\uc758 \uc790\uc5f0\uc218 } n \\text{\uc758 \uac1c\uc218\ub97c } f(m) \\text{\uc774\ub77c \ud560 \ub54c,} \n\\[ \\sum_{m=2}^{9} f(m) \\text{\uc758 \uac12\uc740? [4\uc810]} \\]\n\n\\begin{itemize}\n    \\item[1] 37\n    \\item[2] 42\n    \\item[3] 47\n    \\item[4] 52\n    \\item[5] 57\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":14,"name":"14","problem":"14. \\text{\ub2e4\ud56d\ud568\uc218 } f(x) \\text{\uc5d0 \ub300\ud558\uc5ec \ud568\uc218 } g(x) \\text{\ub97c \ub2e4\uc74c\uacfc \uac19\uc774 \uc815\uc758\ud55c\ub2e4.}\n\n\\[ g(x) = \\begin{cases} x & (x < -1 \\text{ \ub610\ub294 } x > 1) \\\\ f(x) & (-1 \\leq x \\leq 1) \\end{cases} \\]\n\\text{\ud568\uc218 } h(x) = \\lim_{t \\to 0^+} g(x+t) \\times \\lim_{t \\to 2^+} g(x+t) \\text{\uc5d0 \ub300\ud558\uc5ec} \n\\text{\ubcf4\uae30\uc5d0\uc11c \uc633\uc740 \uac83\ub9cc\uc744 \uc788\ub294 \ub300\ub85c \uace0\ub978 \uac83\uc740? [4\uc810]}\n\n\\<\ubcf4\uae30>\n\n\u3131. h(1) = 3 \n\n\u3134. \ud568\uc218 h(x)\ub294 \uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \uc5f0\uc18d\uc774\ub2e4. \n\n\u3137. \ud568\uc218 g(x)\uac00 \ub2eb\ud78c\uad6c\uac04 \\([-1, 1]\\)\uc5d0\uc11c \uac10\uc18c\ud558\uace0 \\(g(-1) = -2\\)\uc774\uba74 \ud568\uc218 h(x)\ub294 \uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \ucd5c\uc19f\uac12\uc744 \uac16\ub294\ub2e4.\n\n\\begin{itemize}\n    \\item[1] \u3131\n    \\item[2] \u3134\n    \\item[3] \u3131, \u3134\n    \\item[4] \u3131, \u3137\n    \\item[5] \u3134, \u3137\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":15,"name":"15","problem":"15. \\text{\ubaa8\ub4e0 \ud56d\uc774 \uc790\uc5f0\uc218\uc774\uace0 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a4\ub294 \ubaa8\ub4e0 \uc218\uc5f4 } \\{a_n\\} \\text{\uc5d0 \ub300\ud558\uc5ec } a_9 \\text{\uc758 \ucd5c\ub300\uac12\uacfc \ucd5c\uc19f\uac12\uc744 \uac01\uac01 } M, m \\text{\uc774\ub77c \ud560 \ub54c, } M + m \\text{\uc758 \uac12\uc740? [4\uc810]}\n\n\\text{(\uac00) } a_7 = 40 \n\n\\text{(\ub098) \ubaa8\ub4e0 \uc790\uc5f0\uc218 } n \\text{\uc5d0 \ub300\ud558\uc5ec}\n\\[ a_{n+2} = \\begin{cases} a_{n+1} + a_n & (a_{n+1}\\text{\uc774 } 3 \\text{\uc758 \ubc30\uc218\uac00 \uc544\ub2cc \uacbd\uc6b0}) \\\\ \\frac{1}{3} a_{n+1} & (a_{n+1}\\text{\uc774 } 3 \\text{\uc758 \ubc30\uc218\uc778 \uacbd\uc6b0}) \\end{cases} \\]\n\n\\begin{itemize}\n    \\item[1] 216\n    \\item[2] 218\n    \\item[3] 220\n    \\item[4] 222\n    \\item[5] 224\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":16,"name":"16","problem":"16. \\text{\ubc29\uc815\uc2dd}\n\n\\[ \\log_2(3x + 2) = 2 + \\log_2(x - 2) \\]\n\\text{\ub97c \ub9cc\uc871\uc2dc\ud0a4\ub294 \uc2e4\uc218 } x \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [3\uc810]}\n","answer":-1,"score":-1,"review":null}
+{"id":17,"name":"17","problem":"17. \\text{\ud568\uc218 } f(x) \\text{\uc5d0 \ub300\ud558\uc5ec } f'(x) = 4x^3 - 2x \\text{\uc774\uace0 } f(0) = 3 \\text{\uc77c \ub54c, } f(2) \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [3\uc810]}\n","answer":-1,"score":-1,"review":null}
+{"id":18,"name":"18","problem":"18. \\text{\ub450 \uc218\uc5f4 } \\{a_n\\}, \\{b_n\\} \\text{\uc5d0 \ub300\ud558\uc5ec}\n\n\\[ \\sum_{k=1}^{5} (3a_k + 5) = 55, \\quad \\sum_{k=1}^{5} (a_k + b_k) = 32 \\]\n\\text{\uc77c \ub54c, } \\sum_{k=1}^{5} b_k \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [3\uc810]}\n","answer":-1,"score":-1,"review":null}
+{"id":19,"name":"19","problem":"19. \\text{\ubc29\uc815\uc2dd } 2x^3 - 6x^2 + k = 0 \\text{\uc758 \uc11c\ub85c \ub2e4\ub978 \uc591\uc758 \uc2e4\uadfc\uc758 \uac1c\uc218\uac00 2\uac00 \ub418\ub3c4\ub85d \ud558\ub294 \uc815\uc218 } k \\text{\uc758 \uac1c\uc218\ub97c \uad6c\ud558\uc2dc\uc624. [3\uc810]}\n","answer":-1,"score":-1,"review":null}
+{"id":20,"name":"20","problem":"20. \\text{\uc218\uc9c1\uc120 \uc704\ub97c \uc6c0\uc9c1\uc774\ub294 \uc810 P\uc758 \uc2dc\uac01 } t(t \\geq 0) \\text{\uc5d0\uc11c\uc758 \uc18d\ub3c4 } v(t) \\text{\uc640 \uac00\uc18d\ub3c4 } a(t) \\text{\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a8\ub2e4.}\n\n\\text{(\uac00) } 0 \\leq t \\leq 2 \\text{\uc77c \ub54c, } v(t) = 2t^3 - 8t \\text{\uc774\ub2e4.}\n\\text{(\ub098) } t \\geq 2 \\text{\uc77c \ub54c, } a(t) = 6t + 4 \\text{\uc774\ub2e4.}\n\n\\text{\uc2dc\uac01 } t = 0 \\text{\uc5d0\uc11c } t = 3 \\text{\uae4c\uc9c0 \uc810 P\uac00 \uc6c0\uc9c1\uc778 \uac70\ub9ac\ub97c \uad6c\ud558\uc2dc\uc624. [4\uc810]}\n","answer":-1,"score":-1,"review":null}
+{"id":21,"name":"21","problem":"21. \\text{\uc790\uc5f0\uc218 } n \\text{\uc5d0 \ub300\ud558\uc5ec \ud568\uc218 } f(x) \\text{\ub97c}\n\n\\[ f(x) = \\begin{cases} |3^x + 2 - n| & (x < 0) \\\\ |\\log_2(x + 4) - n| & (x \\geq 0) \\end{cases} \\]\n\\text{\uc774\ub77c \ud558\uc790. \uc2e4\uc218 } t \\text{\uc5d0 \ub300\ud558\uc5ec } x \\text{\uc5d0 \ub300\ud55c \ubc29\uc815\uc2dd } f(x) = t \\text{\uc758 \uc11c\ub85c \ub2e4\ub978 \uc2e4\uadfc\uc758 \uac1c\uc218\ub97c } g(t) \\text{\ub77c \ud560 \ub54c, \ud568\uc218 } g(t) \\text{\uc758 \ucd5c\ub313\uac12\uc774 4\uac00 \ub418\ub3c4\ub85d \ud558\ub294 \ubaa8\ub4e0 \uc790\uc5f0\uc218 } n \\text{\uc758 \uac12\uc758 \ud569\uc744 \uad6c\ud558\uc2dc\uc624. [4\uc810]}\n","answer":-1,"score":-1,"review":null}
+{"id":22,"name":"22","problem":"22. \\text{\ucd5c\uace0\ucc28\ud56d\uc758 \uacc4\uc218\uac00 1\uc778 \uc0bc\ucc28\ud568\uc218 } f(x) \\text{\uc640 \uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \uc5f0\uc18d\uc778 \ud568\uc218 } g(x) \\text{\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0ac \ub54c, } f(4) \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [4\uc810]}\n\n\\text{(\uac00) \ubaa8\ub4e0 \uc2e4\uc218 } x \\text{\uc5d0 \ub300\ud558\uc5ec } f(x) = f(1) + (x - 1)f'(g(x)) \\text{\uc774\ub2e4.}\n\\text{(\ub098) \ud568\uc218 } g(x) \\text{\uc758 \ucd5c\uc19f\uac12\uc740 } \\frac{5}{2} \\text{\uc774\ub2e4.}\n\\text{(\ub2e4) } f(0) = -3, \\quad f(g(1)) = 6 \n","answer":-1,"score":-1,"review":null}
+{"id":23,"name":"23_prob","problem":"23. \\( (x^3 + 3)^5 \\)\uc758 \uc804\uac1c\uc2dd\uc5d0\uc11c \\(x^9\\)\uc758 \uacc4\uc218\ub294? [2\uc810]\n\\begin{itemize}\n    \\item[1] 30\n    \\item[2] 60\n    \\item[3] 90\n    \\item[4] 120\n    \\item[5] 150\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":24,"name":"24_prob","problem":"24. \\text{\uc22b\uc790 } 1, 2, 3, 4, 5 \\text{ \uc911\uc5d0\uc11c \uc911\ubcf5\uc744 \ud5c8\ub77d\ud558\uc5ec 4\uac1c\ub97c \ud0dd\ud574 \uc77c\ub82c\ub85c \ub098\uc5f4\ud558\uc5ec \ub9cc\ub4e4 \uc218 \uc788\ub294 \ub124 \uc790\ub9ac\uc758 \uc790\uc5f0\uc218 \uc911 4000 \uc774\uc0c1\uc778 \ud640\uc218\uc758 \uac1c\uc218\ub294? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] 125\n    \\item[2] 150\n    \\item[3] 175\n    \\item[4] 200\n    \\item[5] 225\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":25,"name":"25_prob","problem":"25. \\text{\ud770\uc0c9 \ub9c8\uc2a4\ud06c 5\uac1c, \uac80\uc740\uc0c9 \ub9c8\uc2a4\ud06c 9\uac1c\uac00 \ub4e4\uc5b4 \uc788\ub294 \uc0c1\uc790\uac00 \uc788\ub2e4. \uc774 \uc0c1\uc790\uc5d0\uc11c \uc784\uc758\ub85c 3\uac1c\uc758 \ub9c8\uc2a4\ud06c\ub97c \ub3d9\uc2dc\uc5d0 \uaebc\ub0bc \ub54c, \uaebc\ub0b8 3\uac1c\uc758 \ub9c8\uc2a4\ud06c \uc911\uc5d0\uc11c \uc801\uc5b4\ub3c4 \ud55c \uac1c\uac00 \ud770\uc0c9 \ub9c8\uc2a4\ud06c\uc77c \ud655\ub960\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{8}{13}\n    \\item[2] \\frac{17}{26}\n    \\item[3] \\frac{9}{13}\n    \\item[4] \\frac{19}{26}\n    \\item[5] \\frac{10}{13}\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":26,"name":"26_prob","problem":"26. \\text{\uc8fc\uba38\ub2c8\uc5d0 1\uc774 \uc801\ud78c \ud770 \uacf5 1\uac1c, 2\uac00 \uc801\ud78c \ud770 \uacf5 1\uac1c, 1\uc774 \uc801\ud78c \uac80\uc740 \uacf5 1\uac1c, 2\uac00 \uc801\ud78c \uac80\uc740 \uacf5 3\uac1c\uac00 \ub4e4\uc5b4 \uc788\ub2e4. \uc774 \uc8fc\uba38\ub2c8\uc5d0\uc11c \uc784\uc758\ub85c 3\uac1c\uc758 \uacf5\uc744 \ub3d9\uc2dc\uc5d0 \uaebc\ub0b4\ub294 \uc2dc\ud589\uc744 \ud55c\ub2e4. \uc774 \uc2dc\ud589\uc5d0\uc11c \uaebc\ub0b8 3\uac1c\uc758 \uacf5 \uc911\uc5d0\uc11c \ud770 \uacf5\uc774 1\uac1c\uc774\uace0 \uac80\uc740 \uacf5\uc774 2\uac1c\uc778 \uc0ac\uac74\uc744 } A, \\text{ \uaebc\ub0b8 3\uac1c\uc758 \uacf5\uc5d0 \uc801\ud600 \uc788\ub294 \uc218\ub97c \ubaa8\ub450 \uacf1\ud55c \uac12\uc774 8\uc778 \uc0ac\uac74\uc744 } B \\text{\ub77c \ud560 \ub54c, } P(A \\cup B) \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{11}{20}\n    \\item[2] \\frac{3}{5}\n    \\item[3] \\frac{13}{20}\n    \\item[4] \\frac{7}{10}\n    \\item[5] \\frac{3}{4}\n\\end{itemize}\n","answer":-1,"score":-1,"review":1.0}
+{"id":27,"name":"27_prob","problem":"27. \\text{\uc5b4\ub290 \ud68c\uc0ac\uc5d0\uc11c \uc0dd\uc0b0\ud558\ub294 \uc0f4\ud478 1\uac1c\uc758 \uc6a9\ub7c9\uc740 \uc815\uaddc\ubd84\ud3ec } N(m, \\sigma^2) \\text{\uc744 \ub530\ub978\ub2e4\uace0 \ud55c\ub2e4. \uc774 \ud68c\uc0ac\uc5d0\uc11c \uc0dd\uc0b0\ud558\ub294 \uc0f4\ud478 \uc911\uc5d0\uc11c 16\uac1c\ub97c \uc784\uc758\ucd94\ucd9c\ud558\uc5ec \uc5bb\uc740 \ud45c\ubcf8\ud3c9\uade0\uc744 \uc774\uc6a9\ud558\uc5ec \uad6c\ud55c } m \\text{\uc5d0 \ub300\ud55c \uc2e0\ub8b0\ub3c4 95%\uc758 \uc2e0\ub8b0\uad6c\uac04\uc774 } 746.1 \\leq m \\leq 755.9 \\text{\uc774\ub2e4. \uc774 \ud68c\uc0ac\uc5d0\uc11c \uc0dd\uc0b0\ud558\ub294 \uc0f4\ud478 \uc911\uc5d0\uc11c } n \\text{\uac1c\ub97c \uc784\uc758\ucd94\ucd9c\ud558\uc5ec \uc5bb\uc740 \ud45c\ubcf8\ud3c9\uade0\uc744 \uc774\uc6a9\ud558\uc5ec \uad6c\ud558\ub294 } m \\text{\uc5d0 \ub300\ud55c \uc2e0\ub8b0\ub3c4 99%\uc758 \uc2e0\ub8b0\uad6c\uac04\uc774 } a \\leq m \\leq b \\text{\uc77c \ub54c, } b - a \\text{\uc758 \uac12\uc774 6 \uc774\ud558\uac00 \ub418\uae30 \uc704\ud55c \uc790\uc5f0\uc218 } n \\text{\uc758 \ucd5c\uc19f\uac12\uc740? (\ub2e8, \uc6a9\ub7c9\uc758 \ub2e8\uc704\ub294 mL\uc774\uace0, } Z \\text{\uac00 \ud45c\uc900\uc815\uaddc\ubd84\ud3ec\ub97c \ub530\ub974\ub294 \ud655\ub960\ubcc0\uc218\uc77c \ub54c, } P(|Z| \\leq 1.96) = 0.95, P(|Z| \\leq 2.58) = 0.99 \\text{\ub85c \uacc4\uc0b0\ud55c\ub2e4.) [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] 70\n    \\item[2] 74\n    \\item[3] 78\n    \\item[4] 82\n    \\item[5] 86\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":28,"name":"28_prob","problem":"28. \\text{\uc5f0\uc18d\ud655\ub960\ubcc0\uc218 } X \\text{\uac00 \uac16\ub294 \uac12\uc758 \ubc94\uc704\ub294 } 0 \\leq X \\leq a \\text{\uc774\uace0, } X \\text{\uc758 \ud655\ub960\ubc00\ub3c4\ud568\uc218\uc758 \uadf8\ub798\ud504\uac00 \uadf8\ub9bc\uacfc \uac19\ub2e4.}\n\n\\[ P(X \\leq b) - P(X \\geq b) = \\frac{1}{4}, \\quad P(X \\leq \\sqrt{5}) = \\frac{1}{2} \\]\n\\text{\uc77c \ub54c, } a + b + c \\text{\uc758 \uac12\uc740? (\ub2e8, } a, b, c \\text{\ub294 \uc0c1\uc218\uc774\ub2e4.) [4\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{11}{2}\n    \\item[2] 6\n    \\item[3] \\frac{13}{2}\n    \\item[4] 7\n    \\item[5] \\frac{15}{2}\n\\end{itemize}\n","answer":-1,"score":-1,"review":2.0}
+{"id":29,"name":"29_prob","problem":"29. \\text{\uc55e\uba74\uc5d0\ub294 1\ubd80\ud130 6\uae4c\uc9c0\uc758 \uc790\uc5f0\uc218\uac00 \ud558\ub098\uc529 \uc801\ud600 \uc788\uace0, \ub4b7\uba74\uc5d0\ub294 \ubaa8\ub450 0\uc774 \ud558\ub098\uc529 \uc801\ud600 \uc788\ub294 6\uc7a5\uc758 \uce74\ub4dc\uac00 \uc788\ub2e4. \uc774 6\uc7a5\uc758 \uce74\ub4dc\uac00 \uadf8\ub9bc\uacfc \uac19\uc774 6 \uc774\ud558\uc758 \uc790\uc5f0\uc218 } k \\text{\uc5d0 \ub300\ud558\uc5ec } k \\text{\ubc88\uc9f8 \uc790\ub9ac\uc5d0 \uc790\uc5f0\uc218 } k \\text{\uac00 \ubcf4\uc774\ub3c4\ub85d \ub193\uc5ec \uc788\ub2e4.}\n\n\\text{\uc774 6\uc7a5\uc758 \uce74\ub4dc\uc640 \ud55c \uac1c\uc758 \uc8fc\uc0ac\uc704\ub97c \uc0ac\uc6a9\ud558\uc5ec \ub2e4\uc74c \uc2dc\ud589\uc744 \ud55c\ub2e4.}\n\n\\[ \\text{\uc8fc\uc0ac\uc704\ub97c \ud55c \ubc88 \ub358\uc838 \ub098\uc628 \ub208\uc758 \uc218\uac00 } k \\text{\uc774\uba74 } k \\text{\ubc88\uc9f8 \uc790\ub9ac\uc5d0 \ub193\uc5ec \uc788\ub294 \uce74\ub4dc\ub97c \ud55c \ubc88 \ub4a4\uc9d1\uc5b4 \uc81c\uc790\ub9ac\uc5d0 \ub193\ub294\ub2e4.} \\]\n\n\\text{\uc704\uc758 \uc2dc\ud589\uc744 3\ubc88 \ubc18\ubcf5\ud55c \ud6c4 6\uc7a5\uc758 \uce74\ub4dc\uc5d0 \ubcf4\uc774\ub294 \ubaa8\ub4e0 \uc218\uc758 \ud569\uc774 \uc9dd\uc218\uc77c \ub54c, \uc8fc\uc0ac\uc704\uc758 1\uc758 \ub208\uc774 \ud55c \ubc88\ub9cc \ub098\uc654\uc744 \ud655\ub960\uc740 } \\frac{q}{p} \\text{\uc774\ub2e4. } p + q \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. (\ub2e8, } p \\text{\uc640 } q \\text{\ub294 \uc11c\ub85c\uc18c\uc778 \uc790\uc5f0\uc218\uc774\ub2e4.) [4\uc810]}\n","answer":-1,"score":-1,"review":1.0}
+{"id":30,"name":"30_prob","problem":"30. \\text{\uc9d1\ud569 } X = \\{x | x \\text{\ub294 10 \uc774\ud558\uc758 \uc790\uc5f0\uc218}\\} \\text{\uc5d0 \ub300\ud558\uc5ec \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a4\ub294 \ud568\uc218 } f: X \\to X \\text{\uc758 \uac1c\uc218\ub97c \uad6c\ud558\uc2dc\uc624. [4\uc810]}\n\n\\text{(\uac00) 9 \uc774\ud558\uc758 \ubaa8\ub4e0 \uc790\uc5f0\uc218 } x \\text{\uc5d0 \ub300\ud558\uc5ec } f(x) \\leq f(x+1) \\text{\uc774\ub2e4.}\n\\text{(\ub098) } 1 \\leq x \\leq 5 \\text{\uc77c \ub54c } f(x) \\leq x \\text{\uc774\uace0, } 6 \\leq x \\leq 10 \\text{\uc77c \ub54c } f(x) \\geq x \\text{\uc774\ub2e4.}\n\\text{(\ub2e4) } f(6) = f(5) + 6\n","answer":-1,"score":-1,"review":null}
+{"id":31,"name":"23_calc","problem":"23. \\lim_{x \\to 0} \\frac{\\ln(x+1)}{\\sqrt{x+4} - 2} \\text{\uc758 \uac12\uc740? [2\uc810]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":32,"name":"24_calc","problem":"24. \\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=1}^{n} \\sqrt{1 + \\frac{3k}{n}} \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{4}{3}\n    \\item[2] \\frac{13}{9}\n    \\item[3] \\frac{14}{9}\n    \\item[4] \\frac{5}{3}\n    \\item[5] \\frac{16}{9}\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":33,"name":"25_calc","problem":"25. \\text{\ub4f1\ube44\uc218\uc5f4 } \\{a_n\\} \\text{\uc5d0 \ub300\ud558\uc5ec } \\lim_{n \\to \\infty} \\frac{a_n + 1}{3^n + 2^{2n-1}} = 3 \\text{\uc77c \ub54c, } a_2 \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] 16\n    \\item[2] 18\n    \\item[3] 20\n    \\item[4] 22\n    \\item[5] 24\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":34,"name":"26_calc","problem":"26. \\text{\uadf8\ub9bc\uacfc \uac19\uc774 \uace1\uc120 } y = \\sqrt{\\sec^2 x} + \\tan x \\left(0 \\leq x \\leq \\frac{\\pi}{3}\\right) \\text{\uc640 } x \\text{\ucd95, } y \\text{\ucd95 \ubc0f \uc9c1\uc120 } x = \\frac{\\pi}{3} \\text{\ub85c \ub458\ub7ec\uc2f8\uc778 \ubd80\ubd84\uc744 \ubc11\uba74\uc73c\ub85c \ud558\ub294 \uc785\uccb4\ub3c4\ud615\uc774 \uc788\ub2e4. \uc774 \uc785\uccb4\ub3c4\ud615\uc744 } x \\text{\ucd95\uc5d0 \uc218\uc9c1\uc778 \ud3c9\uba74\uc73c\ub85c \uc790\ub978 \ub2e8\uba74\uc774 \ubaa8\ub450 \uc815\uc0ac\uac01\ud615\uc77c \ub54c, \uc774 \uc785\uccb4\ub3c4\ud615\uc758 \ubd80\ud53c\ub294? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{\\sqrt{3}}{2} + \\frac{\\ln 2}{2}\n    \\item[2] \\frac{\\sqrt{3}}{2} + \\ln 2\n    \\item[3] \\sqrt{3} + \\frac{\\ln 2}{2}\n    \\item[4] \\sqrt{3} + \\ln 2\n    \\item[5] \\sqrt{3} + 2\\ln 2\n\\end{itemize}\n","answer":-1,"score":-1,"review":1.0}
+{"id":35,"name":"27_prob","problem":"27. \\text{\uadf8\ub9bc\uacfc \uac19\uc774 \uc911\uc2ec\uc774 } O, \\text{\ubc18\uc9c0\ub984\uc758 \uae38\uc774\uac00 1\uc774\uace0 \uc911\uc2ec\uac01\uc758 \ud06c\uae30\uac00 } \\frac{\\pi}{2} \\text{\uc778 \ubd80\ucc44\uaf34 } OA_1B_1 \\text{\uc774 \uc788\ub2e4. \ud638 } A_1B_1 \\text{ \uc704\uc5d0 \uc810 } P_1, \\text{\uc120\ubd84 } OA_1 \\text{ \uc704\uc5d0 \uc810 } C_1, \\text{\uc120\ubd84 } OB_1 \\text{ \uc704\uc5d0 \uc810 } D_1 \\text{\uc744 \uc0ac\uac01\ud615 } OC_1P_1D_1 \\text{\uc774 } OC_1 : OD_1 = 3:4 \\text{\uc778 \uc9c1\uc0ac\uac01\ud615\uc774 \ub418\ub3c4\ub85d \uc7a1\ub294\ub2e4.}\n\n\\text{\ubd80\ucc44\uaf34 } OA_1B_1 \\text{\uc758 \ub0b4\ubd80\uc5d0 \uc810 } Q_1 \\text{\uc744 } PQ_1 = AQ_1, \\angle PQ_1A_1 = \\frac{\\pi}{2} \\text{\uac00 \ub418\ub3c4\ub85d \uc7a1\uace0, \uc774\ub4f1\ubcc0\uc0bc\uac01\ud615 } P_1Q_1A_1 \\text{\uc5d0 \uc0c9\uce60\ud558\uc5ec \uc5bb\uc740 \uadf8\ub9bc\uc744 } R_1 \\text{\uc774\ub77c \ud558\uc790.}\n\\text{\uadf8\ub9bc } R_1 \\text{\uc5d0\uc11c \uc120\ubd84 } OA_1 \\text{ \uc704\uc758 \uc810 } A_2 \\text{\uc640 \uc120\ubd84 } OB_1 \\text{ \uc704\uc758 \uc810 } B_2 \\text{\ub97c } OQ_1 = OA_2 = OB_2 \\text{\uac00 \ub418\ub3c4\ub85d \uc7a1\uace0, \uc911\uc2ec\uc774 } O, \\text{\ubc18\uc9c0\ub984\uc758 \uae38\uc774\uac00 } OQ_1, \\text{\uc911\uc2ec\uac01\uc758 \ud06c\uae30\uac00 } \\frac{\\pi}{2} \\text{\uc778 \ubd80\ucc44\uaf34 } OA_2B_2 \\text{\ub97c \uadf8\ub9b0\ub2e4. \uadf8\ub9b0 } R_1 \\text{\uc744 \uc5bb\uc740 \uac83\uacfc \uac19\uc740 \ubc29\ubc95\uc73c\ub85c \ub124 \uc810 } P_2, C_2, D_2, Q_2 \\text{\ub97c \uc7a1\uace0, \uc774\ub4f1\ubcc0\uc0bc\uac01\ud615 } P_2Q_2A_2 \\text{\uc5d0 \uc0c9\uce60\ud558\uc5ec \uc5bb\uc740 \uadf8\ub9bc\uc744 } R_2 \\text{\ub77c \ud558\uc790. \uc774\uc640 \uac19\uc740 \uacfc\uc815\uc744 \uacc4\uc18d\ud558\uc5ec } n \\text{\ubc88\uc9f8 \uc5bb\uc740 \uadf8\ub9bc } R_n \\text{\uc5d0 \uc0c9\uce60\ub418\uc5b4 \uc788\ub294 \ubd80\ubd84\uc758 \ub113\uc774\ub97c } S_n \\text{\uc774\ub77c \ud560 \ub54c, } \\lim_{n \\to \\infty} S_n \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{9}{40}\n    \\item[2] \\frac{1}{4}\n    \\item[3] \\frac{11}{40}\n    \\item[4] \\frac{3}{10}\n    \\item[5] \\frac{13}{40}\n\\end{itemize}\n","answer":-1,"score":-1,"review":1.0}
+{"id":36,"name":"28_prob","problem":"28. \\text{\uadf8\ub9bc\uacfc \uac19\uc774 \uc911\uc2ec\uc774 } O \\text{\uc774\uace0 \uae38\uc774\uac00 2\uc778 \uc120\ubd84 } AB \\text{\ub97c \uc9c0\ub984\uc73c\ub85c \ud558\ub294 \ubc18\uc6d0 \uc704\uc5d0 } \\angle AOC = \\frac{\\pi}{2} \\text{\uc778 \uc810 } C \\text{\uac00 \uc788\ub2e4.}\n\\text{\ud638 } BC \\text{ \uc704\uc5d0 \uc810 } P \\text{\uc640 \ud638 } CA \\text{ \uc704\uc5d0 \uc810 } Q \\text{\ub97c } PB = QC \\text{\uac00 \ub418\ub3c4\ub85d \uc7a1\uace0, \uc120\ubd84 } AP \\text{ \uc704\uc5d0 \uc810 } R \\text{\uc744 } \\angle CQR = \\frac{\\pi}{2} \\text{\uac00 \ub418\ub3c4\ub85d \uc7a1\ub294\ub2e4.}\n\\text{\uc120\ubd84 } AP \\text{\uc640 \uc120\ubd84 } CO \\text{\uc758 \uad50\uc810\uc744 } S \\text{\ub77c \ud558\uc790. } \\angle PAB = \\theta \\text{\uc77c \ub54c, \uc0bc\uac01\ud615 } POB \\text{\uc758 \ub113\uc774\ub97c } f(\\theta), \\text{\uc0ac\uac01\ud615 } CQRS \\text{\uc758 \ub113\uc774\ub97c } g(\\theta) \\text{\ub77c \ud558\uc790.}\n\n\\lim_{\\theta \\to 0^+} \\frac{3f(\\theta) - 2g(\\theta)}{\\theta^2} \\text{\uc758 \uac12\uc740? (\ub2e8, } 0 < \\theta < \\frac{\\pi}{4} \\text{) [4\uc810]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n","answer":-1,"score":-1,"review":1.0}
+{"id":37,"name":"29_prob","problem":"29. \\text{\uc138 \uc0c1\uc218 } a, b, c \\text{\uc5d0 \ub300\ud558\uc5ec \ud568\uc218 } f(x) = ae^{2x} + be^x + c \\text{\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a8\ub2e4.}\n\n\\text{(\uac00) } \\lim_{x \\to -\\infty} \\frac{f(x) + 6}{e^x} = 1\n\\text{(\ub098) } f(\\ln 2) = 0\n\n\\text{\ud568\uc218 } f(x) \\text{\uc758 \uc5ed\ud568\uc218\ub97c } g(x) \\text{\ub77c \ud560 \ub54c,}\n\\[ \\int_0^{14} g(x) dx = p + q \\ln 2 \\text{\uc774\ub2e4. } p + q \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624.}\n\\text{(\ub2e8, } p, q \\text{\ub294 \uc720\ub9ac\uc218\uc774\uace0, } \\ln 2 \\text{\ub294 \ubb34\ub9ac\uc218\uc774\ub2e4.) [4\uc810]}\n","answer":-1,"score":-1,"review":null}
+{"id":38,"name":"30_prob","problem":"30. \\text{\ucd5c\uace0\ucc28\ud56d\uc758 \uacc4\uc218\uac00 \uc591\uc218\uc778 \uc0bc\ucc28\ud568\uc218 } f(x) \\text{\uc640 \ud568\uc218 } g(x) = e^{\\sin \\pi x} - 1 \\text{\uc5d0 \ub300\ud558\uc5ec \uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \uc815\uc758\ub41c \ud569\uc131\ud568\uc218 } h(x) = g(f(x)) \\text{\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a8\ub2e4.}\n\n\\text{(\uac00) \ud568\uc218 } h(x) \\text{\ub294 } x = 0 \\text{\uc5d0\uc11c \uadf9\ub313\uac12 0\uc744 \uac16\ub294\ub2e4.}\n\\text{(\ub098) \uc5f4\ub9b0\uad6c\uac04 } (0, 3) \\text{\uc5d0\uc11c \ubc29\uc815\uc2dd } h(x) = 1 \\text{\uc758 \uc11c\ub85c \ub2e4\ub978 \uc2e4\uadfc\uc758 \uac1c\uc218\ub294 7\uc774\ub2e4.}\n\nf(3) = \\frac{1}{2}, f'(3) = 0 \\text{\uc77c \ub54c, } f(2) = \\frac{q}{p} \\text{\uc774\ub2e4. } p + q \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. (\ub2e8, } p \\text{\uc640 } q \\text{\ub294 \uc11c\ub85c\uc18c\uc778 \uc790\uc5f0\uc218\uc774\ub2e4.) [4\uc810]}\n","answer":-1,"score":-1,"review":null}
+{"id":39,"name":"23_geom","problem":"23. \\text{\uc88c\ud45c\uacf5\uac04\uc758 \uc810 } A(2, 2, -1) \\text{\uc744 } x \\text{\ucd95\uc5d0 \ub300\ud558\uc5ec \ub300\uce6d\uc774\ub3d9\ud55c \uc810\uc744 } B \\text{\ub77c \ud558\uc790. \uc810 } C(-2, 1, 1) \\text{\uc5d0 \ub300\ud558\uc5ec \uc120\ubd84 BC\uc758 \uae38\uc774\ub294? [2\uc810]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":40,"name":"24_geom","problem":"24. \\text{\ucd08\uc810\uc774 } F\\left(\\frac{1}{3}, 0\\right) \\text{\uc774\uace0 \uc900\uc120\uc774 } x = -\\frac{1}{3} \\text{\uc778 \ud3ec\ubb3c\uc120\uc774 \uc810 } (a, 2) \\text{\ub97c \uc9c0\ub0a0 \ub54c, } a \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":41,"name":"25_geom","problem":"25. \\text{\ud0c0\uc6d0 } \\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1 \\text{ \uc704\uc758 \uc810 } (2, 1) \\text{\uc5d0\uc11c\uc758 \uc811\uc120\uc758 \uae30\uc6b8\uae30\uac00 } -\\frac{1}{2} \\text{\uc77c \ub54c, \uc774 \ud0c0\uc6d0\uc758 \ub450 \ucd08\uc810 \uc0ac\uc774\uc758 \uac70\ub9ac\ub294? (\ub2e8, } a, b \\text{\ub294 \uc591\uc218\uc774\ub2e4.) [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] 2\\sqrt{3}\n    \\item[2] 4\n    \\item[3] 2\\sqrt{5}\n    \\item[4] 2\\sqrt{6}\n    \\item[5] 2\\sqrt{7}\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":42,"name":"26_geom","problem":"26. \\text{\uc88c\ud45c\ud3c9\uba74\uc5d0\uc11c \uc138 \ubca1\ud130 } \\vec{a} = (2, 4), \\vec{b} = (2, 8), \\vec{c} = (1, 0) \\text{\uc5d0 \ub300\ud558\uc5ec \ub450 \ubca1\ud130 } \\vec{p}, \\vec{q} \\text{\uac00}\n\n(\\vec{p} - \\vec{a}) \\cdot (\\vec{p} - \\vec{b}) = 0, \\quad \\vec{q} = \\frac{1}{2} \\vec{a} + t \\vec{c} \\quad (t \\text{\ub294 \uc2e4\uc218}) \\text{\ub97c \ub9cc\uc871\uc2dc\ud0ac \ub54c, } |\\vec{p} - \\vec{q}| \\text{\uc758 \ucd5c\uc19f\uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{3}{2}\n    \\item[2] 2\n    \\item[3] \\frac{5}{2}\n    \\item[4] 3\n    \\item[5] \\frac{7}{2}\n\\end{itemize}\n","answer":-1,"score":-1,"review":null}
+{"id":43,"name":"27_geom","problem":"27. \\text{\uc88c\ud45c\uacf5\uac04\uc5d0 \uc9c1\uc120 AB\ub97c \ud3ec\ud568\ud558\ub294 \ud3c9\uba74 } \\alpha \\text{\uac00 \uc788\ub2e4. \ud3c9\uba74 } \\alpha \\text{ \uc704\uc5d0 \uc788\uc9c0 \uc54a\uc740 \uc810 C\uc5d0 \ub300\ud558\uc5ec \uc9c1\uc120 AB\uc640 \uc9c1\uc120 AC\uac00 \uc774\ub8e8\ub294 \uc608\uac01\uc758 \ud06c\uae30\ub97c } \\theta_1 \\text{\uc774\ub77c \ud560 \ub54c } \\sin \\theta_1 = \\frac{4}{5} \\text{\uc774\uace0, \uc9c1\uc120 AC\uc640 \ud3c9\uba74 } \\alpha \\text{\uac00 \uc774\ub8e8\ub294 \uc608\uac01\uc758 \ud06c\uae30\ub294 } \\frac{\\pi}{2} - \\theta_1 \\text{\uc774\ub2e4. \ud3c9\uba74 ABC\uc640 \ud3c9\uba74 } \\alpha \\text{\uac00 \uc774\ub8e8\ub294 \uc608\uac01\uc758 \ud06c\uae30\ub97c } \\theta_2 \\text{\ub77c \ud560 \ub54c, } \\cos \\theta_2 \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{\\sqrt{7}}{4}\n    \\item[2] \\frac{\\sqrt{7}}{5}\n    \\item[3] \\frac{\\sqrt{7}}{6}\n    \\item[4] \\frac{\\sqrt{7}}{7}\n    \\item[5] \\frac{\\sqrt{7}}{8}\n\\end{itemize}\n","answer":-1,"score":-1,"review":1.0}
+{"id":44,"name":"28_geom","problem":"28. \\text{\ub450 \ucd08\uc810\uc774 } F(c, 0), F'(-c, 0) \\text{(} c > 0 \\text{)\uc778 \uc30d\uace1\uc120 } C \\text{\uc640 } y \\text{\ucd95 \uc704\uc758 \uc810 } A \\text{\uac00 \uc788\ub2e4. \uc30d\uace1\uc120 } C \\text{\uac00 \uc120\ubd84 } AF \\text{\uc640 \ub9cc\ub098\ub294 \uc810\uc744 } P, \\text{\uc120\ubd84 } AF' \\text{\uacfc \ub9cc\ub098\ub294 \uc810\uc744 } P' \\text{\uc774\ub77c \ud558\uc790. \uc9c1\uc120 } AF \\text{\ub294 \uc30d\uace1\uc120 } C \\text{\uc758 \ud55c \uc811\uadfc\uc120\uacfc \ud3c9\ud589\ud558\uace0 }\n\\frac{AP}{PP'} = \\frac{5}{6}, PF = 1 \\text{\uc77c \ub54c, \uc30d\uace1\uc120 } C \\text{\uc758 \uc8fc\ucd95\uc758 \uae38\uc774\ub294? [4\uc810]}\n\n\\begin{itemize}\n    \\item[1] \\frac{13}{6}\n    \\item[2] \\frac{9}{4}\n    \\item[3] \\frac{7}{3}\n    \\item[4] \\frac{29}{12}\n    \\item[5] \\frac{5}{2}\n\\end{itemize}\n","answer":-1,"score":-1,"review":1.0}
+{"id":45,"name":"29_geom","problem":"29. \\text{\ud3c9\uba74 } \\alpha \\text{ \uc704\uc5d0 } \\overline{AB} = \\overline{CD} = \\overline{AD} = 2, \\quad \\angle ABC = \\angle BCD = \\frac{\\pi}{3} \\text{\uc778 \uc0ac\ub2e4\ub9ac\uaf34 ABCD\uac00 \uc788\ub2e4. \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a4\ub294 \ud3c9\uba74 } \\alpha \\text{ \uc704\uc758 \ub450 \uc810 P, Q\uc5d0 \ub300\ud558\uc5ec } \\overrightarrow{CP} \\cdot \\overrightarrow{DQ} \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [4\uc810]}\n\n\\text{(\uac00) } \\overrightarrow{AC} = 2(\\overrightarrow{AD} + \\overrightarrow{BP})\n\\text{(\ub098) } \\overrightarrow{AC} \\cdot \\overrightarrow{PQ} = 6\n\\text{(\ub2e4) } 2 \\times \\angle BQA = \\angle PBQ < \\frac{\\pi}{2}\n","answer":-1,"score":-1,"review":2.0}
+{"id":46,"name":"30_geom","problem":"30. \\text{\uc88c\ud45c\uacf5\uac04\uc5d0 \uc815\uc0ac\uba74\uccb4 ABCD\uac00 \uc788\ub2e4. \uc815\uc0bc\uac01\ud615 BCD\uc758 \uc678\uc2ec\uc744 \uc911\uc2ec\uc73c\ub85c \ud558\uace0 \uc810 B\ub97c \uc9c0\ub098\ub294 \uad6c\ub97c } S \\text{\ub77c \ud558\uc790.}\n\\text{\uad6c } S \\text{\uc640 \uc120\ubd84 AB\uac00 \ub9cc\ub098\ub294 \uc810 \uc911 B\uac00 \uc544\ub2cc \uc810\uc744 P,}\n\\text{\uad6c } S \\text{\uc640 \uc120\ubd84 AC\uac00 \ub9cc\ub098\ub294 \uc810 \uc911 C\uac00 \uc544\ub2cc \uc810\uc744 Q,}\n\\text{\uad6c } S \\text{\uc640 \uc120\ubd84 AD\uac00 \ub9cc\ub098\ub294 \uc810 \uc911 D\uac00 \uc544\ub2cc \uc810\uc744 R\ub77c \ud558\uace0, \uc810 P\uc5d0\uc11c \uad6c } S \\text{\uc5d0 \uc811\ud558\ub294 \ud3c9\uba74\uc744 } \\alpha \\text{\ub77c \ud558\uc790.}\n\\text{\uad6c } S \\text{\uc758 \ubc18\uc9c0\ub984\uc758 \uae38\uc774\uac00 6\uc77c \ub54c, \uc0bc\uac01\ud615 PQR\uc758 \ud3c9\uba74 } \\alpha \\text{\uc704\ub85c\uc758 \uc815\uc0ac\uc601\uc758 \ub113\uc774\ub294 } k \\alpha \\text{\uc774\ub2e4. } k^2 \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [4\uc810]}\n","answer":-1,"score":-1,"review":null}
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6

data/json/2023/math_temp.json ADDED Viewed

	@@ -0,0 +1,336 @@

+{
+"id": 1,
+"name": "1",
+"problem": "1. \\left( \\frac{4}{2^{\\sqrt{2}}} \\right)^{2 + \\sqrt{2}} \\text{의 값은? [2점]}\n\n\\begin{itemize}\n    \\item[1] $\\frac{1}{4}$\n    \\item[2] $\\frac{1}{2}$\n    \\item[3] $1$\n    \\item[4] $2$\n    \\item[5] $4$\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 2,
+"name": "2",
+"problem": "2. \\lim_{x \\to \\infty} \\frac{\\sqrt{x^2 - 2 + 3x}}{x + 5} \\text{의 값은? [2점]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 3,
+"name": "3",
+"problem": "3. \\text{공비가 양수인 등비수열 } \\{a_n\\}\\text{이}\n\n\\[ a_2 + a_4 = 30, \\quad a_4 + a_6 = \\frac{15}{2} \\]\n\\text{를 만족시킬 때, } a_1 \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] 48\n    \\item[2] 56\n    \\item[3] 64\n    \\item[4] 72\n    \\item[5] 80\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 4,
+"name": "4",
+"problem": "4. \\text{다항함수 } f(x) \\text{에 대하여 함수 } g(x) \\text{를}\n\n\\[ g(x) = x^2 f(x) \\]\n\\text{라 하자. } f(2) = 1, \\ f'(2) = 3 \\text{일 때, } g'(2) \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] 12\n    \\item[2] 14\n    \\item[3] 16\n    \\item[4] 18\n    \\item[5] 20\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 5,
+"name": "5",
+"problem": "5. \\tan \\theta < 0 \\text{이고} \\cos \\left( \\frac{\\pi}{2} + \\theta \\right) = \\frac{\\sqrt{5}}{5} \\text{일 때, } \\cos \\theta \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] - \\frac{2 \\sqrt{5}}{5}\n    \\item[2] - \\frac{\\sqrt{5}}{5}\n    \\item[3] 0\n    \\item[4] \\frac{\\sqrt{5}}{5}\n    \\item[5] \\frac{2 \\sqrt{5}}{5}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 6,
+"name": "6",
+"problem": "6. \\text{함수 } f(x) = 2x^3 - 9x^2 + ax + 5 \\text{는 } x = 1 \\text{에서 극대이고, } x = b \\text{에서 극소이다. } a + b \\text{의 값은? (단, } a, b \\text{는 상수이다.) [3점]}\n\n\\begin{itemize}\n    \\item[1] 12\n    \\item[2] 14\n    \\item[3] 16\n    \\item[4] 18\n    \\item[5] 20\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 7,
+"name": "7",
+"problem": "7. \\text{모든 항이 양수이고 첫째항과 공차가 같은 등차수열 } \\{a_n\\}\\text{이}\n\n\\[ \\sum_{k=1}^{15} \\frac{1}{\\sqrt{a_k} + \\sqrt{a_{k+1}}} = 2 \\]\n\\text{를 만족시킬 때, } a_4 \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] 6\n    \\item[2] 7\n    \\item[3] 8\n    \\item[4] 9\n    \\item[5] 10\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 8,
+"name": "8",
+"problem": "8. \\text{점 } (0, 4) \\text{에서 곡선 } y = x^3 - x + 2 \\text{에 그은 접선의 } x \\text{절편은? [3점]}\n\n\\begin{itemize}\n    \\item[1] -\\frac{1}{2}\n    \\item[2] -1\n    \\item[3] -\\frac{3}{2}\n    \\item[4] -2\n    \\item[5] -\\frac{5}{2}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 9,
+"name": "9",
+"problem": "9. \\text{함수}\n\n\\[ f(x) = a - \\sqrt{3} \\tan 2x \\]\n\\text{가 닫힌구간} \\left[ -\\frac{\\pi}{6}, b \\right] \\text{에서 최대값 7, 최솟값 3을 가질 때, } a \\times b \\text{의 값은? (단, } a, b \\text{는 상수이다.) [4점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{\\pi}{2}\n    \\item[2] \\frac{5\\pi}{12}\n    \\item[3] \\frac{\\pi}{3}\n    \\item[4] \\frac{\\pi}{4}\n    \\item[5] \\frac{\\pi}{6}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 10,
+"name": "10",
+"problem": "10. \\text{두 곡선 } y = x^3 + x^2, \\ y = -x^2 + k \\text{와 } y \\text{축으로 둘러싸인 부분의 넓이를 } A, \\text{ 두 곡선 } y = x^3 + x^2, \\ y = -x^2 + k \\text{와 직선 } x = 2 \\text{로 둘러싸인 부분의 넓이를 } B \\text{라 하자.} A = B \\text{일 때, 상수 } k \\text{의 값은? (단, } 4 < k < 5) [4점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{25}{6}\n    \\item[2] \\frac{13}{3}\n    \\item[3] \\frac{9}{2}\n    \\item[4] \\frac{14}{3}\n    \\item[5] \\frac{29}{6}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1,
+"review": 1
+}
+{
+"id": 11,
+"name": "11",
+"problem": "11. \\text{그림과 같이 사각형 } ABCD \\text{가 한 원에 내접하고}\n\n\\[ \\overline{AB} = 5, \\quad \\overline{AC} = 3 \\sqrt{5}, \\quad \\overline{AD} = 7, \\quad \\angle BAC = \\angle CAD \\]\n\\text{일 때, 이 원의 반지름의 길이는? [4점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{5 \\sqrt{2}}{2}\n    \\item[2] \\frac{8 \\sqrt{5}}{5}\n    \\item[3] \\frac{5 \\sqrt{5}}{3}\n    \\item[4] \\frac{8 \\sqrt{2}}{3}\n    \\item[5] \\frac{9 \\sqrt{3}}{4}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1,
+"review": 1
+}
+{
+"id": 12,
+"name": "12",
+"problem": "12. \\text{실수 전체의 집합에서 연속인 함수 } f(x) \\text{가 다음 조건을 만족시킨다.}\n\n\\[ n - 1 \\leq x < n \\text{일 때, } |f(x)| = |6(x - n + 1)(x - n)| \\text{이다. (단, } n \\text{은 자연수이다.)} \\]\n\n\\text{열린구간 } (0, 4) \\text{에서 정��된 함수} \n\\[ g(x) = \\int_0^x f(t) dt - \\int_x^4 f(t) dt \\]\n\\text{가 } x = 2 \\text{에서 최솟값 0을 가질 때, } \\int_{\\frac{1}{2}}^4 f(x) dx \\text{의 값은? [4점]}\n\n\\begin{itemize}\n    \\item[1] -\\frac{3}{2}\n    \\item[2] -\\frac{1}{2}\n    \\item[3] \\frac{1}{2}\n    \\item[4] \\frac{3}{2}\n    \\item[5] \\frac{5}{2}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 13,
+"name": "13",
+"problem": "13. \\text{자연수 } m(m \\geq 2) \\text{에 대하여 } m^{12} \\text{의 } n \\text{제곱근 중에서 정수가 존재하도록 하는 2 이상의 자연수 } n \\text{의 개수를 } f(m) \\text{이라 할 때,} \n\\[ \\sum_{m=2}^{9} f(m) \\text{의 값은? [4점]} \\]\n\n\\begin{itemize}\n    \\item[1] 37\n    \\item[2] 42\n    \\item[3] 47\n    \\item[4] 52\n    \\item[5] 57\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 14,
+"name": "14",
+"problem": "14. \\text{다항함수 } f(x) \\text{에 대하여 함수 } g(x) \\text{를 다음과 같이 정의한다.}\n\n\\[ g(x) = \\begin{cases} x & (x < -1 \\text{ 또는 } x > 1) \\\\ f(x) & (-1 \\leq x \\leq 1) \\end{cases} \\]\n\\text{함수 } h(x) = \\lim_{t \\to 0^+} g(x+t) \\times \\lim_{t \\to 2^+} g(x+t) \\text{에 대하여} \n\\text{보기에서 옳은 것만을 있는 대로 고른 것은? [4점]}\n\n\\<보기>\n\nㄱ. h(1) = 3 \n\nㄴ. 함수 h(x)는 실수 전체의 집합에서 연속이다. \n\nㄷ. 함수 g(x)가 닫힌구간 \\([-1, 1]\\)에서 감소하고 \\(g(-1) = -2\\)이면 함수 h(x)는 실수 전체의 집합에서 최솟값을 갖는다.\n\n\\begin{itemize}\n    \\item[1] ㄱ\n    \\item[2] ㄴ\n    \\item[3] ㄱ, ㄴ\n    \\item[4] ㄱ, ㄷ\n    \\item[5] ㄴ, ㄷ\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 15,
+"name": "15",
+"problem": "15. \\text{모든 항이 자연수이고 다음 조건을 만족시키는 모든 수열 } \\{a_n\\} \\text{에 대하여 } a_9 \\text{의 최대값과 최솟값을 각각 } M, m \\text{이라 할 때, } M + m \\text{의 값은? [4점]}\n\n\\text{(가) } a_7 = 40 \n\n\\text{(나) 모든 자연수 } n \\text{에 대하여}\n\\[ a_{n+2} = \\begin{cases} a_{n+1} + a_n & (a_{n+1}\\text{이 } 3 \\text{의 배수가 아닌 경우}) \\\\ \\frac{1}{3} a_{n+1} & (a_{n+1}\\text{이 } 3 \\text{의 배수인 경우}) \\end{cases} \\]\n\n\\begin{itemize}\n    \\item[1] 216\n    \\item[2] 218\n    \\item[3] 220\n    \\item[4] 222\n    \\item[5] 224\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 16,
+"name": "16",
+"problem": "16. \\text{방정식}\n\n\\[ \\log_2(3x + 2) = 2 + \\log_2(x - 2) \\]\n\\text{를 만족시키는 실수 } x \\text{의 값을 구하시오. [3점]}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 17,
+"name": "17",
+"problem": "17. \\text{함수 } f(x) \\text{에 대하여 } f'(x) = 4x^3 - 2x \\text{이고 } f(0) = 3 \\text{일 때, } f(2) \\text{의 값을 구하시오. [3점]}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 18,
+"name": "18",
+"problem": "18. \\text{두 수열 } \\{a_n\\}, \\{b_n\\} \\text{에 대하여}\n\n\\[ \\sum_{k=1}^{5} (3a_k + 5) = 55, \\quad \\sum_{k=1}^{5} (a_k + b_k) = 32 \\]\n\\text{일 때, } \\sum_{k=1}^{5} b_k \\text{의 값을 구하시오. [3점]}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 19,
+"name": "19",
+"problem": "19. \\text{방정식 } 2x^3 - 6x^2 + k = 0 \\text{의 서로 다른 양의 실근의 개수가 2가 되도록 하는 정수 } k \\text{의 개수를 구하시오. [3점]}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 20,
+"name": "20",
+"problem": "20. \\text{수직선 위를 움직이는 점 P의 시각 } t(t \\geq 0) \\text{에서의 속도 } v(t) \\text{와 가속도 } a(t) \\text{가 다음 조건을 만족시킨다.}\n\n\\text{(가) } 0 \\leq t \\leq 2 \\text{일 때, } v(t) = 2t^3 - 8t \\text{이다.}\n\\text{(나) } t \\geq 2 \\text{일 때, } a(t) = 6t + 4 \\text{이다.}\n\n\\text{시각 } t = 0 \\text{에서 } t = 3 \\text{까지 점 P가 움직인 거리를 구하시오. [4점]}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 21,
+"name": "21",
+"problem": "21. \\text{자연수 } n \\text{에 대하여 함수 } f(x) \\text{를}\n\n\\[ f(x) = \\begin{cases} |3^x + 2 - n| & (x < 0) \\\\ |\\log_2(x + 4) - n| & (x \\geq 0) \\end{cases} \\]\n\\text{이라 하자. 실수 } t \\text{에 대하여 } x \\text{에 대한 방정식 } f(x) = t \\text{의 서로 다른 실근의 개수를 } g(t) \\text{라 할 때, 함수 } g(t) \\text{의 최댓값이 4가 되도록 하는 모든 자연수 } n \\text{의 값의 합을 구하시오. [4점]}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 22,
+"name": "22",
+"problem": "22. \\text{최고차항의 계수가 1인 삼차함수 } f(x) \\text{와 실수 전체의 집합에서 연속인 함수 } g(x) \\text{가 다음 조건을 만족시킬 때, } f(4) \\text{의 값을 구하시오. [4점]}\n\n\\text{(가) 모든 실수 } x \\text{에 대하여 } f(x) = f(1) + (x - 1)f'(g(x)) \\text{이다.}\n\\text{(나) 함수 } g(x) \\text{의 최솟값은 } \\frac{5}{2} \\text{이다.}\n\\text{(다) } f(0) = -3, \\quad f(g(1)) = 6 \n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 23,
+"name": "23_prob",
+"problem": "23. \\( (x^3 + 3)^5 \\)의 전개식에서 \\(x^9\\)의 계수는? [2점]\n\\begin{itemize}\n    \\item[1] 30\n    \\item[2] 60\n    \\item[3] 90\n    \\item[4] 120\n    \\item[5] 150\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 24,
+"name": "24_prob",
+"problem": "24. \\text{숫자 } 1, 2, 3, 4, 5 \\text{ 중에서 중복을 허락하여 4개를 택해 일렬로 나열하여 만들 수 있는 네 자리의 자연수 중 4000 이상인 홀수의 개수는? [3점]}\n\n\\begin{itemize}\n    \\item[1] 125\n    \\item[2] 150\n    \\item[3] 175\n    \\item[4] 200\n    \\item[5] 225\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 25,
+"name": "25_prob",
+"problem": "25. \\text{흰색 마스크 5개, 검은색 마스크 9개가 들어 있는 상자가 있다. 이 상자에서 임의로 3개의 마스크를 동시에 꺼낼 때, 꺼낸 3개의 마스크 중에서 적어도 한 개가 흰색 마스크일 확률은? [3점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{8}{13}\n    \\item[2] \\frac{17}{26}\n    \\item[3] \\frac{9}{13}\n    \\item[4] \\frac{19}{26}\n    \\item[5] \\frac{10}{13}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 26,
+"name": "26_prob",
+"problem": "26. \\text{주머니에 1이 적힌 흰 공 1개, 2가 적힌 흰 공 1개, 1이 적힌 검은 공 1개, 2가 적힌 검은 공 3개가 들어 있다. 이 주머니에서 임의로 3개의 공을 동시에 꺼내는 시행을 한다. 이 시행에서 꺼낸 3개의 공 중에서 흰 공이 1개이고 검은 공이 2개인 사건을 } A, \\text{ 꺼낸 3개의 공에 적혀 있는 수를 모두 곱한 값이 8인 사건을 } B \\text{라 할 때, } P(A \\cup B) \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{11}{20}\n    \\item[2] \\frac{3}{5}\n    \\item[3] \\frac{13}{20}\n    \\item[4] \\frac{7}{10}\n    \\item[5] \\frac{3}{4}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1,
+"review": 1
+}
+{
+"id": 27,
+"name": "27_prob",
+"problem": "27. \\text{어느 회사에서 생산하는 샴푸 1개의 용량은 정규분포 } N(m, \\sigma^2) \\text{을 따른다고 한다. 이 회사에서 생산하는 샴푸 중에서 16개를 임의추출하여 얻은 표본평균을 이용하여 구한 } m \\text{에 대한 신뢰도 95%의 신뢰구간이 } 746.1 \\leq m \\leq 755.9 \\text{이다. 이 회사에서 생산하는 샴푸 중에서 } n \\text{개를 임의추출하여 얻은 표본평균을 이용하여 구하는 } m \\text{에 대한 신뢰도 99%의 신뢰구간이 } a \\leq m \\leq b \\text{일 때, } b - a \\text{의 값이 6 이하가 되기 위한 자연수 } n \\text{의 최솟값은? (단, 용량의 단위는 mL이고, } Z \\text{가 표준정규분포를 따르는 확률변수일 때, } P(|Z| \\leq 1.96) = 0.95, P(|Z| \\leq 2.58) = 0.99 \\text{로 계산한다.) [3점]}\n\n\\begin{itemize}\n    \\item[1] 70\n    \\item[2] 74\n    \\item[3] 78\n    \\item[4] 82\n    \\item[5] 86\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 28,
+"name": "28_prob",
+"problem": "28. \\text{연속확률변수 } X \\text{가 갖는 값의 범위는 } 0 \\leq X \\leq a \\text{이고, } X \\text{의 확률밀도함수의 그래프가 그림과 같다.}\n\n\\[ P(X \\leq b) - P(X \\geq b) = \\frac{1}{4}, \\quad P(X \\leq \\sqrt{5}) = \\frac{1}{2} \\]\n\\text{일 때, } a + b + c \\text{의 값은? (단, } a, b, c \\text{는 상수이다.) [4점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{11}{2}\n    \\item[2] 6\n    \\item[3] \\frac{13}{2}\n    \\item[4] 7\n    \\item[5] \\frac{15}{2}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1,
+"review": 2
+}
+{
+"id": 29,
+"name": "29_prob",
+"problem": "29. \\text{앞면에는 1부터 6까지의 자연수가 하나씩 적혀 있고, 뒷면에는 모두 0이 하나씩 적혀 있는 6장의 카드가 있다. 이 6장의 카드가 그림과 같이 6 이하의 자연수 } k \\text{에 대하여 } k \\text{번째 자리에 자연수 } k \\text{가 보이도록 놓여 있다.}\n\n\\text{이 6장의 카드와 한 개의 주사위를 사용하여 다음 시행을 한다.}\n\n\\[ \\text{주사위를 한 번 던져 나온 눈의 수가 } k \\text{이면 } k \\text{번째 자리에 놓여 있는 카드를 한 번 뒤집어 제자리에 놓는다.} \\]\n\n\\text{위의 시행을 3번 반복한 후 6장의 카드에 보이는 모든 수의 합이 짝수일 때, 주사위의 1의 눈이 한 번만 나왔을 확률은 } \\frac{q}{p} \\text{이다. } p + q \\text{의 값을 구하시오. (단, } p \\text{와 } q \\text{는 서로소인 자연수이다.) [4점]}\n",
+"answer": -1,
+"score": -1,
+"review": 1
+}
+{
+"id": 30,
+"name": "30_prob",
+"problem": "30. \\text{집합 } X = \\{x | x \\text{는 10 이하의 자연수}\\} \\text{에 대하여 다음 조건을 만족시키는 함수 } f: X \\to X \\text{의 개수를 구하시오. [4점]}\n\n\\text{(가) 9 이하의 모든 자연수 } x \\text{에 대하여 } f(x) \\leq f(x+1) \\text{이다.}\n\\text{(나) } 1 \\leq x \\leq 5 \\text{일 때 } f(x) \\leq x \\text{이고, } 6 \\leq x \\leq 10 \\text{일 때 } f(x) \\geq x \\text{이다.}\n\\text{(다) } f(6) = f(5) + 6\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 31,
+"name": "23_calc",
+"problem": "23. \\lim_{x \\to 0} \\frac{\\ln(x+1)}{\\sqrt{x+4} - 2} \\text{의 값은? [2점]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 32,
+"name": "24_calc",
+"problem": "24. \\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=1}^{n} \\sqrt{1 + \\frac{3k}{n}} \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{4}{3}\n    \\item[2] \\frac{13}{9}\n    \\item[3] \\frac{14}{9}\n    \\item[4] \\frac{5}{3}\n    \\item[5] \\frac{16}{9}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 33,
+"name": "25_calc",
+"problem": "25. \\text{등비수열 } \\{a_n\\} \\text{에 대하여 } \\lim_{n \\to \\infty} \\frac{a_n + 1}{3^n + 2^{2n-1}} = 3 \\text{일 때, } a_2 \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] 16\n    \\item[2] 18\n    \\item[3] 20\n    \\item[4] 22\n    \\item[5] 24\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 34,
+"name": "26_calc",
+"problem": "26. \\text{그림과 같이 곡선 } y = \\sqrt{\\sec^2 x} + \\tan x \\left(0 \\leq x \\leq \\frac{\\pi}{3}\\right) \\text{와 } x \\text{축, } y \\text{축 및 직선 } x = \\frac{\\pi}{3} \\text{로 둘러싸인 부분을 밑면으로 하는 입체도형이 있다. 이 입체도형을 } x \\text{축에 수직인 평면으로 자른 단면이 모두 정사각형일 때, 이 입체도형의 부피는? [3점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{\\sqrt{3}}{2} + \\frac{\\ln 2}{2}\n    \\item[2] \\frac{\\sqrt{3}}{2} + \\ln 2\n    \\item[3] \\sqrt{3} + \\frac{\\ln 2}{2}\n    \\item[4] \\sqrt{3} + \\ln 2\n    \\item[5] \\sqrt{3} + 2\\ln 2\n\\end{itemize}\n",
+"answer": -1,
+"score": -1,
+"review": 1
+}
+{
+"id": 35,
+"name": "27_prob",
+"problem": "27. \\text{그림과 같이 중심이 } O, \\text{반지름의 길이가 1이고 중심각의 크기가 } \\frac{\\pi}{2} \\text{인 부채꼴 } OA_1B_1 \\text{이 있다. 호 } A_1B_1 \\text{ 위에 점 } P_1, \\text{선분 } OA_1 \\text{ 위에 점 } C_1, \\text{선분 } OB_1 \\text{ 위에 점 } D_1 \\text{을 사각형 } OC_1P_1D_1 \\text{이 } OC_1 : OD_1 = 3:4 \\text{인 직사각형이 되도록 잡는다.}\n\n\\text{부채꼴 } OA_1B_1 \\text{의 내부에 점 } Q_1 \\text{을 } PQ_1 = AQ_1, \\angle PQ_1A_1 = \\frac{\\pi}{2} \\text{가 되도록 잡고, 이등변삼각형 } P_1Q_1A_1 \\text{에 색칠하여 얻은 그림을 } R_1 \\text{이라 하자.}\n\\text{그림 } R_1 \\text{에서 선분 } OA_1 \\text{ 위의 점 } A_2 \\text{와 선분 } OB_1 \\text{ 위의 점 } B_2 \\text{를 } OQ_1 = OA_2 = OB_2 \\text{가 되도록 잡고, 중심이 } O, \\text{반지름의 길이가 } OQ_1, \\text{중심각의 크기가 } \\frac{\\pi}{2} \\text{인 부채꼴 } OA_2B_2 \\text{를 그린다. 그린 } R_1 \\text{을 얻은 것과 같은 방법으로 네 점 } P_2, C_2, D_2, Q_2 \\text{를 잡고, 이등변삼각형 } P_2Q_2A_2 \\text{에 색칠하여 얻은 그림을 } R_2 \\text{라 하자. 이와 같은 과정을 계속하여 } n \\text{번째 얻은 그림 } R_n \\text{에 색칠되어 있는 부분의 넓이를 } S_n \\text{이라 할 때, } \\lim_{n \\to \\infty} S_n \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{9}{40}\n    \\item[2] \\frac{1}{4}\n    \\item[3] \\frac{11}{40}\n    \\item[4] \\frac{3}{10}\n    \\item[5] \\frac{13}{40}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1,
+"review": 1
+}
+{
+"id": 36,
+"name": "28_prob",
+"problem": "28. \\text{그림과 같이 중심이 } O \\text{이고 길이가 2인 선분 } AB \\text{를 지름으로 하는 반원 위에 } \\angle AOC = \\frac{\\pi}{2} \\text{인 점 } C \\text{가 있다.}\n\\text{호 } BC \\text{ 위에 점 } P \\text{와 호 } CA \\text{ 위에 점 } Q \\text{를 } PB = QC \\text{가 되도록 잡고, 선분 } AP \\text{ 위에 점 } R \\text{을 } \\angle CQR = \\frac{\\pi}{2} \\text{가 되도록 잡는다.}\n\\text{선분 } AP \\text{와 선분 } CO \\text{의 교점을 } S \\text{라 하자. } \\angle PAB = \\theta \\text{일 때, 삼각형 } POB \\text{의 넓이를 } f(\\theta), \\text{사각형 } CQRS \\text{의 넓이를 } g(\\theta) \\text{라 하자.}\n\n\\lim_{\\theta \\to 0^+} \\frac{3f(\\theta) - 2g(\\theta)}{\\theta^2} \\text{의 값은? (단, } 0 < \\theta < \\frac{\\pi}{4} \\text{) [4점]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n",
+"answer": -1,
+"score": -1,
+"review": 1
+}
+{
+"id": 37,
+"name": "29_prob",
+"problem": "29. \\text{세 상수 } a, b, c \\text{에 대하여 함수 } f(x) = ae^{2x} + be^x + c \\text{가 다음 조건을 만족시킨다.}\n\n\\text{(가) } \\lim_{x \\to -\\infty} \\frac{f(x) + 6}{e^x} = 1\n\\text{(나) } f(\\ln 2) = 0\n\n\\text{함수 } f(x) \\text{의 역함수를 } g(x) \\text{라 할 때,}\n\\[ \\int_0^{14} g(x) dx = p + q \\ln 2 \\text{이다. } p + q \\text{의 값��� 구하시오.}\n\\text{(단, } p, q \\text{는 유리수이고, } \\ln 2 \\text{는 무리수이다.) [4점]}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 38,
+"name": "30_prob",
+"problem": "30. \\text{최고차항의 계수가 양수인 삼차함수 } f(x) \\text{와 함수 } g(x) = e^{\\sin \\pi x} - 1 \\text{에 대하여 실수 전체의 집합에서 정의된 합성함수 } h(x) = g(f(x)) \\text{가 다음 조건을 만족시킨다.}\n\n\\text{(가) 함수 } h(x) \\text{는 } x = 0 \\text{에서 극댓값 0을 갖는다.}\n\\text{(나) 열린구간 } (0, 3) \\text{에서 방정식 } h(x) = 1 \\text{의 서로 다른 실근의 개수는 7이다.}\n\nf(3) = \\frac{1}{2}, f'(3) = 0 \\text{일 때, } f(2) = \\frac{q}{p} \\text{이다. } p + q \\text{의 값을 구하시오. (단, } p \\text{와 } q \\text{는 서로소인 자연수이다.) [4점]}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 39,
+"name": "23_geom",
+"problem": "23. \\text{좌표공간의 점 } A(2, 2, -1) \\text{을 } x \\text{축에 대하여 대칭이동한 점을 } B \\text{라 하자. 점 } C(-2, 1, 1) \\text{에 대하여 선분 BC의 길이는? [2점]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 40,
+"name": "24_geom",
+"problem": "24. \\text{초점이 } F\\left(\\frac{1}{3}, 0\\right) \\text{이고 준선이 } x = -\\frac{1}{3} \\text{인 포물선이 점 } (a, 2) \\text{를 지날 때, } a \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 41,
+"name": "25_geom",
+"problem": "25. \\text{타원 } \\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1 \\text{ 위의 점 } (2, 1) \\text{에서의 접선의 기울기가 } -\\frac{1}{2} \\text{일 때, 이 타원의 두 초점 사이의 거리는? (단, } a, b \\text{는 양수이다.) [3점]}\n\n\\begin{itemize}\n    \\item[1] 2\\sqrt{3}\n    \\item[2] 4\n    \\item[3] 2\\sqrt{5}\n    \\item[4] 2\\sqrt{6}\n    \\item[5] 2\\sqrt{7}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 42,
+"name": "26_geom",
+"problem": "26. \\text{좌표평면에서 세 벡터 } \\vec{a} = (2, 4), \\vec{b} = (2, 8), \\vec{c} = (1, 0) \\text{에 대하여 두 벡터 } \\vec{p}, \\vec{q} \\text{가}\n\n(\\vec{p} - \\vec{a}) \\cdot (\\vec{p} - \\vec{b}) = 0, \\quad \\vec{q} = \\frac{1}{2} \\vec{a} + t \\vec{c} \\quad (t \\text{는 실수}) \\text{를 만족시킬 때, } |\\vec{p} - \\vec{q}| \\text{의 최솟값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{3}{2}\n    \\item[2] 2\n    \\item[3] \\frac{5}{2}\n    \\item[4] 3\n    \\item[5] \\frac{7}{2}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1
+}
+{
+"id": 43,
+"name": "27_geom",
+"problem": "27. \\text{좌표공간에 직선 AB를 포함하는 평면 } \\alpha \\text{가 있다. 평면 } \\alpha \\text{ 위에 있지 않은 점 C에 대하여 직선 AB와 직선 AC가 이루는 예각의 크기를 } \\theta_1 \\text{이라 할 때 } \\sin \\theta_1 = \\frac{4}{5} \\text{이고, 직선 AC와 평면 } \\alpha \\text{가 이루는 예각의 크기는 } \\frac{\\pi}{2} - \\theta_1 \\text{이다. 평면 ABC와 평면 } \\alpha \\text{가 이루는 예각의 크기를 } \\theta_2 \\text{라 할 때, } \\cos \\theta_2 \\text{의 값은? [3점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{\\sqrt{7}}{4}\n    \\item[2] \\frac{\\sqrt{7}}{5}\n    \\item[3] \\frac{\\sqrt{7}}{6}\n    \\item[4] \\frac{\\sqrt{7}}{7}\n    \\item[5] \\frac{\\sqrt{7}}{8}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1,
+"review": 1
+}
+{
+"id": 44,
+"name": "28_geom",
+"problem": "28. \\text{두 초점이 } F(c, 0), F'(-c, 0) \\text{(} c > 0 \\text{)인 쌍곡선 } C \\text{와 } y \\text{축 위의 점 } A \\text{가 있다. 쌍곡선 } C \\text{가 선분 } AF \\text{와 만나는 점을 } P, \\text{선분 } AF' \\text{과 만나는 점을 } P' \\text{이라 하자. 직선 } AF \\text{는 쌍곡선 } C \\text{의 한 접근선과 평행하고 }\n\\frac{AP}{PP'} = \\frac{5}{6}, PF = 1 \\text{일 때, 쌍곡선 } C \\text{의 주축의 길이는? [4점]}\n\n\\begin{itemize}\n    \\item[1] \\frac{13}{6}\n    \\item[2] \\frac{9}{4}\n    \\item[3] \\frac{7}{3}\n    \\item[4] \\frac{29}{12}\n    \\item[5] \\frac{5}{2}\n\\end{itemize}\n",
+"answer": -1,
+"score": -1,
+"review": 1
+}
+{
+"id": 45,
+"name": "29_geom",
+"problem": "29. \\text{평면 } \\alpha \\text{ 위에 } \\overline{AB} = \\overline{CD} = \\overline{AD} = 2, \\quad \\angle ABC = \\angle BCD = \\frac{\\pi}{3} \\text{인 사다리꼴 ABCD가 있다. 다음 조건을 만족시키는 평면 } \\alpha \\text{ 위의 두 점 P, Q에 대하여 } \\overrightarrow{CP} \\cdot \\overrightarrow{DQ} \\text{의 값을 구하시오. [4점]}\n\n\\text{(가) } \\overrightarrow{AC} = 2(\\overrightarrow{AD} + \\overrightarrow{BP})\n\\text{(나) } \\overrightarrow{AC} \\cdot \\overrightarrow{PQ} = 6\n\\text{(다) } 2 \\times \\angle BQA = \\angle PBQ < \\frac{\\pi}{2}\n",
+"answer": -1,
+"score": -1,
+"review": 2
+}
+{
+"id": 46,
+"name": "30_geom",
+"problem": "30. \\text{좌표공간에 정사면체 ABCD가 있다. 정삼각형 BCD의 외심을 중심으로 하고 점 B를 지나는 구를 } S \\text{라 하자.}\n\\text{구 } S \\text{와 선분 AB가 만나는 점 중 B가 아닌 점을 P,}\n\\text{구 } S \\text{와 선분 AC가 만나는 점 중 C가 아닌 점을 Q,}\n\\text{구 } S \\text{와 선분 AD가 만나는 점 중 D가 아닌 점을 R라 하고, 점 P에서 구 } S \\text{에 접하는 평면을 } \\alpha \\text{라 하자.}\n\\text{구 } S \\text{의 반지름의 길이가 6일 때, 삼각형 PQR의 평면 } \\alpha \\text{위로의 정사영의 넓이는 } k \\alpha \\text{이다. } k^2 \\text{의 값을 구하시오. [4점]}\n",
+"answer": -1,
+"score": -1
+}

data/json/2024/math.json ADDED Viewed

	@@ -0,0 +1,49 @@

+{"id":1,"name":"1","problem":"1. \\left( \\frac{4}{2 ^{\\sqrt{2}}} \\right)^{2 + \\sqrt{2}} \\text{\uc758 \uac12\uc740? [2\uc810]}\n\n\\begin{itemize}\n    \\item[1] $\\frac{1}{4}$\n    \\item[2] $\\frac{1}{2}$\n    \\item[3] $1$\n    \\item[4] $2$\n    \\item[5] $4$\n\\end{itemize}\n","answer":"1","score":2}
+{"id":2,"name":"2","problem":"2. \\lim_{x \\to \\infty} \\frac{\\sqrt{x^2 - 2} + 3x}{x+5} \\text{\uc758 \uac12\uc740? [2\uc810]}\n\n\\begin{itemize}\n  \\item[1] 1\n  \\item[2] 2\n  \\item[3] 3\n  \\item[4] 4\n  \\item[5] 5\n\\end{itemize}\n","answer":"4","score":2}
+{"id":3,"name":"3","problem":"3. \uacf5\ube44\uac00 \uc591\uc218\uc778 \ub4f1\ube44\uc218\uc5f4 $\\{a_n\\}$\uc774\n\\[\na_2 + a_4 = 30, \\quad a_4 + a_6 = \\frac{15}{2}\n\\]\n\ub97c \ub9cc\uc871\uc2dc\ud0ac \ub54c, $a_1$\uc758 \uac12\uc740? [3\uc810]\n\\begin{itemize}\n    \\item[1] 48\n    \\item[2] 56\n    \\item[3] 64\n    \\item[4] 72\n    \\item[5] 80\n\\end{itemize}\n","answer":"2","score":3}
+{"id":4,"name":"4","problem":"4. \ub2e4\ud56d\ud568\uc218 $f(x)$\uc5d0 \ub300\ud558\uc5ec \ud568\uc218 $g(x)$\ub97c\n\\[\ng(x) = x^2 f(x)\n\\]\n\ub77c \ud558\uc790. $f(2) = 1$, $f'(2) = 3$\uc77c \ub54c, $g'(2)$\uc758 \uac12\uc740? [3\uc810]\n\\begin{itemize}\n    \\item[1] 12\n    \\item[2] 14\n    \\item[3] 16\n    \\item[4] 18\n    \\item[5] 20\n\\end{itemize}\n","answer":"1","score":3}
+{"id":5,"name":"5","problem":"5. \\(\\tan \\theta < 0\\)\uc774\uace0 \\(\\cos \\left(\\frac{\\pi}{2} + \\theta \\right) = \\frac{\\sqrt{5}}{5}\\)\uc77c \ub54c, \\(\\cos \\theta\\)\uc758 \uac12\uc740? [3\uc810]\n\\begin{itemize}\n    \\item[1] \\(- \\frac{2\\sqrt{5}}{5}\\)\n    \\item[2] \\(- \\frac{\\sqrt{5}}{5}\\)\n    \\item[3] 0\n    \\item[4] \\(\\frac{\\sqrt{5}}{5}\\)\n    \\item[5] \\(\\frac{2\\sqrt{5}}{5}\\)\n\\end{itemize}\n","answer":"4","score":3}
+{"id":6,"name":"6","problem":"6. \ud568\uc218 \\( f(x) = 2x^3 - 9x^2 + ax + 5 \\)\ub294 \\( x = 1 \\)\uc5d0\uc11c \uadf9\ub300\uc774\uace0, \\( x = b \\)\uc5d0\uc11c \uadf9\uc18c\uc774\ub2e4. \\( a + b \\)\uc758 \uac12\uc740? (\ub2e8, \\( a, b \\)\ub294 \uc0c1\uc218\uc774\ub2e4.) [3\uc810]\n\\begin{itemize}\n    \\item[1] 12\n    \\item[2] 14\n    \\item[3] 16\n    \\item[4] 18\n    \\item[5] 20\n\\end{itemize}\n","answer":"4","score":3}
+{"id":7,"name":"7","problem":"7. \ubaa8\ub4e0 \ud56d\uc774 \uc591\uc218\uc774\uace0 \uccab\uc9f8\ud56d\uacfc \uacf5\ucc28\uac00 \uac19\uc740 \ub4f1\ucc28\uc218\uc5f4 $\\{a_n\\}$\uc774 \n\\[\n\\sum_{k=1}^{15} \\frac{1}{\\sqrt{a_k} + \\sqrt{a_{k+1}}} = 2\n\\]\n\ub97c \ub9cc\uc871\uc2dc\ud0ac \ub54c, $a_4$\uc758 \uac12\uc740? \\textbf{[3\uc810]}\n\\begin{itemize}\n    \\item[1] 6\n    \\item[2] 7\n    \\item[3] 8\n    \\item[4] 9\n    \\item[5] 10\n\\end{itemize}\n","answer":"5","score":3}
+{"id":8,"name":"8","problem":"8. \uc810 $(0, 4)$\uc5d0\uc11c \uace1\uc120 $y = x^3 - x + 2$\uc5d0 \uadf8\uc740 \uc811\uc120\uc758 $x$\uc808\ud3b8\uc740? [3\uc810]\n\\begin{itemize}\n    \\item[1] $-\\frac{1}{2}$\n    \\item[2] $-1$\n    \\item[3] $-\\frac{3}{2}$\n    \\item[4] $-2$\n    \\item[5] $-\\frac{5}{2}$\n\\end{itemize}\n","answer":"2","score":3}
+{"id":9,"name":"9","problem":"9. \ud568\uc218\n\\[\nf(x) = a - \\sqrt{3} \\tan 2x\n\\]\n\uac00 \ub2eb\ud78c\uad6c\uac04 \\(\\left[ -\\frac{\\pi}{6}, b \\right]\\) \uc5d0\uc11c \ucd5c\ub313\uac12 7, \ucd5c\uc19f\uac12 3\uc744 \uac00\uc9c8 \ub54c, \\(a \\times b\\)\uc758 \uac12\uc740? (\ub2e8, \\(a, b\\)\ub294 \uc0c1\uc218\uc774\ub2e4.) [4\uc810]\n\\begin{itemize}\n    \\item[1] \\(\\frac{\\pi}{2}\\)\n    \\item[2] \\(\\frac{5\\pi}{12}\\)\n    \\item[3] \\(\\frac{\\pi}{3}\\)\n    \\item[4] \\(\\frac{\\pi}{4}\\)\n    \\item[5] \\(\\frac{\\pi}{6}\\)\n\\end{itemize}\n","answer":"4","score":4}
+{"id":10,"name":"10","problem":"10. \ub450 \uace1\uc120 \\(y = x^3 + x^2\\), \\(y = -x^2 + k\\)\uc640 \\(y\\) \ucd95\uc73c\ub85c \ub458\ub7ec\uc2f8\uc778 \ubd80\ubd84\uc758 \ub113\uc774\ub97c \\(A\\), \ub450 \uace1\uc120 \\(y = x^3 + x^2\\), \\(y = -x^2 + k\\)\uc640 \uc9c1\uc120 \\(x = 2\\)\ub85c \ub458\ub7ec\uc2f8\uc778 \ubd80\ubd84\uc758 \ub113\uc774\ub97c \\(B\\)\ub77c \ud558\uc790. \\(A = B\\)\uc77c \ub54c, \uc0c1\uc218 \\(k\\)\uc758 \uac12\uc740? (\ub2e8, \\(4 < k < 5\\)) [4\uc810]\n\\begin{itemize}\n    \\item[1] \\(\\frac{25}{6}\\)\n    \\item[2] \\(\\frac{13}{3}\\)\n    \\item[3] \\(\\frac{9}{2}\\)\n    \\item[4] \\(\\frac{14}{3}\\)\n    \\item[5] \\(\\frac{29}{6}\\)\n\\end{itemize}\n","answer":"2","score":4}
+{"id":11,"name":"11","problem":"11. \uadf8\ub9bc\uacfc \uac19\uc774 \uc0ac\uac01\ud615 ABCD\uac00 \ud55c \uc6d0\uc5d0 \ub0b4\uc811\ud558\uace0 \\\\\n\\[\n\\overline{AB} = 5, \\quad \\overline{AC} = 3\\sqrt{5}, \\quad \\overline{AD} = 7, \\quad \\angle BAC = \\angle CAD\n\\]\n\uc77c \ub54c, \uc774 \uc6d0\uc758 \ubc18\uc9c0\ub984\uc758 \uae38\uc774\ub294? \\textbf{[4\uc810]}\\\\\n\\begin{itemize}\n    \\item[1] \\(\\frac{5\\sqrt{2}}{2}\\)\n    \\item[2] \\(\\frac{8\\sqrt{5}}{5}\\)\n    \\item[3] \\(\\frac{5\\sqrt{5}}{3}\\)\n    \\item[4] \\(\\frac{8\\sqrt{2}}{3}\\)\n    \\item[5] \\(\\frac{9\\sqrt{3}}{4}\\)\n\\end{itemize}\n","answer":"1","score":4}
+{"id":12,"name":"12","problem":"12. \uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \uc5f0\uc18d\uc778 \ud568\uc218 \\( f(x) \\)\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a8\ub2e4.\n\\[\n\\boxed{\nn-1 \\leq x < n \\text{\uc77c \ub54c}, \\, |f(x)| = |6(x-n+1)(x-n)| \\, \\text{\uc774\ub2e4}. \\, (\\text{\ub2e8}, n \\, \\text{\uc740 \uc790\uc5f0\uc218\uc774\ub2e4.})\n}\n\\]\n\uc5f4\ub9b0\uad6c\uac04 \\( (0, 4) \\)\uc5d0\uc11c \uc815\uc758\ub41c \ud568\uc218\n\\[\ng(x) = \\int_0^x f(t)dt - \\int_x^4 f(t)dt\n\\]\n\uac00 \\( x = 2 \\)\uc5d0\uc11c \ucd5c\uc19f\uac12 0\uc744 \uac00\uc9c8 \ub54c, \\( \\int_\\frac{1}{2}^4 f(x)dx \\)\uc758 \uac12\uc740? [4\uc810]\n\\begin{itemize}\n    \\item[1] \\( -\\frac{3}{2} \\)\n    \\item[2] \\( -\\frac{1}{2} \\)\n    \\item[3] \\( \\frac{1}{2} \\)\n    \\item[4] \\( \\frac{3}{2} \\)\n    \\item[5] \\( \\frac{5}{2} \\)\n\\end{itemize}\n","answer":"3","score":4}
+{"id":13,"name":"13","problem":"13. \uc790\uc5f0\uc218 $m(m \\geq 2)$\uc5d0 \ub300\ud558\uc5ec $m^{12}$\uc758 $n$\uc81c\uacf1\uadfc \uc911\uc5d0\uc11c \uc815\uc218\uac00 \uc874\uc7ac\ud558\ub3c4\ub85d \ud558\ub294 2 \uc774\uc0c1\uc758 \uc790\uc5f0\uc218 $n$\uc758 \uac1c\uc218\ub97c $f(m)$\uc774\ub77c \ud560 \ub54c,\n\\[\n\\sum_{m=2}^{9} f(m) \\text{\uc758 \uac12\uc740? [4\uc810]} \n\\]\n\\begin{itemize}\n    \\item[1] 37\n    \\item[2] 42\n    \\item[3] 47\n    \\item[4] 52\n    \\item[5] 57\n\\end{itemize}\n","answer":"1","score":4}
+{"id":14,"name":"14","problem":"14. \ub2e4\ud56d\ud568\uc218 $f(x)$\uc5d0 \ub300\ud558\uc5ec \ud568\uc218 $g(x)$\ub97c \ub2e4\uc74c\uacfc \uac19\uc774 \uc815\uc758\ud55c\ub2e4.\n\\[\ng(x) = \\begin{cases} \nx & (x < -1 \\text{ \ub610\ub294 } x > 1) \\\\\nf(x) & (-1 \\leq x \\leq 1)\n\\end{cases}\n\\]\n\ud568\uc218 $h(x) = \\lim_{t \\to 0^+} g(x + t) \\times \\lim_{t \\to 2^+} g(x + t)$\uc5d0 \ub300\ud558\uc5ec \\\\\n\\textless \ubcf4\uae30\\textgreater \uc5d0\uc11c \uc633\uc740 \uac83\ub9cc\uc744 \uc788\ub294 \ub300\ub85c \uace0\ub978 \uac83\uc740? [4\uc810]\n\\textless \ubcf4\uae30\\textgreater \\\\\n\\fbox{\n  \\parbox{\\textwidth}{\n    \u3131. $h(1) = 3$ \\\\\n    \u3134. \ud568\uc218 $h(x)$\ub294 \uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \uc5f0\uc18d\uc774\ub2e4. \\\\\n    \u3137. \ud568\uc218 $g(x)$\uac00 \ub2eb\ud78c\uad6c\uac04 $[-1, 1]$\uc5d0\uc11c \uac10\uc18c\ud558\uace0 $g(-1) = -2$\uc774\uba74 \ud568\uc218 $h(x)$\ub294 \uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \ucd5c\uc19f\uac12\uc744 \uac16\ub294\ub2e4.\n  }\n}\n\\begin{itemize}\n\\item[1] \u3131\n\\item[2] \u3134\n\\item[3] \u3131, \u3134\n\\item[4] \u3131, \u3137\n\\item[5] \u3134, \u3137\n\\end{itemize}\n","answer":"1","score":4}
+{"id":15,"name":"15","problem":"15. \ubaa8\ub4e0 \ud56d\uc774 \uc790\uc5f0\uc218\uc774\uace0 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a4\ub294 \ubaa8\ub4e0 \uc218\uc5f4 $\\{a_n\\}$\uc5d0 \ub300\ud558\uc5ec $a_9$\uc758 \ucd5c\ub313\uac12\uacfc \ucd5c\uc19f\uac12\uc744 \uac01\uac01 $M, m$\uc774\ub77c \ud560 \ub54c, $M+m$\uc758 \uac12\uc740? \\textbf{[4\uc810]}\n\\[\n\\text{(\uac00)} \\quad a_7 = 40\n\\]\n\\text{(\ub098)} \\quad \ubaa8\ub4e0 \uc790\uc5f0\uc218 $n$\uc5d0 \ub300\ud558\uc5ec \n\\[\na_{n+2} = \n\\begin{cases} \na_{n+1} + a_n & \\text{(}a_{n+1}\\text{\uc774 3\uc758 \ubc30\uc218\uac00 \uc544\ub2cc \uacbd\uc6b0)}\\\\\n\\frac{1}{3} a_{n+1} & \\text{(}a_{n+1}\\text{\uc774 3\uc758 \ubc30\uc218\uc778 \uacbd\uc6b0)}\n\\end{cases}\n\\]\n\\begin{itemize}\n  \\item[1] 216\n  \\item[2] 218\n  \\item[3] 220\n  \\item[4] 222\n  \\item[5] 224\n\\end{itemize}\n","answer":"3","score":4}
+{"id":16,"name":"16","problem":"16. \ubc29\uc815\uc2dd\n\\[\n\\log_2{(3x+2)} = 2 + \\log_2{(x-2)}\n\\]\n\\text{\ub97c \ub9cc\uc871\uc2dc\ud0a4\ub294 \uc2e4\uc218 } \\( x \\) \\text{\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [3\uc810]}\n","answer":"2","score":3}
+{"id":17,"name":"17","problem":"17. \ud568\uc218 $f(x)$\uc5d0 \ub300\ud558\uc5ec $f'(x) = 4x^3 - 2x$\uc774\uace0 $f(0) = 3$\uc77c \ub54c, $f(2)$\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [3\uc810]\n","answer":"8","score":3}
+{"id":18,"name":"18","problem":"18. \ub450 \uc218\uc5f4 $\\{a_n\\}$, $\\{b_n\\}$\uc5d0 \ub300\ud558\uc5ec\n\\[\n\\sum_{k=1}^{5} (3a_k + 5) = 55, \\quad \\sum_{k=1}^{5} (a_k + b_k) = 32\n\\]\n\uc77c \ub54c, $\\sum_{k=1}^{5} b_k$\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [3\uc810]\n","answer":"9","score":3}
+{"id":19,"name":"19","problem":"19. \ubc29\uc815\uc2dd $2x^3 - 6x^2 + k = 0$\uc758 \uc11c\ub85c \ub2e4\ub978 \uc591\uc758 \uc2e4\uadfc\uc758 \uac1c\uc218\uac00 2\uac00 \ub418\ub3c4\ub85d \ud558\ub294 \uc815\uc218 $k$\uc758 \uac1c\uc218\ub97c \uad6c\ud558\uc2dc\uc624. [3\uc810]\n","answer":"32","score":3}
+{"id":20,"name":"20","problem":"20. \uc218\uc9c1\uc120 \uc704\ub97c \uc6c0\uc9c1\uc774\ub294 \uc810 P\uc758 \uc2dc\uac01 \\(t(t\\geq0)\\)\uc5d0\uc11c\uc758 \uc18d\ub3c4 \\(v(t)\\)\uc640 \uac00\uc18d\ub3c4 \\(a(t)\\)\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a8\ub2e4.\n\\[\n\\text{(\uac00)} \\quad 0 \\leq t \\leq 2 \\text{\uc77c \ub54c}, \\quad v(t) = 2t^3 - 8t \\text{\uc774\ub2e4.}\n\\]\n\\[\n\\text{(\ub098)} \\quad t \\geq 2 \\text{\uc77c \ub54c}, \\quad a(t) = 6t + 4\\text{\uc774\ub2e4.}\n\\]\n\uc2dc\uac01 \\( t=0 \\)\uc5d0\uc11c \\( t=3 \\)\uae4c\uc9c0 \uc810 P\uac00 \uc6c0\uc9c1\uc778 \uac70\ub9ac\ub97c \uad6c\ud558\uc2dc\uc624. \\hfill [4\uc810]\n","answer":"25","score":4}
+{"id":21,"name":"21","problem":"21. \uc790\uc5f0\uc218 \\(n\\)\uc5d0 \ub300\ud558\uc5ec \ud568\uc218 \\(f(x)\\)\ub97c\n\\[\nf(x) =\n\\begin{cases} \n    |3^{x+2}-n| & (x<0) \\\\ \n    | \\log_2 (x+4) -n| & (x \\geq 0)\n\\end{cases}\n\\]\n\uc774\ub77c \ud558\uc790. \uc2e4\uc218 \\(t\\)\uc5d0 \ub300\ud558\uc5ec \\(x\\)\uc5d0 \ub300\ud55c \ubc29\uc815\uc2dd \\(f(x) = t\\)\uc758 \uc11c\ub85c \ub2e4\ub978 \uc2e4\uadfc\uc758 \uac1c\uc218\ub97c \\(g(t)\\)\ub77c \ud560 \ub54c, \ud568\uc218 \\(g(t)\\)\uc758 \ucd5c\ub313\uac12\uc774 4\uac00 \ub418\ub3c4\ub85d \ud558\ub294 \ubaa8\ub4e0 \uc790\uc5f0\uc218 \\(n\\)\uc758 \uac12\uc758 \ud569\uc744 \uad6c\ud558\uc2dc\uc624. [4\uc810]\n","answer":"10","score":4}
+{"id":22,"name":"22","problem":"22. \ucd5c\uace0\ucc28\ud56d\uc758 \uacc4\uc218\uac00 1\uc778 \uc0bc\ucc28\ud568\uc218 \\( f(x) \\)\uc640 \uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \uc5f0\uc18d\uc778 \ud568\uc218 \\( g(x) \\)\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0ac \ub54c, \\( f(4) \\)\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. [4\uc810]\n\\[\n\\begin{aligned}\n\\text{(\uac00)} & \\quad \\text{\ubaa8\ub4e0 \uc2e4\uc218 } x \\text{\uc5d0 \ub300\ud558\uc5ec} \\\\\n& \\quad f(x) = f(1) + (x - 1)f'(g(x)) \\text{\uc774\ub2e4.} \\\\\n\\text{(\ub098)} & \\quad \\text{\ud568\uc218 } g(x) \\text{\uc758 \ucd5c\uc19f\uac12\uc740 } \\frac{5}{2} \\text{\uc774\ub2e4.} \\\\\n\\text{(\ub2e4)} & \\quad f(0) = -3, \\, f(g(1)) = 6\n\\end{aligned}\n\\]\n","answer":"483","score":4}
+{"id":23,"name":"23_prob","problem":"23. \ub2e4\ud56d\uc2dd $(x^3 + 3)^5$ \uc758 \uc804\uac1c\uc2dd\uc5d0\uc11c $x^9$\uc758 \uacc4\uc218\ub294? [2\uc810]\n\\begin{itemize}\n    \\item[1] 30\n    \\item[2] 60\n    \\item[3] 90\n    \\item[4] 120\n    \\item[5] 150\n\\end{itemize}\n","answer":"3","score":2}
+{"id":24,"name":"24_prob","problem":"24. \uc22b\uc790 1, 2, 3, 4, 5 \uc911\uc5d0\uc11c \uc911\ubcf5\uc744 \ud5c8\ub77d\ud558\uc5ec 4\uac1c\ub97c \ud0dd\ud574 \uc77c\ub82c\ub85c \ub098\uc5f4\ud558\uc5ec \ub9cc\ub4e4 \uc218 \uc788\ub294 \ub124 \uc790\ub9ac\uc758 \uc790\uc5f0\uc218 \uc911 4000 \uc774\uc0c1\uc778 \ud640\uc218\uc758 \uac1c\uc218\ub294? [3\uc810]\n\\begin{itemize}\n  \\item[1] 125\n  \\item[2] 150\n  \\item[3] 175\n  \\item[4] 200\n  \\item[5] 225\n\\end{itemize}\n","answer":"4","score":3}
+{"id":25,"name":"25_prob","problem":"25. \ud770\uc0c9 \ub9c8\uc2a4\ud06c 5\uac1c, \uac80\uc740\uc0c9 \ub9c8\uc2a4\ud06c 9\uac1c\uac00 \ub4e4\uc5b4 \uc788\ub294 \uc0c1\uc790\uac00 \uc788\ub2e4. \uc774 \uc0c1\uc790\uc5d0\uc11c \uc784\uc758\ub85c 3\uac1c\uc758 \ub9c8\uc2a4\ud06c\ub97c \ub3d9\uc2dc\uc5d0 \uaebc\ub0bc \ub54c, \uaebc\ub0b8 3\uac1c\uc758 \ub9c8\uc2a4\ud06c \uc911\uc5d0\uc11c \uc801\uc5b4\ub3c4 \ud55c \uac1c\uac00 \ud770\uc0c9 \ub9c8\uc2a4\ud06c\uc77c \ud655\ub960\uc740? [3\uc810]\n\\begin{itemize}\n    \\item[1] $\\frac{8}{13}$\n    \\item[2] $\\frac{17}{26}$\n    \\item[3] $\\frac{9}{13}$\n    \\item[4] $\\frac{19}{26}$\n    \\item[5] $\\frac{10}{13}$\n\\end{itemize}\n","answer":"5","score":3}
+{"id":26,"name":"26_prob","problem":"26. \uc8fc\uba38\ub2c8\uc5d0 1\uc774 \uc801\ud78c \ud770 \uacf5 1\uac1c, 2\uac00 \uc801\ud78c \ud770 \uacf5 1\uac1c, 1\uc774 \uc801\ud78c \uac80\uc740 \uacf5 1\uac1c, 2\uac00 \uc801\ud78c \uac80\uc740 \uacf5 3\uac1c\uac00 \ub4e4\uc5b4 \uc788\ub2e4.  \n\uc774 \uc8fc\uba38\ub2c8\uc5d0\uc11c \uc784\uc758\ub85c 3\uac1c\uc758 \uacf5\uc744 \ub3d9\uc2dc\uc5d0 \uaebc\ub0b4\ub294 \uc2dc\ud589\uc744 \ud55c\ub2e4.  \n\uc774 \uc2dc\ud589\uc5d0\uc11c \uaebc\ub0b8 3\uac1c\uc758 \uacf5 \uc911\uc5d0\uc11c \ud770 \uacf5\uc774 1\uac1c\uc774\uace0 \uac80\uc740 \uacf5\uc774 2\uac1c\uc778 \uc0ac\uac74\uc744 A, \uaebc\ub0b8 3\uac1c\uc758 \uacf5\uc5d0 \uc801\ud600 \uc788\ub294 \uc218\ub97c \ubaa8\ub450 \uacf1\ud55c \uac12\uc774 8\uc778 \uc0ac\uac74\uc744 B\ub77c \ud560 \ub54c, $P(A \\cup B)$\uc758 \uac12\uc740? [3\uc810]\n\\begin{itemize}\n  \\item[1] $\\frac{11}{20}$\n  \\item[2] $\\frac{3}{5}$\n  \\item[3] $\\frac{13}{20}$\n  \\item[4] $\\frac{7}{10}$\n  \\item[5] $\\frac{3}{4}$\n\\end{itemize}\n","answer":"2","score":3}
+{"id":27,"name":"27_prob","problem":"27. \uc5b4\ub290 \ud68c\uc0ac\uc5d0\uc11c \uc0dd\uc0b0\ud558\ub294 \uc0d8\ud50c 1\uac1c\uc758 \uc6a9\ub7c9\uc740 \uc815\uaddc\ubd84\ud3ec \\( N(\\mu, \\sigma^2) \\) \ub97c \ub530\ub978\ub2e4\uace0 \ud55c\ub2e4. \uc774 \ud68c\uc0ac\uc5d0\uc11c \uc0dd\uc0b0\ud558\ub294 \uc0d8\ud50c \uc911\uc5d0\uc11c 16\uac1c\ub97c \uc784\uc758\ucd94\ucd9c\ud558\uc5ec \uc5bb\uc740 \ud45c\ubcf8\ud3c9\uade0\uc744 \uc774\uc6a9\ud558\uc5ec \uad6c\ud55c \\( m \\) \uc5d0 \ub300\ud55c \uc2e0\ub8b0\ub3c4 95\\%\uc758 \uc2e0\ub8b0\uad6c\uac04\uc774 \\( 746.1 \\leq m \\leq 755.9 \\)\uc774\ub2e4. \uc774 \ud68c\uc0ac\uc5d0\uc11c \uc0dd\uc0b0\ud558\ub294 \uc0d8\ud50c \uc911\uc5d0\uc11c \\( n \\) \uac1c\ub97c \uc784\uc758\ucd94\ucd9c\ud558\uc5ec \uc5bb\uc740 \ud45c\ubcf8\ud3c9\uade0\uc744 \uc774\uc6a9\ud558\uc5ec \uad6c\ud558\ub294 \\( m \\) \uc5d0 \ub300\ud55c \uc2e0\ub8b0\ub3c4 99\\%\uc758 \uc2e0\ub8b0\uad6c\uac04\uc774 \\( a \\leq m \\leq b \\)\uc77c \ub54c, \\( b-a \\)\uc758 \uac12\uc774 6 \uc774\ud558\uac00 \ub418\uae30 \uc704\ud55c \uc790\uc5f0\uc218 \\( n \\)\uc758 \ucd5c\uc18c\uac12\uc740? (\ub2e8, \uc6a9\ub7c9\uc758 \ub2e8\uc704\ub294 mL\uc774\uace0, \\( Z \\)\uac00 \ud45c\uc900\uc815\uaddc\ubd84\ud3ec\ub97c \ub530\ub974\ub294 \ud655\ub960\ubcc0\uc218\uc77c \ub54c, \\( P(|Z| \\leq 1.96) = 0.95, P(|Z| \\leq 2.58) = 0.99 \\) \ub85c \uacc4\uc0b0\ud55c\ub2e4.) [3\uc810]\n\\begin{itemize}\n    \\item[1] 70\n    \\item[2] 74\n    \\item[3] 78\n    \\item[4] 82\n    \\item[5] 86\n\\end{itemize}\n","answer":"2","score":3}
+{"id":28,"name":"28_prob","problem":"28. \uc5f0\uc18d\ud655\ub960\ubcc0\uc218 \\( X \\) \uac00 \uac16\ub294 \uac12\uc758 \ubc94\uc704\ub294 \\( 0 \\leq X \\leq a \\) \uc774\uace0, \\( X \\)\uc758 \ud655\ub960\ubc00\ub3c4\ud568\uc218\uc758 \uadf8\ub798\ud504\uac00 \uadf8\ub9bc\uacfc \uac19\ub2e4.\\\\\n\\begin{center}\n\\begin{tikzpicture}\n    % Draw axes\n    \\draw[->] (0,0) -- (5,0) node[right] {$x$};\n    \\draw[->] (0,0) -- (0,4) node[above] {$y$};\n    % Label points\n    \\node at (1,-0.3) {$O$};\n    \\node at (3,-0.3) {$b$};\n    \\node at (5,-0.3) {$a$};\n    \\node at (-0.3,3) {$c$};\n    % Draw the function\n    \\draw[thick] (0,0) -- (3,3) -- (5,0);\n    % Dotted lines for the heights\n    \\draw[dashed] (3,0) -- (3,3);\n    \\draw[dashed] (5,0) -- (5,0);\n\\end{tikzpicture}\n\\end{center}\n\\( P(X \\leq b) - P(X \\geq b) = \\frac{1}{4}, \\quad P(X \\leq \\sqrt{5}) = \\frac{1}{2} \\)\uc77c \ub54c,\\\\\n\\( a + b + c \\)\uc758 \uac12\uc740? (\ub2e8, \\(a, b, c\\)\ub294 \uc0c1\uc218\uc774\ub2e4.) [4\uc810] \n\\begin{itemize}\n  \\item[1] \\(\\frac{11}{2}\\)\n  \\item[2] 6\n  \\item[3] \\(\\frac{13}{2}\\)\n  \\item[4] 7\n  \\item[5] \\(\\frac{15}{2}\\)\n\\end{itemize}\n","answer":"4","score":4}
+{"id":29,"name":"29_prob","problem":"29. \uc55e\uba74\uc5d0\ub294 1\ubd80\ud130 6\uae4c\uc9c0\uc758 \uc790\uc5f0\uc218\uac00 \ud558\ub098\uc529 \uc801\ud600 \uc788\uace0 \ub4b7\uba74\uc5d0\ub294 \ubaa8\ub450 0\uc774 \ud558\ub098\uc529 \uc801\ud600 \uc788\ub294 6\uc7a5\uc758 \uce74\ub4dc\uac00 \uc788\ub2e4. \uc774 6\uc7a5\uc758 \uce74\ub4dc\ub97c \uadf8\ub9bc\uacfc \uac19\uc774 6 \uc774\ud558\uc758 \uc790\uc5f0\uc218 $k$\uc5d0 \ub300\ud558\uc5ec $k$\ubc88\uc9f8 \uc790\ub9ac\uc5d0 \uc790\uc5f0\uc218 $k$\uac00 \ubcf4\uc774\ub3c4\ub85d \ub193\uc5ec \uc788\ub2e4. \\\\\n\\[\n\\begin{array}{|c|c|c|c|c|c|}\n\\hline\n\\text{1\ubc88\uc9f8 \uc790\ub9ac} & \\text{2\ubc88\uc9f8 \uc790\ub9ac} & \\text{3\ubc88\uc9f8 \uc790\ub9ac} & \\text{4\ubc88\uc9f8 \uc790\ub9ac} & \\text{5\ubc88\uc9f8 \uc790\ub9ac} & \\text{6\ubc88\uc9f8 \uc790\ub9ac} \\\\\n\\hline\n1 & 2 & 3 & 4 & 5 & 6 \\\\\n\\hline\n\\end{array}\n\\]\n\uc774 6\uc7a5\uc758 \uce74\ub4dc\uc640 \ud55c \uac1c\uc758 \uc8fc\uc0ac\uc704\ub97c \uc0ac\uc6a9\ud558\uc5ec \ub2e4\uc74c \uc2dc\ud589\uc744 \ud55c\ub2e4. \\\\\n\\framebox{\n\\parbox{\\textwidth}{\n\uc8fc\uc0ac\uc704\ub97c \ud55c \ubc88 \ub358\uc838 \ub098\uc628 \ub208\uc758 \uc218\uac00 $k$\uc774\uba74 $k$\ubc88\uc9f8 \uc790\ub9ac\uc5d0 \ub193\uc5ec \uc788\ub294 \uce74\ub4dc\ub97c \ud55c \ubc88 \ub4a4\uc9d1\uc5b4 \uc81c\uc790\ub9ac\uc5d0 \ub193\ub294\ub2e4.\n}\n} \\\\\n\uc704\uc758 \uc2dc\ud589\uc744 3\ubc88 \ubc18\ubcf5\ud55c \ud6c4 6\uc7a5\uc758 \uce74\ub4dc\uc5d0 \ubcf4\uc774\ub294 \ubaa8\ub4e0 \uc218\uc758 \ud569\uc774 \uc9dd\uc218\uc77c \ub54c, \uc8fc\uc0ac\uc704\uc758 1\uc758 \ub208\uc774 \ud55c \ubc88\ub9cc \ub098\uc654\uc744 \ud655\ub960\uc744 $\\frac{p}{q}$\uc774\ub2e4. $p+q$\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. (\ub2e8, $p$\uc640 $q$\ub294 \uc11c\ub85c\uc18c\uc778 \uc790\uc5f0\uc218\uc774\ub2e4.) [4\uc810]\n","answer":"196","score":4}
+{"id":30,"name":"30_prob","problem":"30. \uc9d1\ud569 $X=\\{x \\mid x \\text{\ub294 10 \uc774\ud558\uc758 \uc790\uc5f0\uc218}\\}$\uc5d0 \ub300\ud558\uc5ec \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a4\ub294 \ud568\uc218 $f: X \\rightarrow X$\uc758 \uac1c\uc218\ub97c \uad6c\ud558\uc2dc\uc624. [4\uc810]\n\\begin{quote}\n\\textbf{(\uac00)} 9 \uc774\ud558\uc758 \ubaa8\ub4e0 \uc790\uc5f0\uc218 $x$\uc5d0 \ub300\ud558\uc5ec $f(x) \\leq f(x+1)$ \uc774\ub2e4.\n\\textbf{(\ub098)} $1 \\leq x \\leq 5$\uc77c \ub54c $f(x) \\leq x$\uc774\uace0, \\\\\n$6 \\leq x \\leq 10$\uc77c \ub54c $f(x) \\geq x$\uc774\ub2e4.\n\\textbf{(\ub2e4)} $f(6) = f(5) + 6$\n\\end{quote}\n","answer":"673","score":4}
+{"id":31,"name":"23_calc","problem":"23. \\lim_{x \\to 0} \\frac{\\ln(x+1)}{\\sqrt{x+4} - 2} \\text{\uc758 \uac12\uc740? [2\uc810]}\n\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n","answer":"3","score":2}
+{"id":32,"name":"24_calc","problem":"24. \\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=1}^{n} \\sqrt{1 + \\frac{3k}{n}} \\text{\uc758 \uac12\uc740? [3\uc810]}\n\n\\begin{itemize}\n    \\item[1] $\\frac{4}{3}$\n    \\item[2] $\\frac{13}{9}$\n    \\item[3] $\\frac{14}{9}$\n    \\item[4] $\\frac{5}{3}$\n    \\item[5] $\\frac{16}{9}$\n\\end{itemize}\n","answer":"2","score":3}
+{"id":33,"name":"25_calc","problem":"25. \ub4f1\ube44\uc218\uc5f4 $\\{a_n\\}$\uc5d0 \ub300\ud558\uc5ec $\\lim_{n \\to \\infty} \\frac{a_{n+1}}{3^n + 2^{2n-1}} = 3$\uc77c \ub54c, $a_2$\uc758 \uac12\uc740? \\hspace{3mm}[3\uc810]\n\\begin{itemize}\n    \\item[1] 16\n    \\item[2] 18\n    \\item[3] 20\n    \\item[4] 22\n    \\item[5] 24\n\\end{itemize}\n","answer":"4","score":3}
+{"id":34,"name":"26_calc","problem":"26. \uadf8\ub9bc\uacfc \uac19\uc774 \uace1\uc120 $y=\\sqrt{\\sec^2x + \\tan x} \\ \\left( 0 \\leq x \\leq \\frac{\\pi}{3} \\right)$ \uc640 $x$\ucd95, $y$\ucd95 \ubc0f \uc9c1\uc120 $x=\\frac{\\pi}{3}$\ub85c \ub458\ub7ec\uc2f8\uc778 \ubd80\ubd84\uc744 \ubc11\uba74\uc73c\ub85c \ud558\ub294 \uc785\uccb4\ub3c4\ud615\uc774 \uc788\ub2e4. \uc774 \uc785\uccb4\ub3c4\ud615\uc744 $x$\ucd95\uc5d0 \uc218\uc9c1\uc778 \ud3c9\uba74\uc73c\ub85c \uc790\ub978 \ub2e8\uba74\uc774 \ubaa8\ub450 \uc815\uc0ac\uac01\ud615\uc77c \ub54c, \uc774 \uc785\uccb4\ub3c4\ud615\uc758 \ubd80\ud53c\ub294? [3\uc810]\n\\begin{itemize}\n    \\item[1] $\\frac{\\sqrt{3}}{2} + \\frac{\\ln 2}{2}$\n    \\item[2] $\\frac{\\sqrt{3}}{2} + \\ln 2$\n    \\item[3] $\\sqrt{3} + \\frac{\\ln 2}{2}$\n    \\item[4] $\\sqrt{3} + \\ln 2$\n    \\item[5] $\\frac{\\sqrt{3}}{2} + 2 \\ln 2$\n\\end{itemize}\n","answer":"3","score":3}
+{"id":35,"name":"27_calc","problem":"27. \uadf8\ub9bc\uacfc \uac19\uc774 \uc911\uc2ec\uc774 $O$, \ubc18\uc9c0\ub984\uc758 \uae38\uc774\uac00 $1$\uc774\uace0 \uc911\uc2ec\uac01\uc758 \ud06c\uae30\uac00 $\\frac{\\pi}{2}$\uc778 \ubd80\ucc44\uaf34 $OA_1B_1$\uc774 \uc788\ub2e4. \ud638 $A_1B_1$ \uc704\uc5d0 \uc810 $P_1$, \uc120\ubd84 $OA_1$ \uc704\uc5d0 \uc810 $C_1$, \uc120\ubd84 $OB_1$ \uc704\uc5d0 \uc810 $D_1$\uc744 \uc0ac\uac01\ud615 $OC_1P_1D_1$\uc774 $OC_1 : OD_1 = 3:4$\uc778 \uc9c1\uc0ac\uac01\ud615\uc774 \ub418\ub3c4\ub85d \uc7a1\ub294\ub2e4.\n\ubd80\ucc44\uaf34 $OA_1B_1$\uc758 \ub0b4\ubd80\uc5d0 \uc810 $Q_1$\uc744 $P_1Q_1 = A_1Q_1$, $\\angle P_1Q_1A_1 = \\frac{\\pi}{2}$\uac00 \ub418\ub3c4\ub85d \uc7a1\uace0, \uc774\ub4f1\ubcc0\uc0bc\uac01\ud615 $P_1Q_1A_1$\uc5d0 \uc0c9\uce60\ud558\uc5ec \uc5bb\uc740 \uadf8\ub9bc\uc744 $R_1$\uc774\ub77c \ud558\uc790.\n\uadf8\ub9bc $R_1$\uc5d0\uc11c \uc120\ubd84 $OA_1$ \uc704\uc758 \uc810 $A_2$\uc640 \uc120\ubd84 $OB_1$ \uc704\uc758 \uc810 $B_2$\ub97c $OQ_1 = OA_2 = OB_2$\uac00 \ub418\ub3c4\ub85d \uc7a1\uace0, \uc911\uc2ec\uc774 $O$, \ubc18\uc9c0\ub984\uc758 \uae38\uc774\uac00 $OQ_1$, \uc911\uc2ec\uac01\uc758 \ud06c\uae30\uac00 $\\frac{\\pi}{2}$\uc778 \ubd80\ucc44\uaf34 $OA_2B_2$\ub97c \uadf8\ub9b0\ub2e4. \uadf8\ub9bc $R_1$\uc744 \uc5bb\uc740 \uac83\uacfc \uac19\uc740 \ubc29\ubc95\uc73c\ub85c \ub124 \uc810 $P_2, C_2, D_2, Q_2$\ub97c \uc7a1\uace0, \uc774\ub4f1\ubcc0\uc0bc\uac01\ud615 $P_2Q_2A_2$\uc5d0 \uc0c9\uce60\ud558\uc5ec \uc5bb\uc740 \uadf8\ub9bc\uc744 $R_2$\ub77c \ud558\uc790.\n\uc774\uc640 \uac19\uc740 \uacfc\uc815\uc744 \uacc4\uc18d\ud558\uc5ec $n$\ubc88\uc9f8 \uc5bb\uc740 \uadf8\ub9bc $R_n$\uc5d0 \uc0c9\uce60\ub418\uc5b4 \uc788\ub294 \ubd80\ubd84\uc758 \ub113\uc774\ub97c $S_n$\uc774\ub77c \ud560 \ub54c, $\\lim_{n \\to \\infty} S_n$\uc758 \uac12\uc740? [3\uc810]\n\\begin{itemize}\n    \\item[1] $\\frac{9}{40}$\n    \\item[2] $\\frac{1}{4}$\n    \\item[3] $\\frac{11}{40}$\n    \\item[4] $\\frac{3}{10}$\n    \\item[5] $\\frac{13}{40}$\n\\end{itemize}\n","answer":"1","score":3}
+{"id":36,"name":"28_calc","problem":"28. \uadf8\ub9bc\uacfc \uac19\uc774 \uc911\uc2ec\uc774 $O$\uc774\uace0 \uae38\uc774\uac00 2\uc778 \uc120\ubd84 $AB$\ub97c \uc9c0\ub984\uc73c\ub85c \ud558\ub294 \ubc18\uc6d0 \uc704\uc5d0 $\\angle AOC = \\frac{\\pi}{2}$\uc778 \uc810 $C$\uac00 \uc788\ub2e4. \ud638 $BC$ \uc704\uc5d0 \uc810 $P$\uc640 \ud638 $CA$ \uc704\uc5d0 \uc810 $Q$\ub97c $PB = QC$\uac00 \ub418\ub3c4\ub85d \uc7a1\uace0, \uc120\ubd84 $AP$ \uc704\uc5d0 \uc810 $R$\uc744 $\\angle CQR = \\frac{\\pi}{2}$\uac00 \ub418\ub3c4\ub85d \uc7a1\ub294\ub2e4.\\\\\n\uc120\ubd84 $AP$\uc640 \uc120\ubd84 $CO$\uc758 \uad50\uc810\uc744 $S$\ub77c \ud558\uc790. $\\angle PAB = \\theta$\uc77c \ub54c, \uc0bc\uac01\ud615 $POB$\uc758 \ub113\uc774\ub97c $f(\\theta)$, \uc0ac\uac01\ud615 $CQRS$\uc758 \ub113\uc774\ub97c $g(\\theta)$\ub77c \ud558\uc790. \\\\\n\\[\n\\lim_{\\theta \\to 0^{+}} \\frac{3f(\\theta) - 2g(\\theta)}{\\theta^2}\n\\]\n\uc758 \uac12\uc740? (\ub2e8, $0 < \\theta < \\frac{\\pi}{4}$) [4\uc810] \n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n","answer":"2","score":4}
+{"id":37,"name":"29_calc","problem":"29. \uc138 \uc0c1\uc218 \\(a, b, c\\)\uc5d0 \ub300\ud558\uc5ec \ud568\uc218 \\(f(x) = ae^{2x} + be^x + c\\)\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a8\ub2e4.\n\\[\n(\uac00)\\ \\lim_{x \\to -\\infty} \\frac{f(x) + 6}{e^x} = 1\n\\]\n\\[\n(\ub098)\\ f(\\ln 2) = 0\n\\]\n\ud568\uc218 \\(f(x)\\)\uc758 \uc5ed\ud568\uc218\ub97c \\(g(x)\\)\ub77c \ud560 \ub54c,\n\\[\n\\int_0^{14} g(x) dx = p + q \\ln 2 \uc774\ub2e4. \\ p+q\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624.\n\\]\n(\ub2e8, \\(p, q\\)\ub294 \uc720\ub9ac\uc218\uc774\uace0, \\(\\ln 2\\)\ub294 \ubb34\ub9ac\uc218\uc774\ub2e4.) [4\uc810]\n","answer":"162","score":4}
+{"id":38,"name":"30_calc","problem":"30. \ucd5c\uace0\ucc28\ud56d\uc758 \uacc4\uc218\uac00 \uc591\uc218\uc778 \uc0bc\ucc28\ud568\uc218 $f(x)$\uc640\\\\\n\ud568\uc218 $g(x) = e^{\\sin \\pi x} - 1$\uc5d0 \ub300\ud558\uc5ec \uc2e4\uc218 \uc804\uccb4\uc758 \uc9d1\ud569\uc5d0\uc11c \uc815\uc758\ub41c \ud569\uc131\ud568\uc218 $h(x) = g(f(x))$\uac00 \ub2e4\uc74c \uc870\uac74\uc744 \ub9cc\uc871\uc2dc\ud0a8\ub2e4.\n\\begin{itemize}\n    \\item[(\uac00)] \ud568\uc218 $h(x)$\ub294 $x = 0$\uc5d0\uc11c \uadf9\ub313\uac12 $0$\uc744 \uac16\ub294\ub2e4.\n    \\item[(\ub098)] \uc5f4\ub9b0\uad6c\uac04 $(0, 3)$\uc5d0\uc11c \ubc29\uc815\uc2dd $h(x) = 1$\uc758 \uc11c\ub85c \ub2e4\ub978 \uc2e4\uadfc\uc758 \uac1c\uc218\ub294 7\uc774\ub2e4.\n\\end{itemize}\n$f(3) = \\frac{1}{2}, f'(3) = 0$\uc77c \ub54c, $f(2) = \\frac{q}{p}$\uc774\ub2e4. $p + q$\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. (\ub2e8, $p$\uc640 $q$\ub294 \uc11c\ub85c\uc18c\uc778 \uc790\uc5f0\uc218\uc774\ub2e4.) [4\uc810]\n","answer":"125","score":4}
+{"id":39,"name":"23_geom","problem":"23. \uc88c\ud45c\uacf5\uac04\uc758 \uc810 A(2, 2, -1)\uc744 \\(x\\)\ucd95\uc5d0 \ub300\ud558\uc5ec \ub300\uce6d\uc774\ub3d9\ud55c \uc810\uc744 B\ub77c \ud558\uc790. \uc810 C(-2, 1, 1)\uc5d0 \ub300\ud558\uc5ec \uc120\ubd84 BC\uc758 \uae38\uc774\ub294? \\hfill [2\uc810]\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n","answer":"4","score":2}
+{"id":40,"name":"24_geom","problem":"24. \ucd08\uc810\uc774 $F\\left(\\frac{1}{3}, 0\\right)$\uc774\uace0 \uc900\uc120\uc774 $x = -\\frac{1}{3}$\uc778 \ud3ec\ubb3c\uc120\uc774 \uc810 $(a, 2)$\ub97c \uc9c0\ub0a0 \ub54c, $a$\uc758 \uac12\uc740? [3\uc810]\n\\begin{itemize}\n    \\item[1] 1\n    \\item[2] 2\n    \\item[3] 3\n    \\item[4] 4\n    \\item[5] 5\n\\end{itemize}\n","answer":"3","score":3}
+{"id":41,"name":"25_geom","problem":"25. \ud0c0\uc6d0 $\\dfrac{x^2}{a^2} + \\dfrac{y^2}{b^2} = 1$ \uc704\uc758 \uc810 $(2, 1)$\uc5d0\uc11c\uc758 \uc811\uc120\uc758 \uae30\uc6b8\uae30\uac00 $-\\dfrac{1}{2}$\uc77c \ub54c, \uc774 \ud0c0\uc6d0\uc758 \ub450 \ucd08\uc810 \uc0ac\uc774\uc758 \uac70\ub9ac\ub294?\\\\\n(\ub2e8, $a$, $b$\ub294 \uc591\uc218\uc774\ub2e4.) [3\uc810]\n\\begin{itemize}\n    \\item[1] $2 \\sqrt{3}$\n    \\item[2] $4$\n    \\item[3] $2 \\sqrt{5}$\n    \\item[4] $2 \\sqrt{6}$\n    \\item[5] $2 \\sqrt{7}$\n\\end{itemize}\n","answer":"2","score":3}
+{"id":42,"name":"26_geom","problem":"26. \uc88c\ud45c\ud3c9\uba74\uc5d0\uc11c \uc138 \ubca1\ud130\n\\[\n\\vec{a} = (2, 4), \\quad \\vec{b} = (2, 8), \\quad \\vec{c} = (1, 0)\n\\]\n\uc5d0 \ub300\ud558\uc5ec \ub450 \ubca1\ud130 \\(\\vec{p}, \\vec{q}\\)\uac00\n\\[\n(\\vec{p} - \\vec{a}) \\cdot (\\vec{p} - \\vec{b}) = 0, \\quad \\vec{q} = \\frac{1}{2} \\vec{a} + t \\vec{c} \\quad (t\ub294 \\, \uc2e4\uc218)\n\\]\n\ub97c \ub9cc\uc871\uc2dc\ud0ac \ub54c, \\(\\left| \\vec{p} - \\vec{q} \\right|\\)\uc758 \ucd5c\uc18c\uac12\uc740? [3\uc810]\n\\begin{itemize}\n    \\item[1] \\(\\frac{3}{2}\\)\n    \\item[2] 2\n    \\item[3] \\(\\frac{5}{2}\\)\n    \\item[4] 3\n    \\item[5] \\(\\frac{7}{2}\\)\n\\end{itemize}\n","answer":"5","score":3}
+{"id":43,"name":"27_geom","problem":"27. \uc88c\ud45c\uacf5\uac04\uc5d0 \uc9c1\uc120 AB\ub97c \ud3ec\ud568\ud558\ub294 \ud3c9\uba74 $\\alpha$\uac00 \uc788\ub2e4. \ud3c9\uba74 $\\alpha$ \uc704\uc5d0 \uc788\uc9c0 \uc54a\uc740 \uc810 C\uc5d0 \ub300\ud558\uc5ec \uc9c1\uc120 AB\uc640 \uc9c1\uc120 AC\uac00 \uc774\ub8e8\ub294 \uc608\uac01\uc758 \ud06c\uae30\ub97c $\\theta_1$\uc774\ub77c \ud560 \ub54c $\\sin \\theta_1 = \\frac{4}{5}$\uc774\uace0, \uc9c1\uc120 AC\uc640 \ud3c9\uba74 $\\alpha$\uac00 \uc774\ub8e8\ub294 \uc608\uac01\uc758 \ud06c\uae30\ub294 $\\frac{\\pi}{2} - \\theta_1$\uc774\ub2e4. \ud3c9\uba74 ABC\uc640 \ud3c9\uba74 $\\alpha$\uac00 \uc774\ub8e8\ub294 \uc608\uac01\uc758 \ud06c\uae30\ub97c $\\theta_2$\ub77c \ud560 \ub54c, $\\cos \\theta_2$\uc758 \uac12\uc740? [3\uc810]\n\\begin{itemize}\n    \\item[1] $\\frac{\\sqrt{7}}{4}$\n    \\item[2] $\\frac{\\sqrt{7}}{5}$\n    \\item[3] $\\frac{\\sqrt{7}}{6}$\n    \\item[4] $\\frac{\\sqrt{7}}{7}$\n    \\item[5] $\\frac{\\sqrt{7}}{8}$\n\\end{itemize}\n","answer":"3","score":3}
+{"id":44,"name":"28_geom","problem":"28. \ub450 \ucd08\uc810\uc774 $F(c,0), F'(-c,0)(c>0)$\uc778 \uc30d\uace1\uc120 $C$\uc640 y\ucd95 \uc704\uc758 \uc810 $A$\uac00 \uc788\ub2e4. \uc30d\uace1\uc120 $C$\uac00 \uc120\ubd84 $AF$\uc640 \ub9cc\ub098\ub294 \uc810\uc744 $P$, \uc120\ubd84 $AF'$\uacfc \ub9cc\ub098\ub294 \uc810\uc744 $P'$\uc774\ub77c \ud558\uc790. \\\\\n\uc9c1\uc120 $AF$\ub294 \uc30d\uace1\uc120 $C$\uc758 \ud55c \uc810\uadfc\uc120\uacfc \ud3c9\ud589\ud558\uace0 \\\\\n\\[\n\\frac{AP}{PP'} = \\frac{5}{6}, \\quad PF = 1\n\\]\n\uc77c \ub54c, \uc30d\uace1\uc120 $C$\uc758 \uc8fc\ucd95\uc758 \uae38\uc774\ub294? \\textbf{[4\uc810]} \\\\\n\\begin{itemize}\n    \\item[1] $\\frac{13}{6}$\n    \\item[2] $9\/4$\n    \\item[3] $7\/3$\n    \\item[4] $\\frac{29}{12}$\n    \\item[5] $\\frac{5}{2}$\n\\end{itemize}\n","answer":"5","score":4}
+{"id":45,"name":"29_geom","problem":"29.\\ \ud3c9\uba74\\ \\(\\alpha\\) \uc704\uc5d0\\ \\(\\overline{AB} = \\overline{CD} = \\overline{AD} = 2\\),\\ \\(\\angle ABC = \\angle BCD = \\frac{\\pi}{3}\\)\\ \uc778\\ \uc0ac\ub2e4\ub9ac\uaf34\\ \\(ABCD\\)\\ \uac00\\ \uc788\ub2e4.\\ \ub2e4\uc74c\\ \uc870\uac74\uc744\\ \ub9cc\uc871\uc2dc\ud0a4\ub294\\ \ud3c9\uba74\\ \\(\\alpha\\) \uc704\uc758\\ \ub450\\ \uc810\\ \\(P, Q\\)\uc5d0\\ \ub300\ud558\uc5ec\\ \\(CP \\cdot DQ\\)\uc758\\ \uac12\uc744\\ \uad6c\ud558\uc2dc\uc624.\\ [4\uc810]\n\\begin{itemize}\n    \\item[(\uac00)] \\(\\overrightarrow{AC} = 2(\\overrightarrow{AD} + \\overrightarrow{BP})\\)\n    \\item[(\ub098)] \\(\\overrightarrow{AC} \\cdot \\overrightarrow{PQ} = 6\\)\n    \\item[(\ub2e4)] \\(2 \\times \\angle BQA = \\angle PBQ < \\frac{\\pi}{2}\\)\n\\end{itemize}\n\\begin{center}\n\\begin{tikzpicture}\n    \\draw (0,0) -- (2,0) -- (2.5,1.5) -- (-0.5,1.5) -- cycle;\n    \\node[below] at (0,0) {B};\n    \\node[below] at (2,0) {C};\n    \\node[above] at (2.5,1.5) {D};\n    \\node[above] at (-0.5,1.5) {A};\n\\end{tikzpicture}\n\\end{center}\n","answer":"11","score":4}
+{"id":46,"name":"30_geom","problem":"30. \uc88c\ud45c\uacf5\uac04\uc5d0 \uc815\uc0ac\uba74\uccb4 $ABCD$ \uac00 \uc788\ub2e4. \uc815\uc0bc\uac01\ud615 $BCD$ \uc758 \uc678\uc2ec\uc744 \uc911\uc2ec\uc73c\ub85c \ud558\uace0 \uc810 $B$\ub97c \uc9c0\ub098\ub294 \uad6c\ub97c $S$\ub77c \ud558\uc790. \\\\\n\uad6c $S$\uc640 \uc120\ubd84 $AB$\uac00 \ub9cc\ub098\ub294 \uc810 \uc911 $B$\uac00 \uc544\ub2cc \uc810\uc744 $P$, \\\\\n\uad6c $S$\uc640 \uc120\ubd84 $AC$\uac00 \ub9cc\ub098\ub294 \uc810 \uc911 $C$\uac00 \uc544\ub2cc \uc810\uc744 $Q$, \\\\\n\uad6c $S$\uc640 \uc120\ubd84 $AD$\uac00 \ub9cc\ub098\ub294 \uc810 \uc911 $D$\uac00 \uc544\ub2cc \uc810\uc744 $R$ \ud558\uace0, \\\\\n\uc810 $P$\uc5d0\uc11c \uad6c $S$\uc5d0 \uc811\ud558\ub294 \ud3c9\uba74\uc744 $\\alpha$\ub77c \ud558\uc790. \\\\\n\uad6c $S$\uc758 \ubc18\uc9c0\ub984\uc758 \uae38\uc774\uac00 6\uc77c \ub54c, \uc0bc\uac01\ud615 $PQR$\uc758 \ud3c9\uba74 $\\alpha$ \uc704\ub85c\uc758 \uc815\uc0ac\uc601\uc758 \ub113\uc774\ub294 $k$\uc774\ub2e4. $k^2$\uc758 \uac12\uc744 \uad6c\ud558\uc2dc\uc624. \\hfill [4\uc810]\n","answer":"147","score":4}

test_dataset.ipynb DELETED Viewed

@@ -1,104 +0,0 @@
-{
- "cells": [
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "from datasets import load_dataset\n",
-    "\n",
-    "ds = load_dataset(\"cfpark00/KoreanSAT\")"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "DatasetDict({\n",
-       "    2023_math: Dataset({\n",
-       "        features: ['id', 'name', 'problem', 'answer', 'score', 'review'],\n",
-       "        num_rows: 46\n",
-       "    })\n",
-       "    2024_math: Dataset({\n",
-       "        features: ['id', 'name', 'problem', 'answer', 'score', 'review'],\n",
-       "        num_rows: 46\n",
-       "    })\n",
-       "})"
-      ]
-     },
-     "execution_count": 9,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "ds"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "{'id': 31,\n",
-       " 'name': '23_calc',\n",
-       " 'problem': '23. \\\\lim_{x \\\\to 0} \\\\frac{\\\\ln(x+1)}{\\\\sqrt{x+4} - 2} \\\\text{의 값은? [2점]}\\n\\n\\\\begin{itemize}\\n    \\\\item[1] 1\\n    \\\\item[2] 2\\n    \\\\item[3] 3\\n    \\\\item[4] 4\\n    \\\\item[5] 5\\n\\\\end{itemize}\\n',\n",
-       " 'answer': 3,\n",
-       " 'score': 2,\n",
-       " 'review': None}"
-      ]
-     },
-     "execution_count": 13,
-     "metadata": {},
-     "output_type": "execute_result"
-    }
-   ],
-   "source": [
-    "ds[\"2024_math\"][30]"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
-  }
- ],
- "metadata": {
-  "kernelspec": {
-   "display_name": "venv1",
-   "language": "python",
-   "name": "python3"
-  },
-  "language_info": {
-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 3
-   },
-   "file_extension": ".py",
-   "mimetype": "text/x-python",
-   "name": "python",
-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython3",
-   "version": "3.10.9"
-  }
- },
- "nbformat": 4,
- "nbformat_minor": 2
-}

to_parquet.ipynb CHANGED Viewed

@@ -2,16 +2,25 @@
  "cells": [
   {
    "cell_type": "code",
    "execution_count": 19,
    "metadata": {},
    "outputs": [],
    "source": [
     "from datasets import load_dataset\n",
     "import json"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": 21,
    "metadata": {},
    "outputs": [],
@@ -32,11 +41,15 @@
   {
    "cell_type": "code",
    "execution_count": 22,
    "metadata": {},
    "outputs": [
     {
      "data": {
       "application/vnd.jupyter.widget-view+json": {
        "model_id": "462231fd51404fd290bffe90a8c0b158",
        "version_major": 2,
        "version_minor": 0
@@ -80,6 +93,9 @@
      "data": {
       "application/vnd.jupyter.widget-view+json": {
        "model_id": "3598e1ab98304d5a98a1cd7c322c7fa6",
        "version_major": 2,
        "version_minor": 0
       },
@@ -89,16 +105,140 @@
      },
      "metadata": {},
      "output_type": "display_data"
     }
    ],
    "source": [
     "dataset=load_dataset(\"json\",data_files={\"2022_math\":\"./data/json/2022/math.json\",\n",
     "                                        \"2023_math\":\"./data/json/2023/math.json\"})"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": 25,
    "metadata": {},
    "outputs": [
     {
@@ -106,6 +246,7 @@
       "text/plain": [
        "{'id': 1,\n",
        " 'name': '1',\n",
        " 'problem': '1. $\\\\left(2^{\\\\sqrt{3}} \\\\times 4\\\\right)^{\\\\sqrt{3} - 2}$ 의 값은? [2점] \\\\begin{itemize} \\\\item[1] \\\\frac{1}{4} \\\\item[2] \\\\frac{1}{2} \\\\item[3] 1 \\\\item[4] 2 \\\\item[5] 4 \\\\end{itemize}',\n",
        " 'answer': 2,\n",
        " 'score': 2,\n",
@@ -114,28 +255,73 @@
       ]
      },
      "execution_count": 25,
      "metadata": {},
      "output_type": "execute_result"
     }
    ],
    "source": [
     "dataset[\"2022_math\"][0]"
    ]
   },
   {
    "cell_type": "code",
    "execution_count": 24,
    "metadata": {},
    "outputs": [
     {
      "data": {
       "application/vnd.jupyter.widget-view+json": {
        "model_id": "f612f040825040c6affa3b897f2633ef",
        "version_major": 2,
        "version_minor": 0
       },
       "text/plain": [
        "Pushing dataset shards to the dataset hub:   0%|          | 0/1 [00:00<?, ?it/s]"
       ]
      },
      "metadata": {},
@@ -143,6 +329,7 @@
     },
     {
      "data": {
       "application/vnd.jupyter.widget-view+json": {
        "model_id": "f3cb1129210b4586bbb81b52357c424e",
        "version_major": 2,
@@ -215,6 +402,49 @@
    "source": [
     "#save as parquet\n",
     "dataset.push_to_hub(\"cfpark00/KoreanSAT\")"
    ]
   },
   {
@@ -224,6 +454,113 @@
    "outputs": [],
    "source": []
   },
   {
    "cell_type": "code",
    "execution_count": null,
@@ -248,7 +585,11 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
    "version": "3.10.9"
   }
  },
  "nbformat": 4,

  "cells": [
   {
    "cell_type": "code",
+<<<<<<< HEAD
    "execution_count": 19,
    "metadata": {},
    "outputs": [],
    "source": [
     "from datasets import load_dataset\n",
     "import json"
+=======
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from datasets import load_dataset"
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
    ]
   },
   {
    "cell_type": "code",
+<<<<<<< HEAD
    "execution_count": 21,
    "metadata": {},
    "outputs": [],
   {
    "cell_type": "code",
    "execution_count": 22,
+=======
+   "execution_count": 4,
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
    "metadata": {},
    "outputs": [
     {
      "data": {
       "application/vnd.jupyter.widget-view+json": {
+<<<<<<< HEAD
        "model_id": "462231fd51404fd290bffe90a8c0b158",
        "version_major": 2,
        "version_minor": 0
      "data": {
       "application/vnd.jupyter.widget-view+json": {
        "model_id": "3598e1ab98304d5a98a1cd7c322c7fa6",
+=======
+       "model_id": "b5d51386a14e407ca2bdd18926ef997c",
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
        "version_major": 2,
        "version_minor": 0
       },
      },
      "metadata": {},
      "output_type": "display_data"
+<<<<<<< HEAD
     }
    ],
    "source": [
     "dataset=load_dataset(\"json\",data_files={\"2022_math\":\"./data/json/2022/math.json\",\n",
     "                                        \"2023_math\":\"./data/json/2023/math.json\"})"
+=======
+    },
+    {
+     "data": {
+      "application/vnd.jupyter.widget-view+json": {
+       "model_id": "a3015bdd603e44c789b8347863bb94c8",
+       "version_major": 2,
+       "version_minor": 0
+      },
+      "text/plain": [
+       "Generating 2024_math split: 0 examples [00:00, ? examples/s]"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    }
+   ],
+   "source": [
+    "dataset=load_dataset(\"json\",data_files={\"2023_math\":\"./data/json/2023/math.json\",\n",
+    "                                        \"2024_math\":\"./data/json/2024/math.json\"})"
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
    ]
   },
   {
    "cell_type": "code",
+<<<<<<< HEAD
    "execution_count": 25,
+=======
+   "execution_count": 5,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "application/vnd.jupyter.widget-view+json": {
+       "model_id": "221c4bba0e78418b9686c1ba0a0fa668",
+       "version_major": 2,
+       "version_minor": 0
+      },
+      "text/plain": [
+       "Uploading the dataset shards:   0%|          | 0/1 [00:00<?, ?it/s]"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "data": {
+      "application/vnd.jupyter.widget-view+json": {
+       "model_id": "1763d5a9fc5c4c8099d0deb716f06c73",
+       "version_major": 2,
+       "version_minor": 0
+      },
+      "text/plain": [
+       "Creating parquet from Arrow format:   0%|          | 0/1 [00:00<?, ?ba/s]"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "data": {
+      "application/vnd.jupyter.widget-view+json": {
+       "model_id": "04f7bfc0e76c438ab84048301cb0e10e",
+       "version_major": 2,
+       "version_minor": 0
+      },
+      "text/plain": [
+       "Uploading the dataset shards:   0%|          | 0/1 [00:00<?, ?it/s]"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "data": {
+      "application/vnd.jupyter.widget-view+json": {
+       "model_id": "8707c527afaf47d0928b5167c509d5f7",
+       "version_major": 2,
+       "version_minor": 0
+      },
+      "text/plain": [
+       "Creating parquet from Arrow format:   0%|          | 0/1 [00:00<?, ?ba/s]"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "data": {
+      "text/plain": [
+       "CommitInfo(commit_url='https://huggingface.co/datasets/cfpark00/KoreanSAT/commit/8fdf990d118abc5511f7fe828a4ae482e4b67d07', commit_message='Upload dataset', commit_description='', oid='8fdf990d118abc5511f7fe828a4ae482e4b67d07', pr_url=None, repo_url=RepoUrl('https://huggingface.co/datasets/cfpark00/KoreanSAT', endpoint='https://huggingface.co', repo_type='dataset', repo_id='cfpark00/KoreanSAT'), pr_revision=None, pr_num=None)"
+      ]
+     },
+     "execution_count": 5,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "#save as parquet\n",
+    "dataset.push_to_hub(\"cfpark00/KoreanSAT\")"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
    "metadata": {},
    "outputs": [
     {
       "text/plain": [
        "{'id': 1,\n",
        " 'name': '1',\n",
+<<<<<<< HEAD
        " 'problem': '1. $\\\\left(2^{\\\\sqrt{3}} \\\\times 4\\\\right)^{\\\\sqrt{3} - 2}$ 의 값은? [2점] \\\\begin{itemize} \\\\item[1] \\\\frac{1}{4} \\\\item[2] \\\\frac{1}{2} \\\\item[3] 1 \\\\item[4] 2 \\\\item[5] 4 \\\\end{itemize}',\n",
        " 'answer': 2,\n",
        " 'score': 2,\n",
       ]
      },
      "execution_count": 25,
+=======
+       " 'problem': '1. \\\\left( \\\\frac{4}{2^{\\\\sqrt{2}}} \\\\right)^{2 + \\\\sqrt{2}} \\\\text{의 값은? [2점]}\\n\\n\\\\begin{itemize}\\n    \\\\item[1] $\\\\frac{1}{4}$\\n    \\\\item[2] $\\\\frac{1}{2}$\\n    \\\\item[3] $1$\\n    \\\\item[4] $2$\\n    \\\\item[5] $4$\\n\\\\end{itemize}\\n',\n",
+       " 'answer': -1,\n",
+       " 'score': -1}"
+      ]
+     },
+     "execution_count": 4,
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
      "metadata": {},
      "output_type": "execute_result"
     }
    ],
    "source": [
+<<<<<<< HEAD
     "dataset[\"2022_math\"][0]"
+=======
+    "dataset[\"2023_math\"][0]"
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
    ]
   },
   {
    "cell_type": "code",
+<<<<<<< HEAD
    "execution_count": 24,
+=======
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
    "metadata": {},
    "outputs": [
     {
      "data": {
       "application/vnd.jupyter.widget-view+json": {
+<<<<<<< HEAD
        "model_id": "f612f040825040c6affa3b897f2633ef",
+=======
+       "model_id": "ed6870fcaa5d4641ae911c05f8d5e1a3",
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
        "version_major": 2,
        "version_minor": 0
       },
       "text/plain": [
+<<<<<<< HEAD
        "Pushing dataset shards to the dataset hub:   0%|          | 0/1 [00:00<?, ?it/s]"
+=======
+       "Creating json from Arrow format:   0%|          | 0/1 [00:00<?, ?ba/s]"
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
       ]
      },
      "metadata": {},
     },
     {
      "data": {
+<<<<<<< HEAD
       "application/vnd.jupyter.widget-view+json": {
        "model_id": "f3cb1129210b4586bbb81b52357c424e",
        "version_major": 2,
    "source": [
     "#save as parquet\n",
     "dataset.push_to_hub(\"cfpark00/KoreanSAT\")"
+=======
+      "text/plain": [
+       "33057"
+      ]
+     },
+     "execution_count": 19,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "from datasets import load_dataset\n",
+    "\n",
+    "ds = load_dataset(\"cfpark00/KoreanSAT\")\n",
+    "ds[\"2024_math\"].to_json(\"./data/json/2024/math.json\")"
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
+   ]
+  },
+  {
+   "cell_type": "code",
+<<<<<<< HEAD
+=======
+   "execution_count": 46,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "DatasetDict({\n",
+       "    2024_math: Dataset({\n",
+       "        features: ['id', 'name', 'problem', 'answer', 'score'],\n",
+       "        num_rows: 46\n",
+       "    })\n",
+       "})"
+      ]
+     },
+     "execution_count": 46,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "ds"
    ]
   },
   {
    "outputs": [],
    "source": []
   },
+  {
+   "cell_type": "code",
+   "execution_count": 37,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "application/vnd.jupyter.widget-view+json": {
+       "model_id": "56b0357800c24f42af02e56bcd9f9133",
+       "version_major": 2,
+       "version_minor": 0
+      },
+      "text/plain": [
+       "Creating json from Arrow format:   0%|          | 0/1 [00:00<?, ?ba/s]"
+      ]
+     },
+     "metadata": {},
+     "output_type": "display_data"
+    },
+    {
+     "data": {
+      "text/plain": [
+       "33057"
+      ]
+     },
+     "execution_count": 37,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "dataset[\"2024_math\"].to_json(\"./data/json/2024/math.json\")"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 29,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "import json\n",
+    "jsons=[]\n",
+    "for line in open(\"./data/json/2024/math.json\"):\n",
+    "    jsons.append(json.loads(line))"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 31,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "16. 방정식\n",
+      "\\[\n",
+      "\\log_2{(3x+2)} = 2 + \\log_2{(x-2)}\n",
+      "\\]\n",
+      "\\text{를 만족시키는 실수 } \\( x \\) \\text{의 값을 ��하시오. [3점]}\n",
+      "\n"
+     ]
+    }
+   ],
+   "source": [
+    "print(jsons[15][\"problem\"])"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 30,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "{'id': 16,\n",
+       " 'name': '16',\n",
+       " 'problem': '16. 방정식\\n\\\\[\\n\\\\log_2{(3x+2)} = 2 + \\\\log_2{(x-2)}\\n\\\\]\\n\\\\text{를 만족시키는 실수 } \\\\( x \\\\) \\\\text{의 값을 구하시오. [3점]}\\n',\n",
+       " 'answer': '2',\n",
+       " 'score': 3}"
+      ]
+     },
+     "execution_count": 30,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "jsons[15]"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
+  {
+   "cell_type": "code",
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
+  },
   {
    "cell_type": "code",
    "execution_count": null,
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
+<<<<<<< HEAD
    "version": "3.10.9"
+=======
+   "version": "3.8.9"
+>>>>>>> 41af016e94a20b80de932855d2fd5110dfdd4df6
   }
  },
  "nbformat": 4,