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- data/json/2022/math.json +46 -0
- data/json/2022/math/answer_score_comment.json +278 -0
- data/json/2022/math/math_1.txt +9 -0
- data/json/2022/math/math_10.txt +9 -0
- data/json/2022/math/math_11.txt +13 -0
- data/json/2022/math/math_12.txt +21 -0
- data/json/2022/math/math_13.txt +9 -0
- data/json/2022/math/math_14.txt +19 -0
- data/json/2022/math/math_15.txt +31 -0
- data/json/2022/math/math_16.txt +1 -0
- data/json/2022/math/math_17.txt +1 -0
- data/json/2022/math/math_18.txt +7 -0
- data/json/2022/math/math_19.txt +1 -0
- data/json/2022/math/math_2.txt +9 -0
- data/json/2022/math/math_20.txt +8 -0
- data/json/2022/math/math_21.txt +9 -0
- data/json/2022/math/math_22.txt +8 -0
- data/json/2022/math/math_23_calc.txt +12 -0
- data/json/2022/math/math_23_geom.txt +9 -0
- data/json/2022/math/math_23_prob.txt +9 -0
- data/json/2022/math/math_24_calc.txt +13 -0
- data/json/2022/math/math_24_geom.txt +9 -0
- data/json/2022/math/math_24_prob.txt +9 -0
- data/json/2022/math/math_25_calc.txt +13 -0
- data/json/2022/math/math_25_geom.txt +15 -0
- data/json/2022/math/math_25_prob.txt +14 -0
- data/json/2022/math/math_26_calc.txt +12 -0
- data/json/2022/math/math_26_geom.txt +9 -0
- data/json/2022/math/math_26_prob.txt +9 -0
- data/json/2022/math/math_27_calc.txt +9 -0
- data/json/2022/math/math_27_geom.txt +9 -0
- data/json/2022/math/math_27_prob.txt +15 -0
- data/json/2022/math/math_28_calc.txt +15 -0
- data/json/2022/math/math_28_geom.txt +9 -0
- data/json/2022/math/math_28_prob.txt +14 -0
- data/json/2022/math/math_29_calc.txt +7 -0
- data/json/2022/math/math_29_geom.txt +14 -0
- data/json/2022/math/math_29_prob.txt +12 -0
- data/json/2022/math/math_3.txt +13 -0
- data/json/2022/math/math_30_calc.txt +11 -0
- data/json/2022/math/math_30_geom.txt +11 -0
- data/json/2022/math/math_30_prob.txt +15 -0
- data/json/2022/math/math_4.txt +13 -0
- data/json/2022/math/math_5.txt +16 -0
- data/json/2022/math/math_6.txt +9 -0
- data/json/2022/math/math_7.txt +9 -0
- data/json/2022/math/math_8.txt +9 -0
- data/json/2022/math/math_9.txt +15 -0
- data/json/2022/math/prompt.txt +52 -0
- data/json/2022/math_v1.json +46 -0
data/json/2022/math.json
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{"id": 1, "name": "1", "problem": "1. $\\left(2^{\\sqrt{3}} \\times 4\\right)^{\\sqrt{3} - 2}$ μ κ°μ? [2μ ] \\begin{itemize} \\item[1] \\frac{1}{4} \\item[2] \\frac{1}{2} \\item[3] 1 \\item[4] 2 \\item[5] 4 \\end{itemize}", "answer": 2, "score": 2, "review": null, "incomplete": false}
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{"id": 2, "name": "2", "problem": "2. ν¨μ $f(x) = x^3 + 3x^2 + x - 1$ μ λνμ¬ $f'(1)$μ κ°μ? [2μ ] \\begin{itemize} \\item[1] 6 \\item[2] 7 \\item[3] 8 \\item[4] 9 \\item[5] 10 \\end{itemize}", "answer": 5, "score": 2, "review": null, "incomplete": false}
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{"id": 3, "name": "3", "problem": "3. λ±μ°¨μμ΄ $\\{a_n\\}$μ λνμ¬ \\[ a_2 = 6, \\quad a_4 + a_6 = 36 \\] μΌ λ, $a_{10}$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 30 \\item[2] 32 \\item[3] 34 \\item[4] 36 \\item[5] 38 \\end{itemize}", "answer": 5, "score": 3, "review": null, "incomplete": false}
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{"id": 4, "name": "4", "problem": "4. ν¨μ $( y = f(x) )$μ κ·Έλνκ° κ·Έλ¦Όκ³Ό κ°λ€.\n\n\\[ \\lim_{x \\to -1-} f(x) + \\lim_{x \\to 2} f(x) \\text{μ κ°μ? [3μ ]} \\]\n\n\\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}", "answer": 4, "score": 3, "review": "Removed figure and the statement referring to the figure. The figure is needed to solve the problem.", "incomplete": true}
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{"id": 5, "name": "5", "problem": "5. 첫째νμ΄ 1μΈ μμ΄ $\\{a_n\\}$μ΄ λͺ¨λ μμ°μ $n$μ λνμ¬ \\[ a_{n+1} = \\begin{cases} 2a_n & (a_n < 7) \\\\ a_n - 7 & (a_n \\geq 7) \\end{cases} \\] μΌ λ, $\\sum_{k=1}^{8} a_k$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 30 \\item[2] 32 \\item[3] 34 \\item[4] 36 \\item[5] 38 \\end{itemize}", "answer": 1, "score": 3, "review": null, "incomplete": false}
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{"id": 6, "name": "6", "problem": "6. λ°©μ μ $( 2x^3 - 3x^2 - 12x + k = 0 )$μ΄ μλ‘ λ€λ₯Έ μΈ μ€κ·Όμ κ°λλ‘ νλ μ μ $k$μ κ°μλ? [3μ ] \\begin{itemize} \\item[1] 20 \\item[2] 23 \\item[3] 26 \\item[4] 29 \\item[5] 32 \\end{itemize}", "answer": 3, "score": 3, "review": null, "incomplete": false}
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{"id": 7, "name": "7", "problem": "7. $( \\pi < \\theta < \\frac{3}{2}\\pi )$μΈ $\\theta$μ λνμ¬ $\\tan \\theta - \\frac{6}{\\tan \\theta} = 1$μΌ λ, $ \\sin \\theta + \\cos \\theta $μ κ°μ? [3μ ] \\begin{itemize} \\item[1] -\\frac{2\\sqrt{10}}{5} \\item[2] -\\frac{\\sqrt{10}}{5} \\item[3] 0 \\item[4] \\frac{\\sqrt{10}}{5} \\item[5] \\frac{2\\sqrt{10}}{5} \\end{itemize}", "answer": 1, "score": 3, "review": null, "incomplete": false}
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{"id": 8, "name": "8", "problem": "8. 곑μ $( y = x^2 - 5x )$μ μ§μ $( y = x )$λ‘ λλ¬μΈμΈ λΆλΆμ λμ΄λ₯Ό μ§μ $( x = k )$κ° μ΄λ±λΆν λ, μμ $k$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 3 \\item[2] \\frac{13}{4} \\item[3] \\frac{7}{2} \\item[4] \\frac{15}{4} \\item[5] 4 \\end{itemize}", "answer": 1, "score": 3, "review": null, "incomplete": false}
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{"id": 9, "name": "9", "problem": "9. μ§μ $( y = 2x + k )$ κ° λ ν¨μ \\[ y = \\left( \\frac{2}{3} \\right)^{x+3} + 1, \\quad y = \\left( \\frac{2}{3} \\right)^{x+1} + \\frac{8}{3} \\] μ κ·Έλνμ λ§λλ μ μ κ°κ° $( \\mathrm{P} )$, $( \\mathrm{Q} )$λΌ νμ. $\\overline{\\mathrm{PQ}} = \\sqrt{5}$μΌ λ, μμ $k$μ κ°μ? [4μ ] \\begin{itemize} \\item[1] \\frac{31}{6} \\item[2] \\frac{16}{3} \\item[3] \\frac{11}{2} \\item[4] \\frac{17}{3} \\item[5] \\frac{35}{6} \\end{itemize}", "answer": 4, "score": 4, "review": "Removed figure.", "incomplete": false}
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{"id": 10, "name": "10", "problem": "10. μΌμ°¨ν¨μ $( f(x) )$μ λνμ¬ κ³‘μ $( y = f(x) )$ μμ μ $( 0, 0 )$μμμ μ μ κ³Ό 곑μ $( y = x f(x) )$ μμ μ $( 1, 2 )$μμμ μ μ μ΄ μΌμΉν λ, $f'(2)$μ κ°μ? [4μ ] \\begin{itemize} \\item[1] -18 \\item[2] -17 \\item[3] -16 \\item[4] -15 \\item[5] -14 \\end{itemize}", "answer": 5, "score": 4, "review": null, "incomplete": false}
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{"id": 11, "name": "11", "problem": "11. μμ $a$μ λνμ¬ μ§ν© $\\left\\{ x \\ \\middle| \\ -\\frac{a}{2} < x \\leq a, \\ x \\neq \\frac{a}{2} \\right\\}$ μμ μ μλ ν¨μ \\[ f(x) = \\tan \\frac{\\pi x}{a} \\] κ° μλ€. κ·Έλ¦Όκ³Ό κ°μ΄ ν¨μ $y = f(x)$μ κ·Έλν μμ μΈ μ $( \\mathrm{O, A, B} )$λ₯Ό μ§λλ μ§μ μ΄ μλ€. μ $( \\mathrm{A} )$λ₯Ό μ§λκ³ $x$μΆμ ννν μ§μ μ΄ ν¨μ $y = f(x)$μ κ·Έλνμ λ§λλ μ μ€ $( \\mathrm{A} )$κ° μλ μ μ $( \\mathrm{C} )$λΌ νμ. μΌκ°ν $( \\mathrm{ABC} )$κ° μ μΌκ°νμΌ λ, μΌκ°ν $( \\mathrm{ABC} )$μ λμ΄λ? (λ¨, $( \\mathrm{O} )$λ μμ μ΄λ€.) [4μ ] \\begin{itemize} \\item[1] \\frac{3\\sqrt{3}}{2} \\item[2] \\frac{17\\sqrt{3}}{12} \\item[3] \\frac{4\\sqrt{3}}{3} \\item[4] \\frac{5\\sqrt{3}}{4} \\item[5] \\frac{7\\sqrt{3}}{6} \\end{itemize}", "answer": 3, "score": 4, "review": "Removed figure and the statement referring to the figure.", "incomplete": false}
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{"id": 12, "name": "12", "problem": "12. μ€μ μ 체μ μ§ν©μμ μ°μμΈ ν¨μ $f(x)$κ° λͺ¨λ μ€μ $x$μ λνμ¬ \\[ \\{f(x)\\}^3 - \\{f(x)\\}^2 - x^2 f(x) + x^2 = 0 \\] μ λ§μ‘±μν¨λ€. ν¨μ $f(x)$μ μ΅λκ°μ΄ 1μ΄κ³ μ΅μκ°μ΄ 0μΌ λ, \\[ f\\left( -\\frac{4}{3} \\right) + f(0) + f\\left( \\frac{1}{2} \\right) \\] μ κ°μ? [4μ ] \\begin{itemize} \\item[1] \\frac{1}{2} \\item[2] 1 \\item[3] \\frac{3}{2} \\item[4] 2 \\item[5] \\frac{5}{2} \\end{itemize}", "answer": 3, "score": 4, "review": null, "incomplete": false}
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{"id": 13, "name": "13", "problem": "13. λ μμ $( a, b \\ (1 < a < b) )$μ λνμ¬ μ’ννλ©΄ μμ λ μ $(a, \\log_2 a), \\ (b, \\log_2 b)$λ₯Ό μ§λλ μ§μ μ $y$μ νΈκ³Ό λ μ $(a, \\log_4 a), \\ (b, \\log_4 b)$λ₯Ό μ§λλ μ§μ μ $y$μ νΈμ΄ κ°λ€. ν¨μ $f(x) = a^{bx} + b^{ax}$μ λνμ¬ $f(1) = 40$μΌ λ, $f(2)$μ κ°μ? [4μ ] \\begin{itemize} \\item[1] 760 \\item[2] 800 \\item[3] 840 \\item[4] 880 \\item[5] 920 \\end{itemize}", "answer": 2, "score": 4, "review": null, "incomplete": false}
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{"id": 14, "name": "14", "problem": "14. μμ§μ μλ₯Ό μμ§μ΄λ μ $\\mathrm{P}$μ μκ° $t$μμμ μμΉ $x(t)$κ° λ μμ $a$, $b$μ λνμ¬ \\[ x(t) = t(t - 1)(at + b) \\quad (a \\neq 0) \\] μ΄λ€. μ $\\mathrm{P}$μ μκ° $t$μμμ μλ $v(t)$κ° $\\int_0^1 |v(t)| \\, dt = 2$λ₯Ό λ§μ‘±μν¬ λ, μλ γ±, γ΄, γ· μ€μμ μ³μ κ²λ§μ μλ λλ‘ κ³ λ₯Έ κ²μ? [4μ ]\n\n\\begin{itemize} \\item[γ±.] $\\int_0^1 v(t) \\, dt = 0$ \\item[γ΄.] $|x(t_1)| > 1$μΈ $t_1$μ΄ μ΄λ¦°κ΅¬κ° $(0, 1)$μ μ‘΄μ¬νλ€. \\item[γ·.] $0 \\leq t \\leq 1$μΈ λͺ¨λ $t$μ λνμ¬ $|x(t)| < 1$μ΄λ©΄ $x(t_2) = 0$μΈ $t_2$κ° μ΄λ¦°κ΅¬κ° $(0, 1)$μ μ‘΄μ¬νλ€. \\end{itemize}\n\n\\begin{itemize} \\item[1] γ± \\item[2] γ±, γ΄ \\item[3] γ±, γ· \\item[4] γ΄, γ· \\item[5] γ±, γ΄, γ· \\end{itemize}", "answer": 3, "score": 4, "review": "<보기> changed to 'μλ γ±,γ΄,γ·, μ€'", "incomplete": false}
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{"id": 15, "name": "15", "problem": "15. λ μ $( \\mathrm{O}_1, \\mathrm{O}_2 )$λ₯Ό κ°κ° μ€μ¬μΌλ‘ νκ³ λ°μ§λ¦μ κΈΈμ΄κ° $(\\overline{\\mathrm{O}_1\\mathrm{O}_2} )$μΈ λ μ $( C_1, C_2 )$κ° μλ€. κ·Έλ¦Όκ³Ό κ°μ΄ μ $( C_1 )$ μμ μλ‘ λ€λ₯Έ μΈ μ $( \\mathrm{A}, \\mathrm{B}, \\mathrm{C} )$μ μ $( C_2 )$ μμ μ $( \\mathrm{D} )$κ° μ£Όμ΄μ Έ μκ³ , μΈ μ $( \\mathrm{A}, \\mathrm{O}_1, \\mathrm{O}_2 )$μ μΈ μ $( \\mathrm{C}, \\mathrm{O}_2, \\mathrm{D} )$κ° κ°κ° ν μ§μ μμ μλ€.\n\nμ΄λ $(\\angle \\mathrm{B}\\mathrm{O}_1\\mathrm{A} = \\theta_1)$, $(\\angle \\mathrm{O}_2\\mathrm{O}_1\\mathrm{C} = \\theta_2)$, $(\\angle \\mathrm{O}_1\\mathrm{O}_2\\mathrm{D} = \\theta_3)$μ΄λΌ νμ.\n\nλ€μμ $( \\overline{\\mathrm{A}\\mathrm{B}} : \\overline{\\mathrm{O}_1\\mathrm{D}} = 1 : 2\\sqrt{2} )$μ΄κ³ $( \\theta_3 = \\theta_1 + \\theta_2 )$μΌ λ, μ λΆ $( \\mathrm{A}\\mathrm{B} )$μ μ λΆ $( \\mathrm{C}\\mathrm{D} )$μ κΈΈμ΄μ λΉλ₯Ό ꡬνλ κ³Όμ μ΄λ€.\n\n\\[ \\begin{aligned} &\\angle \\mathrm{C}\\mathrm{O}_2\\mathrm{O}_1 + \\angle \\mathrm{O}_1\\mathrm{O}_2\\mathrm{D} = \\pi \\text{μ΄λ―λ‘ } \\theta_3 = \\frac{\\pi}{2} + \\frac{\\theta_2}{2} \\text{μ΄κ³ } \\\\ &\\theta_3 = \\theta_1 + \\theta_2 \\text{μμ } 2\\theta_1 + \\theta_2 = \\pi \\text{μ΄λ―λ‘ } \\angle \\mathrm{C}\\mathrm{O}_1\\mathrm{B} = \\theta_1 \\text{μ΄λ€.} \\\\ &\\text{μ΄λ } \\angle \\mathrm{O}_2\\mathrm{O}_1\\mathrm{B} = \\theta_1 + \\theta_2 = \\theta_3 \\text{μ΄λ―λ‘ μΌκ°ν } \\mathrm{O}_1\\mathrm{O}_2\\mathrm{B} \\text{μ μΌκ°ν } \\mathrm{O}_2\\mathrm{O}_1\\mathrm{D} \\text{λ ν©λμ΄λ€.} \\\\ &\\overline{\\mathrm{A}\\mathrm{B}} = k \\text{λΌ ν λ} \\\\ &\\overline{\\mathrm{B}\\mathrm{O}_2} = \\overline{\\mathrm{O}_1\\mathrm{D}}= 2\\sqrt{2}k \\text{μ΄λ―λ‘ } \\overline{\\mathrm{A}\\mathrm{O}_2} = \\text{(κ°)μ΄κ³ ,} \\\\ &\\angle \\mathrm{B}\\mathrm{O}_2\\mathrm{A} = \\frac{\\theta_1}{2} \\text{μ΄λ―λ‘ } \\cos \\frac{\\theta_1}{2} = \\text{(λ) μ΄λ€.} \\\\ &\\text{μΌκ°ν } \\mathrm{O}_2\\mathrm{B}\\mathrm{C} \\text{μμ} \\\\ &\\overline{\\mathrm{B}\\mathrm{C}} = k, \\overline{\\mathrm{B}\\mathrm{O}_2} = 2\\sqrt{2}k, \\angle \\mathrm{C}\\mathrm{O}_2\\mathrm{B} = \\frac{\\theta_1}{2} \\text{μ΄λ―λ‘} \\\\ &\\text{μ½μ¬μΈλ²μΉμ μνμ¬ } \\overline{\\mathrm{O}_2\\mathrm{C}} = \\text{(λ€) μ΄λ€.} \\\\ &\\overline{\\mathrm{C}\\mathrm{D}} = \\overline{\\mathrm{O}_2\\mathrm{D}} + \\overline{\\mathrm{O}_2\\mathrm{C}} = \\overline{\\mathrm{O}_1\\mathrm{O}_2} + \\overline{\\mathrm{O}_2\\mathrm{C}} \\text{μ΄λ―λ‘} \\\\ &\\overline{\\mathrm{A}\\mathrm{B}} : \\overline{\\mathrm{C}\\mathrm{D}} = k : \\left(\\frac{\\text{(κ°)}}{2} + \\text{(λ€)}\\right) \\text{μ΄λ€.} \\end{aligned} \\]\n\nμμ (κ°), (λ€)μ μλ§μ μμ κ°κ° $( f(k), g(k) )$λΌ νκ³ , (λ)μ μλ§μ μλ₯Ό $( p )$λΌ ν λ, $( f(p) \\times g(p) )$μ κ°μ? [4μ ]\n\n\\begin{itemize} \\item[1] \\frac{169}{27} \\item[2] \\frac{56}{9} \\item[3] \\frac{167}{27} \\item[4] \\frac{166}{27} \\item[5] \\frac{55}{9} \\end{itemize}", "answer": 2, "score": 4, "review": "Removed figure and the statement referring to the figure.", "incomplete": false}
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{"id": 16, "name": "16", "problem": "16. $\\log_2 120 - \\frac{1}{\\log_{15} 2}$ μ κ°μ ꡬνμμ€. [3μ ]", "answer": 3, "score": 3, "review": null, "incomplete": false}
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{"id": 17, "name": "17", "problem": "17. ν¨μ $f(x)$μ λνμ¬ $f'(x) = 3x^2 + 2x$μ΄κ³ $f(0) = 2$μΌ λ, $f(1)$μ κ°μ ꡬνμμ€. [3μ ]", "answer": 4, "score": 3, "review": null, "incomplete": false}
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{"id": 18, "name": "18", "problem": "18. μμ΄ $\\{a_n\\}$μ λνμ¬ \\[ \\sum_{k=1}^{10} a_k - \\sum_{k=1}^{7} \\frac{a_k}{2} = 56, \\quad \\sum_{k=1}^{10} 2a_k - \\sum_{k=1}^{8} a_k = 100 \\] μΌ λ, $a_8$μ κ°μ ꡬνμμ€. [3μ ]", "answer": 12, "score": 3, "review": null, "incomplete": false}
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{"id": 19, "name": "19", "problem": "19. ν¨μ $f(x) = x^3 + ax^2 - (a^2 - 8a)x + 3$μ΄ μ€μ μ 체μ μ§ν©μμ μ¦κ°νλλ‘ νλ μ€μ $a$μ μ΅λκ°μ ꡬνμμ€. [3μ ]", "answer": 6, "score": 3, "review": null, "incomplete": false}
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{"id": 20, "name": "20", "problem": "20. μ€μ μ 체μ μ§ν©μμ λ―ΈλΆκ°λ₯ν ν¨μ $( f(x) )$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] λ«νκ΅¬κ° $[0, 1]$μμ $f(x) = x$μ΄λ€. \\item[(λ)] μ΄λ€ μμ $a, b$μ λνμ¬ κ΅¬κ° $[0, \\infty)$μμ $f(x+1) - x f(x) = ax + b$μ΄λ€. \\end{itemize}\n\n\\[ 60 \\times \\int_1^2 f(x) \\, dx \\] μ κ°μ ꡬνμμ€. [4μ ]", "answer": 110, "score": 4, "review": null, "incomplete": false}
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{"id": 21, "name": "21", "problem": "21. μμ΄ $\\{a_n\\}$μ΄ λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] $( |a_1| = 2 )$ \\item[(λ)] λͺ¨λ μμ°μ $( n )$μ λνμ¬ $( |a_{n+1}| = 2|a_n| )$μ΄λ€. \\item[(λ€)] $\\sum_{n=1}^{10} a_n = -14$ \\end{itemize}\n\n$a_1 + a_3 + a_5 + a_7 + a_9$μ κ°μ ꡬνμμ€. [4μ ]", "answer": 678, "score": 4, "review": null, "incomplete": false}
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{"id": 22, "name": "22", "problem": "22. μ΅κ³ μ°¨νμ κ³μκ° $\\frac{1}{2}$ μΈ μΌμ°¨ν¨μ $f(x)$μ μ€μ $t$μ λνμ¬ λ°©μ μ $f'(x) = 0$μ΄ λ«νκ΅¬κ° $[t, t+2]$μμ κ°λ μ€κ·Όμ κ°μλ₯Ό $g(t)$λΌ ν λ, ν¨μ $g(t)$λ λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] λͺ¨λ μ€μ $( a )$μ λνμ¬ $( \\lim_{t \\to a+} g(t) + \\lim_{t \\to a-} g(t) \\leq 2 )$μ΄λ€. \\item[(λ)] $( g(f(1)) = g(f(4)) = 2, \\ g(f(0)) = 1 )$ \\end{itemize}\n\n$f(5)$μ κ°μ ꡬνμμ€. [4μ ]", "answer": 9, "score": 4, "review": null, "incomplete": false}
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{"id": 23, "name": "23_prob", "problem": "23. λ€νμ $(x+2)^7$μ μ κ°μμμ $x^5$μ κ³μλ? [2μ ] \\begin{itemize} \\item[1] 42 \\item[2] 56 \\item[3] 70 \\item[4] 84 \\item[5] 98 \\end{itemize}", "answer": 4, "score": 2, "review": null, "incomplete": false}
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{"id": 24, "name": "24_prob", "problem": "24. νλ₯ λ³μ $X$κ° μ΄νλΆν¬ $\\mathrm{B}\\left(n, \\frac{1}{3}\\right)$μ λ°λ₯΄κ³ $\\mathrm{V}(2X) = 40$μΌ λ, $n$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 30 \\item[2] 35 \\item[3] 40 \\item[4] 45 \\item[5] 50 \\end{itemize}", "answer": 4, "score": 3, "review": null, "incomplete": false}
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{"id": 25, "name": "25_prob", "problem": "25. λ€μ 쑰건μ λ§μ‘±μν€λ μμ°μ $a, \\ b, \\ c, \\ d, \\ e$μ λͺ¨λ μμμ $(a, b, c, d, e)$μ κ°μλ? [3μ ]\n\n\\begin{itemize} \\item[(κ°)] $a + b + c + d + e = 12$ \\item[(λ)] $\\left| a^2 - b^2 \\right| = 5$ \\end{itemize}\n\n\\begin{itemize} \\item[1] 30 \\item[2] 32 \\item[3] 34 \\item[4] 36 \\item[5] 38 \\end{itemize}", "answer": 1, "score": 3, "review": null, "incomplete": false}
|
26 |
+
{"id": 26, "name": "26_prob", "problem": "26. $( 1 )$λΆν° $( 10 )$κΉμ§ μμ°μκ° νλμ© μ ν μλ $( 10 )$μ₯μ μΉ΄λκ° λ€μ΄ μλ μ£Όλ¨Έλκ° μλ€. μ΄ μ£Όλ¨Έλμμ μμλ‘ μΉ΄λ $( 3 )$μ₯μ λμμ κΊΌλΌ λ, κΊΌλΈ μΉ΄λμ μ ν μλ μΈ μμ°μ μ€μμ κ°μ₯ μμ μκ° $( 4 )$ μ΄νμ΄κ±°λ $( 7 )$ μ΄μμΌ νλ₯ μ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{4}{5} \\item[2] \\frac{5}{6} \\item[3] \\frac{13}{15} \\item[4] \\frac{9}{10} \\item[5] \\frac{14}{15} \\end{itemize}", "answer": 3, "score": 3, "review": "Removed figure.", "incomplete": false}
|
27 |
+
{"id": 27, "name": "27_prob", "problem": "27. μ΄λ μλμ°¨ νμ¬μμ μμ°νλ μ κΈ° μλμ°¨μ 1ν μΆ©μ μ£Όν 거리λ νκ· μ΄ $m$μ΄κ³ νμ€νΈμ°¨κ° $\\sigma$μΈ μ κ·λΆν¬λ₯Ό λ°λ₯Έλ€κ³ νλ€.\n\nμ΄ μλμ°¨ νμ¬μμ μμ°ν μ κΈ° μλμ°¨ 100λλ₯Ό μμμΆμΆνμ¬ μ»μ 1ν μΆ©μ μ£Όν 거리μ νλ³Ένκ· μ΄ $\\overline{x_1}$μΌ λ, λͺ¨νκ· $m$μ λν μ λ’°λ 95\\%μ μ 뒰ꡬκ°μ΄ $a \\le m \\le b$μ΄λ€.\n\nμ΄ μλμ°¨ νμ¬μμ μμ°ν μ κΈ° μλμ°¨ 400λλ₯Ό μμμΆμΆνμ¬ μ»μ 1ν μΆ©μ μ£Όν 거리μ νλ³Ένκ· μ΄ $\\overline{x_2}$μΌ λ, λͺ¨νκ· $m$μ λν μ λ’°λ 99\\%μ μ 뒰ꡬκ°μ΄ $c \\le m \\le d$μ΄λ€.\n\n$\\overline{x_1} - \\overline{x_2} = 1.34$μ΄κ³ $a = c$μΌ λ, $b - a$μ κ°μ? (λ¨, μ£Όν 거리μ λ¨μλ kmμ΄κ³ , $Z$κ° νμ€μ κ·λΆν¬λ₯Ό λ°λ₯΄λ νλ₯ λ³μμΌ λ $\\mathrm{P}(|Z| \\le 1.96) = 0.95$, $\\mathrm{P}(|Z| \\le 2.58) = 0.99$λ‘ κ³μ°νλ€.) [3μ ]\n\n\\begin{itemize} \\item[1] 5.88 \\item[2] 7.84 \\item[3] 9.80 \\item[4] 11.76 \\item[5] 13.72 \\end{itemize}", "answer": 2, "score": 3, "review": null, "incomplete": false}
|
28 |
+
{"id": 28, "name": "28_prob", "problem": "28. λ μ§ν© $X = \\{1, 2, 3, 4, 5\\}$, $Y = \\{1, 2, 3, 4\\}$μ λνμ¬ λ€μ 쑰건μ λ§μ‘±μν€λ $X$μμ $Y$λ‘μ ν¨μ $f$μ κ°μλ? [4μ ]\n\n\\begin{itemize} \\item[(κ°)] μ§ν© $X$μ λͺ¨λ μμ $x$μ λνμ¬ $f(x) \\geq \\sqrt{x}$μ΄λ€. \\item[(λ)] ν¨μ $f$μ μΉμμ μμμ κ°μλ 3μ΄λ€. \\end{itemize}\n\n\\begin{itemize} \\item[1] 128 \\item[2] 138 \\item[3] 148 \\item[4] 158 \\item[5] 168 \\end{itemize}", "answer": 1, "score": 4, "review": null, "incomplete": false}
|
29 |
+
{"id": 29, "name": "29_prob", "problem": "29. λ μ°μνλ₯ λ³μ $( X )$μ $( Y )$κ° κ°λ κ°μ λ²μλ $( 0 \\leq X \\leq 6 )$, $( 0 \\leq Y \\leq 6 )$μ΄κ³ , $( X )$μ $( Y )$μ νλ₯ λ°λν¨μλ κ°κ° $( f(x), g(x) )$μ΄λ€. νλ₯ λ³μ $( X )$μ νλ₯ λ°λν¨μ $( f(x) )$μ κ·Έλνλ κ·Έλ¦Όκ³Ό κ°λ€.\n\n\\[ 0 \\leq x \\leq 6\\ \\text{μΈ λͺ¨λ } x \\text{μ λνμ¬} \\]\n\\[ f(x) + g(x) = k \\quad (k \\text{λ μμ}) \\]\nλ₯Ό λ§μ‘±μν¬ λ, $( \\mathrm{P}(6k \\leq Y \\leq 15k) = \\frac{q}{p} )$μ΄λ€. $( p + q )$μ κ°μ ꡬνμμ€. (λ¨, $( p )$μ $( q )$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]", "answer": 31, "score": 4, "review": "Removed figure and the statement referring to the figure. The figure is needed to solve the problem.", "incomplete": true}
|
30 |
+
{"id": 30, "name": "30_prob", "problem": "30. ν° κ³΅κ³Ό κ²μ κ³΅μ΄ κ°κ° 10κ° μ΄μ λ€μ΄ μλ λ°κ΅¬λμ λΉμ΄ μλ μ£Όλ¨Έλκ° μλ€. ν κ°μ μ£Όμ¬μλ₯Ό μ¬μ©νμ¬ λ€μ μνμ νλ€.\n\n\\[ \\begin{array}{|c|} \\hline \\text{μ£Όμ¬μλ₯Ό ν λ² λμ Έ} \\\\ \\text{λμ¨ λμ μκ° 5 μ΄μμ΄λ©΄} \\\\ \\text{λ°κ΅¬λμ μλ ν° κ³΅ 2κ°λ₯Ό μ£Όλ¨Έλμ λ£κ³ ,} \\\\ \\text{λμ¨ λμ μκ° 4 μ΄νμ΄λ©΄} \\\\ \\text{λ°κ΅¬λμ μλ κ²μ 곡 1κ°λ₯Ό μ£Όλ¨Έλμ λ£λλ€.} \\\\ \\hline \\end{array} \\]\n\nμμ μνμ 5λ² λ°λ³΅ν λ, $( n(1 \\leq n \\leq 5) )$λ²μ§Έ μν ν μ£Όλ¨Έλμ λ€μ΄ μλ ν° κ³΅κ³Ό κ²μ 곡μ κ°μλ₯Ό κ°κ° $( a_n )$, $( b_n )$μ΄λΌ νμ. $( a_5 + b_5 \\geq 7 )$μΌ λ, $( a_k = b_k )$μΈ μμ°μ $( k(1 \\leq k \\leq 5) )$κ° μ‘΄μ¬ν νλ₯ μ $( \\frac{q}{p} )$μ΄λ€. $( p + q )$μ κ°μ ꡬνμμ€. (λ¨, $(p)$μ $(q)$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]", "answer": 191, "score": 4, "review": null, "incomplete": false}
|
31 |
+
{"id": 31, "name": "23_calc", "problem": "23. \\[ \\lim_{n \\to \\infty} \\frac{\\frac{5}{n} + \\frac{3}{n^2}}{\\frac{1}{n} - \\frac{2}{n^3}} \\text{μ κ°μ? [2μ ]} \\] \\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}", "answer": 5, "score": 2, "review": null, "incomplete": false}
|
32 |
+
{"id": 32, "name": "24_calc", "problem": "24. μ€μ μ 체μ μ§ν©μμ λ―ΈλΆκ°λ₯ν ν¨μ $f(x)$κ° λͺ¨λ μ€μ $x$μ λνμ¬ \\[ f(x^3 + x) = e^x \\] μ λ§μ‘±μν¬ λ, $f'(2)$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] e \\item[2] \\frac{e}{2} \\item[3] \\frac{e}{3} \\item[4] \\frac{e}{4} \\item[5] \\frac{e}{5} \\end{itemize}", "answer": 4, "score": 3, "review": null, "incomplete": false}
|
33 |
+
{"id": 33, "name": "25_calc", "problem": "25. λ±λΉμμ΄ $\\{a_n\\}$μ λνμ¬ \\[ \\sum_{n=1}^{\\infty} (a_{2n-1} - a_{2n}) = 3, \\quad \\sum_{n=1}^{\\infty} a_n^2 = 6 \\] μΌ λ, $\\sum_{n=1}^{\\infty} a_n$ μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}", "answer": 2, "score": 3, "review": null, "incomplete": false}
|
34 |
+
{"id": 34, "name": "26_calc", "problem": "26. \\[ \\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\frac{k^2 + 2kn}{k^3 + 3k^2 n + n^3} \\text{μ κ°μ?} \\quad [3 \\text{μ }] \\] \\begin{itemize} \\item[1] \\ln 5 \\item[2] \\frac{\\ln 5}{2} \\item[3] \\frac{\\ln 5}{3} \\item[4] \\frac{\\ln 5}{4} \\item[5] \\frac{\\ln 5}{5} \\end{itemize}", "answer": 3, "score": 3, "review": null, "incomplete": false}
|
35 |
+
{"id": 35, "name": "27_calc", "problem": "27. μ’ννλ©΄ μλ₯Ό μμ§μ΄λ μ $\\mathrm{P}$μ μκ° $t \\ (t>0)$μμμ μμΉκ° 곑μ $y = x^2$κ³Ό μ§μ $y = t^2 x - \\frac{\\ln t}{8}$κ° λ§λλ μλ‘ λ€λ₯Έ λ μ μ μ€μ μΌ λ, μκ° $t=1$μμ $t=e$κΉμ§ μ $\\mathrm{P}$κ° μμ§μΈ 거리λ? [3μ ] \\begin{itemize} \\item[1] \\frac{e^4}{2} - \\frac{3}{8} \\item[2] \\frac{e^4}{2} - \\frac{5}{16} \\item[3] \\frac{e^4}{2} - \\frac{1}{4} \\item[4] \\frac{e^4}{2} - \\frac{3}{16} \\item[5] \\frac{e^4}{2} - \\frac{1}{8} \\end{itemize}", "answer": 1, "score": 3, "review": null, "incomplete": false}
|
36 |
+
{"id": 36, "name": "28_calc", "problem": "28. ν¨μ $( f(x) = 6\\pi (x - 1)^2 )$μ λνμ¬ ν¨μ $( g(x) )$λ₯Ό \\[ g(x) = 3f(x) + 4\\cos f(x) \\] λΌ νμ. $( 0 < x < 2 )$μμ ν¨μ $( g(x) )$κ° κ·Ήμκ° λλ $( x )$μ κ°μλ? [4μ ] \\begin{itemize} \\item[1] 6 \\item[2] 7 \\item[3] 8 \\item[4] 9 \\item[5] 10 \\end{itemize}", "answer": 2, "score": 4, "review": null, "incomplete": false}
|
37 |
+
{"id": 37, "name": "29_calc", "problem": "29. κ·Έλ¦Όκ³Ό κ°μ΄ κΈΈμ΄κ° 2μΈ μ λΆ $(\\mathrm{AB})$λ₯Ό μ§λ¦μΌλ‘ νλ λ°μμ΄ μλ€. νΈ $(\\mathrm{AB})$ μμ λ μ $(\\mathrm{P})$, $(\\mathrm{Q})$λ₯Ό $(\\angle \\mathrm{PAB} = \\theta)$, $(\\angle \\mathrm{QBA} = 2\\theta)$κ° λλλ‘ μ‘κ³ , λ μ λΆ $(\\mathrm{AP})$, $(\\mathrm{BQ})$μ κ΅μ μ $(\\mathrm{R})$λΌ νμ. μ λΆ $(\\mathrm{AB})$ μμ μ $(\\mathrm{S})$, μ λΆ $(\\mathrm{BR})$ μμ μ $(\\mathrm{T})$, μ λΆ $(\\mathrm{AR})$ μμ μ $(\\mathrm{U})$λ₯Ό μ λΆ $(\\mathrm{UT})$κ° μ λΆ $(\\mathrm{AB})$μ νννκ³ μΌκ°ν $(\\mathrm{STU})$κ° μ μΌκ°νμ΄ λλλ‘ μ‘λλ€. λ μ λΆ $(\\mathrm{AR})$, $(\\mathrm{QR})$μ νΈ $(\\mathrm{AQ})$λ‘ λλ¬μΈμΈ λΆλΆμ λμ΄λ₯Ό $(f(\\theta))$, μΌκ°ν $(\\mathrm{STU})$μ λμ΄λ₯Ό $(g(\\theta))$λΌ ν λ,\n\\[ \\lim_{\\theta \\to 0+} \\frac{g(\\theta)}{\\theta \\times f(\\theta)} = \\frac{q}{p} \\sqrt{3} \\]\nμ΄λ€. $(p + q)$μ κ°μ ꡬνμμ€.\n\n(λ¨, $(0 < \\theta < \\frac{\\pi}{6})$μ΄κ³ , $(p)$μ $(q)$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]", "answer": 11, "score": 4, "review": "Removed figure and the statement referring to the figure.", "incomplete": false}
|
38 |
+
{"id": 38, "name": "30_calc", "problem": "30. μ€μ μ 체μ μ§ν©μμ μ¦κ°νκ³ λ―ΈλΆκ°λ₯ν ν¨μ $f(x)$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] $f(1) = 1$, \\quad $\\int_{1}^{2} f(x) \\, dx = \\frac{5}{4}$ \\item[(λ)] ν¨μ $f(x)$μ μν¨μλ₯Ό $g(x)$λΌ ν λ, $x \\geq 1$μΈ λͺ¨λ μ€μ $x$μ λνμ¬ $g(2x) = 2f(x)$μ΄λ€. \\end{itemize}\n\n\\[ \\int_{1}^{8} x f'(x) \\, dx = \\frac{q}{p} \\text{μΌ λ, } p+q \\text{μ κ°μ ꡬνμμ€.} \\]\n(λ¨, $p$μ $q$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]", "answer": 143, "score": 4, "review": null, "incomplete": false}
|
39 |
+
{"id": 39, "name": "23_geom", "problem": "23. μ’ν곡κ°μ μ $\\mathrm{A}(2, 1, 3)$μ $xy$ νλ©΄μ λνμ¬ λμΉμ΄λν μ μ $\\mathrm{P}$λΌ νκ³ , μ $\\mathrm{A}$λ₯Ό $yz$ νλ©΄μ λνμ¬ λμΉμ΄λν μ μ $\\mathrm{Q}$λΌ ν λ, μ λΆ $\\mathrm{PQ}$μ κΈΈμ΄λ? [2μ ]\n\n\\begin{itemize} \\item[1] 5 \\sqrt{2} \\item[2] 2 \\sqrt{13} \\item[3] 3 \\sqrt{6} \\item[4] 2 \\sqrt{14} \\item[5] 2 \\sqrt{15} \\end{itemize}", "answer": 2, "score": 2, "review": null, "incomplete": false}
|
40 |
+
{"id": 40, "name": "24_geom", "problem": "24. ν μ΄μ μ μ’νκ° $\\left( 3\\sqrt{2}, 0 \\right)$ μΈ μ곑μ $\\frac{x^2}{a^2} - \\frac{y^2}{6} = 1$ μ μ£ΌμΆμ κΈΈμ΄λ? (λ¨, $a$ λ μμμ΄λ€.) [3μ ]\n\n\\begin{itemize} \\item[1] 3\\sqrt{3} \\item[2] \\frac{7\\sqrt{3}}{2} \\item[3] 4\\sqrt{3} \\item[4] \\frac{9\\sqrt{3}}{2} \\item[5] 5\\sqrt{3} \\end{itemize}", "answer": 3, "score": 3, "review": null, "incomplete": false}
|
41 |
+
{"id": 41, "name": "25_geom", "problem": "25. μ’ννλ©΄μμ λ μ§μ \\[ \\frac{x+1}{2} = y - 3, \\quad x - 2 = \\frac{y - 5}{3} \\] κ° μ΄λ£¨λ μκ°μ ν¬κΈ°λ₯Ό $\\theta$λΌ ν λ, $\\cos \\theta$μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{1}{2} \\item[2] \\frac{\\sqrt{5}}{4} \\item[3] \\frac{\\sqrt{6}}{4} \\item[4] \\frac{\\sqrt{7}}{4} \\item[5] \\frac{\\sqrt{2}}{2} \\end{itemize}", "answer": 5, "score": 3, "review": null, "incomplete": false}
|
42 |
+
{"id": 42, "name": "26_geom", "problem": "26. λ μ΄μ μ΄ $( \\mathrm{F}, \\mathrm{F'} )$μΈ νμ $\\frac{x^2}{64} + \\frac{y^2}{16} = 1$ μμ μ μ€ μ 1μ¬λΆλ©΄μ μλ μ $( \\mathrm{A} )$κ° μλ€. λ μ§μ $( \\mathrm{AF}, \\mathrm{AF'} )$μ λμμ μ νκ³ μ€μ¬μ΄ $y$μΆ μμ μλ μ μ€ μ€μ¬μ $y$μ’νκ° μμμΈ κ²μ $( C )$λΌ νμ. μ $( C )$μ μ€μ¬μ $( \\mathrm{B} )$λΌ ν λ μ¬κ°ν $( \\mathrm{AFBF'} )$μ λμ΄κ° 72μ΄λ€. μ $( C )$μ λ°μ§λ¦μ κΈΈμ΄λ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{17}{2} \\item[2] 9 \\item[3] \\frac{19}{2} \\item[4] 10 \\item[5] \\frac{21}{2} \\end{itemize}", "answer": 2, "score": 3, "review": "Removed figure.", "incomplete": false}
|
43 |
+
{"id": 43, "name": "27_geom", "problem": "27. κ·Έλ¦Όκ³Ό κ°μ΄ ν λͺ¨μ리μ κΈΈμ΄κ° 4μΈ μ μ‘면체 $\\mathrm{ABCD - EFGH}$ κ° μλ€. μ λΆ $\\mathrm{AD}$ μ μ€μ μ $\\mathrm{M}$μ΄λΌ ν λ, μΌκ°ν $\\mathrm{MEG}$ μ λμ΄λ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{21}{2} \\item[2] 11 \\item[3] \\frac{23}{2} \\item[4] 12 \\item[5] \\frac{25}{2} \\end{itemize}", "answer": 4, "score": 3, "review": "Removed figure and the statement referring to the figure.", "incomplete": false}
|
44 |
+
{"id": 44, "name": "28_geom", "problem": "28. λ μμ $( a )$, $( p )$μ λνμ¬ ν¬λ¬Όμ $( (y - a)^2 = 4px )$μ μ΄μ μ $( \\mathrm{F}_1 )$μ΄λΌ νκ³ , ν¬λ¬Όμ $( y^2 = -4x )$μ μ΄μ μ $( \\mathrm{F}_2 )$λΌ νμ. μ λΆ $( \\mathrm{F}_1 \\mathrm{F}_2 )$κ° λ ν¬λ¬Όμ κ³Ό λ§λλ μ μ κ°κ° $( \\mathrm{P} )$, $( \\mathrm{Q} )$λΌ ν λ, $( \\overline{\\mathrm{F}_1 \\mathrm{F}_2} = 3 )$, $( \\overline{\\mathrm{P}\\mathrm{Q}} = 1 )$μ΄λ€. $( a^2 + p^2 )$μ κ°μ? [4μ ]\n\n\\begin{itemize} \\item[1] 6 \\item[2] \\frac{25}{4} \\item[3] \\frac{13}{2} \\item[4] \\frac{27}{4} \\item[5] 7 \\end{itemize}", "answer": 5, "score": 4, "review": "Removed figure.", "incomplete": false}
|
45 |
+
{"id": 45, "name": "29_geom", "problem": "29. μ’ννλ©΄μμ $\\overline{\\mathrm{OA}} = \\sqrt{2}$, $\\overline{\\mathrm{OB}} = 2\\sqrt{2}$μ΄κ³ \n\\[ \\cos(\\angle \\mathrm{AOB}) = \\frac{1}{4} \\]\nμΈ ννμ¬λ³ν $\\mathrm{OACB}$μ λνμ¬ μ $\\mathrm{P}$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] $\\overrightarrow{\\mathrm{OP}} = s \\overrightarrow{\\mathrm{OA}} + t \\overrightarrow{\\mathrm{OB}} \\quad (0 \\leq s \\leq 1, \\ 0 \\leq t \\leq 1)$ \\item[(λ)] $\\overrightarrow{\\mathrm{OP}} \\cdot \\overrightarrow{\\mathrm{OB}} + \\overrightarrow{\\mathrm{BP}} \\cdot \\overrightarrow{\\mathrm{BC}} = 2$ \\end{itemize}\n\nμ $\\mathrm{O}$λ₯Ό μ€μ¬μΌλ‘ νκ³ μ $\\mathrm{A}$λ₯Ό μ§λλ μ μλ₯Ό μμ§μ΄λ μ $\\mathrm{X}$μ λνμ¬ $|3\\overrightarrow{\\mathrm{OP}} - \\overrightarrow{\\mathrm{OX}}|$μ μ΅λκ°κ³Ό μ΅μκ°μ κ°κ° $M$, $m$μ΄λΌ νμ. $M \\times m = a\\sqrt{6} + b$μΌ λ, $a^2 + b^2$μ κ°μ ꡬνμμ€. (λ¨, $a$μ $b$λ μ 리μμ΄λ€.) [4μ ]", "answer": 100, "score": 4, "review": "Removed figure.", "incomplete": false}
|
46 |
+
{"id": 46, "name": "30_geom", "problem": "30. μ’ν곡κ°μ μ€μ¬μ΄ $\\mathrm{C}(2, \\sqrt{5}, 5)$μ΄κ³ μ $\\mathrm{P}(0, 0, 1)$μ μ§λλ ꡬ \\[ S: (x - 2)^2 + (y - \\sqrt{5})^2 + (z - 5)^2 = 25 \\] κ° μλ€. ꡬ $S$κ° νλ©΄ $\\mathrm{OPC}$μ λ§λμ μκΈ°λ μ μλ₯Ό μμ§μ΄λ μ $\\mathrm{Q}$, ꡬ $S$ μλ₯Ό μμ§μ΄λ μ $\\mathrm{R}$μ λνμ¬ λ μ $\\mathrm{Q}, \\mathrm{R}$μ $xy$νλ©΄ μλ‘μ μ μ¬μμ κ°κ° $\\mathrm{Q}_1, \\mathrm{R}_1$μ΄λΌ νμ.\n\nμΌκ°ν $\\mathrm{O}\\mathrm{Q}_1\\mathrm{R}_1$μ λμ΄κ° μ΅λκ° λλλ‘ νλ λ μ $\\mathrm{Q}, \\mathrm{R}$μ λνμ¬ μΌκ°ν $\\mathrm{O}\\mathrm{Q}_1\\mathrm{R}_1$μ νλ©΄ $\\mathrm{PQR}$ μλ‘μ μ μ¬μμ λμ΄λ $\\frac{q}{p} \\sqrt{6}$μ΄λ€. $p+q$μ κ°μ ꡬνμμ€.\n\n(λ¨, $\\mathrm{O}$λ μμ μ΄κ³ μΈ μ $\\mathrm{O}, \\mathrm{Q}_1, \\mathrm{R}_1$μ ν μ§μ μμ μμ§ μμΌλ©°, $p$μ $q$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]", "answer": 23, "score": 4, "review": "Removed figure.", "incomplete": false}
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data/json/2022/math/answer_score_comment.json
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|
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|
|
|
|
|
|
|
|
|
|
|
1 |
+
{
|
2 |
+
"1":{
|
3 |
+
"name":"1",
|
4 |
+
"answer":"2",
|
5 |
+
"score":"2",
|
6 |
+
"comment":""
|
7 |
+
},
|
8 |
+
"2":{
|
9 |
+
"name":"2",
|
10 |
+
"answer":"5",
|
11 |
+
"score":"2",
|
12 |
+
"comment":""
|
13 |
+
},
|
14 |
+
"3":{
|
15 |
+
"name":"3",
|
16 |
+
"answer":"5",
|
17 |
+
"score":"3",
|
18 |
+
"comment":""
|
19 |
+
},
|
20 |
+
"4":{
|
21 |
+
"name":"4",
|
22 |
+
"answer":"4",
|
23 |
+
"score":"3",
|
24 |
+
"comment":"Removed figure and the statement referring to the figure. Need the figure to solve the problem"
|
25 |
+
},
|
26 |
+
"5":{
|
27 |
+
"name":"5",
|
28 |
+
"answer":"1",
|
29 |
+
"score":"3",
|
30 |
+
"comment":""
|
31 |
+
},
|
32 |
+
"6":{
|
33 |
+
"name":"6",
|
34 |
+
"answer":"3",
|
35 |
+
"score":"3",
|
36 |
+
"comment":""
|
37 |
+
},
|
38 |
+
"7":{
|
39 |
+
"name":"7",
|
40 |
+
"answer":"1",
|
41 |
+
"score":"3",
|
42 |
+
"comment":""
|
43 |
+
},
|
44 |
+
"8":{
|
45 |
+
"name":"8",
|
46 |
+
"answer":"1",
|
47 |
+
"score":"3",
|
48 |
+
"comment":""
|
49 |
+
},
|
50 |
+
"9":{
|
51 |
+
"name":"9",
|
52 |
+
"answer":"4",
|
53 |
+
"score":"4",
|
54 |
+
"comment":"Removed figure"
|
55 |
+
},
|
56 |
+
"10":{
|
57 |
+
"name":"10",
|
58 |
+
"answer":"5",
|
59 |
+
"score":"4",
|
60 |
+
"comment":""
|
61 |
+
},
|
62 |
+
"11":{
|
63 |
+
"name":"11",
|
64 |
+
"answer":"3",
|
65 |
+
"score":"4",
|
66 |
+
"comment":"Removed figure and the statement referring to the figure"
|
67 |
+
},
|
68 |
+
"12":{
|
69 |
+
"name":"12",
|
70 |
+
"answer":"3",
|
71 |
+
"score":"4",
|
72 |
+
"comment":""
|
73 |
+
},
|
74 |
+
"13":{
|
75 |
+
"name":"13",
|
76 |
+
"answer":"2",
|
77 |
+
"score":"4",
|
78 |
+
"comment":""
|
79 |
+
},
|
80 |
+
"14":{
|
81 |
+
"name":"14",
|
82 |
+
"answer":"3",
|
83 |
+
"score":"4",
|
84 |
+
"comment":"<보기> changed to 'μλ γ±,γ΄,γ·, μ€'"
|
85 |
+
},
|
86 |
+
"15":{
|
87 |
+
"name":"15",
|
88 |
+
"answer":"2",
|
89 |
+
"score":"4",
|
90 |
+
"comment":"Removed figure and the statement referring to the figure"
|
91 |
+
},
|
92 |
+
"16":{
|
93 |
+
"name":"16",
|
94 |
+
"answer":"3",
|
95 |
+
"score":"3",
|
96 |
+
"comment":""
|
97 |
+
},
|
98 |
+
"17":{
|
99 |
+
"name":"17",
|
100 |
+
"answer":"4",
|
101 |
+
"score":"3",
|
102 |
+
"comment":""
|
103 |
+
},
|
104 |
+
"18":{
|
105 |
+
"name":"18",
|
106 |
+
"answer":"12",
|
107 |
+
"score":"3",
|
108 |
+
"comment":""
|
109 |
+
},
|
110 |
+
"19":{
|
111 |
+
"name":"19",
|
112 |
+
"answer":"6",
|
113 |
+
"score":"3",
|
114 |
+
"comment":""
|
115 |
+
},
|
116 |
+
"20":{
|
117 |
+
"name":"20",
|
118 |
+
"answer":"110",
|
119 |
+
"score":"4",
|
120 |
+
"comment":""
|
121 |
+
},
|
122 |
+
"21":{
|
123 |
+
"name":"21",
|
124 |
+
"answer":"678",
|
125 |
+
"score":"4",
|
126 |
+
"comment":""
|
127 |
+
},
|
128 |
+
"22":{
|
129 |
+
"name":"22",
|
130 |
+
"answer":"9",
|
131 |
+
"score":"4",
|
132 |
+
"comment":""
|
133 |
+
},
|
134 |
+
"23":{
|
135 |
+
"name":"23_prob",
|
136 |
+
"answer":"4",
|
137 |
+
"score":"2",
|
138 |
+
"comment":""
|
139 |
+
},
|
140 |
+
"24":{
|
141 |
+
"name":"24_prob",
|
142 |
+
"answer":"4",
|
143 |
+
"score":"3",
|
144 |
+
"comment":""
|
145 |
+
},
|
146 |
+
"25":{
|
147 |
+
"name":"25_prob",
|
148 |
+
"answer":"1",
|
149 |
+
"score":"3",
|
150 |
+
"comment":""
|
151 |
+
},
|
152 |
+
"26":{
|
153 |
+
"name":"26_prob",
|
154 |
+
"answer":"3",
|
155 |
+
"score":"3",
|
156 |
+
"comment":"Removed figure"
|
157 |
+
},
|
158 |
+
"27":{
|
159 |
+
"name":"27_prob",
|
160 |
+
"answer":"2",
|
161 |
+
"score":"3",
|
162 |
+
"comment":""
|
163 |
+
},
|
164 |
+
"28":{
|
165 |
+
"name":"28_prob",
|
166 |
+
"answer":"1",
|
167 |
+
"score":"4",
|
168 |
+
"comment":""
|
169 |
+
},
|
170 |
+
"29":{
|
171 |
+
"name":"29_prob",
|
172 |
+
"answer":"31",
|
173 |
+
"score":"4",
|
174 |
+
"comment":"Removed figure and the statement referring to the figure. Need the figure to solve the problem"
|
175 |
+
},
|
176 |
+
"30":{
|
177 |
+
"name":"30_prob",
|
178 |
+
"answer":"191",
|
179 |
+
"score":"4",
|
180 |
+
"comment":""
|
181 |
+
},
|
182 |
+
"31":{
|
183 |
+
"name":"23_calc",
|
184 |
+
"answer":"5",
|
185 |
+
"score":"2",
|
186 |
+
"comment":""
|
187 |
+
},
|
188 |
+
"32":{
|
189 |
+
"name":"24_calc",
|
190 |
+
"answer":"4",
|
191 |
+
"score":"3",
|
192 |
+
"comment":""
|
193 |
+
},
|
194 |
+
"33":{
|
195 |
+
"name":"25_calc",
|
196 |
+
"answer":"2",
|
197 |
+
"score":"3",
|
198 |
+
"comment":""
|
199 |
+
},
|
200 |
+
"34":{
|
201 |
+
"name":"26_calc",
|
202 |
+
"answer":"3",
|
203 |
+
"score":"3",
|
204 |
+
"comment":""
|
205 |
+
},
|
206 |
+
"35":{
|
207 |
+
"name":"27_calc",
|
208 |
+
"answer":"1",
|
209 |
+
"score":"3",
|
210 |
+
"comment":""
|
211 |
+
},
|
212 |
+
"36":{
|
213 |
+
"name":"28_calc",
|
214 |
+
"answer":"2",
|
215 |
+
"score":"4",
|
216 |
+
"comment":""
|
217 |
+
},
|
218 |
+
"37":{
|
219 |
+
"name":"29_calc",
|
220 |
+
"answer":"11",
|
221 |
+
"score":"4",
|
222 |
+
"comment":"Removed figure and the statement referring to the figure"
|
223 |
+
},
|
224 |
+
"38":{
|
225 |
+
"name":"30_calc",
|
226 |
+
"answer":"143",
|
227 |
+
"score":"4",
|
228 |
+
"comment":""
|
229 |
+
},
|
230 |
+
"39":{
|
231 |
+
"name":"23_geom",
|
232 |
+
"answer":"2",
|
233 |
+
"score":"2",
|
234 |
+
"comment":""
|
235 |
+
},
|
236 |
+
"40":{
|
237 |
+
"name":"24_geom",
|
238 |
+
"answer":"3",
|
239 |
+
"score":"3",
|
240 |
+
"comment":""
|
241 |
+
},
|
242 |
+
"41":{
|
243 |
+
"name":"25_geom",
|
244 |
+
"answer":"5",
|
245 |
+
"score":"3",
|
246 |
+
"comment":""
|
247 |
+
},
|
248 |
+
"42":{
|
249 |
+
"name":"26_geom",
|
250 |
+
"answer":"2",
|
251 |
+
"score":"3",
|
252 |
+
"comment":"Removed figure"
|
253 |
+
},
|
254 |
+
"43":{
|
255 |
+
"name":"27_geom",
|
256 |
+
"answer":"4",
|
257 |
+
"score":"3",
|
258 |
+
"comment":"Removed figure and the statement referring to the figure"
|
259 |
+
},
|
260 |
+
"44":{
|
261 |
+
"name":"28_geom",
|
262 |
+
"answer":"5",
|
263 |
+
"score":"4",
|
264 |
+
"comment":"Removed figure"
|
265 |
+
},
|
266 |
+
"45":{
|
267 |
+
"name":"29_geom",
|
268 |
+
"answer":"100",
|
269 |
+
"score":"4",
|
270 |
+
"comment":"Removed figure"
|
271 |
+
},
|
272 |
+
"46":{
|
273 |
+
"name":"30_geom",
|
274 |
+
"answer":"23",
|
275 |
+
"score":"4",
|
276 |
+
"comment":"Removed figure"
|
277 |
+
}
|
278 |
+
}
|
data/json/2022/math/math_1.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
1. $\left(2^{\sqrt{3}} \times 4\right)^{\sqrt{3} - 2}$ μ κ°μ? [2μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] $\frac{1}{4}$
|
5 |
+
\item[2] $\frac{1}{2}$
|
6 |
+
\item[3] $1$
|
7 |
+
\item[4] $2$
|
8 |
+
\item[5] $4$
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_10.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
10. μΌμ°¨ν¨μ \( f(x) \)μ λνμ¬ κ³‘μ \( y = f(x) \) μμ μ \( (0, 0) \)μμμ μ μ κ³Ό 곑μ \( y = x f(x) \) μμ μ \( (1, 2) \)μμμ μ μ μ΄ μΌμΉν λ, \( f'(2) \)μ κ°μ? [4μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] $-18$
|
5 |
+
\item[2] $-17$
|
6 |
+
\item[3] $-16$
|
7 |
+
\item[4] $-15$
|
8 |
+
\item[5] $-14$
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_11.txt
ADDED
@@ -0,0 +1,13 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
11. μμ \( a \)μ λνμ¬ μ§ν© \( \left\{ x \ \middle| \ -\frac{a}{2} < x \leq a, \ x \neq \frac{a}{2} \right\} \) μμ μ μλ ν¨μ
|
2 |
+
\[
|
3 |
+
f(x) = \tan \frac{\pi x}{a}
|
4 |
+
\]
|
5 |
+
κ° μλ€. κ·Έλ¦Όκ³Ό κ°μ΄ ν¨μ \( y = f(x) \)μ κ·Έλν μμ μΈ μ \( \mathrm{O, A, B} \)λ₯Ό μ§λλ μ§μ μ΄ μλ€. μ \( \mathrm{A} \)λ₯Ό μ§λκ³ \( x \)μΆμ ννν μ§μ μ΄ ν¨μ \( y = f(x) \)μ κ·Έλνμ λ§λλ μ μ€ \( \mathrm{A} \)κ° μλ μ μ \( \mathrm{C} \)λΌ νμ. μΌκ°ν \( \mathrm{ABC} \)κ° μ μΌκ°νμΌ λ, μΌκ°ν \( \mathrm{ABC} \)μ λμ΄λ? (λ¨, \( \mathrm{O} \)λ μμ μ΄λ€.) [4μ ]
|
6 |
+
|
7 |
+
\begin{itemize}
|
8 |
+
\item[1] \( \frac{3\sqrt{3}}{2} \)
|
9 |
+
\item[2] \( \frac{17\sqrt{3}}{12} \)
|
10 |
+
\item[3] \( \frac{4\sqrt{3}}{3} \)
|
11 |
+
\item[4] \( \frac{5\sqrt{3}}{4} \)
|
12 |
+
\item[5] \( \frac{7\sqrt{3}}{6} \)
|
13 |
+
\end{itemize}
|
data/json/2022/math/math_12.txt
ADDED
@@ -0,0 +1,21 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
12. μ€μ μ 체μ μ§ν©μμ μ°μμΈ ν¨μ $f(x)$κ° λͺ¨λ μ€μ $x$μ λνμ¬
|
2 |
+
|
3 |
+
\[
|
4 |
+
\{f(x)\}^3 - \{f(x)\}^2 - x^2 f(x) + x^2 = 0
|
5 |
+
\]
|
6 |
+
|
7 |
+
μ λ§μ‘±μν¨λ€. ν¨μ $f(x)$μ μ΅λκ°μ΄ 1μ΄κ³ μ΅μκ°μ΄ 0μΌ λ,
|
8 |
+
|
9 |
+
\[
|
10 |
+
f\left( -\frac{4}{3} \right) + f(0) + f\left( \frac{1}{2} \right)
|
11 |
+
\]
|
12 |
+
|
13 |
+
μ κ°μ? [4μ ]
|
14 |
+
|
15 |
+
\begin{itemize}
|
16 |
+
\item[1] $\frac{1}{2}$
|
17 |
+
\item[2] $1$
|
18 |
+
\item[3] $\frac{3}{2}$
|
19 |
+
\item[4] $2$
|
20 |
+
\item[5] $\frac{5}{2}$
|
21 |
+
\end{itemize}
|
data/json/2022/math/math_13.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
13. λ μμ \( a, b \ (1 < a < b) \)μ λνμ¬ μ’ννλ©΄ μμ λ μ \((a, \log_2 a), \ (b, \log_2 b)\)λ₯Ό μ§λλ μ§μ μ \(y\)μ νΈκ³Ό λ μ \((a, \log_4 a), \ (b, \log_4 b)\)λ₯Ό μ§λλ μ§μ μ \(y\)μ νΈμ΄ κ°λ€. ν¨μ \( f(x) = a^{bx} + b^{ax} \)μ λνμ¬ \( f(1) = 40 \)μΌ λ, \( f(2) \)μ κ°μ? [4μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] 760
|
5 |
+
\item[2] 800
|
6 |
+
\item[3] 840
|
7 |
+
\item[4] 880
|
8 |
+
\item[5] 920
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_14.txt
ADDED
@@ -0,0 +1,19 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
14. μμ§μ μλ₯Ό μμ§μ΄λ μ $\mathrm{P}$μ μκ° $t$μμμ μμΉ $x(t)$κ° λ μμ $a$, $b$μ λνμ¬
|
2 |
+
\[
|
3 |
+
x(t) = t(t - 1)(at + b) \quad (a \neq 0)
|
4 |
+
\]
|
5 |
+
μ΄λ€. μ $\mathrm{P}$μ μκ° $t$μμμ μλ $v(t)$κ° $\int_0^1 |v(t)| \, dt = 2$λ₯Ό λ§μ‘±μν¬ λ, μλ γ±, γ΄, γ· μ€μμ μ³μ κ²λ§μ μλ λλ‘ κ³ λ₯Έ κ²μ? [4μ ]
|
6 |
+
|
7 |
+
\begin{itemize}
|
8 |
+
\item[γ±.] $\int_0^1 v(t) \, dt = 0$
|
9 |
+
\item[γ΄.] $|x(t_1)| > 1$μΈ $t_1$μ΄ μ΄λ¦°κ΅¬κ° $(0, 1)$μ μ‘΄μ¬νλ€.
|
10 |
+
\item[γ·.] $0 \leq t \leq 1$μΈ λͺ¨λ $t$μ λνμ¬ $|x(t)| < 1$μ΄λ©΄ $x(t_2) = 0$μΈ $t_2$κ° μ΄λ¦°κ΅¬κ° $(0, 1)$μ μ‘΄μ¬νλ€.
|
11 |
+
\end{itemize}
|
12 |
+
|
13 |
+
\begin{itemize}
|
14 |
+
\item[1] γ±
|
15 |
+
\item[2] γ±, γ΄
|
16 |
+
\item[3] γ±, γ·
|
17 |
+
\item[4] γ΄, γ·
|
18 |
+
\item[5] γ±, γ΄, γ·
|
19 |
+
\end{itemize}
|
data/json/2022/math/math_15.txt
ADDED
@@ -0,0 +1,31 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
15. λ μ \( \mathrm{O}_1, \mathrm{O}_2 \)λ₯Ό κ°κ° μ€μ¬μΌλ‘ νκ³ λ°μ§λ¦μ κΈΈμ΄κ° \(\overline{\mathrm{O}_1\mathrm{O}_2} \)μΈ λ μ \( C_1, C_2 \)κ° μλ€. κ·Έλ¦Όκ³Ό κ°μ΄ μ \( C_1 \) μμ μλ‘ λ€λ₯Έ μΈ μ \( \mathrm{A}, \mathrm{B}, \mathrm{C} \)μ μ \( C_2 \) μμ μ \( \mathrm{D} \)κ° μ£Όμ΄μ Έ μκ³ , μΈ μ \( \mathrm{A}, \mathrm{O}_1, \mathrm{O}_2 \)μ μΈ μ \( \mathrm{C}, \mathrm{O}_2, \mathrm{D} \)κ° κ°κ° ν μ§μ μμ μλ€.
|
2 |
+
|
3 |
+
μ΄λ \(\angle \mathrm{B}\mathrm{O}_1\mathrm{A} = \theta_1\), \(\angle \mathrm{O}_2\mathrm{O}_1\mathrm{C} = \theta_2\), \(\angle \mathrm{O}_1\mathrm{O}_2\mathrm{D} = \theta_3\)μ΄λΌ νμ.
|
4 |
+
|
5 |
+
λ€μμ \( \overline{\mathrm{A}\mathrm{B}} : \overline{\mathrm{O}_1\mathrm{D}} = 1 : 2\sqrt{2} \)μ΄κ³ \( \theta_3 = \theta_1 + \theta_2 \)μΌ λ, μ λΆ \( \mathrm{A}\mathrm{B} \)μ μ λΆ \( \mathrm{C}\mathrm{D} \)μ κΈΈμ΄μ λΉλ₯Ό ꡬνλ κ³Όμ μ΄λ€.
|
6 |
+
|
7 |
+
\[
|
8 |
+
\begin{aligned}
|
9 |
+
&\angle \mathrm{C}\mathrm{O}_2\mathrm{O}_1 + \angle \mathrm{O}_1\mathrm{O}_2\mathrm{D} = \pi \text{μ΄λ―λ‘ } \theta_3 = \frac{\pi}{2} + \frac{\theta_2}{2} \text{μ΄κ³ } \\
|
10 |
+
&\theta_3 = \theta_1 + \theta_2 \text{μμ } 2\theta_1 + \theta_2 = \pi \text{μ΄λ―λ‘ } \angle \mathrm{C}\mathrm{O}_1\mathrm{B} = \theta_1 \text{μ΄λ€.} \\
|
11 |
+
&\text{μ΄λ } \angle \mathrm{O}_2\mathrm{O}_1\mathrm{B} = \theta_1 + \theta_2 = \theta_3 \text{μ΄λ―λ‘ μΌκ°ν } \mathrm{O}_1\mathrm{O}_2\mathrm{B} \text{μ μΌκ°ν } \mathrm{O}_2\mathrm{O}_1\mathrm{D} \text{λ ν©λμ΄λ€.} \\
|
12 |
+
&\overline{\mathrm{A}\mathrm{B}} = k \text{λΌ ν λ} \\
|
13 |
+
&\overline{\mathrm{B}\mathrm{O}_2} = \overline{\mathrm{O}_1\mathrm{D}}= 2\sqrt{2}k \text{μ΄λ―λ‘ } \overline{\mathrm{A}\mathrm{O}_2} = \text{(κ°)μ΄κ³ ,} \\
|
14 |
+
&\angle \mathrm{B}\mathrm{O}_2\mathrm{A} = \frac{\theta_1}{2} \text{μ΄λ―λ‘ } \cos \frac{\theta_1}{2} = \text{(λ) μ΄λ€.} \\
|
15 |
+
&\text{μΌκ°ν } \mathrm{O}_2\mathrm{B}\mathrm{C} \text{μμ} \\
|
16 |
+
&\overline{\mathrm{B}\mathrm{C}} = k, \overline{\mathrm{B}\mathrm{O}_2} = 2\sqrt{2}k, \angle \mathrm{C}\mathrm{O}_2\mathrm{B} = \frac{\theta_1}{2} \text{μ΄λ―λ‘} \\
|
17 |
+
&\text{μ½μ¬μΈλ²μΉμ μνμ¬ } \overline{\mathrm{O}_2\mathrm{C}} = \text{(λ€) μ΄λ€.} \\
|
18 |
+
&\overline{\mathrm{C}\mathrm{D}} = \overline{\mathrm{O}_2\mathrm{D}} + \overline{\mathrm{O}_2\mathrm{C}} = \overline{\mathrm{O}_1\mathrm{O}_2} + \overline{\mathrm{O}_2\mathrm{C}} \text{μ΄λ―λ‘} \\
|
19 |
+
&\overline{\mathrm{A}\mathrm{B}} : \overline{\mathrm{C}\mathrm{D}} = k : \left(\frac{\text{(κ°)}}{2} + \text{(λ€)}\right) \text{μ΄λ€.}
|
20 |
+
\end{aligned}
|
21 |
+
\]
|
22 |
+
|
23 |
+
μμ (κ°), (λ€)μ μλ§μ μμ κ°κ° \( f(k), g(k) \)λΌ νκ³ , (λ)μ μλ§μ μλ₯Ό \( p \)λΌ ν λ, \( f(p) \times g(p) \)μ κ°μ? [4μ ]
|
24 |
+
|
25 |
+
\begin{itemize}
|
26 |
+
\item[1] \(\frac{169}{27}\)
|
27 |
+
\item[2] \(\frac{56}{9}\)
|
28 |
+
\item[3] \(\frac{167}{27}\)
|
29 |
+
\item[4] \(\frac{166}{27}\)
|
30 |
+
\item[5] \(\frac{55}{9}\)
|
31 |
+
\end{itemize}
|
data/json/2022/math/math_16.txt
ADDED
@@ -0,0 +1 @@
|
|
|
|
|
1 |
+
16. $\log_2 120 - \frac{1}{\log_{15} 2}$ μ κ°μ ꡬνμμ€. [3μ ]
|
data/json/2022/math/math_17.txt
ADDED
@@ -0,0 +1 @@
|
|
|
|
|
1 |
+
17. ν¨μ $f(x)$μ λνμ¬ $f'(x) = 3x^2 + 2x$μ΄κ³ $f(0) = 2$μΌ λ, $f(1)$μ κ°μ ꡬνμμ€. [3μ ]
|
data/json/2022/math/math_18.txt
ADDED
@@ -0,0 +1,7 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
18. μμ΄ $\{a_n\}$μ λνμ¬
|
2 |
+
|
3 |
+
\[
|
4 |
+
\sum_{k=1}^{10} a_k - \sum_{k=1}^{7} \frac{a_k}{2} = 56, \quad \sum_{k=1}^{10} 2a_k - \sum_{k=1}^{8} a_k = 100
|
5 |
+
\]
|
6 |
+
|
7 |
+
μΌ λ, $a_8$μ κ°μ ꡬνμμ€. [3μ ]
|
data/json/2022/math/math_19.txt
ADDED
@@ -0,0 +1 @@
|
|
|
|
|
1 |
+
19. ν¨μ $f(x) = x^3 + ax^2 - (a^2 - 8a)x + 3$μ΄ μ€μ μ 체μ μ§ν©μμ μ¦κ°νλλ‘ νλ μ€μ $a$μ μ΅λκ°μ ꡬνμμ€. [3μ ]
|
data/json/2022/math/math_2.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
2. ν¨μ \( f(x) = x^3 + 3x^2 + x - 1 \) μ λνμ¬ \( f'(1) \)μ κ°μ? [2μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] 6
|
5 |
+
\item[2] 7
|
6 |
+
\item[3] 8
|
7 |
+
\item[4] 9
|
8 |
+
\item[5] 10
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_20.txt
ADDED
@@ -0,0 +1,8 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
20. μ€μ μ 체μ μ§ν©μμ λ―ΈλΆκ°λ₯ν ν¨μ \( f(x) \)κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[(κ°)] λ«νκ΅¬κ° \([0, 1]\)μμ \( f(x) = x \)μ΄λ€.
|
5 |
+
\item[(λ)] μ΄λ€ μμ \( a, b \)μ λνμ¬ κ΅¬κ° \([0, \infty)\)μμ \( f(x+1) - x f(x) = ax + b \)μ΄λ€.
|
6 |
+
\end{itemize}
|
7 |
+
|
8 |
+
\[ 60 \times \int_1^2 f(x) \, dx \] μ κ°μ ꡬνμμ€. [4μ ]
|
data/json/2022/math/math_21.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
21. μμ΄ $\{a_n\}$μ΄ λ€μ 쑰건μ λ§μ‘±μν¨λ€.
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[(κ°)] \( |a_1| = 2 \)
|
5 |
+
\item[(λ)] λͺ¨λ μμ°μ \( n \)μ λνμ¬ \( |a_{n+1}| = 2|a_n| \)μ΄λ€.
|
6 |
+
\item[(λ€)] \(\sum_{n=1}^{10} a_n = -14\)
|
7 |
+
\end{itemize}
|
8 |
+
|
9 |
+
$a_1 + a_3 + a_5 + a_7 + a_9$μ κ°μ ꡬνμμ€. [4μ ]
|
data/json/2022/math/math_22.txt
ADDED
@@ -0,0 +1,8 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
22. μ΅κ³ μ°¨νμ κ³μκ° $\frac{1}{2}$ μΈ μΌμ°¨ν¨μ $f(x)$μ μ€μ $t$μ λνμ¬ λ°©μ μ $f'(x) = 0$μ΄ λ«νκ΅¬κ° $[t, t+2]$μμ κ°λ μ€κ·Όμ κ°μλ₯Ό $g(t)$λΌ ν λ, ν¨μ $g(t)$λ λ€μ 쑰건μ λ§μ‘±μν¨λ€.
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[(κ°)] λͺ¨λ μ€μ \( a \)μ λνμ¬ \(\lim_{t \to a+} g(t) + \lim_{t \to a-} g(t) \leq 2\)μ΄λ€.
|
5 |
+
\item[(λ)] \( g(f(1)) = g(f(4)) = 2, \ g(f(0)) = 1 \)
|
6 |
+
\end{itemize}
|
7 |
+
|
8 |
+
$f(5)$μ κ°μ ꡬνμμ€. [4μ ]
|
data/json/2022/math/math_23_calc.txt
ADDED
@@ -0,0 +1,12 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
23.
|
2 |
+
\[
|
3 |
+
\lim_{n \to \infty} \frac{\frac{5}{n} + \frac{3}{n^2}}{\frac{1}{n} - \frac{2}{n^3}} \text{μ κ°μ? [2μ ]}
|
4 |
+
\]
|
5 |
+
|
6 |
+
\begin{itemize}
|
7 |
+
\item[1] 1
|
8 |
+
\item[2] 2
|
9 |
+
\item[3] 3
|
10 |
+
\item[4] 4
|
11 |
+
\item[5] 5
|
12 |
+
\end{itemize}
|
data/json/2022/math/math_23_geom.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
23. μ’ν곡κ°μ μ $\mathrm{A}(2, 1, 3)$μ $xy$ νλ©΄μ λνμ¬ λμΉμ΄λν μ μ $\mathrm{P}$λΌ νκ³ , μ $\mathrm{A}$λ₯Ό $yz$ νλ©΄μ λνμ¬ λμΉμ΄λν μ μ $\mathrm{Q}$λΌ ν λ, μ λΆ $\mathrm{PQ}$μ κΈΈμ΄λ? [2μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] $5 \sqrt{2}$
|
5 |
+
\item[2] $2 \sqrt{13}$
|
6 |
+
\item[3] $3 \sqrt{6}$
|
7 |
+
\item[4] $2 \sqrt{14}$
|
8 |
+
\item[5] $2 \sqrt{15}$
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_23_prob.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
23. λ€νμ $(x+2)^7$μ μ κ°μμμ $x^5$μ κ³μλ? [2μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] 42
|
5 |
+
\item[2] 56
|
6 |
+
\item[3] 70
|
7 |
+
\item[4] 84
|
8 |
+
\item[5] 98
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_24_calc.txt
ADDED
@@ -0,0 +1,13 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
24. μ€μ μ 체μ μ§ν©μμ λ―ΈλΆκ°λ₯ν ν¨μ $f(x)$κ° λͺ¨λ μ€μ $x$μ λνμ¬
|
2 |
+
\[
|
3 |
+
f(x^3 + x) = e^x
|
4 |
+
\]
|
5 |
+
μ λ§μ‘±μν¬ λ, $f'(2)$μ κ°μ? [3μ ]
|
6 |
+
|
7 |
+
\begin{itemize}
|
8 |
+
\item[1] $e$
|
9 |
+
\item[2] $\frac{e}{2}$
|
10 |
+
\item[3] $\frac{e}{3}$
|
11 |
+
\item[4] $\frac{e}{4}$
|
12 |
+
\item[5] $\frac{e}{5}$
|
13 |
+
\end{itemize}
|
data/json/2022/math/math_24_geom.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
24. ν μ΄μ μ μ’νκ° $\left( 3\sqrt{2}, 0 \right)$ μΈ μ곑μ $\frac{x^2}{a^2} - \frac{y^2}{6} = 1$ μ μ£ΌμΆμ κΈΈμ΄λ? (λ¨, $a$ λ μμμ΄λ€.) [3μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] $3\sqrt{3}$
|
5 |
+
\item[2] $\frac{7\sqrt{3}}{2}$
|
6 |
+
\item[3] $4\sqrt{3}$
|
7 |
+
\item[4] $\frac{9\sqrt{3}}{2}$
|
8 |
+
\item[5] $5\sqrt{3}$
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_24_prob.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
24. νλ₯ λ³μ \( X \)κ° μ΄νλΆν¬ \( \mathrm{B}\left(n, \frac{1}{3}\right) \)μ λ°λ₯΄κ³ \( \mathrm{V}(2X) = 40 \)μΌ λ, \( n \)μ κ°μ? [3μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] 30
|
5 |
+
\item[2] 35
|
6 |
+
\item[3] 40
|
7 |
+
\item[4] 45
|
8 |
+
\item[5] 50
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_25_calc.txt
ADDED
@@ -0,0 +1,13 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
25. λ±λΉμμ΄ $\{a_n\}$μ λνμ¬
|
2 |
+
\[
|
3 |
+
\sum_{n=1}^{\infty} (a_{2n-1} - a_{2n}) = 3, \quad \sum_{n=1}^{\infty} a_n^2 = 6
|
4 |
+
\]
|
5 |
+
μΌ λ, $\sum_{n=1}^{\infty} a_n$ μ κ°μ? [3μ ]
|
6 |
+
|
7 |
+
\begin{itemize}
|
8 |
+
\item[1] 1
|
9 |
+
\item[2] 2
|
10 |
+
\item[3] 3
|
11 |
+
\item[4] 4
|
12 |
+
\item[5] 5
|
13 |
+
\end{itemize}
|
data/json/2022/math/math_25_geom.txt
ADDED
@@ -0,0 +1,15 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
25. μ’ννλ©΄μμ λ μ§μ
|
2 |
+
|
3 |
+
\[
|
4 |
+
\frac{x+1}{2} = y - 3, \quad x - 2 = \frac{y - 5}{3}
|
5 |
+
\]
|
6 |
+
|
7 |
+
κ° μ΄λ£¨λ μκ°μ ν¬κΈ°λ₯Ό $\theta$λΌ ν λ, $\cos \theta$μ κ°μ? [3μ ]
|
8 |
+
|
9 |
+
\begin{itemize}
|
10 |
+
\item[1] $\frac{1}{2}$
|
11 |
+
\item[2] $\frac{\sqrt{5}}{4}$
|
12 |
+
\item[3] $\frac{\sqrt{6}}{4}$
|
13 |
+
\item[4] $\frac{\sqrt{7}}{4}$
|
14 |
+
\item[5] $\frac{\sqrt{2}}{2}$
|
15 |
+
\end{itemize}
|
data/json/2022/math/math_25_prob.txt
ADDED
@@ -0,0 +1,14 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
25. λ€μ 쑰건μ λ§μ‘±μν€λ μμ°μ $a, \ b, \ c, \ d, \ e$μ λͺ¨λ μμμ $(a, b, c, d, e)$μ κ°μλ? [3μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[(κ°)] $a + b + c + d + e = 12$
|
5 |
+
\item[(λ)] $\left| a^2 - b^2 \right| = 5$
|
6 |
+
\end{itemize}
|
7 |
+
|
8 |
+
\begin{itemize}
|
9 |
+
\item[1] 30
|
10 |
+
\item[2] 32
|
11 |
+
\item[3] 34
|
12 |
+
\item[4] 36
|
13 |
+
\item[5] 38
|
14 |
+
\end{itemize}
|
data/json/2022/math/math_26_calc.txt
ADDED
@@ -0,0 +1,12 @@
|
|
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|
|
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|
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|
|
|
|
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|
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|
|
1 |
+
26.
|
2 |
+
\[
|
3 |
+
\lim_{n \to \infty} \sum_{k=1}^{n} \frac{k^2 + 2kn}{k^3 + 3k^2 n + n^3} \text{μ κ°μ?} \quad [3 \text{μ }]
|
4 |
+
\]
|
5 |
+
|
6 |
+
\begin{itemize}
|
7 |
+
\item[1] $\ln 5$
|
8 |
+
\item[2] $\frac{\ln 5}{2}$
|
9 |
+
\item[3] $\frac{\ln 5}{3}$
|
10 |
+
\item[4] $\frac{\ln 5}{4}$
|
11 |
+
\item[5] $\frac{\ln 5}{5}$
|
12 |
+
\end{itemize}
|
data/json/2022/math/math_26_geom.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
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|
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|
|
|
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|
1 |
+
26. λ μ΄μ μ΄ \( \mathrm{F}, \mathrm{F'} \)μΈ νμ \( \frac{x^2}{64} + \frac{y^2}{16} = 1 \) μμ μ μ€ μ 1μ¬λΆλ©΄μ μλ μ \( \mathrm{A} \)κ° μλ€. λ μ§μ \( \mathrm{AF}, \mathrm{AF'} \)μ λμμ μ νκ³ μ€μ¬μ΄ \( y \)μΆ μμ μλ μ μ€ μ€μ¬μ \( y \)μ’νκ° μμμΈ κ²μ \( C \)λΌ νμ. μ \( C \)μ μ€μ¬μ \( \mathrm{B} \)λΌ ν λ μ¬κ°ν \( \mathrm{AFBF'} \)μ λμ΄κ° 72μ΄λ€. μ \( C \)μ λ°μ§λ¦μ κΈΈμ΄λ? [3μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] \( \frac{17}{2} \)
|
5 |
+
\item[2] 9
|
6 |
+
\item[3] \( \frac{19}{2} \)
|
7 |
+
\item[4] 10
|
8 |
+
\item[5] \( \frac{21}{2} \)
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_26_prob.txt
ADDED
@@ -0,0 +1,9 @@
|
|
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|
1 |
+
26. \( 1 \)λΆν° \( 10 \)κΉμ§ μμ°μκ° νλμ© μ ν μλ \( 10 \)μ₯μ μΉ΄λκ° λ€μ΄ μλ μ£Όλ¨Έλκ° μλ€. μ΄ μ£Όλ¨Έλμμ μμλ‘ μΉ΄λ \( 3 \)μ₯μ λμμ κΊΌλΌ λ, κΊΌλΈ μΉ΄λμ μ ν μλ μΈ μμ°μ μ€μμ κ°μ₯ μμ μκ° \( 4 \) μ΄νμ΄κ±°λ \( 7 \) μ΄μμΌ νλ₯ μ? [3μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] \( \frac{4}{5} \)
|
5 |
+
\item[2] \( \frac{5}{6} \)
|
6 |
+
\item[3] \( \frac{13}{15} \)
|
7 |
+
\item[4] \( \frac{9}{10} \)
|
8 |
+
\item[5] \( \frac{14}{15} \)
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_27_calc.txt
ADDED
@@ -0,0 +1,9 @@
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|
1 |
+
27. μ’ννλ©΄ μλ₯Ό μμ§μ΄λ μ $\mathrm{P}$μ μκ° $t \ (t>0)$μμμ μμΉκ° 곑μ $y = x^2$κ³Ό μ§μ $y = t^2 x - \frac{\ln t}{8}$κ° λ§λλ μλ‘ λ€λ₯Έ λ μ μ μ€μ μΌ λ, μκ° $t=1$μμ $t=e$κΉμ§ μ $\mathrm{P}$κ° μμ§μΈ 거리λ? [3μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] $\frac{e^4}{2} - \frac{3}{8}$
|
5 |
+
\item[2] $\frac{e^4}{2} - \frac{5}{16}$
|
6 |
+
\item[3] $\frac{e^4}{2} - \frac{1}{4}$
|
7 |
+
\item[4] $\frac{e^4}{2} - \frac{3}{16}$
|
8 |
+
\item[5] $\frac{e^4}{2} - \frac{1}{8}$
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_27_geom.txt
ADDED
@@ -0,0 +1,9 @@
|
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|
1 |
+
27. κ·Έλ¦Όκ³Ό κ°μ΄ ν λͺ¨μ리μ κΈΈμ΄κ° 4μΈ μ μ‘면체 $\mathrm{ABCD - EFGH}$ κ° μλ€. μ λΆ $\mathrm{AD}$ μ μ€μ μ $\mathrm{M}$μ΄λΌ ν λ, μΌκ°ν $\mathrm{MEG}$ μ λμ΄λ? [3μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] $\frac{21}{2}$
|
5 |
+
\item[2] 11
|
6 |
+
\item[3] $\frac{23}{2}$
|
7 |
+
\item[4] 12
|
8 |
+
\item[5] $\frac{25}{2}$
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_27_prob.txt
ADDED
@@ -0,0 +1,15 @@
|
|
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|
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|
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|
1 |
+
27. μ΄λ μλμ°¨ νμ¬μμ μμ°νλ μ κΈ° μλμ°¨μ 1ν μΆ©μ μ£Όν 거리λ νκ· μ΄ $m$μ΄κ³ νμ€νΈμ°¨κ° $\sigma$μΈ μ κ·λΆν¬λ₯Ό λ°λ₯Έλ€κ³ νλ€.
|
2 |
+
|
3 |
+
μ΄ μλμ°¨ νμ¬μμ μμ°ν μ κΈ° μλμ°¨ 100λλ₯Ό μμμΆμΆνμ¬ μ»μ 1ν μΆ©μ μ£Όν 거리μ νλ³Ένκ· μ΄ $\overline{x_1}$μΌ λ, λͺ¨νκ· $m$μ λν μ λ’°λ 95\%μ μ 뒰ꡬκ°μ΄ $a \le m \le b$μ΄λ€.
|
4 |
+
|
5 |
+
μ΄ μλμ°¨ νμ¬μμ μμ°ν μ κΈ° μλμ°¨ 400λλ₯Ό μμμΆμΆνμ¬ μ»μ 1ν μΆ©μ μ£Όν 거리μ νλ³Ένκ· μ΄ $\overline{x_2}$μΌ λ, λͺ¨νκ· $m$μ λν μ λ’°λ 99\%μ μ 뒰ꡬκ°μ΄ $c \le m \le d$μ΄λ€.
|
6 |
+
|
7 |
+
$\overline{x_1} - \overline{x_2} = 1.34$μ΄κ³ $a = c$μΌ λ, $b - a$μ κ°μ? (λ¨, μ£Όν 거리μ λ¨μλ kmμ΄κ³ , $Z$κ° νμ€μ κ·λΆν¬λ₯Ό λ°λ₯΄λ νλ₯ λ³μμΌ λ $\mathrm{P}(|Z| \le 1.96) = 0.95$, $\mathrm{P}(|Z| \le 2.58) = 0.99$λ‘ κ³μ°νλ€.) [3μ ]
|
8 |
+
|
9 |
+
\begin{itemize}
|
10 |
+
\item[1] 5.88
|
11 |
+
\item[2] 7.84
|
12 |
+
\item[3] 9.80
|
13 |
+
\item[4] 11.76
|
14 |
+
\item[5] 13.72
|
15 |
+
\end{itemize}
|
data/json/2022/math/math_28_calc.txt
ADDED
@@ -0,0 +1,15 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
28. ν¨μ \( f(x) = 6\pi (x - 1)^2 \)μ λνμ¬ ν¨μ \( g(x) \)λ₯Ό
|
2 |
+
|
3 |
+
\[
|
4 |
+
g(x) = 3f(x) + 4\cos f(x)
|
5 |
+
\]
|
6 |
+
|
7 |
+
λΌ νμ. \( 0 < x < 2 \)μμ ν¨μ \( g(x) \)κ° κ·Ήμκ° λλ \( x \)μ κ°μλ? [4μ ]
|
8 |
+
|
9 |
+
\begin{itemize}
|
10 |
+
\item[1] 6
|
11 |
+
\item[2] 7
|
12 |
+
\item[3] 8
|
13 |
+
\item[4] 9
|
14 |
+
\item[5] 10
|
15 |
+
\end{itemize}
|
data/json/2022/math/math_28_geom.txt
ADDED
@@ -0,0 +1,9 @@
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
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|
|
|
|
1 |
+
28. λ μμ \( a \), \( p \)μ λνμ¬ ν¬λ¬Όμ \( (y - a)^2 = 4px \)μ μ΄μ μ \( \mathrm{F}_1 \)μ΄λΌ νκ³ , ν¬λ¬Όμ \( y^2 = -4x \)μ μ΄μ μ \( \mathrm{F}_2 \)λΌ νμ. μ λΆ \( \mathrm{F}_1 \mathrm{F}_2 \)κ° λ ν¬λ¬Όμ κ³Ό λ§λλ μ μ κ°κ° \( \mathrm{P} \), \( \mathrm{Q} \)λΌ ν λ, \( \overline{\mathrm{F}_1 \mathrm{F}_2} = 3 \), \( \overline{\mathrm{P}\mathrm{Q}} = 1 \)μ΄λ€. \( a^2 + p^2 \)μ κ°μ? [4μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[1] 6
|
5 |
+
\item[2] \(\frac{25}{4}\)
|
6 |
+
\item[3] \(\frac{13}{2}\)
|
7 |
+
\item[4] \(\frac{27}{4}\)
|
8 |
+
\item[5] 7
|
9 |
+
\end{itemize}
|
data/json/2022/math/math_28_prob.txt
ADDED
@@ -0,0 +1,14 @@
|
|
|
|
|
|
|
|
|
|
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|
|
|
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|
|
|
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|
|
|
|
|
|
|
|
|
|
|
1 |
+
28. λ μ§ν© $X = \{1, 2, 3, 4, 5\}$, $Y = \{1, 2, 3, 4\}$μ λνμ¬ λ€μ 쑰건μ λ§μ‘±μν€λ $X$μμ $Y$λ‘μ ν¨μ $f$μ κ°μλ? [4μ ]
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[(κ°)] μ§ν© $X$μ λͺ¨λ μμ $x$μ λνμ¬ $f(x) \geq \sqrt{x}$μ΄λ€.
|
5 |
+
\item[(λ)] ν¨μ $f$μ μΉμμ μμμ κ°μλ 3μ΄λ€.
|
6 |
+
\end{itemize}
|
7 |
+
|
8 |
+
\begin{itemize}
|
9 |
+
\item[1] 128
|
10 |
+
\item[2] 138
|
11 |
+
\item[3] 148
|
12 |
+
\item[4] 158
|
13 |
+
\item[5] 168
|
14 |
+
\end{itemize}
|
data/json/2022/math/math_29_calc.txt
ADDED
@@ -0,0 +1,7 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
29. κ·Έλ¦Όκ³Ό κ°μ΄ κΈΈμ΄κ° 2μΈ μ λΆ \(\mathrm{AB}\)λ₯Ό μ§λ¦μΌλ‘ νλ λ°μμ΄ μλ€. νΈ \(\mathrm{AB}\) μμ λ μ \(\mathrm{P}\), \(\mathrm{Q}\)λ₯Ό \(\angle \mathrm{PAB} = \theta\), \(\angle \mathrm{QBA} = 2\theta\)κ° λλλ‘ μ‘κ³ , λ μ λΆ \(\mathrm{AP}\), \(\mathrm{BQ}\)μ κ΅μ μ \(\mathrm{R}\)λΌ νμ. μ λΆ \(\mathrm{AB}\) μμ μ \(\mathrm{S}\), μ λΆ \(\mathrm{BR}\) μμ μ \(\mathrm{T}\), μ λΆ \(\mathrm{AR}\) μμ μ \(\mathrm{U}\)λ₯Ό μ λΆ \(\mathrm{UT}\)κ° μ λΆ \(\mathrm{AB}\)μ νννκ³ μΌκ°ν \(\mathrm{STU}\)κ° μ μΌκ°νμ΄ λλλ‘ μ‘λλ€. λ μ λΆ \(\mathrm{AR}\), \(\mathrm{QR}\)μ νΈ \(\mathrm{AQ}\)λ‘ λλ¬μΈμΈ λΆλΆμ λμ΄λ₯Ό \(f(\theta)\), μΌκ°ν \(\mathrm{STU}\)μ λμ΄λ₯Ό \(g(\theta)\)λΌ ν λ,
|
2 |
+
\[
|
3 |
+
\lim_{\theta \to 0+} \frac{g(\theta)}{\theta \times f(\theta)} = \frac{q}{p} \sqrt{3}
|
4 |
+
\]
|
5 |
+
μ΄λ€. \(p + q\)μ κ°μ ꡬνμμ€.
|
6 |
+
|
7 |
+
(λ¨, \(0 < \theta < \frac{\pi}{6}\)μ΄κ³ , \(p\)μ \(q\)λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]
|
data/json/2022/math/math_29_geom.txt
ADDED
@@ -0,0 +1,14 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
29. μ’ννλ©΄μμ $\overline{\mathrm{OA}} = \sqrt{2}$, $\overline{\mathrm{OB}} = 2\sqrt{2}$μ΄κ³
|
2 |
+
|
3 |
+
\[
|
4 |
+
\cos(\angle \mathrm{AOB}) = \frac{1}{4}
|
5 |
+
\]
|
6 |
+
|
7 |
+
μΈ ννμ¬λ³ν $\mathrm{OACB}$μ λνμ¬ μ $\mathrm{P}$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.
|
8 |
+
|
9 |
+
\begin{itemize}
|
10 |
+
\item[(κ°)] $\overrightarrow{\mathrm{OP}} = s \overrightarrow{\mathrm{OA}} + t \overrightarrow{\mathrm{OB}} \quad (0 \leq s \leq 1, \ 0 \leq t \leq 1)$
|
11 |
+
\item[(λ)] $\overrightarrow{\mathrm{OP}} \cdot \overrightarrow{\mathrm{OB}} + \overrightarrow{\mathrm{BP}} \cdot \overrightarrow{\mathrm{BC}} = 2$
|
12 |
+
\end{itemize}
|
13 |
+
|
14 |
+
μ $\mathrm{O}$λ₯Ό μ€μ¬μΌλ‘ νκ³ μ $\mathrm{A}$λ₯Ό μ§λλ μ μλ₯Ό μμ§μ΄λ μ $\mathrm{X}$μ λνμ¬ $|3\overrightarrow{\mathrm{OP}} - \overrightarrow{\mathrm{OX}}|$μ μ΅λκ°κ³Ό μ΅μκ°μ κ°κ° $M$, $m$μ΄λΌ νμ. $M \times m = a\sqrt{6} + b$μΌ λ, $a^2 + b^2$μ κ°μ ꡬνμμ€. (λ¨, $a$μ $b$λ μ 리μμ΄λ€.) [4μ ]
|
data/json/2022/math/math_29_prob.txt
ADDED
@@ -0,0 +1,12 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
29. λ μ°μνλ₯ λ³μ \( X \)μ \( Y \)κ° κ°λ κ°μ λ²μλ \( 0 \leq X \leq 6 \),
|
2 |
+
\( 0 \leq Y \leq 6 \)μ΄κ³ , \( X \)μ \( Y \)μ νλ₯ λ°λν¨μλ κ°κ° \( f(x), g(x) \)μ΄λ€.
|
3 |
+
νλ₯ λ³μ \( X \)μ νλ₯ λ°λν¨μ \( f(x) \)μ κ·Έλνλ κ·Έλ¦Όκ³Ό κ°λ€.
|
4 |
+
|
5 |
+
\[
|
6 |
+
0 \leq x \leq 6\ \text{μΈ λͺ¨λ } x \text{μ λνμ¬}
|
7 |
+
\]
|
8 |
+
\[
|
9 |
+
f(x) + g(x) = k \quad (k \text{λ μμ})
|
10 |
+
\]
|
11 |
+
λ₯Ό λ§μ‘±μν¬ λ, \( \mathrm{P}(6k \leq Y \leq 15k) = \frac{q}{p} \) μ΄λ€. \( p + q \)μ κ°μ ꡬνμμ€.
|
12 |
+
(λ¨, \( p \)μ \( q \)λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]
|
data/json/2022/math/math_3.txt
ADDED
@@ -0,0 +1,13 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
3. λ±μ°¨μμ΄ $\{a_n\}$μ λνμ¬
|
2 |
+
\[
|
3 |
+
a_2 = 6, \quad a_4 + a_6 = 36
|
4 |
+
\]
|
5 |
+
μΌ λ, $a_{10}$μ κ°μ? [3μ ]
|
6 |
+
|
7 |
+
\begin{itemize}
|
8 |
+
\item[1] 30
|
9 |
+
\item[2] 32
|
10 |
+
\item[3] 34
|
11 |
+
\item[4] 36
|
12 |
+
\item[5] 38
|
13 |
+
\end{itemize}
|
data/json/2022/math/math_30_calc.txt
ADDED
@@ -0,0 +1,11 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
30. μ€μ μ 체μ μ§ν©μμ μ¦κ°νκ³ λ―ΈλΆκ°λ₯ν ν¨μ $f(x)$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.
|
2 |
+
|
3 |
+
\begin{itemize}
|
4 |
+
\item[(κ°)] $f(1) = 1$, \quad $\int_{1}^{2} f(x) \, dx = \frac{5}{4}$
|
5 |
+
\item[(λ)] ν¨μ $f(x)$μ μν¨μλ₯Ό $g(x)$λΌ ν λ, $x \geq 1$μΈ λͺ¨λ μ€μ $x$μ λνμ¬ $g(2x) = 2f(x)$μ΄λ€.
|
6 |
+
\end{itemize}
|
7 |
+
|
8 |
+
\[
|
9 |
+
\int_{1}^{8} x f'(x) \, dx = \frac{q}{p} \text{μΌ λ, } p+q \text{μ κ°μ ꡬνμμ€.}
|
10 |
+
\]
|
11 |
+
(λ¨, $p$μ $q$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]
|
data/json/2022/math/math_30_geom.txt
ADDED
@@ -0,0 +1,11 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
30. μ’ν곡κ°μ μ€μ¬μ΄ $\mathrm{C}(2, \sqrt{5}, 5)$μ΄κ³ μ $\mathrm{P}(0, 0, 1)$μ μ§λλ ꡬ
|
2 |
+
|
3 |
+
\[
|
4 |
+
S: (x - 2)^2 + (y - \sqrt{5})^2 + (z - 5)^2 = 25
|
5 |
+
\]
|
6 |
+
|
7 |
+
κ° μλ€. ꡬ $S$κ° νλ©΄ $\mathrm{OPC}$μ λ§λμ μκΈ°λ μ μλ₯Ό μμ§μ΄λ μ $\mathrm{Q}$, ꡬ $S$ μλ₯Ό μμ§μ΄λ μ $\mathrm{R}$μ λνμ¬ λ μ $\mathrm{Q}, \mathrm{R}$μ $xy$νλ©΄ μλ‘μ μ μ¬μμ κ°κ° $\mathrm{Q}_1, \mathrm{R}_1$μ΄λΌ νμ.
|
8 |
+
|
9 |
+
μΌκ°ν $\mathrm{O}\mathrm{Q}_1\mathrm{R}_1$μ λμ΄κ° μ΅λκ° λλλ‘ νλ λ μ $\mathrm{Q}, \mathrm{R}$μ λνμ¬ μΌκ°ν $\mathrm{O}\mathrm{Q}_1\mathrm{R}_1$μ νλ©΄ $\mathrm{PQR}$ μλ‘μ μ μ¬μμ λμ΄λ $\frac{q}{p} \sqrt{6}$μ΄λ€. $p+q$μ κ°μ ꡬνμμ€.
|
10 |
+
|
11 |
+
(λ¨, $\mathrm{O}$λ μμ μ΄κ³ μΈ μ $\mathrm{O}, \mathrm{Q}_1, \mathrm{R}_1$μ ν μ§μ μμ μμ§ μμΌλ©°, $p$μ $q$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]
|
data/json/2022/math/math_30_prob.txt
ADDED
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30. ν° κ³΅κ³Ό κ²μ κ³΅μ΄ κ°κ° 10κ° μ΄μ λ€μ΄ μλ λ°κ΅¬λμ λΉμ΄ μλ μ£Όλ¨Έλκ° μλ€. ν κ°μ μ£Όμ¬μλ₯Ό μ¬μ©νμ¬ λ€μ μνμ νλ€.
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\[
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\begin{array}{|c|}
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\hline
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\text{μ£Όμ¬μλ₯Ό ν λ² λμ Έ} \\
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\text{λμ¨ λμ μκ° 5 μ΄μμ΄λ©΄} \\
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\text{λ°κ΅¬λμ μλ ν° κ³΅ 2κ°λ₯Ό μ£Όλ¨Έλμ λ£κ³ ,} \\
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\text{λμ¨ λμ μκ° 4 μ΄νμ΄λ©΄} \\
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\text{λ°κ΅¬λμ μλ κ²μ 곡 1κ°λ₯Ό μ£Όλ¨Έλμ λ£λλ€.} \\
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\hline
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\end{array}
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\]
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μμ μνμ 5λ² λ°λ³΅ν λ, \( n(1 \leq n \leq 5) \)λ²μ§Έ μν ν μ£Όλ¨Έλμ λ€μ΄ μλ ν° κ³΅κ³Ό κ²μ 곡μ κ°μλ₯Ό κ°κ° \( a_n \), \( b_n \)μ΄λΌ νμ. \( a_5 + b_5 \geq 7 \)μΌ λ, \( a_k = b_k \)μΈ μμ°μ \( k(1 \leq k \leq 5) \)κ° μ‘΄μ¬ν νλ₯ μ \( \frac{q}{p} \)μ΄λ€. \( p + q \)μ κ°μ ꡬνμμ€. (λ¨, \(p\)μ \(q\)λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]
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data/json/2022/math/math_4.txt
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4. ν¨μ \( y = f(x) \)μ κ·Έλνκ° κ·Έλ¦Όκ³Ό κ°λ€.
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\[
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\lim_{x \to -1-} f(x) + \lim_{x \to 2} f(x) \text{μ κ°μ? [3μ ]}
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\]
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\begin{itemize}
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\item[1] 1
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\item[2] 2
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\item[3] 3
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\item[4] 4
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\item[5] 5
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\end{itemize}
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data/json/2022/math/math_5.txt
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5. 첫째νμ΄ 1μΈ μμ΄ $\{a_n\}$μ΄ λͺ¨λ μμ°μ $n$μ λνμ¬
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\[
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a_{n+1} = \begin{cases}
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2a_n & (a_n < 7) \\
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a_n - 7 & (a_n \geq 7)
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\end{cases}
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\]
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μΌ λ, $\sum_{k=1}^{8} a_k$μ κ°μ? [3μ ]
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\begin{itemize}
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\item[1] 30
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\item[2] 32
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\item[3] 34
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\item[4] 36
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\item[5] 38
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\end{itemize}
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data/json/2022/math/math_6.txt
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6. λ°©μ μ \( 2x^3 - 3x^2 - 12x + k = 0 \)μ΄ μλ‘ λ€λ₯Έ μΈ μ€κ·Όμ κ°λλ‘ νλ μ μ \( k \)μ κ°μλ? [3μ ]
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\begin{itemize}
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\item[1] 20
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\item[2] 23
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\item[3] 26
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\item[4] 29
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\item[5] 32
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\end{itemize}
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data/json/2022/math/math_7.txt
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7. \(\pi < \theta < \frac{3}{2}\pi\)μΈ \(\theta\)μ λνμ¬ \(\tan \theta - \frac{6}{\tan \theta} = 1\)μΌ λ, \( \sin \theta + \cos \theta \)μ κ°μ? [3μ ]
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\begin{itemize}
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\item[1] \(-\frac{2\sqrt{10}}{5}\)
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\item[2] \(-\frac{\sqrt{10}}{5}\)
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\item[3] \(0\)
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\item[4] \(\frac{\sqrt{10}}{5}\)
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\item[5] \(\frac{2\sqrt{10}}{5}\)
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\end{itemize}
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data/json/2022/math/math_8.txt
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8. 곑μ \( y = x^2 - 5x \)μ μ§μ \( y = x \)λ‘ λλ¬μΈμΈ λΆλΆμ λμ΄λ₯Ό μ§μ \( x = k \)κ° μ΄λ±λΆν λ, μμ \( k \)μ κ°μ? [3μ ]
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\begin{itemize}
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\item[1] \( 3 \)
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\item[2] \( \frac{13}{4} \)
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\item[3] \( \frac{7}{2} \)
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\item[4] \( \frac{15}{4} \)
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\item[5] \( 4 \)
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\end{itemize}
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data/json/2022/math/math_9.txt
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9. μ§μ \( y = 2x + k \) κ° λ ν¨μ
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\[
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y = \left( \frac{2}{3} \right)^{x+3} + 1, \quad y = \left( \frac{2}{3} \right)^{x+1} + \frac{8}{3}
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\]
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μ κ·Έλνμ λ§λλ μ μ κ°κ° \( \mathrm{P} \), \( \mathrm{Q} \)λΌ νμ. \( \overline{\mathrm{PQ}} = \sqrt{5} \)μΌ λ, μμ \( k \)μ κ°μ? [4μ ]
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\begin{itemize}
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\item[1] \( \frac{31}{6} \)
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\item[2] \( \frac{16}{3} \)
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\item[3] \( \frac{11}{2} \)
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\item[4] \( \frac{17}{3} \)
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\item[5] \( \frac{35}{6} \)
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\end{itemize}
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data/json/2022/math/prompt.txt
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1. $\left( \frac{4}{2^{\sqrt{2}}} \right)^{2 + \sqrt{2}}$ μ κ°μ? [2μ ]
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\begin{itemize}
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\item[1] $\frac{1}{4}$
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\item[2] $\frac{1}{2}$
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\item[3] $1$
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\item[4] $2$
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\item[5] $4$
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\end{itemize}
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#############
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2. $\lim_{x \to \infty} \frac{\sqrt{x^2 - 2 + 3x}}{x + 5}$ μ κ°μ? [2μ ]
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\begin{itemize}
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\item[1] 1
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\item[2] 2
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\item[3] 3
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\item[4] 4
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\item[5] 5
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\end{itemize}
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#############
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3. 곡λΉκ° μμμΈ λ±λΉμμ΄$\{a_n\}$μ΄
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\[ a_2 + a_4 = 30, \quad a_4 + a_6 = \frac{15}{2} \]
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λ₯Ό λ§μ‘±μν¬ λ, $a_1$ μ κ°μ? [3μ ]
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\begin{itemize}
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\item[1] 48
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\item[2] 56
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\item[3] 64
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\item[4] 72
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\item[5] 80
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\end{itemize}
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#############
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4. λ€νν¨μ $f(x)$μ λνμ¬ ν¨μ $g(x)$ λ₯Ό
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\[ g(x) = x^2 f(x) \]
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λΌ νμ. $f(2) = 1, \ f'(2) = 3$ μΌ λ, $g'(2)$ μ κ°μ? [3μ ]
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\begin{itemize}
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\item[1] 12
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\item[2] 14
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\item[3] 16
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\item[4] 18
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\item[5] 20
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\end{itemize}
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#############
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Give the latex code like the examples for the problem in the image
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data/json/2022/math_v1.json
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{"id":1,"name":"1","problem":"1. $\\left(2^{\\sqrt{3}} \\times 4\\right)^{\\sqrt{3} - 2}$ μ κ°μ? [2μ ] \\begin{itemize} \\item[1] \\frac{1}{4} \\item[2] \\frac{1}{2} \\item[3] 1 \\item[4] 2 \\item[5] 4 \\end{itemize}","answer":2,"score":2,"review":null}
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{"id":2,"name":"2","problem":"2. ν¨μ $f(x) = x^3 + 3x^2 + x - 1$ μ λνμ¬ $f'(1)$μ κ°μ? [2μ ] \\begin{itemize} \\item[1] 6 \\item[2] 7 \\item[3] 8 \\item[4] 9 \\item[5] 10 \\end{itemize}","answer":5,"score":2,"review":null}
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{"id":3,"name":"3","problem":"3. λ±μ°¨μμ΄ $\\{a_n\\}$μ λνμ¬ \\[ a_2 = 6, \\quad a_4 + a_6 = 36 \\] μΌ λ, $a_{10}$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 30 \\item[2] 32 \\item[3] 34 \\item[4] 36 \\item[5] 38 \\end{itemize}","answer":5,"score":3,"review":null}
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{"id":4,"name":"4","problem":"4. ν¨μ $( y = f(x) )$μ κ·Έλνκ° κ·Έλ¦Όκ³Ό κ°λ€.\n\n\\[ \\lim_{x \\to -1-} f(x) + \\lim_{x \\to 2} f(x) \\text{μ κ°μ? [3μ ]} \\]\n\n\\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}","answer":4,"score":3,"review":"Removed figure and the statement referring to the figure. The figure is needed to solve the problem.","incomplete":true}
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{"id":5,"name":"5","problem":"5. 첫째νμ΄ 1μΈ μμ΄ $\\{a_n\\}$μ΄ λͺ¨λ μμ°μ $n$μ λνμ¬ \\[ a_{n+1} = \\begin{cases} 2a_n & (a_n < 7) \\\\ a_n - 7 & (a_n \\geq 7) \\end{cases} \\] μΌ λ, $\\sum_{k=1}^{8} a_k$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 30 \\item[2] 32 \\item[3] 34 \\item[4] 36 \\item[5] 38 \\end{itemize}","answer":1,"score":3,"review":null}
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{"id":6,"name":"6","problem":"6. λ°©μ μ $( 2x^3 - 3x^2 - 12x + k = 0 )$μ΄ μλ‘ λ€λ₯Έ μΈ μ€κ·Όμ κ°λλ‘ νλ μ μ $k$μ κ°μλ? [3μ ] \\begin{itemize} \\item[1] 20 \\item[2] 23 \\item[3] 26 \\item[4] 29 \\item[5] 32 \\end{itemize}","answer":3,"score":3,"review":null}
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{"id":7,"name":"7","problem":"7. $( \\pi < \\theta < \\frac{3}{2}\\pi )$μΈ $\\theta$μ λνμ¬ $\\tan \\theta - \\frac{6}{\\tan \\theta} = 1$μΌ λ, $ \\sin \\theta + \\cos \\theta $μ κ°μ? [3μ ] \\begin{itemize} \\item[1] -\\frac{2\\sqrt{10}}{5} \\item[2] -\\frac{\\sqrt{10}}{5} \\item[3] 0 \\item[4] \\frac{\\sqrt{10}}{5} \\item[5] \\frac{2\\sqrt{10}}{5} \\end{itemize}","answer":1,"score":3,"review":null}
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{"id":8,"name":"8","problem":"8. 곑μ $( y = x^2 - 5x )$μ μ§μ $( y = x )$λ‘ λλ¬μΈμΈ λΆλΆμ λμ΄λ₯Ό μ§μ $( x = k )$κ° μ΄λ±λΆν λ, μμ $k$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 3 \\item[2] \\frac{13}{4} \\item[3] \\frac{7}{2} \\item[4] \\frac{15}{4} \\item[5] 4 \\end{itemize}","answer":1,"score":3,"review":null}
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{"id":9,"name":"9","problem":"9. μ§μ $( y = 2x + k )$ κ° λ ν¨μ \\[ y = \\left( \\frac{2}{3} \\right)^{x+3} + 1, \\quad y = \\left( \\frac{2}{3} \\right)^{x+1} + \\frac{8}{3} \\] μ κ·Έλνμ λ§λλ μ μ κ°κ° $( \\mathrm{P} )$, $( \\mathrm{Q} )$λΌ νμ. $\\overline{\\mathrm{PQ}} = \\sqrt{5}$μΌ λ, μμ $k$μ κ°μ? [4μ ] \\begin{itemize} \\item[1] \\frac{31}{6} \\item[2] \\frac{16}{3} \\item[3] \\frac{11}{2} \\item[4] \\frac{17}{3} \\item[5] \\frac{35}{6} \\end{itemize}","answer":4,"score":4,"review":"Removed figure."}
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{"id":10,"name":"10","problem":"10. μΌμ°¨ν¨μ $( f(x) )$μ λνμ¬ κ³‘μ $( y = f(x) )$ μμ μ $( 0, 0 )$μμμ μ μ κ³Ό 곑μ $( y = x f(x) )$ μμ μ $( 1, 2 )$μμμ μ μ μ΄ μΌμΉν λ, $f'(2)$μ κ°μ? [4μ ] \\begin{itemize} \\item[1] -18 \\item[2] -17 \\item[3] -16 \\item[4] -15 \\item[5] -14 \\end{itemize}","answer":5,"score":4,"review":null}
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{"id":11,"name":"11","problem":"11. μμ $a$μ λνμ¬ μ§ν© $\\left\\{ x \\ \\middle| \\ -\\frac{a}{2} < x \\leq a, \\ x \\neq \\frac{a}{2} \\right\\}$ μμ μ μλ ν¨μ \\[ f(x) = \\tan \\frac{\\pi x}{a} \\] κ° μλ€. κ·Έλ¦Όκ³Ό κ°μ΄ ν¨μ $y = f(x)$μ κ·Έλν μμ μΈ μ $( \\mathrm{O, A, B} )$λ₯Ό μ§λλ μ§μ μ΄ μλ€. μ $( \\mathrm{A} )$λ₯Ό μ§λκ³ $x$μΆμ ννν μ§μ μ΄ ν¨μ $y = f(x)$μ κ·Έλνμ λ§λλ μ μ€ $( \\mathrm{A} )$κ° μλ μ μ $( \\mathrm{C} )$λΌ νμ. μΌκ°ν $( \\mathrm{ABC} )$κ° μ μΌκ°νμΌ λ, μΌκ°ν $( \\mathrm{ABC} )$μ λμ΄λ? (λ¨, $( \\mathrm{O} )$λ μμ μ΄λ€.) [4μ ] \\begin{itemize} \\item[1] \\frac{3\\sqrt{3}}{2} \\item[2] \\frac{17\\sqrt{3}}{12} \\item[3] \\frac{4\\sqrt{3}}{3} \\item[4] \\frac{5\\sqrt{3}}{4} \\item[5] \\frac{7\\sqrt{3}}{6} \\end{itemize}","answer":3,"score":4,"review":"Removed figure and the statement referring to the figure."}
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{"id":12,"name":"12","problem":"12. μ€μ μ 체μ μ§ν©μμ μ°μμΈ ν¨μ $f(x)$κ° λͺ¨λ μ€μ $x$μ λνμ¬ \\[ \\{f(x)\\}^3 - \\{f(x)\\}^2 - x^2 f(x) + x^2 = 0 \\] μ λ§μ‘±μν¨λ€. ν¨μ $f(x)$μ μ΅λκ°μ΄ 1μ΄κ³ μ΅μκ°μ΄ 0μΌ λ, \\[ f\\left( -\\frac{4}{3} \\right) + f(0) + f\\left( \\frac{1}{2} \\right) \\] μ κ°μ? [4μ ] \\begin{itemize} \\item[1] \\frac{1}{2} \\item[2] 1 \\item[3] \\frac{3}{2} \\item[4] 2 \\item[5] \\frac{5}{2} \\end{itemize}","answer":3,"score":4,"review":null}
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13 |
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{"id":13,"name":"13","problem":"13. λ μμ $( a, b \\ (1 < a < b) )$μ λνμ¬ μ’ννλ©΄ μμ λ μ $(a, \\log_2 a), \\ (b, \\log_2 b)$λ₯Ό μ§λλ μ§μ μ $y$μ νΈκ³Ό λ μ $(a, \\log_4 a), \\ (b, \\log_4 b)$λ₯Ό μ§λλ μ§μ μ $y$μ νΈμ΄ κ°λ€. ν¨μ $f(x) = a^{bx} + b^{ax}$μ λνμ¬ $f(1) = 40$μΌ λ, $f(2)$μ κ°μ? [4μ ] \\begin{itemize} \\item[1] 760 \\item[2] 800 \\item[3] 840 \\item[4] 880 \\item[5] 920 \\end{itemize}","answer":2,"score":4,"review":null}
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{"id":14,"name":"14","problem":"14. μμ§μ μλ₯Ό μμ§μ΄λ μ $\\mathrm{P}$μ μκ° $t$μμμ μμΉ $x(t)$κ° λ μμ $a$, $b$μ λνμ¬ \\[ x(t) = t(t - 1)(at + b) \\quad (a \\neq 0) \\] μ΄λ€. μ $\\mathrm{P}$μ μκ° $t$μμμ μλ $v(t)$κ° $\\int_0^1 |v(t)| \\, dt = 2$λ₯Ό λ§μ‘±μν¬ λ, μλ γ±, γ΄, γ· μ€μμ μ³μ κ²λ§μ μλ λλ‘ κ³ λ₯Έ κ²μ? [4μ ]\n\n\\begin{itemize} \\item[γ±.] $\\int_0^1 v(t) \\, dt = 0$ \\item[γ΄.] $|x(t_1)| > 1$μΈ $t_1$μ΄ μ΄λ¦°κ΅¬κ° $(0, 1)$μ μ‘΄μ¬νλ€. \\item[γ·.] $0 \\leq t \\leq 1$μΈ λͺ¨λ $t$μ λνμ¬ $|x(t)| < 1$μ΄λ©΄ $x(t_2) = 0$μΈ $t_2$κ° μ΄λ¦°κ΅¬κ° $(0, 1)$μ μ‘΄μ¬νλ€. \\end{itemize}\n\n\\begin{itemize} \\item[1] γ± \\item[2] γ±, γ΄ \\item[3] γ±, γ· \\item[4] γ΄, γ· \\item[5] γ±, γ΄, γ· \\end{itemize}","answer":3,"score":4,"review":"<보기> changed to 'μλ γ±,γ΄,γ·, μ€'"}
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{"id":15,"name":"15","problem":"15. λ μ $( \\mathrm{O}_1, \\mathrm{O}_2 )$λ₯Ό κ°κ° μ€μ¬μΌλ‘ νκ³ λ°μ§λ¦μ κΈΈμ΄κ° $(\\overline{\\mathrm{O}_1\\mathrm{O}_2} )$μΈ λ μ $( C_1, C_2 )$κ° μλ€. κ·Έλ¦Όκ³Ό κ°μ΄ μ $( C_1 )$ μμ μλ‘ λ€λ₯Έ μΈ μ $( \\mathrm{A}, \\mathrm{B}, \\mathrm{C} )$μ μ $( C_2 )$ μμ μ $( \\mathrm{D} )$κ° μ£Όμ΄μ Έ μκ³ , μΈ μ $( \\mathrm{A}, \\mathrm{O}_1, \\mathrm{O}_2 )$μ μΈ μ $( \\mathrm{C}, \\mathrm{O}_2, \\mathrm{D} )$κ° κ°κ° ν μ§μ μμ μλ€.\n\nμ΄λ $(\\angle \\mathrm{B}\\mathrm{O}_1\\mathrm{A} = \\theta_1)$, $(\\angle \\mathrm{O}_2\\mathrm{O}_1\\mathrm{C} = \\theta_2)$, $(\\angle \\mathrm{O}_1\\mathrm{O}_2\\mathrm{D} = \\theta_3)$μ΄λΌ νμ.\n\nλ€μμ $( \\overline{\\mathrm{A}\\mathrm{B}} : \\overline{\\mathrm{O}_1\\mathrm{D}} = 1 : 2\\sqrt{2} )$μ΄κ³ $( \\theta_3 = \\theta_1 + \\theta_2 )$μΌ λ, μ λΆ $( \\mathrm{A}\\mathrm{B} )$μ μ λΆ $( \\mathrm{C}\\mathrm{D} )$μ κΈΈμ΄μ λΉλ₯Ό ꡬνλ κ³Όμ μ΄λ€.\n\n\\[ \\begin{aligned} &\\angle \\mathrm{C}\\mathrm{O}_2\\mathrm{O}_1 + \\angle \\mathrm{O}_1\\mathrm{O}_2\\mathrm{D} = \\pi \\text{μ΄λ―λ‘ } \\theta_3 = \\frac{\\pi}{2} + \\frac{\\theta_2}{2} \\text{μ΄κ³ } \\\\ &\\theta_3 = \\theta_1 + \\theta_2 \\text{μμ } 2\\theta_1 + \\theta_2 = \\pi \\text{μ΄λ―λ‘ } \\angle \\mathrm{C}\\mathrm{O}_1\\mathrm{B} = \\theta_1 \\text{μ΄λ€.} \\\\ &\\text{μ΄λ } \\angle \\mathrm{O}_2\\mathrm{O}_1\\mathrm{B} = \\theta_1 + \\theta_2 = \\theta_3 \\text{μ΄λ―λ‘ μΌκ°ν } \\mathrm{O}_1\\mathrm{O}_2\\mathrm{B} \\text{μ μΌκ°ν } \\mathrm{O}_2\\mathrm{O}_1\\mathrm{D} \\text{λ ν©λμ΄λ€.} \\\\ &\\overline{\\mathrm{A}\\mathrm{B}} = k \\text{λΌ ν λ} \\\\ &\\overline{\\mathrm{B}\\mathrm{O}_2} = \\overline{\\mathrm{O}_1\\mathrm{D}}= 2\\sqrt{2}k \\text{μ΄λ―λ‘ } \\overline{\\mathrm{A}\\mathrm{O}_2} = \\text{(κ°)μ΄κ³ ,} \\\\ &\\angle \\mathrm{B}\\mathrm{O}_2\\mathrm{A} = \\frac{\\theta_1}{2} \\text{μ΄λ―λ‘ } \\cos \\frac{\\theta_1}{2} = \\text{(λ) μ΄λ€.} \\\\ &\\text{μΌκ°ν } \\mathrm{O}_2\\mathrm{B}\\mathrm{C} \\text{μμ} \\\\ &\\overline{\\mathrm{B}\\mathrm{C}} = k, \\overline{\\mathrm{B}\\mathrm{O}_2} = 2\\sqrt{2}k, \\angle \\mathrm{C}\\mathrm{O}_2\\mathrm{B} = \\frac{\\theta_1}{2} \\text{μ΄λ―λ‘} \\\\ &\\text{μ½μ¬μΈλ²μΉμ μνμ¬ } \\overline{\\mathrm{O}_2\\mathrm{C}} = \\text{(λ€) μ΄λ€.} \\\\ &\\overline{\\mathrm{C}\\mathrm{D}} = \\overline{\\mathrm{O}_2\\mathrm{D}} + \\overline{\\mathrm{O}_2\\mathrm{C}} = \\overline{\\mathrm{O}_1\\mathrm{O}_2} + \\overline{\\mathrm{O}_2\\mathrm{C}} \\text{μ΄λ―λ‘} \\\\ &\\overline{\\mathrm{A}\\mathrm{B}} : \\overline{\\mathrm{C}\\mathrm{D}} = k : \\left(\\frac{\\text{(κ°)}}{2} + \\text{(λ€)}\\right) \\text{μ΄λ€.} \\end{aligned} \\]\n\nμμ (κ°), (λ€)μ μλ§μ μμ κ°κ° $( f(k), g(k) )$λΌ νκ³ , (λ)μ μλ§μ μλ₯Ό $( p )$λΌ ν λ, $( f(p) \\times g(p) )$μ κ°μ? [4μ ]\n\n\\begin{itemize} \\item[1] \\frac{169}{27} \\item[2] \\frac{56}{9} \\item[3] \\frac{167}{27} \\item[4] \\frac{166}{27} \\item[5] \\frac{55}{9} \\end{itemize}","answer":2,"score":4,"review":"Removed figure and the statement referring to the figure."}
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{"id":16,"name":"16","problem":"16. $\\log_2 120 - \\frac{1}{\\log_{15} 2}$ μ κ°μ ꡬνμμ€. [3μ ]","answer":3,"score":3,"review":null}
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{"id":17,"name":"17","problem":"17. ν¨μ $f(x)$μ λνμ¬ $f'(x) = 3x^2 + 2x$μ΄κ³ $f(0) = 2$μΌ λ, $f(1)$μ κ°μ ꡬνμμ€. [3μ ]","answer":4,"score":3,"review":null}
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{"id":18,"name":"18","problem":"18. μμ΄ $\\{a_n\\}$μ λνμ¬ \\[ \\sum_{k=1}^{10} a_k - \\sum_{k=1}^{7} \\frac{a_k}{2} = 56, \\quad \\sum_{k=1}^{10} 2a_k - \\sum_{k=1}^{8} a_k = 100 \\] μΌ λ, $a_8$μ κ°μ ꡬνμμ€. [3μ ]","answer":12,"score":3,"review":null}
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{"id":19,"name":"19","problem":"19. ν¨μ $f(x) = x^3 + ax^2 - (a^2 - 8a)x + 3$μ΄ μ€μ μ 체μ μ§ν©μμ μ¦κ°νλλ‘ νλ μ€μ $a$μ μ΅λκ°μ ꡬνμμ€. [3μ ]","answer":6,"score":3,"review":null}
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{"id":20,"name":"20","problem":"20. μ€μ μ 체μ μ§ν©μμ λ―ΈλΆκ°λ₯ν ν¨μ $( f(x) )$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] λ«νκ΅¬κ° $[0, 1]$μμ $f(x) = x$μ΄λ€. \\item[(λ)] μ΄λ€ μμ $a, b$μ λνμ¬ κ΅¬κ° $[0, \\infty)$μμ $f(x+1) - x f(x) = ax + b$μ΄λ€. \\end{itemize}\n\n\\[ 60 \\times \\int_1^2 f(x) \\, dx \\] μ κ°μ ꡬνμμ€. [4μ ]","answer":110,"score":4,"review":null}
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{"id":21,"name":"21","problem":"21. μμ΄ $\\{a_n\\}$μ΄ λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] $( |a_1| = 2 )$ \\item[(λ)] λͺ¨λ μμ°μ $( n )$μ λνμ¬ $( |a_{n+1}| = 2|a_n| )$μ΄λ€. \\item[(λ€)] $\\sum_{n=1}^{10} a_n = -14$ \\end{itemize}\n\n$a_1 + a_3 + a_5 + a_7 + a_9$μ κ°μ ꡬνμμ€. [4μ ]","answer":678,"score":4,"review":null}
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{"id":22,"name":"22","problem":"22. μ΅κ³ μ°¨νμ κ³μκ° $\\frac{1}{2}$ μΈ μΌμ°¨ν¨μ $f(x)$μ μ€μ $t$μ λνμ¬ λ°©μ μ $f'(x) = 0$μ΄ λ«νκ΅¬κ° $[t, t+2]$μμ κ°λ μ€κ·Όμ κ°μλ₯Ό $g(t)$λΌ ν λ, ν¨μ $g(t)$λ λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] λͺ¨λ μ€μ $( a )$μ λνμ¬ $( \\lim_{t \\to a+} g(t) + \\lim_{t \\to a-} g(t) \\leq 2 )$μ΄λ€. \\item[(λ)] $( g(f(1)) = g(f(4)) = 2, \\ g(f(0)) = 1 )$ \\end{itemize}\n\n$f(5)$μ κ°μ ꡬνμμ€. [4μ ]","answer":9,"score":4,"review":null}
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{"id":23,"name":"23_prob","problem":"23. λ€νμ $(x+2)^7$μ μ κ°μμμ $x^5$μ κ³μλ? [2μ ] \\begin{itemize} \\item[1] 42 \\item[2] 56 \\item[3] 70 \\item[4] 84 \\item[5] 98 \\end{itemize}","answer":4,"score":2,"review":null}
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{"id":24,"name":"24_prob","problem":"24. νλ₯ λ³μ $X$κ° μ΄νλΆν¬ $\\mathrm{B}\\left(n, \\frac{1}{3}\\right)$μ λ°λ₯΄κ³ $\\mathrm{V}(2X) = 40$μΌ λ, $n$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 30 \\item[2] 35 \\item[3] 40 \\item[4] 45 \\item[5] 50 \\end{itemize}","answer":4,"score":3,"review":null}
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{"id":25,"name":"25_prob","problem":"25. λ€μ 쑰건μ λ§μ‘±μν€λ μμ°μ $a, \\ b, \\ c, \\ d, \\ e$μ λͺ¨λ μμμ $(a, b, c, d, e)$μ κ°μλ? [3μ ]\n\n\\begin{itemize} \\item[(κ°)] $a + b + c + d + e = 12$ \\item[(λ)] $\\left| a^2 - b^2 \\right| = 5$ \\end{itemize}\n\n\\begin{itemize} \\item[1] 30 \\item[2] 32 \\item[3] 34 \\item[4] 36 \\item[5] 38 \\end{itemize}","answer":1,"score":3,"review":null}
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{"id":26,"name":"26_prob","problem":"26. $( 1 )$λΆν° $( 10 )$κΉμ§ μμ°μκ° νλμ© μ ν μλ $( 10 )$μ₯μ μΉ΄λκ° λ€μ΄ μλ μ£Όλ¨Έλκ° μλ€. μ΄ μ£Όλ¨Έλμμ μμλ‘ μΉ΄λ $( 3 )$μ₯μ λμμ κΊΌλΌ λ, κΊΌλΈ μΉ΄λμ μ ν μλ μΈ μμ°μ μ€μμ κ°μ₯ μμ μκ° $( 4 )$ μ΄νμ΄κ±°λ $( 7 )$ μ΄μμΌ νλ₯ μ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{4}{5} \\item[2] \\frac{5}{6} \\item[3] \\frac{13}{15} \\item[4] \\frac{9}{10} \\item[5] \\frac{14}{15} \\end{itemize}","answer":3,"score":3,"review":"Removed figure."}
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{"id":27,"name":"27_prob","problem":"27. μ΄λ μλμ°¨ νμ¬μμ μμ°νλ μ κΈ° μλμ°¨μ 1ν μΆ©μ μ£Όν 거리λ νκ· μ΄ $m$μ΄κ³ νμ€νΈμ°¨κ° $\\sigma$μΈ μ κ·λΆν¬λ₯Ό λ°λ₯Έλ€κ³ νλ€.\n\nμ΄ μλμ°¨ νμ¬μμ μμ°ν μ κΈ° μλμ°¨ 100λλ₯Ό μμμΆμΆνμ¬ μ»μ 1ν μΆ©μ μ£Όν 거리μ νλ³Ένκ· μ΄ $\\overline{x_1}$μΌ λ, λͺ¨νκ· $m$μ λν μ λ’°λ 95\\%μ μ 뒰ꡬκ°μ΄ $a \\le m \\le b$μ΄λ€.\n\nμ΄ μλμ°¨ νμ¬μμ μμ°ν μ κΈ° μλμ°¨ 400λλ₯Ό μμμΆμΆνμ¬ μ»μ 1ν μΆ©μ μ£Όν 거리μ νλ³Ένκ· μ΄ $\\overline{x_2}$μΌ λ, λͺ¨νκ· $m$μ λν μ λ’°λ 99\\%μ μ 뒰ꡬκ°μ΄ $c \\le m \\le d$μ΄λ€.\n\n$\\overline{x_1} - \\overline{x_2} = 1.34$μ΄κ³ $a = c$μΌ λ, $b - a$μ κ°μ? (λ¨, μ£Όν 거리μ λ¨μλ kmμ΄κ³ , $Z$κ° νμ€μ κ·λΆν¬λ₯Ό λ°λ₯΄λ νλ₯ λ³μμΌ λ $\\mathrm{P}(|Z| \\le 1.96) = 0.95$, $\\mathrm{P}(|Z| \\le 2.58) = 0.99$λ‘ κ³μ°νλ€.) [3μ ]\n\n\\begin{itemize} \\item[1] 5.88 \\item[2] 7.84 \\item[3] 9.80 \\item[4] 11.76 \\item[5] 13.72 \\end{itemize}","answer":2,"score":3,"review":null}
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{"id":28,"name":"28_prob","problem":"28. λ μ§ν© $X = \\{1, 2, 3, 4, 5\\}$, $Y = \\{1, 2, 3, 4\\}$μ λνμ¬ λ€μ 쑰건μ λ§μ‘±μν€λ $X$μμ $Y$λ‘μ ν¨μ $f$μ κ°μλ? [4μ ]\n\n\\begin{itemize} \\item[(κ°)] μ§ν© $X$μ λͺ¨λ μμ $x$μ λνμ¬ $f(x) \\geq \\sqrt{x}$μ΄λ€. \\item[(λ)] ν¨μ $f$μ μΉμμ μμμ κ°μλ 3μ΄λ€. \\end{itemize}\n\n\\begin{itemize} \\item[1] 128 \\item[2] 138 \\item[3] 148 \\item[4] 158 \\item[5] 168 \\end{itemize}","answer":1,"score":4,"review":null}
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{"id":29,"name":"29_prob","problem":"29. λ μ°μνλ₯ λ³μ $( X )$μ $( Y )$κ° κ°λ κ°μ λ²μλ $( 0 \\leq X \\leq 6 )$, $( 0 \\leq Y \\leq 6 )$μ΄κ³ , $( X )$μ $( Y )$μ νλ₯ λ°λν¨μλ κ°κ° $( f(x), g(x) )$μ΄λ€. νλ₯ λ³μ $( X )$μ νλ₯ λ°λν¨μ $( f(x) )$μ κ·Έλνλ κ·Έλ¦Όκ³Ό κ°λ€.\n\n\\[ 0 \\leq x \\leq 6\\ \\text{μΈ λͺ¨λ } x \\text{μ λνμ¬} \\]\n\\[ f(x) + g(x) = k \\quad (k \\text{λ μμ}) \\]\nλ₯Ό λ§μ‘±μν¬ λ, $( \\mathrm{P}(6k \\leq Y \\leq 15k) = \\frac{q}{p} )$μ΄λ€. $( p + q )$μ κ°μ ꡬνμμ€. (λ¨, $( p )$μ $( q )$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]","answer":31,"score":4,"review":"Removed figure and the statement referring to the figure. The figure is needed to solve the problem.","incomplete":true}
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{"id":30,"name":"30_prob","problem":"30. ν° κ³΅κ³Ό κ²μ κ³΅μ΄ κ°κ° 10κ° μ΄μ λ€μ΄ μλ λ°κ΅¬λμ λΉμ΄ μλ μ£Όλ¨Έλκ° μλ€. ν κ°μ μ£Όμ¬μλ₯Ό μ¬μ©νμ¬ λ€μ μνμ νλ€.\n\n\\[ \\begin{array}{|c|} \\hline \\text{μ£Όμ¬μλ₯Ό ν λ² λμ Έ} \\\\ \\text{λμ¨ λμ μκ° 5 μ΄μμ΄λ©΄} \\\\ \\text{λ°κ΅¬λμ μλ ν° κ³΅ 2κ°λ₯Ό μ£Όλ¨Έλμ λ£κ³ ,} \\\\ \\text{λμ¨ λμ μκ° 4 μ΄νμ΄λ©΄} \\\\ \\text{λ°κ΅¬λμ μλ κ²μ 곡 1κ°λ₯Ό μ£Όλ¨Έλμ λ£λλ€.} \\\\ \\hline \\end{array} \\]\n\nμμ μνμ 5λ² λ°λ³΅ν λ, $( n(1 \\leq n \\leq 5) )$λ²μ§Έ μν ν μ£Όλ¨Έλμ λ€μ΄ μλ ν° κ³΅κ³Ό κ²μ 곡μ κ°μλ₯Ό κ°κ° $( a_n )$, $( b_n )$μ΄λΌ νμ. $( a_5 + b_5 \\geq 7 )$μΌ λ, $( a_k = b_k )$μΈ μμ°μ $( k(1 \\leq k \\leq 5) )$κ° μ‘΄μ¬ν νλ₯ μ $( \\frac{q}{p} )$μ΄λ€. $( p + q )$μ κ°μ ꡬνμμ€. (λ¨, $(p)$μ $(q)$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]","answer":191,"score":4,"review":null}
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{"id":31,"name":"23_calc","problem":"23. \\[ \\lim_{n \\to \\infty} \\frac{\\frac{5}{n} + \\frac{3}{n^2}}{\\frac{1}{n} - \\frac{2}{n^3}} \\text{μ κ°μ? [2μ ]} \\] \\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}","answer":5,"score":2,"review":null}
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{"id":32,"name":"24_calc","problem":"24. μ€μ μ 체μ μ§ν©μμ λ―ΈλΆκ°λ₯ν ν¨μ $f(x)$κ° λͺ¨λ μ€μ $x$μ λνμ¬ \\[ f(x^3 + x) = e^x \\] μ λ§μ‘±μν¬ λ, $f'(2)$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] e \\item[2] \\frac{e}{2} \\item[3] \\frac{e}{3} \\item[4] \\frac{e}{4} \\item[5] \\frac{e}{5} \\end{itemize}","answer":4,"score":3,"review":null}
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{"id":33,"name":"25_calc","problem":"25. λ±λΉμμ΄ $\\{a_n\\}$μ λνμ¬ \\[ \\sum_{n=1}^{\\infty} (a_{2n-1} - a_{2n}) = 3, \\quad \\sum_{n=1}^{\\infty} a_n^2 = 6 \\] μΌ λ, $\\sum_{n=1}^{\\infty} a_n$ μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}","answer":2,"score":3,"review":null}
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{"id":34,"name":"26_calc","problem":"26. \\[ \\lim_{n \\to \\infty} \\sum_{k=1}^{n} \\frac{k^2 + 2kn}{k^3 + 3k^2 n + n^3} \\text{μ κ°μ?} \\quad [3 \\text{μ }] \\] \\begin{itemize} \\item[1] \\ln 5 \\item[2] \\frac{\\ln 5}{2} \\item[3] \\frac{\\ln 5}{3} \\item[4] \\frac{\\ln 5}{4} \\item[5] \\frac{\\ln 5}{5} \\end{itemize}","answer":3,"score":3,"review":null}
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{"id":35,"name":"27_calc","problem":"27. μ’ννλ©΄ μλ₯Ό μμ§μ΄λ μ $\\mathrm{P}$μ μκ° $t \\ (t>0)$μμμ μμΉκ° 곑μ $y = x^2$κ³Ό μ§μ $y = t^2 x - \\frac{\\ln t}{8}$κ° λ§λλ μλ‘ λ€λ₯Έ λ μ μ μ€μ μΌ λ, μκ° $t=1$μμ $t=e$κΉμ§ μ $\\mathrm{P}$κ° μμ§μΈ 거리λ? [3μ ] \\begin{itemize} \\item[1] \\frac{e^4}{2} - \\frac{3}{8} \\item[2] \\frac{e^4}{2} - \\frac{5}{16} \\item[3] \\frac{e^4}{2} - \\frac{1}{4} \\item[4] \\frac{e^4}{2} - \\frac{3}{16} \\item[5] \\frac{e^4}{2} - \\frac{1}{8} \\end{itemize}","answer":1,"score":3,"review":null}
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{"id":36,"name":"28_calc","problem":"28. ν¨μ $( f(x) = 6\\pi (x - 1)^2 )$μ λνμ¬ ν¨μ $( g(x) )$λ₯Ό \\[ g(x) = 3f(x) + 4\\cos f(x) \\] λΌ νμ. $( 0 < x < 2 )$μμ ν¨μ $( g(x) )$κ° κ·Ήμκ° λλ $( x )$μ κ°μλ? [4μ ] \\begin{itemize} \\item[1] 6 \\item[2] 7 \\item[3] 8 \\item[4] 9 \\item[5] 10 \\end{itemize}","answer":2,"score":4,"review":null}
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{"id":37,"name":"29_calc","problem":"29. κ·Έλ¦Όκ³Ό κ°μ΄ κΈΈμ΄κ° 2μΈ μ λΆ $(\\mathrm{AB})$λ₯Ό μ§λ¦μΌλ‘ νλ λ°μμ΄ μλ€. νΈ $(\\mathrm{AB})$ μμ λ μ $(\\mathrm{P})$, $(\\mathrm{Q})$λ₯Ό $(\\angle \\mathrm{PAB} = \\theta)$, $(\\angle \\mathrm{QBA} = 2\\theta)$κ° λλλ‘ μ‘κ³ , λ μ λΆ $(\\mathrm{AP})$, $(\\mathrm{BQ})$μ κ΅μ μ $(\\mathrm{R})$λΌ νμ. μ λΆ $(\\mathrm{AB})$ μμ μ $(\\mathrm{S})$, μ λΆ $(\\mathrm{BR})$ μμ μ $(\\mathrm{T})$, μ λΆ $(\\mathrm{AR})$ μμ μ $(\\mathrm{U})$λ₯Ό μ λΆ $(\\mathrm{UT})$κ° μ λΆ $(\\mathrm{AB})$μ νννκ³ μΌκ°ν $(\\mathrm{STU})$κ° μ μΌκ°νμ΄ λλλ‘ μ‘λλ€. λ μ λΆ $(\\mathrm{AR})$, $(\\mathrm{QR})$μ νΈ $(\\mathrm{AQ})$λ‘ λλ¬μΈμΈ λΆλΆμ λμ΄λ₯Ό $(f(\\theta))$, μΌκ°ν $(\\mathrm{STU})$μ λμ΄λ₯Ό $(g(\\theta))$λΌ ν λ,\n\\[ \\lim_{\\theta \\to 0+} \\frac{g(\\theta)}{\\theta \\times f(\\theta)} = \\frac{q}{p} \\sqrt{3} \\]\nμ΄λ€. $(p + q)$μ κ°μ ꡬνμμ€.\n\n(λ¨, $(0 < \\theta < \\frac{\\pi}{6})$μ΄κ³ , $(p)$μ $(q)$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]","answer":11,"score":4,"review":"Removed figure and the statement referring to the figure."}
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38 |
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{"id":38,"name":"30_calc","problem":"30. μ€μ μ 체μ μ§ν©μμ μ¦κ°νκ³ λ―ΈλΆκ°λ₯ν ν¨μ $f(x)$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] $f(1) = 1$, \\quad $\\int_{1}^{2} f(x) \\, dx = \\frac{5}{4}$ \\item[(λ)] ν¨μ $f(x)$μ μν¨μλ₯Ό $g(x)$λΌ ν λ, $x \\geq 1$μΈ λͺ¨λ μ€μ $x$μ λνμ¬ $g(2x) = 2f(x)$μ΄λ€. \\end{itemize}\n\n\\[ \\int_{1}^{8} x f'(x) \\, dx = \\frac{q}{p} \\text{μΌ λ, } p+q \\text{μ κ°μ ꡬνμμ€.} \\]\n(λ¨, $p$μ $q$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]","answer":143,"score":4,"review":null}
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{"id":39,"name":"23_geom","problem":"23. μ’ν곡κ°μ μ $\\mathrm{A}(2, 1, 3)$μ $xy$ νλ©΄μ λνμ¬ λμΉμ΄λν μ μ $\\mathrm{P}$λΌ νκ³ , μ $\\mathrm{A}$λ₯Ό $yz$ νλ©΄μ λνμ¬ λμΉμ΄λν μ μ $\\mathrm{Q}$λΌ ν λ, μ λΆ $\\mathrm{PQ}$μ κΈΈμ΄λ? [2μ ]\n\n\\begin{itemize} \\item[1] 5 \\sqrt{2} \\item[2] 2 \\sqrt{13} \\item[3] 3 \\sqrt{6} \\item[4] 2 \\sqrt{14} \\item[5] 2 \\sqrt{15} \\end{itemize}","answer":2,"score":2,"review":null}
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{"id":40,"name":"24_geom","problem":"24. ν μ΄μ μ μ’νκ° $\\left( 3\\sqrt{2}, 0 \\right)$ μΈ μ곑μ $\\frac{x^2}{a^2} - \\frac{y^2}{6} = 1$ μ μ£ΌμΆμ κΈΈμ΄λ? (λ¨, $a$ λ μμμ΄λ€.) [3μ ]\n\n\\begin{itemize} \\item[1] 3\\sqrt{3} \\item[2] \\frac{7\\sqrt{3}}{2} \\item[3] 4\\sqrt{3} \\item[4] \\frac{9\\sqrt{3}}{2} \\item[5] 5\\sqrt{3} \\end{itemize}","answer":3,"score":3,"review":null}
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{"id":41,"name":"25_geom","problem":"25. μ’ννλ©΄μμ λ μ§μ \\[ \\frac{x+1}{2} = y - 3, \\quad x - 2 = \\frac{y - 5}{3} \\] κ° μ΄λ£¨λ μκ°μ ν¬κΈ°λ₯Ό $\\theta$λΌ ν λ, $\\cos \\theta$μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{1}{2} \\item[2] \\frac{\\sqrt{5}}{4} \\item[3] \\frac{\\sqrt{6}}{4} \\item[4] \\frac{\\sqrt{7}}{4} \\item[5] \\frac{\\sqrt{2}}{2} \\end{itemize}","answer":5,"score":3,"review":null}
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{"id":42,"name":"26_geom","problem":"26. λ μ΄μ μ΄ $( \\mathrm{F}, \\mathrm{F'} )$μΈ νμ $\\frac{x^2}{64} + \\frac{y^2}{16} = 1$ μμ μ μ€ μ 1μ¬λΆλ©΄μ μλ μ $( \\mathrm{A} )$κ° μλ€. λ μ§μ $( \\mathrm{AF}, \\mathrm{AF'} )$μ λμμ μ νκ³ μ€μ¬μ΄ $y$μΆ μμ μλ μ μ€ μ€μ¬μ $y$μ’νκ° μμμΈ κ²μ $( C )$λΌ νμ. μ $( C )$μ μ€μ¬μ $( \\mathrm{B} )$λΌ ν λ μ¬κ°ν $( \\mathrm{AFBF'} )$μ λμ΄κ° 72μ΄λ€. μ $( C )$μ λ°μ§λ¦μ κΈΈμ΄λ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{17}{2} \\item[2] 9 \\item[3] \\frac{19}{2} \\item[4] 10 \\item[5] \\frac{21}{2} \\end{itemize}","answer":2,"score":3,"review":"Removed figure."}
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{"id":43,"name":"27_geom","problem":"27. κ·Έλ¦Όκ³Ό κ°μ΄ ν λͺ¨μ리μ κΈΈμ΄κ° 4μΈ μ μ‘면체 $\\mathrm{ABCD - EFGH}$ κ° μλ€. μ λΆ $\\mathrm{AD}$ μ μ€μ μ $\\mathrm{M}$μ΄λΌ ν λ, μΌκ°ν $\\mathrm{MEG}$ μ λμ΄λ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{21}{2} \\item[2] 11 \\item[3] \\frac{23}{2} \\item[4] 12 \\item[5] \\frac{25}{2} \\end{itemize}","answer":4,"score":3,"review":"Removed figure and the statement referring to the figure."}
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{"id":44,"name":"28_geom","problem":"28. λ μμ $( a )$, $( p )$μ λνμ¬ ν¬λ¬Όμ $( (y - a)^2 = 4px )$μ μ΄μ μ $( \\mathrm{F}_1 )$μ΄λΌ νκ³ , ν¬λ¬Όμ $( y^2 = -4x )$μ μ΄μ μ $( \\mathrm{F}_2 )$λΌ νμ. μ λΆ $( \\mathrm{F}_1 \\mathrm{F}_2 )$κ° λ ν¬λ¬Όμ κ³Ό λ§λλ μ μ κ°κ° $( \\mathrm{P} )$, $( \\mathrm{Q} )$λΌ ν λ, $( \\overline{\\mathrm{F}_1 \\mathrm{F}_2} = 3 )$, $( \\overline{\\mathrm{P}\\mathrm{Q}} = 1 )$μ΄λ€. $( a^2 + p^2 )$μ κ°μ? [4μ ]\n\n\\begin{itemize} \\item[1] 6 \\item[2] \\frac{25}{4} \\item[3] \\frac{13}{2} \\item[4] \\frac{27}{4} \\item[5] 7 \\end{itemize}","answer":5,"score":4,"review":"Removed figure."}
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{"id":45,"name":"29_geom","problem":"29. μ’ννλ©΄μμ $\\overline{\\mathrm{OA}} = \\sqrt{2}$, $\\overline{\\mathrm{OB}} = 2\\sqrt{2}$μ΄κ³ \n\\[ \\cos(\\angle \\mathrm{AOB}) = \\frac{1}{4} \\]\nμΈ ννμ¬λ³ν $\\mathrm{OACB}$μ λνμ¬ μ $\\mathrm{P}$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] $\\overrightarrow{\\mathrm{OP}} = s \\overrightarrow{\\mathrm{OA}} + t \\overrightarrow{\\mathrm{OB}} \\quad (0 \\leq s \\leq 1, \\ 0 \\leq t \\leq 1)$ \\item[(λ)] $\\overrightarrow{\\mathrm{OP}} \\cdot \\overrightarrow{\\mathrm{OB}} + \\overrightarrow{\\mathrm{BP}} \\cdot \\overrightarrow{\\mathrm{BC}} = 2$ \\end{itemize}\n\nμ $\\mathrm{O}$λ₯Ό μ€μ¬μΌλ‘ νκ³ μ $\\mathrm{A}$λ₯Ό μ§λλ μ μλ₯Ό μμ§μ΄λ μ $\\mathrm{X}$μ λνμ¬ $|3\\overrightarrow{\\mathrm{OP}} - \\overrightarrow{\\mathrm{OX}}|$μ μ΅λκ°κ³Ό μ΅μκ°μ κ°κ° $M$, $m$μ΄λΌ νμ. $M \\times m = a\\sqrt{6} + b$μΌ λ, $a^2 + b^2$μ κ°μ ꡬνμμ€. (λ¨, $a$μ $b$λ μ 리μμ΄λ€.) [4μ ]","answer":100,"score":4,"review":"Removed figure."}
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{"id":46,"name":"30_geom","problem":"30. μ’ν곡κ°μ μ€μ¬μ΄ $\\mathrm{C}(2, \\sqrt{5}, 5)$μ΄κ³ μ $\\mathrm{P}(0, 0, 1)$μ μ§λλ ꡬ \\[ S: (x - 2)^2 + (y - \\sqrt{5})^2 + (z - 5)^2 = 25 \\] κ° μλ€. ꡬ $S$κ° νλ©΄ $\\mathrm{OPC}$μ λ§λμ μκΈ°λ μ μλ₯Ό μμ§μ΄λ μ $\\mathrm{Q}$, ꡬ $S$ μλ₯Ό μμ§μ΄λ μ $\\mathrm{R}$μ λνμ¬ λ μ $\\mathrm{Q}, \\mathrm{R}$μ $xy$νλ©΄ μλ‘μ μ μ¬μμ κ°κ° $\\mathrm{Q}_1, \\mathrm{R}_1$μ΄λΌ νμ.\n\nμΌκ°ν $\\mathrm{O}\\mathrm{Q}_1\\mathrm{R}_1$μ λμ΄κ° μ΅λκ° λλλ‘ νλ λ μ $\\mathrm{Q}, \\mathrm{R}$μ λνμ¬ μΌκ°ν $\\mathrm{O}\\mathrm{Q}_1\\mathrm{R}_1$μ νλ©΄ $\\mathrm{PQR}$ μλ‘μ μ μ¬μμ λμ΄λ $\\frac{q}{p} \\sqrt{6}$μ΄λ€. $p+q$μ κ°μ ꡬνμμ€.\n\n(λ¨, $\\mathrm{O}$λ μμ μ΄κ³ μΈ μ $\\mathrm{O}, \\mathrm{Q}_1, \\mathrm{R}_1$μ ν μ§μ μμ μμ§ μμΌλ©°, $p$μ $q$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]","answer":23,"score":4,"review":"Removed figure."}
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