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{"id":1,"name":"1","problem":"1. \\( \\sqrt[3]{24} \\times 3^{\\frac{2}{3}} \\) μ κ°μ? [2μ ]\n\n\\begin{itemize} \\item[1] 6 \\item[2] 7 \\item[3] 8 \\item[4] 9 \\item[5] 10 \\end{itemize}","answer":1,"score":2,"review":null} |
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{"id":2,"name":"2","problem":"2. ν¨μ $f(x) = 2x^3 - 5x^2 + 3$μ λνμ¬\n\n\\[ \\lim_{h \\to 0} \\frac{f(2 + h) - f(2)}{h} \\]\n\nμ κ°μ? [2μ ]\n\n\\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}","answer":4,"score":2,"review":null} |
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{"id":3,"name":"3","problem":"3. $(\\frac{3}{2}\\pi < \\theta < 2\\pi)$ μΈ $\\theta$μ λνμ¬ $\\sin(-\\theta) = \\frac{1}{3}$ μΌ λ,\n\n$\\tan\\theta$μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] -\\frac{\\sqrt{2}}{2} \\item[2] -\\frac{\\sqrt{2}}{4} \\item[3] -\\frac{1}{4} \\item[4] \\frac{1}{4} \\item[5] \\frac{\\sqrt{2}}{4} \\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":4,"name":"4","problem":"4. ν¨μ\n\n\\[ f(x) = \\begin{cases} 3x - a & (x < 2) \\\\ x^2 + a & (x \\geq 2) \\end{cases} \\]\n\nκ° μ€μ μ 체μ μ§ν©μμ μ°μμΌ λ, μμ $a$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":1,"score":3,"review":null} |
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{"id":5,"name":"5","problem":"5. λ€νν¨μ $f(x)$κ°\n\n\\[ f'(x) = 3x(x-2), \\quad f(1) = 6 \\]\n\nμ λ§μ‘±μν¬ λ, $f(2)$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":null} |
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{"id":6,"name":"6","problem":"6. λ±λΉμμ΄ $\\{a_n\\}$μ 첫째νλΆν° μ $n$νκΉμ§μ ν©μ $S_n$μ΄λΌ νμ.\n\n\\[ S_4 - S_2 = 3a_4, \\quad a_5 = \\frac{3}{4} \\]\n\nμΌ λ, $a_1 + a_2$μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] 27 \\item[2] 24 \\item[3] 21 \\item[4] 18 \\item[5] 15 \\end{itemize}","answer":4,"score":3,"review":null} |
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{"id":7,"name":"7","problem":"7. ν¨μ $f(x) = \\frac{1}{3}x^3 - 2x^2 - 12x + 4$κ° $x = \\alpha$μμ κ·Ήλμ΄κ³ \n\n$x = \\beta$μμ κ·ΉμμΌ λ, $\\beta - \\alpha$μ κ°μ? (λ¨, $\\alpha$μ $\\beta$λ μμμ΄λ€.) [3μ ]\n\n\\begin{itemize} \\item[1] -4 \\item[2] -1 \\item[3] 2 \\item[4] 5 \\item[5] 8 \\end{itemize}","answer":5,"score":3,"review":null} |
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{"id":8,"name":"8","problem":"8. μΌμ°¨ν¨μ $f(x)$κ° λͺ¨λ μ€μ $x$μ λνμ¬\n\n\\[ x f(x) - f(x) = 3x^4 - 3x \\]\n\nλ₯Ό λ§μ‘±μν¬ λ, $\\int_{-2}^{2} f(x) \\, dx$μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] 12 \\item[2] 16 \\item[3] 20 \\item[4] 24 \\item[5] 28 \\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":9,"name":"9","problem":"9. μμ§μ μμ λ μ $\\mathrm{P}(\\log_{5} 3), \\ \\mathrm{Q}(\\log_{5} 12)$μ λνμ¬ μ λΆ $\\mathrm{PQ}$λ₯Ό $m : (1 - m)$μΌλ‘ λ΄λΆνλ μ μ μ’νκ° 1μΌ λ, $4^m$μ κ°μ? (λ¨, $m$μ $0 < m < 1$μΈ μμμ΄λ€.) [4μ ]\n\n\\begin{itemize} \\item[1] \\frac{7}{6} \\item[2] \\frac{4}{3} \\item[3] \\frac{3}{2} \\item[4] \\frac{5}{3} \\item[5] \\frac{11}{6} \\end{itemize}","answer":4,"score":4,"review":null} |
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{"id":10,"name":"10","problem":"10. μκ° $( t = 0 )$μΌ λ λμμ μμ μ μΆλ°νμ¬ μμ§μ μλ₯Ό μμ§μ΄λ λ μ $( \\mathrm{P}, \\mathrm{Q} )$μ μκ° $( t (t \\geq 0) )$μμμ μλκ° κ°κ°\n\\[ v_1(t) = t^2 - 6t + 5, \\quad v_2(t) = 2t - 7 \\]\nμ΄λ€. μκ° $t$μμμ λ μ $\\mathrm{P}, \\mathrm{Q}$ μ¬μ΄μ 거리λ₯Ό $f(t)$λΌ ν λ, ν¨μ $f(t)$λ κ΅¬κ° $[0, a]$μμ μ¦κ°νκ³ , κ΅¬κ° $[a, b]$μμ κ°μνκ³ , κ΅¬κ° $[b, \\infty)$μμ μ¦κ°νλ€. μκ° $t = a$μμ $t = b$κΉμ§ μ $\\mathrm{Q}$κ° μμ§μΈ 거리λ? (λ¨, $0 < a < b$) [4μ ]\n\n\\begin{itemize} \\item[1] \\frac{15}{2} \\item[2] \\frac{17}{2} \\item[3] \\frac{19}{2} \\item[4] \\frac{21}{2} \\item[5] \\frac{23}{2} \\end{itemize}","answer":2,"score":4,"review":null} |
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{"id":11,"name":"11","problem":"11. κ³΅μ°¨κ° 0μ΄ μλ λ±μ°¨μμ΄ $\\{a_n\\}$μ λνμ¬\n\\[ |a_6| = a_8, \\quad \\sum_{k=1}^{5} \\frac{1}{a_k a_{k+1}} = \\frac{5}{96} \\]\n\nμΌ λ, $\\sum_{k=1}^{15} a_k$μ κ°μ? [4μ ]\n\n\\begin{itemize} \\item[1] 60 \\item[2] 65 \\item[3] 70 \\item[4] 75 \\item[5] 80 \\end{itemize}","answer":1,"score":4,"review":null} |
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{"id":12,"name":"12","problem":"12. ν¨μ $( f(x) = \\frac{1}{9} x (x - 6)(x - 9) )$μ μ€μ $( t \\ (0 < t < 6) )$μ λνμ¬ ν¨μ $( g(x) )$λ\n\n\\[ g(x) = \\begin{cases} f(x) & (x < t) \\\\ -(x - t) + f(t) & (x \\geq t) \\end{cases} \\]\n\nμ΄λ€. ν¨μ $y = g(x)$μ κ·Έλνμ $x$μΆμΌλ‘ λλ¬μΈμΈ μμμ λμ΄μ μ΅λκ°μ? [4μ ]\n\n\\begin{itemize} \\item[1] \\frac{125}{4} \\item[2] \\frac{127}{4} \\item[3] \\frac{129}{4} \\item[4] \\frac{131}{4} \\item[5] \\frac{133}{4} \\end{itemize}","answer":3,"score":4,"review":null} |
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{"id":13,"name":"13","problem":"13. \n\n\\[ \\overline{\\mathrm{AB}} = 3, \\quad \\overline{\\mathrm{BC}} = \\sqrt{13}, \\quad \\overline{\\mathrm{AD}} \\times \\overline{\\mathrm{CD}} = 9, \\quad \\angle \\mathrm{BAC} = \\frac{\\pi}{3} \\]\n\nμΈ μ¬κ°ν $\\mathrm{ABCD}$κ° μλ€. μΌκ°ν $\\mathrm{ABC}$μ λμ΄λ₯Ό $S_1$, μΌκ°ν $\\mathrm{ACD}$μ λμ΄λ₯Ό $S_2$λΌ νκ³ , μΌκ°ν $\\mathrm{ACD}$μ μΈμ μμ λ°μ§λ¦μ κΈΈμ΄λ₯Ό $R$μ΄λΌ νμ.\n$S_2 = \\frac{5}{6} S_1$μΌ λ, $\\frac{R}{\\sin(\\angle \\mathrm{ADC})}$μ κ°μ? [4μ ]\n\n\\begin{itemize} \\item[1] \\frac{54}{25} \\item[2] \\frac{117}{50} \\item[3] \\frac{63}{25} \\item[4] \\frac{27}{10} \\item[5] \\frac{72}{25} \\end{itemize}","answer":1,"score":4,"review":"Removed figure and the statement referring to the figure."} |
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{"id":14,"name":"14","problem":"14. λ μμ°μ $a, b$μ λνμ¬ ν¨μ $f(x)$λ\n\\[ f(x) = \\begin{cases} 2x^3 - 6x + 1 & (x \\leq 2) \\\\ a(x-2)(x-b) + 9 & (x > 2) \\end{cases} \\]\nμ΄λ€. μ€μ $t$μ λνμ¬ ν¨μ $y = f(x)$μ κ·Έλνμ μ§μ $y = t$κ° λ§λλ μ μ κ°μλ₯Ό $g(t)$λΌ νμ.\n\n\\[ g(k) + \\lim_{t \\to k^-} g(t) + \\lim_{t \\to k^+} g(t) = 9 \\]\n\nλ₯Ό λ§μ‘±μν€λ μ€μ $k$μ κ°μκ° 1μ΄ λλλ‘ νλ λ μμ°μ $a, b$μ μμμ $(a, b)$μ λνμ¬ $a + b$μ μ΅λκ°μ? [4μ ]\n\n\\begin{itemize} \\item[1] 51 \\item[2] 52 \\item[3] 53 \\item[4] 54 \\item[5] 55 \\end{itemize}","answer":1,"score":4,"review":null} |
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{"id":15,"name":"15","problem":"15. 첫째νμ΄ μμ°μμΈ μμ΄ $\\{a_n\\}$μ΄ λͺ¨λ μμ°μ $n$μ λνμ¬\n\n\\[ a_{n+1} = \\begin{cases} 2^{a_n} & (a_n \\text{μ΄ νμμΈ κ²½μ°}) \\\\ \\frac{1}{2} a_n & (a_n \\text{μ΄ μ§μμΈ κ²½μ°}) \\end{cases} \\]\n\nλ₯Ό λ§μ‘±μν¬ λ, $a_6 + a_7 = 3$μ΄ λλλ‘ νλ λͺ¨λ $a_1$μ κ°μ ν©μ? [4μ ]\n\n\\begin{itemize} \\item[1] 139 \\item[2] 146 \\item[3] 153 \\item[4] 160 \\item[5] 167 \\end{itemize}","answer":3,"score":4,"review":null} |
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{"id":16,"name":"16","problem":"16. λ°©μ μ $3^{x-8} = \\left(\\frac{1}{27}\\right)^x$μ λ§μ‘±μν€λ μ€μ $x$μ κ°μ ꡬνμμ€. [3μ ]","answer":2,"score":3,"review":null} |
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{"id":17,"name":"17","problem":"17. ν¨μ $f(x) = (x+1)(x^2 + 3)$μ λνμ¬ $f'(1)$μ κ°μ ꡬνμμ€. [3μ ]","answer":8,"score":3,"review":null} |
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{"id":18,"name":"18","problem":"18. λ μμ΄ $\\{a_n\\}, \\{b_n\\}$μ λνμ¬\n\n\\[ \\sum_{k=1}^{10} a_k = \\sum_{k=1}^{10} (2b_k - 1), \\quad \\sum_{k=1}^{10} (3a_k + b_k) = 33 \\]\n\nμΌ λ, $\\sum_{k=1}^{10} b_k$μ κ°μ ꡬνμμ€. [3μ ]","answer":9,"score":3,"review":null} |
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{"id":19,"name":"19","problem":"19. ν¨μ $f(x) = \\sin \\frac{\\pi}{4} x$λΌ ν λ, $0 < x < 16$μμ λΆλ±μ\n\n\\[ f(2+x) f(2-x) < \\frac{1}{4} \\]\n\nμ λ§μ‘±μν€λ λͺ¨λ μμ°μ $x$μ κ°μ ν©μ ꡬνμμ€. [3μ ]","answer":32,"score":3,"review":null} |
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{"id":20,"name":"20","problem":"20. $a > \\sqrt{2}$ μΈ μ€μ $a$μ λνμ¬ ν¨μ $f(x)$λ₯Ό\n\n\\[ f(x) = -x^3 + ax^2 + 2x \\]\n\nλΌ νμ. 곑μ $y = f(x)$ μμ μ $\\mathrm{O}(0, 0)$μμμ μ μ μ΄ κ³‘μ $y = f(x)$μ λ§λλ μ μ€ $\\mathrm{O}$κ° μλ μ μ $\\mathrm{A}$λΌ νκ³ , 곑μ $y = f(x)$ μμ μ $\\mathrm{A}$μμμ μ μ μ΄ $x$μΆκ³Ό λ§λλ μ μ $\\mathrm{B}$λΌ νμ. μ $\\mathrm{A}$κ° μ λΆ $\\mathrm{OB}$λ₯Ό μ§λ¦μΌλ‘ νλ μ μμ μ μΌ λ, $\\overline{\\mathrm{OA}} \\times \\overline{\\mathrm{AB}}$μ κ°μ ꡬνμμ€. [4μ ]","answer":25,"score":4,"review":null} |
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{"id":21,"name":"21","problem":"21. μμ $a$μ λνμ¬ $x \\geq -1$μμ μ μλ ν¨μ $f(x)$λ\n\\[ f(x) = \\begin{cases} -x^2 + 6x, & (-1 \\leq x < 6) \\\\ a \\log_4 (x - 5) & (x \\geq 6) \\end{cases} \\]\nμ΄λ€. $t \\geq 0$μΈ μ€μ $t$μ λνμ¬ λ«νκ΅¬κ° $[t-1, t+1]$μμμ $f(x)$μ μ΅λκ°μ $g(t)$λΌ νμ. κ΅¬κ° $[0, \\infty)$μμ ν¨μ $g(t)$μ μ΅μκ°μ΄ 5κ° λλλ‘ νλ μμ $a$μ μ΅μκ°μ ꡬνμμ€. [4μ ]","answer":10,"score":4,"review":null} |
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{"id":22,"name":"22","problem":"22. μ΅κ³ μ°¨νμ κ³μκ° 1μΈ μΌμ°¨ν¨μ $( f(x) )$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\[ \\boxed{\\begin{array}{c}\\text{ν¨μ } f(x) \\text{μ λνμ¬} \\\\ f(k-1)f(k+1) < 0 \\\\\\text{μ λ§μ‘±μν€λ μ μ } k \\text{λ }\\underline{\\text{μ‘΄μ¬νμ§ μλλ€.}}\\end{array}} \\]\n\n\\[ f'\\left( -\\frac{1}{4} \\right) = -\\frac{1}{4}, \\quad f'\\left( \\frac{1}{4} \\right) < 0 \\text{μΌ λ,} \\quad f(8) \\text{μ κ°μ ꡬνμμ€. [4μ ]} \\]","answer":483,"score":4,"review":null} |
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{"id":23,"name":"23_prob","problem":"23. 5κ°μ λ¬Έμ $(x, x, y, y, z)$λ₯Ό λͺ¨λ μΌλ ¬λ‘ λμ΄νλ κ²½μ°μ μλ? [2μ ]\n\n\\begin{itemize} \\item[1] 10 \\item[2] 20 \\item[3] 30 \\item[4] 40 \\item[5] 50 \\end{itemize}","answer":3,"score":2,"review":null} |
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{"id":24,"name":"24_prob","problem":"24. λ μ¬κ±΄ $( A, B )$λ μλ‘ λ
립μ΄κ³ \n\n\\[ \\mathrm{P}(A \\cap B) = \\frac{1}{4}, \\quad \\mathrm{P}(A^C) = 2\\mathrm{P}(A) \\]\n\nμΌ λ, $( \\mathrm{P}(B) )$μ κ°μ? (λ¨, $A^C$μ $A$μ μ¬μ¬κ±΄μ΄λ€.) [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{3}{8} \\item[2] \\frac{1}{2} \\item[3] \\frac{5}{8} \\item[4] \\frac{3}{4} \\item[5] \\frac{7}{8} \\end{itemize}","answer":4,"score":3,"review":null} |
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{"id":25,"name":"25_prob","problem":"25. μ«μ $1, 2, 3, 4, 5, 6$μ΄ νλμ© μ ν μλ 6μ₯μ μΉ΄λκ° μλ€. μ΄ 6μ₯μ μΉ΄λλ₯Ό λͺ¨λ ν λ²μ© μ¬μ©νμ¬ μΌλ ¬λ‘ μμλ‘ λμ΄ν λ, μ λμ λμΈ μΉ΄λμ μ ν λ μμ ν©μ΄ 10 μ΄νκ° λλλ‘ μΉ΄λκ° λμΌ νλ₯ μ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{8}{15} \\item[2] \\frac{19}{30} \\item[3] \\frac{11}{15} \\item[4] \\frac{5}{6} \\item[5] \\frac{14}{15} \\end{itemize}","answer":5,"score":3,"review":"Removed figure."} |
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{"id":26,"name":"26_prob","problem":"26. 4κ°μ λμ μ λμμ λμ Έμ μλ©΄μ΄ λμ€λ λμ μ κ°μλ₯Ό νλ₯ λ³μ $X$λΌ νκ³ , μ΄μ°νλ₯ λ³μ $Y$λ₯Ό\n\n\\[ Y = \\begin{cases} X & (X\\text{κ°} \\ 0 \\ \\text{λλ} \\ 1\\text{μ κ°μ κ°μ§λ κ²½μ°}) \\\\ 2 & (X\\text{κ°} \\ 2 \\ \\text{μ΄μμ κ°μ κ°μ§λ κ²½μ°}) \\end{cases} \\]\n\nλΌ νμ. $\\mathrm{E}(Y)$μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{25}{16} \\item[2] \\frac{13}{8} \\item[3] \\frac{27}{16} \\item[4] \\frac{7}{4} \\item[5] \\frac{29}{16} \\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":27,"name":"27_prob","problem":"27. μ κ·λΆν¬ $\\mathrm{N}(m, 5^2)$μ λ°λ₯΄λ λͺ¨μ§λ¨μμ ν¬κΈ°κ° 49μΈ νλ³Έμ μμμΆμΆνμ¬ μ»μ νλ³Ένκ· μ΄ $\\bar{x}$μΌ λ, λͺ¨νκ· $m$μ λν μ λ’°λ 95\\%μ μ 뒰ꡬκ°μ΄ $a \\leq m \\leq \\frac{6}{5} a$μ΄λ€. $\\bar{x}$μ κ°μ?\n\n(λ¨, $Z$κ° νμ€μ κ·λΆν¬λ₯Ό λ°λ₯΄λ νλ₯ λ³μμΌ λ, $\\mathrm{P}(|Z| \\leq 1.96) = 0.95$λ‘ κ³μ°νλ€.) [3μ ]\n\n\\begin{itemize} \\item[1] 15.2 \\item[2] 15.4 \\item[3] 15.6 \\item[4] 15.8 \\item[5] 16.0 \\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":28,"name":"28_prob","problem":"28. νλμ μ£Όλ¨Έλμ λ μμ $\\mathrm{A}$, $\\mathrm{B}$κ° μλ€. μ£Όλ¨Έλμλ μ«μ $1, 2, 3, 4$κ° νλμ© μ ν $4$μ₯μ μΉ΄λκ° λ€μ΄ μκ³ , μμ $\\mathrm{A}$μλ ν° κ³΅κ³Ό κ²μ κ³΅μ΄ κ°κ° $8$κ° μ΄μ λ€μ΄ μκ³ , μμ $\\mathrm{B}$λ λΉμ΄ μλ€. μ΄ μ£Όλ¨Έλμ λ μμ $\\mathrm{A}$, $\\mathrm{B}$λ₯Ό μ¬μ©νμ¬ λ€μ μνμ νλ€.\n\n\\[\\begin{tabular}{|l|} \\hline μ£Όλ¨Έλμμ μμλ‘ ν μ₯μ μΉ΄λλ₯Ό κΊΌλ΄μ΄ μΉ΄λμ μ ν μλ₯Ό νμΈν ν λ€μ μ£Όλ¨Έλμ λ£λλ€. \\\\ νμΈν μκ° $1$μ΄λ©΄ μμ $\\mathrm{A}$μ μλ ν° κ³΅ $1$κ°λ₯Ό μμ $\\mathrm{B}$μ λ£κ³ , \\\\ νμΈν μκ° $2$ λλ $3$μ΄λ©΄ μμ $\\mathrm{A}$μ μλ ν° κ³΅ $1$κ°μ κ²μ 곡 $1$κ°λ₯Ό μμ $\\mathrm{B}$μ λ£κ³ , \\\\ νμΈν μκ° $4$μ΄λ©΄ μμ $\\mathrm{A}$μ μλ ν° κ³΅ $2$κ°μ κ²μ 곡 $1$κ°λ₯Ό μμ $\\mathrm{B}$μ λ£λλ€. \\\\ \\hline \\end{tabular}\\]\n\nμ΄ μνμ $4$λ² λ°λ³΅ν ν μμ $\\mathrm{B}$μ λ€μ΄ μλ 곡μ κ°μκ° $8$μΌ λ, μμ $\\mathrm{B}$μ λ€μ΄ μλ κ²μ 곡μ κ°μκ° $2$μΌ νλ₯ μ? [4μ ]\n\n\\begin{itemize} \\item[1] \\frac{3}{70} \\item[2] \\frac{2}{35} \\item[3] \\frac{1}{14} \\item[4] \\frac{3}{35} \\item[5] \\frac{1}{10} \\end{itemize}","answer":4,"score":4,"review":"Removed figure."} |
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{"id":29,"name":"29_prob","problem":"29. λ€μ 쑰건μ λ§μ‘±μν€λ 6 μ΄νμ μμ°μ $a, b, c, d$μ λͺ¨λ μμμ $(a, b, c, d)$μ κ°μλ₯Ό ꡬνμμ€. [4μ ]\n\n\\[ a \\leq c \\leq d \\quad \\text{μ΄κ³ } \\quad b \\leq c \\leq d \\text{μ΄λ€.} \\]","answer":196,"score":4,"review":null} |
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{"id":30,"name":"30_prob","problem":"30. μμ $t$μ λνμ¬ νλ₯ λ³μ $X$κ° μ κ·λΆν¬ $\\mathrm{N}(1, t^2)$μ λ°λ₯Έλ€.\n\\[ \\mathrm{P}(X \\leq 5t) \\geq \\frac{1}{2} \\]\nμ΄ λλλ‘ νλ λͺ¨λ μμ $t$μ λνμ¬\n\\[ \\mathrm{P}(t^2 - t + 1 \\leq X \\leq t^2 + t + 1) \\]\nμ μ΅λκ°μ λ€μ νμ€μ κ·λΆν¬νλ₯Ό μ΄μ©νμ¬ κ΅¬ν κ°μ $k$λΌ νμ. \\\\\n1000$\\times k$μ κ°μ ꡬνμμ€. [4μ ]\n\n\\begin{table}[h!] \\centering \\begin{tabular}{|c|c|} \\hline $z$ & $\\mathrm{P}(0 \\leq Z \\leq z)$ \\\\ \\hline 0.6 & 0.226 \\\\ 0.8 & 0.288 \\\\ 1.0 & 0.341 \\\\ 1.2 & 0.385 \\\\ 1.4 & 0.419 \\\\ \\hline \\end{tabular} \\end{table}","answer":673,"score":4,"review":"'μ€λ₯Έμͺ½' changed to 'λ€μ'."} |
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{"id":31,"name":"23_calc","problem":"23. $\\lim_{x \\to 0} \\frac{\\ln(1+3x)}{\\ln(1+5x)}$ μ κ°μ? [2μ ]\n\n\\begin{itemize} \\item[1] \\frac{1}{5} \\item[2] \\frac{2}{5} \\item[3] \\frac{3}{5} \\item[4] \\frac{4}{5} \\item[5] 1 \\end{itemize}","answer":3,"score":2,"review":null} |
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{"id":32,"name":"24_calc","problem":"24. 맀κ°λ³μ $ t(t > 0) $λ‘ λνλ΄μ΄μ§ 곑μ \n\n\\[ x = \\ln(t^3 + 1), \\quad y = \\sin \\pi t \\]\n\nμμ $ t = 1 $μΌ λ, $ \\frac{dy}{dx} $μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] -\\frac{1}{3}\\pi \\item[2] -\\frac{2}{3}\\pi \\item[3] -\\pi \\item[4] -\\frac{4}{3}\\pi \\item[5] -\\frac{5}{3}\\pi \\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":33,"name":"25_calc","problem":"25. μμ μ€μ μ 체μ μ§ν©μμ μ μλκ³ λ―ΈλΆκ°λ₯ν λ ν¨μ $ f(x), g(x) $κ° μλ€. $ g(x) $λ $ f(x) $μ μν¨μμ΄κ³ , $ g'(x) $λ μμ μ€μ μ 체μ μ§ν©μμ μ°μμ΄λ€. λͺ¨λ μμ $ a $μ λνμ¬\n\\[ \\int_{1}^{a} \\frac{1}{g'(f(x)) f(x)} \\, dx = 2 \\ln a + \\ln (a+1) - \\ln 2 \\]\nμ΄κ³ $ f(1) = 8 $μΌ λ, $ f(2) $μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] 36 \\item[2] 40 \\item[3] 44 \\item[4] 48 \\item[5] 52 \\end{itemize}","answer":4,"score":3,"review":null} |
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{"id":34,"name":"26_calc","problem":"26. 곑μ $ y = \\sqrt{(1 - 2x) \\cos x} \\left( \\frac{3}{4} \\pi \\leq x \\leq \\frac{5}{4} \\pi \\right) $μ\n\n$ x $μΆ λ° λ μ§μ $ x = \\frac{3}{4} \\pi, \\ x = \\frac{5}{4} \\pi $λ‘ λλ¬μΈμΈ λΆλΆμ λ°λ©΄μΌλ‘ νλ μ
체λνμ΄ μλ€. μ΄ μ
체λνμ $ x $μΆμ μμ§μΈ νλ©΄μΌλ‘ μλ₯Έ λ¨λ©΄μ΄ λͺ¨λ μ μ¬κ°νμΌ λ, μ΄ μ
체λνμ λΆνΌλ? [3μ ]\n\n\\begin{itemize} \\item[1] \\sqrt{2}\\pi - \\sqrt{2} \\item[2] \\sqrt{2}\\pi - 1 \\item[3] 2 \\sqrt{2}\\pi - \\sqrt{2} \\item[4] 2 \\sqrt{2}\\pi - 1 \\item[5] 2 \\sqrt{2}\\pi \\end{itemize}","answer":3,"score":3,"review":"Removed figure and the statement referring to the figure."} |
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{"id":35,"name":"27_calc","problem":"27. μ€μ $ t $μ λνμ¬ μμ μ μ§λκ³ κ³‘μ $ y = \\frac{1}{e^x} + e^t $μ μ νλ μ§μ μ κΈ°μΈκΈ°λ₯Ό $ f(t) $λΌ νμ. $ f(a) = -e \\sqrt{e} $λ₯Ό λ§μ‘±μν€λ μμ $ a $μ λνμ¬ $ f'(a) $μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] -\\frac{1}{3} e \\sqrt{e} \\item[2] -\\frac{1}{2} e \\sqrt{e} \\item[3] -\\frac{2}{3} e \\sqrt{e} \\item[4] -\\frac{5}{6} e \\sqrt{e} \\item[5] -e \\sqrt{e} \\end{itemize}","answer":1,"score":3,"review":null} |
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{"id":36,"name":"28_calc","problem":"28. μ€μ μ 체μ μ§ν©μμ μ°μμΈ ν¨μ $f(x)$κ° λͺ¨λ μ€μ $x$μ λνμ¬ $f(x) \\geq 0$μ΄κ³ , $x < 0$μΌ λ $f(x) = -4xe^{4x^2}$μ΄λ€.\nλͺ¨λ μμ $t$μ λνμ¬ $x$μ λν λ°©μ μ $f(x) = t$μ μλ‘ λ€λ₯Έ μ€κ·Όμ κ°μλ 2μ΄κ³ , μ΄ λ°©μ μμ λ μ€κ·Ό μ€ μμ κ°μ $g(t)$, ν° κ°μ $h(t)$λΌ νμ.\nλ ν¨μ $g(t), h(t)$λ λͺ¨λ μμ $t$μ λνμ¬ \n\\[ 2g(t) + h(t) = k \\quad (k \\text{λ μμ}) \\]\nλ₯Ό λ§μ‘±μν¨λ€. $\\int_0^7 f(x) \\, dx = e^4 - 1$μΌ λ, $\\frac{f(9)}{f(8)}$μ κ°μ? [4μ ]\n\n\\begin{itemize}\n \\item[1] $\\frac{3}{2} e^5$\n \\item[2] $\\frac{4}{3} e^7$\n \\item[3] $\\frac{5}{4} e^9$\n \\item[4] $\\frac{6}{5} e^{11}$\n \\item[5] $\\frac{7}{6} e^{13}$\n\\end{itemize}","answer":2,"score":4,"review":null} |
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{"id":37,"name":"29_calc","problem":"29. 첫째νκ³Ό 곡λΉκ° κ°κ° 0μ΄ μλ λ λ±λΉμμ΄ $\\{a_n\\}, \\{b_n\\}$μ λνμ¬ λ κΈμ $\\sum_{n=1}^{\\infty} a_n$, $\\sum_{n=1}^{\\infty} b_n$μ΄ κ°κ° μλ ΄νκ³ \n\n\\[\\sum_{n=1}^{\\infty} a_n b_n = \\left( \\sum_{n=1}^{\\infty} a_n \\right) \\times \\left( \\sum_{n=1}^{\\infty} b_n \\right),\\]\n\n\\[3 \\times \\sum_{n=1}^{\\infty} |a_{2n}| = 7 \\times \\sum_{n=1}^{\\infty} |a_{3n}|\\]\n\nμ΄ μ±λ¦½νλ€. $\\sum_{n=1}^{\\infty} \\frac{b_{2n-1} + b_{3n+1}}{b_n} = S$μΌ λ, $120S$μ κ°μ ꡬνμμ€. [4μ ]","answer":162,"score":4,"review":null} |
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{"id":38,"name":"30_calc","problem":"30. μ€μ μ 체μ μ§ν©μμ λ―ΈλΆκ°λ₯ν ν¨μ $f(x)$μ λν¨μ $f'(x)$κ°\n\n\\[ f'(x) = |\\sin x| \\cos x \\]\n\nμ΄λ€. μμ $a$μ λνμ¬ κ³‘μ $y=f(x)$ μμ μ $(a, f(a))$μμμ μ μ μ λ°©μ μμ $y = g(x)$λΌ νμ. ν¨μ\n\n\\[ h(x) = \\int_{0}^{x} \\{f(t) - g(t)\\} \\, dt \\]\n\nκ° $x = a$μμ κ·Ήλ λλ κ·Ήμκ° λλλ‘ νλ λͺ¨λ μμ $a$λ₯Ό μμ μλΆν° ν¬κΈ°μμΌλ‘ λμ΄ν λ, $n$λ²μ§Έ μλ₯Ό $a_n$μ΄λΌ νμ.\n\n\\[ \\frac{100}{\\pi} \\times (a_6 - a_2) \\]\n\nμ κ°μ ꡬνμμ€. [4μ ]","answer":125,"score":4,"review":null} |
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{"id":39,"name":"23_geom","problem":"23. μ’ν곡κ°μ λ μ \\(\\mathrm{A}(a, -2, 6)\\), \\(\\mathrm{B}(9, 2, b)\\)μ λνμ¬\n\nμ λΆ \\(\\mathrm{AB}\\)μ μ€μ μ μ’νκ° \\((4, 0, 7)\\)μΌ λ, \\(a + b\\)μ κ°μ? [2μ ]\n\n\\begin{itemize}\n \\item[1] 1\n \\item[2] 3\n \\item[3] 5\n \\item[4] 7\n \\item[5] 9\n\\end{itemize}","answer":4,"score":2,"review":null} |
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{"id":40,"name":"24_geom","problem":"24. νμ $\\frac{x^2}{a^2} + \\frac{y^2}{6} = 1$ μμ μ $\\left( \\sqrt{3}, -2 \\right)$ μμμ μ μ μ κΈ°μΈκΈ°λ? (λ¨, $a$λ μμμ΄λ€.) [3μ ]\n\n\\begin{itemize}\n \\item[1] $\\sqrt{3}$\n \\item[2] $\\frac{\\sqrt{3}}{2}$\n \\item[3] $\\frac{\\sqrt{3}}{3}$\n \\item[4] $\\frac{\\sqrt{3}}{4}$\n \\item[5] $\\frac{\\sqrt{3}}{5}$\n\\end{itemize}","answer":3,"score":3,"review":null} |
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{"id":41,"name":"25_geom","problem":"25. λ λ²‘ν° $\\vec{a}$, $\\vec{b}$ μ λνμ¬\n\n\\[|\\vec{a}| = \\sqrt{11}, \\quad |\\vec{b}| = 3, \\quad |2\\vec{a} - \\vec{b}| = \\sqrt{17}\\]\n\nμΌ λ, $|\\vec{a} - \\vec{b}|$ μ κ°μ? [3μ ]\n\n\\begin{itemize}\n \\item[1] $\\frac{\\sqrt{2}}{2}$\n \\item[2] $\\sqrt{2}$\n \\item[3] $\\frac{3\\sqrt{2}}{2}$\n \\item[4] $2\\sqrt{2}$\n \\item[5] $\\frac{5\\sqrt{2}}{2}$\n\\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":42,"name":"26_geom","problem":"26. μ’ν곡κ°μ νλ©΄ $\\alpha$κ° μλ€. νλ©΄ $\\alpha$ μμ μμ§ μμ μλ‘ λ€λ₯Έ λ μ $\\mathrm{A}$, $\\mathrm{B}$μ νλ©΄ $\\alpha$ μλ‘μ μ μ¬μμ κ°κ° $\\mathrm{A'}$, $\\mathrm{B'}$μ΄λΌ ν λ,\n\n\\[\\overline{\\mathrm{AB}} = \\overline{\\mathrm{A'B'}} = 6\\]\n\nμ΄λ€. μ λΆ $\\mathrm{AB}$μ μ€μ $\\mathrm{M}$μ νλ©΄ $\\alpha$ μλ‘μ μ μ¬μμ $\\mathrm{M'}$μ΄λΌ ν λ,\n\n\\[\\overline{\\mathrm{PM'}} \\perp \\overline{\\mathrm{A'B'}}, \\quad \\overline{\\mathrm{PM'}} = 6\\]\n\nμ΄ λλλ‘ νλ©΄ $\\alpha$ μμ μ $\\mathrm{P}$λ₯Ό μ‘λλ€.\n\nμΌκ°ν $\\mathrm{A'B'P}$μ νλ©΄ $\\mathrm{ABP}$ μλ‘μ μ μ¬μμ λμ΄κ° $\\frac{9}{2}$μΌ λ, μ λΆ $\\mathrm{PM}$μ κΈΈμ΄λ? [3μ ]\n\n\\begin{itemize}\n \\item[1] 12\n \\item[2] 15\n \\item[3] 18\n \\item[4] 21\n \\item[5] 24\n\\end{itemize}","answer":5,"score":3,"review":null} |
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{"id":43,"name":"27_geom","problem":"27. μ΄μ μ΄ \\( \\mathrm{F} \\)μΈ ν¬λ¬Όμ \\( y^2 = 8x \\) μμ ν μ \\( \\mathrm{A} \\)μμ ν¬λ¬Όμ μ μ€μ μ λ΄λ¦° μμ μ λ°μ \\( \\mathrm{B} \\)λΌ νκ³ , μ§μ \\( \\mathrm{BF} \\)μ ν¬λ¬Όμ μ΄ λ§λλ λ μ μ κ°κ° \\( \\mathrm{C}, \\mathrm{D} \\)λΌ νμ. \\( \\overline{\\mathrm{BC}} = \\overline{\\mathrm{CD}} \\)μΌ λ, μΌκ°ν \\( \\mathrm{ABD} \\)μ λμ΄λ? (λ¨, \\( \\overline{\\mathrm{CF}} < \\overline{\\mathrm{DF}} \\)μ΄κ³ , μ \\( \\mathrm{A} \\)λ μμ μ΄ μλλ€.) [3μ ]\n\n\\begin{itemize}\n \\item[1] \\( 100 \\sqrt{2} \\)\n \\item[2] \\( 104 \\sqrt{2} \\)\n \\item[3] \\( 108 \\sqrt{2} \\)\n \\item[4] \\( 112 \\sqrt{2} \\)\n \\item[5] \\( 116 \\sqrt{2} \\)\n\\end{itemize}","answer":3,"score":3,"review":null} |
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{"id":44,"name":"28_geom","problem":"28. μλ‘ λ€λ₯Έ λ νλ©΄ $\\alpha$, $\\beta$μ κ΅μ μμ $\\overline{\\mathrm{AB}} = 18$μΈ λ μ $\\mathrm{A}$, $\\mathrm{B}$κ° μλ€. μ λΆ $\\mathrm{AB}$λ₯Ό μ§λ¦μΌλ‘ νλ μ $C_1$μ΄ νλ©΄ $\\alpha$ μμ μκ³ , μ λΆ $\\mathrm{AB}$λ₯Ό μ₯μΆμΌλ‘ νκ³ λ μ $\\mathrm{F}$, $\\mathrm{F'}$λ₯Ό μ΄μ μΌλ‘ νλ νμ $C_2$κ° νλ©΄ $\\beta$ μμ μλ€. μ $C_1$ μμ ν μ $\\mathrm{P}$μμ νλ©΄ $\\beta$μ λ΄λ¦° μμ μ λ°μ $\\mathrm{H}$λΌ ν λ,\n\\[\n\\overline{\\mathrm{HF'}} < \\overline{\\mathrm{HF}} \\quad \\text{μ΄κ³ } \\quad \\angle \\mathrm{HFF'} = \\frac{\\pi}{6}\n\\]\nμ΄λ€. μ§μ $\\mathrm{HF}$μ νμ $C_2$κ° λ§λλ μ μ€ μ $\\mathrm{H}$μ κ°κΉμ΄ μ μ $\\mathrm{Q}$λΌ νλ©΄, $\\overline{\\mathrm{FH}} < \\overline{\\mathrm{FQ}}$μ΄λ€. μ $\\mathrm{H}$λ₯Ό μ€μ¬μΌλ‘ νκ³ μ $\\mathrm{Q}$λ₯Ό μ§λλ νλ©΄ $\\beta$ μμ μμ λ°μ§λ¦μ κΈΈμ΄κ° 4μ΄κ³ μ§μ $\\mathrm{AB}$μ μ νλ€. λ νλ©΄ $\\alpha$, $\\beta$κ° μ΄λ£¨λ κ°μ ν¬κΈ°λ₯Ό $\\theta$λΌ ν λ, $\\cos \\theta$μ κ°μ?\n(λ¨, μ $\\mathrm{P}$λ νλ©΄ $\\beta$ μμ μμ§ μλ€.) [4μ ]\n\n\\begin{itemize} \\item[1] $\\frac{2 \\sqrt{66}}{33}$ \\item[2] $\\frac{4 \\sqrt{69}}{69}$ \\item[3] $\\frac{\\sqrt{2}}{3}$ \\item[4] $\\frac{4 \\sqrt{3}}{15}$ \\item[5] $\\frac{2 \\sqrt{78}}{39}$ \\end{itemize}","answer":5,"score":4,"review":"Removed figure and the statement referring to the figure."} |
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{"id":45,"name":"29_geom","problem":"29. μμ $c$μ λνμ¬ λ μ $\\mathrm{F}(c, 0), \\ \\mathrm{F'}(-c, 0)$μ μ΄μ μΌλ‘ νκ³ , μ£ΌμΆμ κΈΈμ΄κ° 6μΈ μ곑μ μ΄ μλ€. μ΄ μ곑μ μμ λ€μ 쑰건μ λ§μ‘±μν€λ μλ‘ λ€λ₯Έ λ μ $\\mathrm{P}, \\ \\mathrm{Q}$κ° μ‘΄μ¬νλλ‘ νλ λͺ¨λ $c$μ κ°μ ν©μ ꡬνμμ€. [4μ ]\n\n\\begin{enumerate}\n \\item[(κ°)] μ $\\mathrm{P}$λ μ 1μ¬λΆλ©΄ μμ μκ³ , μ $\\mathrm{Q}$λ μ§μ $\\mathrm{PF'}$ μμ μλ€.\n \\item[(λ)] μΌκ°ν $\\mathrm{PF'F}$λ μ΄λ±λ³μΌκ°νμ΄λ€.\n \\item[(λ€)] μΌκ°ν $\\mathrm{PQF}$μ λλ μ κΈΈμ΄λ 28μ΄λ€.\n\\end{enumerate}","answer":11,"score":4,"review":null} |
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{"id":46,"name":"30_geom","problem":"30. μ’ννλ©΄μ ν λ³μ κΈΈμ΄κ° 4μΈ μ μΌκ°ν $\\mathrm{ABC}$κ° μλ€. μ λΆ $\\mathrm{AB}$λ₯Ό 1:3μΌλ‘ λ΄λΆνλ μ μ $\\mathrm{D}$, μ λΆ $\\mathrm{BC}$λ₯Ό 1:3μΌλ‘ λ΄λΆνλ μ μ $\\mathrm{E}$, μ λΆ $\\mathrm{CA}$λ₯Ό 1:3μΌλ‘ λ΄λΆνλ μ μ $\\mathrm{F}$λΌ νμ. λ€ μ $\\mathrm{P}, \\mathrm{Q}, \\mathrm{R}, \\mathrm{X}$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize}\n \\item[(κ°)] $|\\overrightarrow{\\mathrm{DP}}| = |\\overrightarrow{\\mathrm{EQ}}| = |\\overrightarrow{\\mathrm{FR}}| = 1$\n \\item[(λ)] $\\overrightarrow{\\mathrm{AX}} = \\overrightarrow{\\mathrm{PB}} + \\overrightarrow{\\mathrm{QC}} + \\overrightarrow{\\mathrm{RA}}$\n\\end{itemize}\n\n$|\\overrightarrow{\\mathrm{AX}}|$μ κ°μ΄ μ΅λμΌ λ, μΌκ°ν $\\mathrm{PQR}$μ λμ΄λ₯Ό $S$λΌ νμ. $16S^2$μ κ°μ ꡬνμμ€. [4μ ]","answer":147,"score":4,"review":null} |