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{"id":1,"name":"1","problem":"1. $\\left( \\frac{4}{2^{\\sqrt{2}}} \\right)^{2 + \\sqrt{2}}$ 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":5,"score":2,"review":null} |
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{"id":2,"name":"2","problem":"2. $\\lim_{x \\to \\infty} \\frac{\\sqrt{x^2 - 2} + 3x}{x + 5}$ μ κ°μ? [2μ ] \\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. 곡λΉκ° μμμΈ λ±λΉμμ΄$\\{a_n\\}$μ΄ \\[ a_2 + a_4 = 30, \\quad a_4 + a_6 = \\frac{15}{2} \\] λ₯Ό λ§μ‘±μν¬ λ, $a_1$ μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 48 \\item[2] 56 \\item[3] 64 \\item[4] 72 \\item[5] 80 \\end{itemize}","answer":1,"score":3,"review":null} |
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{"id":4,"name":"4","problem":"4. λ€νν¨μ $f(x)$μ λνμ¬ ν¨μ $g(x)$ λ₯Ό \\[ g(x) = x^2 f(x) \\] λΌ νμ. $f(2) = 1, \\ f'(2) = 3$ μΌ λ, $g'(2)$ μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 12 \\item[2] 14 \\item[3] 16 \\item[4] 18 \\item[5] 20 \\end{itemize}","answer":3,"score":3,"review":null} |
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{"id":5,"name":"5","problem":"5. $\\tan \\theta < 0$μ΄κ³ $\\cos\\left(\\frac{\\pi}{2} + \\theta\\right) = \\frac{\\sqrt{5}}{5}$μΌ λ, $\\cos \\theta$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] -\\frac{2\\sqrt{5}}{5} \\item[2] -\\frac{\\sqrt{5}}{5} \\item[3] 0 \\item[4] \\frac{\\sqrt{5}}{5} \\item[5] \\frac{2\\sqrt{5}}{5} \\end{itemize}","answer":5,"score":3,"review":null} |
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{"id":6,"name":"6","problem":"6. ν¨μ $f(x) = 2x^3 - 9x^2 + ax + 5$ λ $x=1$ μμ κ·Ήλμ΄κ³ , $x=b$ μμ κ·Ήμμ΄λ€. $a + b$μ κ°μ? (λ¨, $a$, $b$λ μμμ΄λ€.) [3μ ] \\begin{itemize} \\item[1] 12 \\item[2] 14 \\item[3] 16 \\item[4] 18 \\item[5] 20 \\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":7,"name":"7","problem":"7. λͺ¨λ νμ΄ μμμ΄κ³ 첫째νκ³Ό κ³΅μ°¨κ° κ°μ λ±μ°¨μμ΄ $\\{a_n\\}$μ΄ \\[ \\sum_{k=1}^{15} \\frac{1}{\\sqrt{a_k} + \\sqrt{a_{k+1}}} = 2 \\] λ₯Ό λ§μ‘±μν¬ λ, $a_4$ μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 6 \\item[2] 7 \\item[3] 8 \\item[4] 9 \\item[5] 10 \\end{itemize}","answer":4,"score":3,"review":null} |
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{"id":8,"name":"8","problem":"8. μ $(0, 4)$μμ 곑μ $y = x^3 - x + 2$μ κ·Έμ μ μ μ $x$μ νΈμ? [3μ ] \\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":4,"score":3,"review":null} |
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{"id":9,"name":"9","problem":"9. ν¨μ \\[ f(x) = a - \\sqrt{3} \\tan 2x \\] κ° λ«νκ΅¬κ° \\left[ -\\frac{\\pi}{6}, b \\right] μμ μ΅λκ° 7, μ΅μκ° 3μ κ°μ§ λ, $a \\times b$μ κ°μ? (λ¨, $a$, $b$λ μμμ΄λ€.) [4μ ] \\begin{itemize} \\item[1] \\frac{\\pi}{2} \\item[2] \\frac{5\\pi}{12} \\item[3] \\frac{\\pi}{3} \\item[4] \\frac{\\pi}{4} \\item[5] \\frac{\\pi}{6} \\end{itemize}","answer":3,"score":4,"review":"Removed figure."} |
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{"id":10,"name":"10","problem":"10. λ 곑μ $y = x^3 + x^2$, $y = -x^2 + k$μ $y$μΆμΌλ‘ λλ¬μΈμΈ λΆλΆμ λμ΄λ₯Ό $A$, λ 곑μ $y = x^3 + x^2$, $y = -x^2 + k$μ μ§μ $x = 2$λ‘ λλ¬μΈμΈ λΆλΆμ λμ΄λ₯Ό $B$λΌ νμ. $A = B$μΌ λ, μμ $k$μ κ°μ? (λ¨, $4 < k < 5$) [4μ ] \\begin{itemize} \\item[1] \\frac{25}{6} \\item[2] \\frac{13}{3} \\item[3] \\frac{9}{2} \\item[4] \\frac{14}{3} \\item[5] \\frac{29}{6} \\end{itemize}","answer":4,"score":4,"review":null} |
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{"id":11,"name":"11","problem":"11. μ¬κ°ν $\\mathrm{ABCD}$κ° ν μμ λ΄μ νκ³ \\[ \\overline{\\mathrm{AB}} = 5, \\quad \\overline{\\mathrm{AC}} = 3\\sqrt{5}, \\quad \\overline{\\mathrm{AD}} = 7, \\quad \\angle \\mathrm{BAC} = \\angle \\mathrm{CAD} \\] μΌ λ, μ΄ μμ λ°μ§λ¦μ κΈΈμ΄λ? [4μ ] \\begin{itemize} \\item[1] \\frac{5\\sqrt{2}}{2} \\item[2] \\frac{8\\sqrt{5}}{5} \\item[3] \\frac{5\\sqrt{5}}{3} \\item[4] \\frac{8\\sqrt{2}}{3} \\item[5] \\frac{9\\sqrt{3}}{4} \\end{itemize}","answer":1,"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° λ€μ 쑰건μ λ§μ‘±μν¨λ€. \\[ n-1 \\leq x < n \\text{μΌ λ}, \\ |f(x)| = |6(x-n+1)(x-n)| \\text{μ΄λ€.} \\ (\\text{λ¨}, \\ n \\text{μ μμ°μμ΄λ€.}) \\] μ΄λ¦°κ΅¬κ° $(0, 4)$μμ μ μλ ν¨μ \\[ g(x) = \\int_0^x f(t) \\, dt - \\int_x^4 f(t) \\, dt \\] κ° $x = 2$μμ μ΅μκ° 0μ κ°μ§ λ, $\\int_{\\frac{1}{2}}^{4} f(x) \\, dx$ μ κ°μ? [4μ ] \\begin{itemize} \\item[1] -\\frac{3}{2} \\item[2] -\\frac{1}{2} \\item[3] \\frac{1}{2} \\item[4] \\frac{3}{2} \\item[5] \\frac{5}{2} \\end{itemize}","answer":2,"score":4,"review":null} |
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{"id":13,"name":"13","problem":"13. μμ°μ $m(m \\geq 2)$μ λνμ¬ $m^{12}$μ $n$μ κ³±κ·Ό μ€μμ μ μκ° μ‘΄μ¬νλλ‘ νλ 2 μ΄μμ μμ°μ $n$μ κ°μλ₯Ό $f(m)$μ΄λΌ ν λ, \\[ \\sum_{m=2}^{9} f(m) \\ \\text{μ κ°μ? [4μ ]} \\] \\begin{itemize} \\item[1] 37 \\item[2] 42 \\item[3] 47 \\item[4] 52 \\item[5] 57 \\end{itemize}","answer":3,"score":4,"review":null} |
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{"id":14,"name":"14","problem":"14. λ€νν¨μ $f(x)$μ λνμ¬ ν¨μ $g(x)$λ₯Ό λ€μκ³Ό κ°μ΄ μ μνλ€. \\[ g(x) = \\begin{cases} x & (x < -1 \\text{ λλ } x > 1) \\\\ f(x) & (-1 \\leq x \\leq 1) \\end{cases} \\] ν¨μ $h(x) = \\lim_{t \\to 0+} g(x+t) \\times \\lim_{t \\to 2+} g(x+t)$ μ λνμ¬ μλ γ±, γ΄, γ· μ€μμ μ³μ κ²λ§μ μλ λλ‘ κ³ λ₯Έ κ²μ? [4μ ] \\begin{itemize} \\item[γ±.] $h(1) = 3$ \\item[γ΄.] ν¨μ $h(x)$λ μ€μ μ 체μ μ§ν©μμ μ°μμ΄λ€. \\item[γ·.] ν¨μ $g(x)$κ° λ«νκ΅¬κ° $[-1, 1]$μμ κ°μνκ³ $g(-1) = -2$μ΄λ©΄ ν¨μ $h(x)$λ μ€μ μ 체μ μ§ν©μμ μ΅μκ°μ κ°λλ€. \\end{itemize} \\begin{itemize} \\item[1] γ± \\item[2] γ΄ \\item[3] γ±, γ΄ \\item[4] γ±, γ· \\item[5] γ΄, γ· \\end{itemize}","answer":1,"score":4,"review":"<보기> changed to 'μλ γ±, γ΄, γ· μ€'."} |
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{"id":15,"name":"15","problem":"15. λͺ¨λ νμ΄ μμ°μμ΄κ³ λ€μ 쑰건μ λ§μ‘±μν€λ λͺ¨λ μμ΄ $\\{a_n\\}$μ λνμ¬ $a_9$μ μ΅λκ°κ³Ό μ΅μκ°μ κ°κ° $M, m$μ΄λΌ ν λ, $M + m$μ κ°μ? [4μ ] \\\\ (κ°) $a_7 = 40$ \\\\ (λ) λͺ¨λ μμ°μ $n$μ λνμ¬ \\[ a_{n+2} = \\begin{cases} a_{n+1} + a_n & (a_{n+1}\\text{μ΄ 3μ λ°°μκ° μλ κ²½μ°}) \\\\ \\frac{1}{3} a_{n+1} & (a_{n+1}\\text{μ΄ 3μ λ°°μμΈ κ²½μ°}) \\end{cases} \\] μ΄λ€. \\begin{itemize} \\item[1] 216 \\item[2] 218 \\item[3] 220 \\item[4] 222 \\item[5] 224 \\end{itemize}","answer":5,"score":4,"review":null} |
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{"id":16,"name":"16","problem":"16. λ°©μ μ \\[ \\log_2 (3x + 2) = 2 + \\log_2 (x - 2) \\] λ₯Ό λ§μ‘±μν€λ μ€μ $x$μ κ°μ ꡬνμμ€. [3μ ]","answer":10,"score":3,"review":null} |
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{"id":17,"name":"17","problem":"17. ν¨μ $f(x)$μ λνμ¬ $f'(x) = 4x^3 - 2x$μ΄κ³ $f(0) = 3$μΌ λ, $f(2)$μ κ°μ ꡬνμμ€. [3μ ]","answer":15,"score":3,"review":null} |
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{"id":18,"name":"18","problem":"18. λ μμ΄ $\\{a_n\\}$, $\\{b_n\\}$μ λνμ¬ \\[ \\sum_{k=1}^{5} (3a_k + 5) = 55, \\quad \\sum_{k=1}^{5} (a_k + b_k) = 32 \\] μΌ λ, $\\sum_{k=1}^{5} b_k$μ κ°μ ꡬνμμ€. [3μ ]","answer":22,"score":3,"review":null} |
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{"id":19,"name":"19","problem":"19. λ°©μ μ $2x^3 - 6x^2 + k = 0$ μ μλ‘ λ€λ₯Έ μμ μ€κ·Όμ κ°μκ° 2κ° λλλ‘ νλ μ μ $k$ μ κ°μλ₯Ό ꡬνμμ€. [3μ ]","answer":7,"score":3,"review":null} |
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{"id":20,"name":"20","problem":"20. μμ§μ μλ₯Ό μμ§μ΄λ μ Pμ μκ° $t(t \\geq 0)$μμμ μλ $v(t)$μ κ°μλ $a(t)$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€. \\[ \\text{(κ°)} \\ 0 \\leq t \\leq 2 \\ \\text{μΌ λ}, \\ v(t) = 2t^3 - 8t \\text{μ΄λ€.} \\] \\[ \\text{(λ)} \\ t \\geq 2 \\ \\text{μΌ λ}, \\ a(t) = 6t + 4 \\text{μ΄λ€.} \\] μκ° $t = 0$μμ $t = 3$κΉμ§ μ Pκ° μμ§μΈ 거리λ₯Ό ꡬνμμ€. [4μ ]","answer":17,"score":4,"review":null} |
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{"id":21,"name":"21","problem":"21. μμ°μ $n$μ λνμ¬ ν¨μ $f(x)$λ₯Ό \\[ f(x) = \\begin{cases} | 3^{x + 2} - n | & (x < 0) \\\\ | \\log_2(x + 4) - n | & (x \\geq 0) \\end{cases} \\] μ΄λΌ νμ. μ€μ $t$μ λνμ¬ $x$μ λν λ°©μ μ $f(x) = t$μ μλ‘ λ€λ₯Έ μ€κ·Όμ κ°μλ₯Ό $g(t)$λΌ ν λ, ν¨μ $g(t)$μ μ΅λκ°μ΄ 4κ° λλλ‘ νλ λͺ¨λ μμ°μ $n$μ κ°μ ν©μ ꡬνμμ€. [4μ ]","answer":33,"score":4,"review":null} |
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{"id":22,"name":"22","problem":"22. μ΅κ³ μ°¨νμ κ³μκ° 1μΈ μΌμ°¨ν¨μ $f(x)$μ μ€μ μ 체μ μ§ν©μμ μ°μμΈ ν¨μ $g(x)$κ° λ€μ 쑰건μ λ§μ‘±μν¬ λ, $f(4)$μ κ°μ ꡬνμμ€. [4μ ] \\\\ (κ°) λͺ¨λ μ€μ $x$μ λνμ¬ $f(x) = f(1) + (x-1)f'(g(x))$μ΄λ€. \\\\ (λ) ν¨μ $g(x)$μ μ΅μκ°μ $\\frac{5}{2}$μ΄λ€. \\\\ (λ€) $f(0) = -3,\\ f(g(1)) = 6$","answer":13,"score":4,"review":null} |
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{"id":23,"name":"23_prob","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":3,"score":2,"review":null} |
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{"id":24,"name":"24_prob","problem":"24. μ«μ 1, 2, 3, 4, 5 μ€μμ μ€λ³΅μ νλ½νμ¬ 4κ°λ₯Ό νν΄ μΌλ ¬λ‘ λμ΄νμ¬ λ§λ€ μ μλ λ€ μ리μ μμ°μ μ€ 4000 μ΄μμΈ νμμ κ°μλ? [3μ ] \\begin{itemize} \\item[1] 125 \\item[2] 150 \\item[3] 175 \\item[4] 200 \\item[5] 225 \\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":25,"name":"25_prob","problem":"25. ν°μ λ§μ€ν¬ 5κ°, κ²μμ λ§μ€ν¬ 9κ°κ° λ€μ΄ μλ μμκ° μλ€. μ΄ μμμμ μμλ‘ 3κ°μ λ§μ€ν¬λ₯Ό λμμ κΊΌλΌ λ, κΊΌλΈ 3κ°μ λ§μ€ν¬ μ€μμ μ μ΄λ ν κ°κ° ν°μ λ§μ€ν¬μΌ νλ₯ μ? [3μ ] \\begin{itemize} \\item[1] \\frac{8}{13} \\item[2] \\frac{17}{26} \\item[3] \\frac{9}{13} \\item[4] \\frac{19}{26} \\item[5] \\frac{10}{13} \\end{itemize}","answer":5,"score":3,"review":null} |
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{"id":26,"name":"26_prob","problem":"26. μ£Όλ¨Έλμ 1μ΄ μ ν ν° κ³΅ 1κ°, 2κ° μ ν ν° κ³΅ 1κ°, 1μ΄ μ ν κ²μ 곡 1κ°, 2κ° μ ν κ²μ 곡 3κ°κ° λ€μ΄ μλ€. μ΄ μ£Όλ¨Έλμμ μμλ‘ 3κ°μ 곡μ λμμ κΊΌλ΄λ μνμ νλ€. μ΄ μνμμ κΊΌλΈ 3κ°μ 곡 μ€μμ ν° κ³΅μ΄ 1κ°μ΄κ³ κ²μ κ³΅μ΄ 2κ°μΈ μ¬κ±΄μ $A$, κΊΌλΈ 3κ°μ 곡μ μ ν μλ μλ₯Ό λͺ¨λ κ³±ν κ°μ΄ 8μΈ μ¬κ±΄μ $B$λΌ ν λ, $\\mathrm{P}(A \\cup B)$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] \\frac{11}{20} \\item[2] \\frac{3}{5} \\item[3] \\frac{13}{20} \\item[4] \\frac{7}{10} \\item[5] \\frac{3}{4} \\end{itemize}","answer":3,"score":3,"review":"Removed figure."} |
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{"id":27,"name":"27_prob","problem":"27. μ΄λ νμ¬μμ μμ°νλ μ΄νΈ 1κ°μ μ©λμ μ κ·λΆν¬ $N(m, \\sigma^2)$μ λ°λ₯Έλ€κ³ νλ€. μ΄ νμ¬μμ μμ°νλ μ΄νΈ μ€μμ 16κ°λ₯Ό μμμΆμΆνμ¬ μ»μ νλ³Ένκ· μ μ΄μ©νμ¬ κ΅¬ν $m$μ λν μ λ’°λ 95\\%μ μ 뒰ꡬκ°μ΄ $746.1 \\leq m \\leq 755.9$μ΄λ€. μ΄ νμ¬μμ μμ°νλ μ΄νΈ μ€μμ $n$κ°λ₯Ό μμμΆμΆνμ¬ μ»μ νλ³Ένκ· μ μ΄μ©νμ¬ κ΅¬νλ $m$μ λν μ λ’°λ 99\\%μ μ 뒰ꡬκ°μ΄ $a \\leq m \\leq b$μΌ λ, $b - a$μ κ°μ΄ 6 μ΄νκ° λκΈ° μν μμ°μ $n$μ μ΅μκ°μ? (λ¨, μ©λμ λ¨μλ mLμ΄κ³ , $Z$κ° νμ€μ κ·λΆν¬λ₯Ό λ°λ₯΄λ νλ₯ λ³μμΌ λ, $\\mathrm{P}(|Z| \\leq 1.96) = 0.95$, $\\mathrm{P}(|Z| \\leq 2.58) = 0.99$λ‘ κ³μ°νλ€.) [3μ ] \\begin{itemize} \\item[1] 70 \\item[2] 74 \\item[3] 78 \\item[4] 82 \\item[5] 86 \\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":28,"name":"28_prob","problem":"28. μ°μνλ₯ λ³μ $X$κ° κ°λ κ°μ λ²μλ $0 \\leq X \\leq a$μ΄κ³ , $X$μ νλ₯ λ°λν¨μ f(x)κ° λ€μκ³Ό κ°μ΄ μ μλμ΄ μλ€.\n\n\\[\nf(x) =\n\\begin{cases}\n0, & x < 0, \\\\\n\\frac{c}{b}x, & 0 \\leq x < b, \\\\\nc\\frac{(a-x)}{a-b}, & \\leq x < a, \\\\\n0, & a < x\n\\end{cases}\n\\]\n(λ¨, $a>b$ μ΄λ€.)\n\n $\\mathrm{P}(X \\leq b) - \\mathrm{P}(X \\geq b) = \\frac{1}{4}, \\mathrm{P}(X \\leq \\sqrt{5}) = \\frac{1}{2}$ μΌ λ, $a + b + c$μ κ°μ? (λ¨, $a, b, c$λ μμμ΄λ€.) [4μ ] \\begin{itemize} \\item[1] $\\frac{11}{2}$ \\item[2] $6$ \\item[3] $\\frac{13}{2}$ \\item[4] $7$ \\item[5] $\\frac{15}{2}$ \\end{itemize}","answer":4,"score":4,"review":"Removed figure and the statement referring to the figure. The figure is needed to solve the problem, so we paraphrased the figure into text."} |
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{"id":29,"name":"29_prob","problem":"29. μλ©΄μλ 1λΆν° 6κΉμ§μ μμ°μκ° νλμ© μ ν μκ³ λ·λ©΄μλ λͺ¨λ 0μ΄ νλμ© μ ν μλ 6μ₯μ μΉ΄λκ° μλ€. μ΄ 6μ₯μ μΉ΄λκ° 6 μ΄νμ μμ°μ $k$μ λνμ¬ $k$λ²μ§Έ μ리μ μμ°μ $k$κ° λ³΄μ΄λλ‘ λμ¬ μλ€. μ΄ 6μ₯μ μΉ΄λμ ν κ°μ μ£Όμ¬μλ₯Ό μ¬μ©νμ¬ λ€μ μνμ νλ€. \\[ \\text{μ£Όμ¬μλ₯Ό ν λ² λμ Έ λμ¨ λμ μκ° } k \\text{μ΄λ©΄ } k\\text{λ²μ§Έ μ리μ λμ¬ μλ μΉ΄λλ₯Ό ν λ² λ€μ§μ΄ μ μ리μ λλλ€.} \\] μμ μνμ 3λ² λ°λ³΅ν ν 6μ₯μ μΉ΄λμ 보μ΄λ λͺ¨λ μμ ν©μ΄ μ§μμΌ λ, μ£Όμ¬μμ 1μ λμ΄ ν λ²λ§ λμμ 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":49,"score":4,"review":"Removed figure and the statement referring to the figure."} |
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{"id":30,"name":"30_prob","problem":"30. μ§ν© $X=\\{x \\mid x\\text{λ} \\ 10 \\ \\text{μ΄νμ μμ°μ}\\}$μ λνμ¬ λ€μ 쑰건μ λ§μ‘±μν€λ ν¨μ $f: X \\to X$μ κ°μλ₯Ό ꡬνμμ€. [4μ ] \\begin{itemize} \\item[(κ°)] 9 μ΄νμ λͺ¨λ μμ°μ $x$μ λνμ¬ $f(x) \\leq f(x+1)$μ΄λ€. \\item[(λ)] $1 \\leq x \\leq 5$μΌ λ $f(x) \\leq x$μ΄κ³ , 6 $\\leq x \\leq 10$μΌ λ $f(x) \\geq x$μ΄λ€. \\item[(λ€)] $f(6) = f(5) + 6$ \\end{itemize}","answer":100,"score":4,"review":null} |
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{"id":31,"name":"23_calc","problem":"23. \\[ \\lim_{x \\to 0} \\frac{\\ln(x+1)}{\\sqrt{x+4} - 2} \\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":4,"score":2,"review":null} |
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{"id":32,"name":"24_calc","problem":"24. $\\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=1}^{n} \\sqrt{1 + \\frac{3k}{n}}$ μ κ°μ? [3μ ] \\begin{itemize} \\item[1] \\frac{4}{3} \\item[2] \\frac{13}{9} \\item[3] \\frac{14}{9} \\item[4] \\frac{5}{3} \\item[5] \\frac{16}{9} \\end{itemize}","answer":3,"score":3,"review":null} |
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{"id":33,"name":"25_calc","problem":"25. λ±λΉμμ΄ $\\{a_n\\}$μ λνμ¬ \\[ \\lim_{n \\to \\infty} \\frac{a_{n}+1}{3^n + 2^{2n-1}} = 3 \\] μΌ λ, $a_2$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 16 \\item[2] 18 \\item[3] 20 \\item[4] 22 \\item[5] 24 \\end{itemize}","answer":5,"score":3,"review":null} |
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{"id":34,"name":"26_calc","problem":"26. 곑μ $y = \\sqrt{\\sec^2 x + \\tan x} \\left( 0 \\leq x \\leq \\frac{\\pi}{3} \\right)$μ $x$μΆ, $y$μΆ λ° μ§μ $x = \\frac{\\pi}{3}$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 ] \\begin{itemize} \\item[1] \\frac{\\sqrt{3}}{2} + \\frac{\\ln 2}{2} \\item[2] \\frac{\\sqrt{3}}{2} + \\ln 2 \\item[3] \\sqrt{3} + \\frac{\\ln 2}{2} \\item[4] \\sqrt{3} + \\ln 2 \\item[5] \\sqrt{3} + 2 \\ln 2 \\end{itemize}","answer":4,"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¬μ΄ $\\mathrm{O}$, λ°μ§λ¦μ κΈΈμ΄κ° 1μ΄κ³ , μ€μ¬κ°μ ν¬κΈ°κ° $\\frac{\\pi}{2}$μΈ λΆμ±κΌ΄ $\\mathrm{O}\\mathrm{A}_1\\mathrm{B}_1$μ΄ μλ€. νΈ $\\mathrm{A}_1\\mathrm{B}_1$ μμ μ $\\mathrm{P}_1$, μ λΆ $\\mathrm{O}\\mathrm{A}_1$ μμ μ $\\mathrm{C}_1$, μ λΆ $\\mathrm{O}\\mathrm{B}_1$ μμ μ $\\mathrm{D}_1$μ μ¬κ°ν $\\mathrm{O}\\mathrm{C}_1\\mathrm{P}_1\\mathrm{D}_1$μ΄ $\\overline{\\mathrm{O}\\mathrm{C}_1}:\\overline{\\mathrm{O}\\mathrm{D}_1}=3:4$μΈ μ§μ¬κ°νμ΄ λλλ‘ μ‘λλ€.\n\nλΆμ±κΌ΄ $\\mathrm{O}\\mathrm{A}_1\\mathrm{B}_1$μ λ΄λΆμ μ $\\mathrm{Q}_1$μ $\\overline{\\mathrm{P}_1\\mathrm{Q}_1} = \\overline{\\mathrm{A}_1\\mathrm{Q}_1}$, $\\angle \\mathrm{P}_1\\mathrm{Q}_1\\mathrm{A}_1 = \\frac{\\pi}{2}$κ° λλλ‘ μ‘κ³ , μ΄λ±λ³μΌκ°ν $\\mathrm{P}_1\\mathrm{Q}_1\\mathrm{A}_1$μ μμΉ νμ¬ μ»μ κ·Έλ¦Όμ $R_1$μ΄λΌ νμ. κ·Έλ¦Ό $R_1$μμ μ λΆ $\\mathrm{O}\\mathrm{A}_1$ μμ μ $\\mathrm{A}_2$μ μ λΆ $\\mathrm{O}\\mathrm{B}_1$ μμ μ $\\mathrm{B}_2$λ₯Ό $\\overline{\\mathrm{O}\\mathrm{Q}_1} = \\overline{\\mathrm{O}\\mathrm{A}_2} = \\overline{\\mathrm{O}\\mathrm{B}_2}$κ° λλλ‘ μ‘κ³ , μ€μ¬μ΄ $\\mathrm{O}$, λ°μ§λ¦μ κΈΈμ΄κ° $\\overline{\\mathrm{O}\\mathrm{Q}_1}$, μ€μ¬κ°μ ν¬κΈ°κ° $\\frac{\\pi}{2}$μΈ λΆμ±κΌ΄ $\\mathrm{O}\\mathrm{A}_2\\mathrm{B}_2$λ₯Ό κ·Έλ¦°λ€.\n\nκ·Έλ¦Ό $R_1$μ μ»μ κ²κ³Ό κ°μ λ°©λ²μΌλ‘ λ€ μ $\\mathrm{P}_2, \\mathrm{C}_2, \\mathrm{D}_2, \\mathrm{Q}_2$λ₯Ό μ‘κ³ , μ΄λ±λ³μΌκ°ν $\\mathrm{P}_2\\mathrm{Q}_2\\mathrm{A}_2$μ μμΉ νμ¬ μ»μ κ·Έλ¦Όμ $R_2$λΌ νμ. μ΄μ κ°μ κ³Όμ μ κ³μνμ¬ $n$λ²μ§Έ μ»μ κ·Έλ¦Ό $R_n$μ μμΉ λμ΄ μλ λΆλΆμ λμ΄λ₯Ό $S_n$μ΄λΌ ν λ, \\[ \\lim_{n \\to \\infty} S_n \\text{μ κ°μ? [3μ ]} \\]\n\n\\begin{itemize} \\item[1] \\frac{9}{40} \\item[2] \\frac{1}{4} \\item[3] \\frac{11}{40} \\item[4] \\frac{3}{10} \\item[5] \\frac{13}{40} \\end{itemize}","answer":2,"score":3,"review":"Removed figure and the statement referring to the figure."} |
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{"id":36,"name":"28_calc","problem":"28. μ€μ¬μ΄ $(\\mathrm{O})$μ΄κ³ κΈΈμ΄κ° 2μΈ μ λΆ $(\\mathrm{AB})$λ₯Ό μ§λ¦μΌλ‘ νλ λ°μ μμ $(\\angle \\mathrm{AOC} = \\frac{\\pi}{2})$μΈ μ $(\\mathrm{C})$κ° μλ€. νΈ $(\\mathrm{BC})$ μμ μ $(\\mathrm{P})$μ νΈ $(\\mathrm{CA})$ μμ μ $(\\mathrm{Q})$λ₯Ό $(\\overline{\\mathrm{PB}} = \\overline{\\mathrm{QC}})$κ° λλλ‘ μ‘κ³ , μ λΆ $(\\mathrm{AP})$ μμ μ $(\\mathrm{R})$λ₯Ό $(\\angle \\mathrm{CQR} = \\frac{\\pi}{2})$κ° λλλ‘ μ‘λλ€. μ λΆ $(\\mathrm{AP})$μ μ λΆ $(\\mathrm{CO})$μ κ΅μ μ $(\\mathrm{S})$λΌ νμ. $(\\angle \\mathrm{PAB} = \\theta)$μΌ λ, μΌκ°ν $(\\mathrm{POB})$μ λμ΄λ₯Ό $(f(\\theta))$, μ¬κ°ν $(\\mathrm{CQRS})$μ λμ΄λ₯Ό $(g(\\theta))$λΌ νμ.\n\n\\[ \\lim_{\\theta \\to 0+} \\frac{3f(\\theta) - 2g(\\theta)}{\\theta^2} \\] μ κ°μ? (λ¨, $0 < \\theta < \\frac{\\pi}{4}$) [4μ ]\n\n\\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}","answer":2,"score":4,"review":"Removed figure and the statement referring to the figure."} |
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{"id":37,"name":"29_calc","problem":"29. μΈ μμ $(a)$, $(b)$, $(c)$μ λνμ¬ ν¨μ $f(x) = ae^{2x} + be^x + c$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\[ \\begin{aligned} &\\text{(κ°)} \\quad \\lim_{x \\to -\\infty} \\frac{f(x) + 6}{e^x} = 1 \\\\ &\\text{(λ)} \\quad f(\\ln 2) = 0 \\end{aligned} \\]\n\nν¨μ $f(x)$μ μν¨μλ₯Ό $g(x)$λΌ ν λ,\n\n\\[ \\int_0^{14} g(x) \\ dx = p + q \\ln 2 \\ \\text{μ΄λ€}. \\ p+q \\ \\text{μ κ°μ ꡬνμμ€.} \\]\n\n$(\\text{λ¨, } p, q \\text{λ μ 리μμ΄κ³ , } \\ln 2 \\text{λ 무리μμ΄λ€.)}$ [4μ ]","answer":26,"score":4,"review":null} |
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{"id":38,"name":"30_calc","problem":"30. μ΅κ³ μ°¨νμ κ³μκ° μμμΈ μΌμ°¨ν¨μ $f(x)$μ ν¨μ $g(x) = e^{\\sin{\\pi x}} - 1$μ λνμ¬ μ€μ μ 체μ μ§ν©μμ μ μλ ν©μ±ν¨μ $h(x) = g(f(x))$κ° λ€μ 쑰건μ λ§μ‘±μν¨λ€.\n\n\\begin{itemize} \\item[(κ°)] ν¨μ $( h(x) )$λ $( x = 0 )$μμ κ·Ήλκ° $0$μ κ°λλ€. \\item[(λ)] μ΄λ¦°κ΅¬κ° $( 0,3 )$μμ λ°©μ μ $( h(x) = 1 )$μ μλ‘ λ€λ₯Έ μ€κ·Όμ κ°μλ 7μ΄λ€. \\end{itemize}\n\n$f(3) = \\frac{1}{2}, \\ f'(3) = 0$μΌ λ, $f(2) = \\frac{q}{p}$μ΄λ€. $p + q$μ κ°μ ꡬνμμ€. (λ¨, $p$μ $q$λ μλ‘μμΈ μμ°μμ΄λ€.) [4μ ]","answer":31,"score":4,"review":null} |
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{"id":39,"name":"23_geom","problem":"23. μ’ν곡κ°μ μ $\\mathrm{A}(2, 2, -1)$μ $x$μΆμ λνμ¬ λμΉμ΄λν μ μ $\\mathrm{B}$λΌ νμ. μ $\\mathrm{C}(-2, 1, 1)$μ λνμ¬ μ λΆ $\\mathrm{BC}$μ κΈΈμ΄λ? [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":40,"name":"24_geom","problem":"24. μ΄μ μ΄ $\\mathrm{F}\\left( \\frac{1}{3}, 0 \\right)$μ΄κ³ μ€μ μ΄ $x = -\\frac{1}{3}$μΈ ν¬λ¬Όμ μ΄ μ $(a, 2)$λ₯Ό μ§λ λ, $a$μ κ°μ? [3μ ] \\begin{itemize} \\item[1] 1 \\item[2] 2 \\item[3] 3 \\item[4] 4 \\item[5] 5 \\end{itemize}","answer":3,"score":3,"review":null} |
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{"id":41,"name":"25_geom","problem":"25. νμ $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ μμ μ $(2, 1)$μμμ μ μ μ κΈ°μΈκΈ°κ° $-\\frac{1}{2}$μΌ λ, μ΄ νμμ λ μ΄μ μ¬μ΄μ 거리λ? (λ¨, $a, b$λ μμμ΄λ€.) [3μ ] \\begin{itemize} \\item[1] 2\\sqrt{3} \\item[2] 4 \\item[3] 2\\sqrt{5} \\item[4] 2\\sqrt{6} \\item[5] 2\\sqrt{7} \\end{itemize}","answer":4,"score":3,"review":null} |
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{"id":42,"name":"26_geom","problem":"26. μ’ννλ©΄μμ μΈ λ²‘ν° \\[ \\vec{a} = (2, 4), \\quad \\vec{b} = (2, 8), \\quad \\vec{c} = (1, 0) \\] μ λνμ¬ λ λ²‘ν° $\\vec{p}, \\vec{q}$ κ° \\[ (\\vec{p} - \\vec{a}) \\cdot (\\vec{p} - \\vec{b}) = 0, \\quad \\vec{q} = \\frac{1}{2} \\vec{a} + t \\vec{c} \\quad (t \\text{λ μ€μ}) \\] λ₯Ό λ§μ‘±μν¬ λ, $\\left| \\vec{p} - \\vec{q} \\right|$ μ μ΅μκ°μ? [3μ ] \\begin{itemize} \\item[1] \\frac{3}{2} \\item[2] 2 \\item[3] \\frac{5}{2} \\item[4] 3 \\item[5] \\frac{7}{2} \\end{itemize}","answer":2,"score":3,"review":null} |
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{"id":43,"name":"27_geom","problem":"27. μ’ν곡κ°μ μ§μ $( \\mathrm{AB} )$λ₯Ό ν¬ν¨νλ νλ©΄ $( \\alpha )$κ° μλ€. νλ©΄ $( \\alpha )$ μμ μμ§ μμ μ $( \\mathrm{C} )$μ λνμ¬ μ§μ $( \\mathrm{AB} )$μ μ§μ $( \\mathrm{AC} )$κ° μ΄λ£¨λ μκ°μ ν¬κΈ°λ₯Ό $( \\theta_1 )$μ΄λΌ ν λ $\\sin \\theta_1 = \\frac{4}{5}$μ΄κ³ , μ§μ $( \\mathrm{AC} )$μ νλ©΄ $( \\alpha )$κ° μ΄λ£¨λ μκ°μ ν¬κΈ°λ $( \\frac{\\pi}{2} - \\theta_1 )$μ΄λ€. νλ©΄ $( \\mathrm{ABC} )$μ νλ©΄ $( \\alpha )$κ° μ΄λ£¨λ μκ°μ ν¬κΈ°λ₯Ό $( \\theta_2 )$λΌ ν λ, $\\cos \\theta_2$μ κ°μ? [3μ ]\n\n\\begin{itemize} \\item[1] \\frac{\\sqrt{7}}{4} \\item[2] \\frac{\\sqrt{7}}{5} \\item[3] \\frac{\\sqrt{7}}{6} \\item[4] \\frac{\\sqrt{7}}{7} \\item[5] \\frac{\\sqrt{7}}{8} \\end{itemize}","answer":1,"score":3,"review":"Removed figure."} |
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{"id":44,"name":"28_geom","problem":"28. λ μ΄μ μ΄ $( \\mathrm{F}(c, 0) )$, $( \\mathrm{F'}(-c, 0) \\ (c > 0) )$μΈ μ곑μ $( C )$μ $( y )$μΆ μμ μ $( \\mathrm{A} )$κ° μλ€. μ곑μ $( C )$κ° μ λΆ $( \\mathrm{AF} )$μ λ§λλ μ μ $( \\mathrm{P} )$, μ λΆ $( \\mathrm{AF'} )$μ λ§λλ μ μ $( \\mathrm{P'} )$μ΄λΌ νμ. μ§μ $( \\mathrm{AF} )$λ μ곑μ $( C )$μ ν μ κ·Όμ κ³Ό νννκ³ \n\n\\[ \\overline{\\mathrm{AP}}:\\overline{\\mathrm{PP'}} = 5:6, \\quad \\overline{\\mathrm{PF}} = 1 \\]\n\nμΌ λ, μ곑μ $( C )$μ μ£ΌμΆμ κΈΈμ΄λ? [4μ ]\n\n\\begin{itemize} \\item[1] \\frac{13}{6} \\item[2] \\frac{9}{4} \\item[3] \\frac{7}{3} \\item[4] \\frac{29}{12} \\item[5] \\frac{5}{2} \\end{itemize}","answer":2,"score":4,"review":"Removed figure."} |
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{"id":45,"name":"29_geom","problem":"29. νλ©΄ $\\alpha$ μμ $\\overline{\\mathrm{AB}} = \\overline{\\mathrm{CD}} = \\overline{\\mathrm{AD}} = 2$, $\\angle \\mathrm{ABC} = \\angle \\mathrm{BCD} = \\frac{\\pi}{3}$ μΈ μ¬λ€λ¦¬κΌ΄ $\\mathrm{ABCD}$κ° μλ€. λ€μ 쑰건μ λ§μ‘±μν€λ νλ©΄ $\\alpha$ μμ λ μ $\\mathrm{P}$, $\\mathrm{Q}$μ λνμ¬ $\\overrightarrow{\\mathrm{CP}} \\cdot \\overrightarrow{\\mathrm{DQ}}$μ κ°μ ꡬνμμ€. [4μ ]\n\n\\begin{itemize} \\item[(κ°)] $\\overrightarrow{\\mathrm{AC}} = 2 \\left( \\overrightarrow{\\mathrm{AD}} + \\overrightarrow{\\mathrm{BP}} \\right)$ \\item[(λ)] $\\overrightarrow{\\mathrm{AC}} \\cdot \\overrightarrow{\\mathrm{PQ}} = 6$ \\item[(λ€)] $2 \\times \\angle \\mathrm{BQA} = \\angle \\mathrm{PBQ} < \\frac{\\pi}{2}$ \\end{itemize}","answer":12,"score":4,"review":"Removed figure."} |
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{"id":46,"name":"30_geom","problem":"30. μ’ν곡κ°μ μ μ¬λ©΄μ²΄ $\\mathrm{ABCD}$κ° μλ€. μ μΌκ°ν $\\mathrm{BCD}$μ μΈμ¬μ μ€μ¬μΌλ‘ νκ³ μ $\\mathrm{B}$λ₯Ό μ§λλ ꡬλ₯Ό $S$λΌ νμ.\n\nꡬ $S$μ μ λΆ $\\mathrm{AB}$κ° λ§λλ μ μ€ $\\mathrm{B}$κ° μλ μ μ $\\mathrm{P}$, ꡬ $S$μ μ λΆ $\\mathrm{AC}$κ° λ§λλ μ μ€ $\\mathrm{C}$κ° μλ μ μ $\\mathrm{Q}$, ꡬ $S$μ μ λΆ $\\mathrm{AD}$κ° λ§λλ μ μ€ $\\mathrm{D}$κ° μλ μ μ $\\mathrm{R}$λΌ νκ³ , μ $\\mathrm{P}$μμ ꡬ $S$μ μ νλ νλ©΄μ $\\alpha$λΌ νμ.\n\nꡬ $S$μ λ°μ§λ¦μ κΈΈμ΄κ° $6$μΌ λ, μΌκ°ν $\\mathrm{PQR}$μ νλ©΄ $\\alpha$ μλ‘μ μ μ¬μμ λμ΄λ $k$μ΄λ€. $k^2$μ κ°μ ꡬνμμ€. [4μ ]","answer":24,"score":4,"review":"Removed figure."} |