id int64 | question string | solution list | final_answer list | context string | image_1 image | image_2 image | image_3 image | image_4 image | image_5 image | modality string | difficulty string | is_multiple_answer bool | unit string | answer_type string | error string | question_type string | subfield string | subject string | language string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
530 | 求最小的正整数 $n$, 使得当正整数 $k \geq n$ 时, 在前 $k$ 个正整数构成的集合 $M=\{1,2, \cdots, k\}$ 中, 对任何 $x \in M$, 总存在另一数 $y \in M(y \neq x)$, 满足 $x+y$ 为平方数. | [
"易知当 $n \\leq 6$ 时, 在 $M=\\{1,2,3,4,5,6\\}$ 中, 数 2 与其它任何数之和皆不是平方数.\n\n以下证明, $n$ 的最小值为 7 .\n\n如果正整数 $x, y(x \\neq y)$ 满足: $x+y=$ 平方数, 就称 $\\{x, y\\}$ 是一个 “平方对”, 显然在 $M=\\{1,2, \\cdots, 7\\}$ 中, $\\{1,3\\},\\{2,7\\},\\{3,6\\},\\{4,5\\}$ 为平方对; 在 $M=\\{1,2, \\cdots, 7,8\\}$ 中，增加了平方对 $\\{1,8\\}$; 在 $M=\\{1,2, \\cdots, 7... | [
"$7$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Set Theory | Math | Chinese |
531 | 若椭圆 $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ 上不同的三点 $A\left(x_{1}, y_{1}\right), B\left(4, \frac{9}{5}\right), C\left(x_{2}, y_{2}\right)$, 到椭圆右焦点的距离顺次成等差数列, 线段 $A C$ 的中垂线 $l$ 交 $X$ 轴于点 $T$, 求直线 $B T$ 的方程. | [
"用 $a, b, c$ 分别表示椭圆的半长轴, 半短轴及半焦距之长度, 则 $a=5, b=3, c=4$, 右焦点为 $F(4,0)$,且准线方程为 $x=\\frac{25}{4}$, 由\n\n$$\n\\frac{A F}{\\frac{25}{4}-x_{1}}=\\frac{c}{a}, \\quad \\frac{C F}{\\frac{25}{4}-x_{2}}=\\frac{4}{5}\n$$\n\n得\n\n$$\nA F=5-\\frac{4}{5} x_{1}, C F=5-\\frac{4}{5} x_{2}\n$$\n\n据等差, $A F+C F=2 B F$, 而 $B F=\\frac{9... | [
"$25 x-20 y=64$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Equation | null | Open-ended | Conic Sections | Math | Chinese |
532 | 将前 12 个正整数构成的集合 $M=\{1,2, \cdots, 12\}$ 中的元素分成 $M_{1}, M_{2}, M_{3}, M_{4}$ 四个三元子集, 使得每个三元子集中的三数都满足: 其中一数等于另外两数之和, 试求不同的分法的种数. | [
"设四个子集为 $M_{i}=\\left(a_{i}, b_{i}, c_{i}\\right), i=1,2,3,4$ ，其中 $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}, i=1,2,3,4$, 设 $a_{1}<a_{2}<$ $a_{3}<a_{4}$, 则 $a_{4}=12 . a_{1}+a_{2}+a_{3}+a_{4}=\\frac{1}{2}(1+2+\\cdots+12)=\\frac{78}{2}=39$, 所以 $a_{1}+a_{2}+a_{3}=27$, 故 $3 a_{3}>27$,由此, $10 \\leq a_{3} \\leq 11$, 若 $a_{3}=10$, ... | [
"$8$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Set Theory | Math | Chinese |
536 | 棋盘上标有第 $0,1,2, \cdots, 100$ 站, 棋子开始时位于第 0 站, 棋手抛郑均匀硬币走跳棋游戏, 若掷出正面,棋子向前跳出一站; 若郑出反面, 棋子向前跳出两站, 直到跳到第 99 站（胜利大本营）或第 100 站（失败集中营) 时, 游戏结束, 设棋子跳到第 $n$ 站的概率为 $P_{n}$.
求 $P_{3}$ 的值; | [
"棋子跳到第 3 站有以下三种途径: 连续三次掷出正面, 其概率为 $\\frac{1}{8}$; 第一次掷出反面, 第二次掷出正面, 其概率为 $\\frac{1}{4}$; 第一次掷出正面, 第二次掷出反面, 其概率为 $\\frac{1}{4}$, 因此 $P_{3}=\\frac{5}{8}$."
] | [
"$\\frac{5}{8}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
538 | 棋盘上标有第 $0,1,2, \cdots, 100$ 站, 棋子开始时位于第 0 站, 棋手抛郑均匀硬币走跳棋游戏, 若掷出正面,棋子向前跳出一站; 若郑出反面, 棋子向前跳出两站, 直到跳到第 99 站（胜利大本营）或第 100 站（失败集中营) 时, 游戏结束, 设棋子跳到第 $n$ 站的概率为 $P_{n}$.
求 $P_{99}, P_{100}$ 的值. | [
"易知棋子先跳到第 $n-2$ 站, 再掷出反面, 其概率为 $\\frac{1}{2} P_{n-2}$; 棋子先跳到第 $n-1$ 站, 再掷出正面,其概率为 $\\frac{1}{2} P_{n-1}$, 因此有 $P_{n}=\\frac{1}{2}\\left(P_{n-1}+P_{n-2}\\right)$, 即\n\n$$\nP_{n}-P_{n-1}=-\\frac{1}{2}\\left(P_{n-1}-P_{n-2}\\right)\n$$\n\n或即\n\n$$\nP_{n+1}-P_{n}=-\\frac{1}{2}\\left(P_{n}-P_{n-1}\\right) \\quad(2 \\leq... | [
"$\\frac{2}{3}\\left(1-\\frac{1}{2^{100}}\\right),\\frac{1}{3}\\left(1+\\frac{1}{2^{99}}\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
542 | 设曲线 $C:\left|x^{2}-16 y\right|=256-16|y|$ 所围成的封闭区域为 $D$.
求区域 $D$ 的面积; | [
"由题设，有 $256-16|y| \\geq 0$, 因此 $-16 \\leq y \\leq 16$.\n\n若 $\\left|x^{2}-16 y\\right|=x^{2}-16 y$, 则当 $0 \\leq y \\leq 16$ 时, $\\left|x^{2}-16 y\\right|=x^{2}-16 y=256-16 y, x^{2}=256$, 此时 $x= \\pm 16(0 \\leq y \\leq 16)$, 图像是两条直线段;\n\n当 $-16 \\leq y \\leq 0$ 时, $\\left|x^{2}-16 y\\right|=x^{2}-16 y=256+16 y, y=\\... | [
"$512$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Conic Sections | Math | Chinese |
543 | 设曲线 $C:\left|x^{2}-16 y\right|=256-16|y|$ 所围成的封闭区域为 $D$.
设过点 $M(0,-16)$ 的直线与曲线 $C$ 交于两点 $P, Q$, 求 $|P Q|$ 的最大值. | [
"设过点 $M(0,-16)$ 的直线为 $l$, 为了求 $|P Q|$ 的最大值, 由区域 $D$ 的对称性, 只需考虑直线 $l$ 与 $D$在 $y$ 轴右侧图像相交部分即可.\n\n设过点 $M(0,-16)$ 的直线 $l$ 方程为 $y=k x-16$, 易知此时 $l$ 与 $D$ 相交时有 $1 \\leq k<\\infty$.\n\n(1) 当 $2 \\leq k<\\infty$ 时, $l$ 与 $D$ 分别相交于二次函数 $y=\\frac{x^{2}}{32}-8$ 以及 $y=\\frac{x^{2}}{32}+8$, 两个交点分别为 $P\\left(16\\left(k-\\sqrt{k... | [
"$16 \\sqrt{20-10 \\sqrt{3}}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Conic Sections | Math | Chinese |
544 | 已知函数 $f(x)=\frac{\sqrt{3}}{2} \sin 2 x-\frac{1}{2}\left(\cos ^{2} x-\sin ^{2} x\right)-1, x \in \mathbf{R}$, 将函数 $f(x)$ 向左平移 $\frac{\pi}{6}$ 个单位后得函数 $g(x)$,设 $\triangle A B C$ 三个角 $A, B, C$ 的对边分别为 $a, b, c$.
若 $c=\sqrt{7}, f(C)=0, \sin B=3 \sin A$, 求 $a, b$ 的值; | [
"$f(x)=\\frac{\\sqrt{3}}{2} \\sin 2 x-\\frac{1}{2}\\left(\\cos ^{2} x-\\sin ^{2} x\\right)-1=\\frac{\\sqrt{3}}{2} \\sin 2 x-\\frac{1}{2} \\cos 2 x-1=\\sin \\left(2 x-\\frac{\\pi}{6}\\right)-1$.\n\n$f(C)=\\sin \\left(2 C-\\frac{\\pi}{6}\\right)-1=0$, 所以 $\\sin \\left(2 C-\\frac{\\pi}{6}\\right)=1$.\n\n因为 $2 C-\\frac... | [
"$1,3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
545 | 已知函数 $f(x)=\frac{\sqrt{3}}{2} \sin 2 x-\frac{1}{2}\left(\cos ^{2} x-\sin ^{2} x\right)-1, x \in \mathbf{R}$, 将函数 $f(x)$ 向左平移 $\frac{\pi}{6}$ 个单位后得函数 $g(x)$,设 $\triangle A B C$ 三个角 $A, B, C$ 的对边分别为 $a, b, c$.
若 $g(B)=0, \vec{m}=(\cos A, \cos B), \vec{n}=(1, \sin A-\cos A \tan B)$, 求 $\vec{m} \cdot \vec{n}$ 的取值范围. | [
"$g(x)=\\sin \\left(2 x+\\frac{\\pi}{6}\\right)-1$, 所以 $g(B)=\\sin \\left(2 B+\\frac{\\pi}{6}\\right)-1=0$, 所以 $\\sin \\left(2 B+\\frac{\\pi}{6}\\right)=1$.\n\n因为 $2 B+\\frac{\\pi}{6} \\in\\left(\\frac{\\pi}{6}, \\frac{13 \\pi}{6}\\right)$, 所以 $2 B+\\frac{\\pi}{6}=\\frac{\\pi}{2}$, 即 $B=\\frac{\\pi}{6}, \\vec{m}=\\... | [
"$(0,1]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Trigonometric Functions | Math | Chinese |
546 | 设等比数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 且 $a_{n+1}=2 S_{n}+1\left(n \in \mathbf{N}^{*}\right)$.
求数列 $\left\{a_{n}\right\}$ 的通项公式; | [
"由 $\\left\\{\\begin{array}{l}a_{n+1}=2 S_{n}+1 \\\\ a_{n}=2 S_{n-1}+1(n \\geq 2)\\end{array}\\right.$ 相减得\n\n$$\na_{n+1}-a_{n}=2\\left(S_{n}-S_{n-1}\\right)=2 a_{n}\n$$\n\n所以 $a_{n+1}=3 a_{n}(n \\geq 2)$.\n\n因为 $\\left\\{a_{n}\\right\\}$ 等比, 且 $a_{2}=2 a_{1}+1$, 所以 $3 a_{1}=2 a_{1}+1$, 所以 $a_{1}=1$, 所以 $a_{n}=3^{n... | [
"$a_{n}=3^{n-1}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Sequence | Math | Chinese |
550 | 已知由甲、乙两位男生和丙、丁两位女生组成的四人冲关小组，参加由某电视台举办的知识类答题闯关活动, 活动共有四关, 设男生闯过一至四关的概率依次是 $\frac{5}{6}, \frac{4}{5}, \frac{3}{4}, \frac{2}{3}$; 女生闯过一至四关的概率依次是 $\frac{4}{5}, \frac{3}{4}, \frac{2}{3}, \frac{1}{2}$.
求男生闯过四关的概率; | [
"记男生四关都闯过为事件 $A$, 则 $P(A)=\\frac{5}{6} \\times \\frac{4}{5} \\times \\frac{3}{4} \\times \\frac{2}{3}=\\frac{1}{3}$;"
] | [
"$\\frac{1}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
551 | 已知由甲、乙两位男生和丙、丁两位女生组成的四人冲关小组，参加由某电视台举办的知识类答题闯关活动, 活动共有四关, 设男生闯过一至四关的概率依次是 $\frac{5}{6}, \frac{4}{5}, \frac{3}{4}, \frac{2}{3}$; 女生闯过一至四关的概率依次是 $\frac{4}{5}, \frac{3}{4}, \frac{2}{3}, \frac{1}{2}$.
设 $\xi$ 表示四人冲关小组闯过四关的人数，求随机变量 $\xi$ 的期望. | [
"记女生四关都闯过为事件 $B$, 则 $P(B)=\\frac{4}{5} \\times \\frac{3}{4} \\times \\frac{2}{3} \\times \\frac{1}{2}=\\frac{1}{5}$, 因为\n\n$$\n\\begin{aligned}\n& P(\\varepsilon=0)=\\left(\\frac{2}{3}\\right)^{2}\\left(\\frac{4}{5}\\right)^{2}=\\frac{64}{225} ; \\\\\n& P(\\varepsilon=1)=C_{2}^{1} \\cdot \\frac{1}{3} \\cdot \\frac{... | [
"$\\frac{16}{15}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
552 | 已知椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ 过点 $M(0,2)$, 且右焦点为 $F(2,0)$.
写出椭圆 $C$ 的方程; | [
"由题意: $b=2, c=2, \\therefore a^{2}=8$, 椭圆 $C$ 的方程 $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$."
] | [
"$\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Equation | null | Open-ended | Conic Sections | Math | Chinese |
553 | 已知椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ 过点 $M(0,2)$, 且右焦点为 $F(2,0)$.
过点 $F$ 的直线 $l$ 与椭圆 $C$ 交于 $A, B$ 两点, 交 $y$ 轴于点 $P$; 若 $\overrightarrow{P A}=m \overrightarrow{A F}, \overrightarrow{P B}=n \overrightarrow{B F}$. 其中 $m+n$ 为定值，求该定值; | [
"由题意: $b=2, c=2, \\therefore a^{2}=8$, 椭圆 $C$ 的方程 $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$;\n\n设 $A, B, P$ 的坐标分别为 $\\left(x_{1}, y_{1}\\right),\\left(x_{2}, y_{2}\\right),(0, t)$, 由 $\\overrightarrow{P A}=m \\overrightarrow{A F}$ 知\n\n$$\nx_{1}=\\frac{2 m}{1+m}, \\quad y_{1}=\\frac{t}{1+m}\n$$\n\n又点 $A$ 在椭圆 $C$ 上, 则\n... | [
"$-4$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Conic Sections | Math | Chinese |
554 | 已知椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ 过点 $M(0,2)$, 且右焦点为 $F(2,0)$.
过点 $F$ 的直线 $l$ 与椭圆 $C$ 交于 $A, B$ 两点, 交 $y$ 轴于点 $P$; 若 $\overrightarrow{P A}=m \overrightarrow{A F}, \overrightarrow{P B}=n \overrightarrow{B F}$,点 $P$ 不在椭圆 $C$ 的内部, 点 $Q$ 是点 $P$ 关于原点 $O$ 的对称点, 试求三角形 $Q A B$ 面积的最小值. | [
"由题意: $b=2, c=2, \\therefore a^{2}=8$, 椭圆 $C$ 的方程 $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$;\n\n设 $A, B, P$ 的坐标分别为 $\\left(x_{1}, y_{1}\\right),\\left(x_{2}, y_{2}\\right),(0, t)$, 由 $\\overrightarrow{P A}=m \\overrightarrow{A F}$ 知\n\n$$\nx_{1}=\\frac{2 m}{1+m}, \\quad y_{1}=\\frac{t}{1+m}\n$$\n\n又点 $A$ 在椭圆 $C$ 上, 则\n... | [
"$\\frac{16}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Conic Sections | Math | Chinese |
557 | 设 $x_{1}, x_{2}, x_{3}$ 是方程 $x^{3}-17 x-18=0$ 的三个根, $-4<x_{1}<-3$, 且 $4<x_{3}<5$.
求 $x_{2}$ 的整数部分. | [
"由于 $x_{1}, x_{2}, x_{3}$ 是方程的根, 我们有\n\n$$\nx^{3}-17 x-18=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right)\\left(x-x_{3}\\right)\n$$\n\n比较两端的系数可得\n\n$$\nx_{1}+x_{2}+x_{3}=0, x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=-17, x_{1} x_{2} x_{3}=18\n$$\n\n由 $x_{1} \\in(-4,-3)$ 和 $x_{3} \\in(4,5)$ 可知 $x_{2}=-x_{1}-x_{3} \\in(-2,0)$.... | [
"$-2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Elementary Functions | Math | Chinese |
558 | 设 $x_{1}, x_{2}, x_{3}$ 是方程 $x^{3}-17 x-18=0$ 的三个根, $-4<x_{1}<-3$, 且 $4<x_{3}<5$.
求 $\arctan x_{1}+\arctan x_{2}+\arctan x_{3}$ 的值. | [
"由于 $x_{1}, x_{2}, x_{3}$ 是方程的根, 我们有\n\n$$\nx^{3}-17 x-18=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right)\\left(x-x_{3}\\right)\n$$\n\n比较两端的系数可得\n\n$$\nx_{1}+x_{2}+x_{3}=0, x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=-17, x_{1} x_{2} x_{3}=18\n$$\n\n设 $\\arctan x_{i}=\\theta_{i}, i=1,2,3$, 由 (1) 知 $\\theta_{1}, \\theta_{2} \\... | [
"$-\\frac{\\pi}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Elementary Functions | Math | Chinese |
569 | 已知 $n$ 元正整数集 $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$. 对任意的 $i \in\{1,2, \cdots, n\}$, 由集合 $A$ 去掉元素 $a_{i}$ 后得到的集合 $A_{i}$ 可分成两个不交的子集之并，且两子集元素之和相等，称这样的数集 $A$ 为“好数集”.
求好数集元素个数 $n$ 的最小值 | [
"设 $M=a_{1}+a_{2}+\\cdots+a_{n}$.\n\n由题设, 知集合 $A_{i}(i=1,2, \\cdots, n)$ 中的元素之和为 $M-a_{i}$, 且 $M-a_{i}$ 均为偶数.\n\n从而, $a_{i}$ 与 $M$ 的奇偶性相同.\n\n(i）若 $M$ 为奇数，则 $a_{i}(i=1,2, \\cdots, n)$ 也为奇数， $n$ 为奇数;\n\n(ii) 若 $M$ 为偶数, 则 $a_{i}(i=1,2, \\cdots, n)$ 为偶数.\n\n记\n\n$$\na_{i}=2 b_{i}(i=1,2, \\cdots, n), \\quad M^{\\prime}... | [
"$7$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
570 | 已知动直线 $l$ 与圆 $O: x^{2}+y^{2}=1$ 相切, 与椭圆 $\frac{x^{2}}{9}+y^{2}=1$ 相交于不同的两点 $A, B$, 求原点到 $A B$ 的中垂线的最大距离. | [
"依题意可设 $l: y=k x+m(k \\neq 0)$.\n\n因为直线 $l$ 与圆 $O$ 相切, 所以, $O$ 到直线 $l$ 的距离为 1 , 即 $\\frac{|m|}{\\sqrt{1+k^{2}}}=1$.\n\n这样的直线必与椭圆交于不同的两点 $A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right)$, 联立 $\\left\\{\\begin{array}{l}y=k x+m \\\\ x^{2}+9 y^{2}-9=0\\end{array}\\right.$ 得\n\n$$\n\\left(1+9 k^{2}\\right) x^... | [
"$\\frac{4}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
571 | 设 $a \in \mathbf{R}$, 且对任意实数 $b$ 均有 $\max _{x \in[0,1]}\left|x^{2}+a x+b\right| \geq 1$, 求 $a$ 的取值范围. | [
"设 $f(x)=x^{2}+a x+b$, 对于 $|b| \\geq 1 \\Rightarrow|f(0)| \\geq 1$, 所以只要考虑 $|b|<1$.\n\n(1) 当 $-\\frac{a}{2} \\leq 0$ 时, 即 $a \\geq 0$, 此时函数 $f(x)$ 的最值在抛物线的左右端点取得, 对任意 $|b|<1$ 有 $f(1)=1+a+b>f(0)=b$, 所以 $f(1)=1+a+b \\geq 1$, 解得 $a \\geq 1$;\n\n( 2 ) 当 $0<-\\frac{a}{2} \\leq \\frac{1}{2}$ 时, 即 $-1 \\leq a<0$, 此时函数 $f(... | [
"$(-\\infty, -3] \\cup [1, +\\infty)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Inequality | Math | Chinese |
575 | 设 $\frac{\sqrt{5}+1}{\sqrt{5}-1}$ 的整数部分为 $a$, 小数部分为 $b$.
求 $a, b$. | [
"$\\frac{\\sqrt{5}+1}{\\sqrt{5}-1}=\\frac{6+2 \\sqrt{5}}{4}, \\because 4<2 \\sqrt{5}<5$,\n\n$\\therefore \\frac{10}{4}<\\frac{6+2 \\sqrt{5}}{4}<\\frac{11}{4}$,\n\n$\\therefore \\frac{\\sqrt{5}+1}{\\sqrt{5}-1}$ 的整数部分 $a=2$, 小数部分 $b=\\frac{\\sqrt{5}+1}{\\sqrt{5}-1}$ $-2=\\frac{\\sqrt{5}-1}{2}$."
] | [
"$2, \\frac{\\sqrt{5}-1}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
576 | 设 $\frac{\sqrt{5}+1}{\sqrt{5}-1}$ 的整数部分为 $a$, 小数部分为 $b$.
求 $a^{2}+b^{2}+\frac{a b}{2}$ | [
"$\\frac{\\sqrt{5}+1}{\\sqrt{5}-1}=\\frac{6+2 \\sqrt{5}}{4}, \\because 4<2 \\sqrt{5}<5$,\n\n$\\therefore \\frac{10}{4}<\\frac{6+2 \\sqrt{5}}{4}<\\frac{11}{4}$,\n\n$\\therefore \\frac{\\sqrt{5}+1}{\\sqrt{5}-1}$ 的整数部分 $a=2$, 小数部分 $b=\\frac{\\sqrt{5}+1}{\\sqrt{5}-1}$ $-2=\\frac{\\sqrt{5}-1}{2}$.\n\n$a^{2}+b^{2}+\\frac... | [
"$5$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
577 | 设 $\frac{\sqrt{5}+1}{\sqrt{5}-1}$ 的整数部分为 $a$, 小数部分为 $b$.
求 $\lim _{n \rightarrow \infty}(b$ $\left.+b^{2}+\cdots+b^{n}\right)$. | [
"$\\frac{\\sqrt{5}+1}{\\sqrt{5}-1}=\\frac{6+2 \\sqrt{5}}{4}, \\because 4<2 \\sqrt{5}<5$,\n\n$\\therefore \\frac{10}{4}<\\frac{6+2 \\sqrt{5}}{4}<\\frac{11}{4}$,\n\n$\\therefore \\frac{\\sqrt{5}+1}{\\sqrt{5}-1}$ 的整数部分 $a=2$, 小数部分 $b=\\frac{\\sqrt{5}+1}{\\sqrt{5}-1}$ $-2=\\frac{\\sqrt{5}-1}{2}$.\n\n$\\because b, b^{2}... | [
"$\\frac{\\sqrt{5}+1}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
582 | 随机挑选一个三位数 $I$.
求 $I$ 含有因子 5 的概率; | [
"三位数的个数为 $9 \\times 10 \\times 10=900$.\n能被 5 整除的三位数可以分为两类:\n\n①个位数字是 0 的三位数共有 $9 \\times 10=90$ 个;\n\n②个位数字是 5 的三位数共有 $9 \\times 10=90$ 个.\n\n所以 $I$ 含有因子 5 的三位数共有 180 ,含有因子 5 的概率 $\\frac{180}{900}=0.2$."
] | [
"$0.2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
583 | 随机挑选一个三位数 $I$.
求 $I$ 中恰有两个数码相等的概率. | [
"三位数的个数为 $9 \\times 10 \\times 10=900$.\n恰有两个数码相同的三位数可以分为:\n\n①三位数中数码含有 0 , 恰好 0 是重复数码, 这样的三位数共有 9 个;\n\n②三位数中数码含有 0 ,但 0 不是重复的数码，这样的三位数共有 $2 \\times 9=18$ 个;\n\n③三位数中数码不含 0 , 这样的三位数共有 $\\mathrm{C}_{9}^{2}$ $\\times 3 \\times 2=216$ 个;恰有两个数码相等的三位数共有 $9+18+216$ $=243$ 个.\n\n所以 $I$ 中恰有两个数码相等的概率 $\\frac{243}{900}=\\fr... | [
"$\\frac{27}{100}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
585 | 已知数列 $\left\{a_{n}\right\}$, 且 $S_{n}=n a+n$ $(n-1)$.
求 $\left(a_{n}, \frac{S_{n}}{n}\right)$ 所在的直线方程. | [
"将 $n=1$ 代人 $S_{n}=n a+n(n-1)$ 得 $a_{1}=S_{1}=a, a_{n}=S_{n}-S_{n-1}=n a+n(n-1)-(n$ $-1) a-(n-1)(n-2)=2 n+a-2(n \\geqslant 2)$, 经验证当 $n=1$ 时也符合上式,故数列 $\\left\\{a_{n}\\right\\}$ 的通项公式为 $a_{n}=2 n+a-2$.\n\n$\\because S_{n}=(n+a-1) n, \\therefore \\frac{S_{n}}{n}=n+a-1$, $\\therefore \\frac{2 S_{n}}{n}=2 n+2 a-2$. 又 $... | [
"$2 y-x=a$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Equation | null | Open-ended | Sequence | Math | Chinese |
586 | 一元三次函数 $f(x)$ 的三次项数为 $\frac{a}{3}, f^{\prime}(x)+9 x>0$ 的解集为 $(1,2)$.
若 $f^{\prime}(x)+7 a=0$ 有两个相等实根, 求 $f^{\prime}(x)$ 的解析式. | [
"设 $f(x)=\\frac{a}{3} x^{3}+b x^{2}+c x+d$, 则 $f^{\\prime}(x)$ $=a x^{2}+2 b x+c, f^{\\prime}(x)+9 x=a x^{2}+2 b x+c+9 x$,\n\n$\\because f^{\\prime}(x)+9 x>0$ 的解集为 $(1,2)$, 故有 $a<0$, 且 $\\left\\{\\begin{array}{l}a+2 b+c+9=0, \\\\ 4 a+4 b+c+18=0 .\\end{array}\\right.$ 则 $2 b=-3 a-9, c=2 a$.\n\n由 $f^{\\prime}(x)+7 a=... | [
"$-x^{2}-6 x-2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Elementary Functions | Math | Chinese |
587 | 一元三次函数 $f(x)$ 的三次项数为 $\frac{a}{3}, f^{\prime}(x)+9 x>0$ 的解集为 $(1,2)$.
若 $f(x)$ 在 $\mathbf{R}$ 上单调递减, 求 $a$ 的范围. | [
"设 $f(x)=\\frac{a}{3} x^{3}+b x^{2}+c x+d$, 则 $f^{\\prime}(x)$ $=a x^{2}+2 b x+c, f^{\\prime}(x)+9 x=a x^{2}+2 b x+c+9 x$,\n\n$\\because f^{\\prime}(x)+9 x>0$ 的解集为 $(1,2)$, 故有 $a<0$, 且 $\\left\\{\\begin{array}{l}a+2 b+c+9=0, \\\\ 4 a+4 b+c+18=0 .\\end{array}\\right.$ 则 $2 b=-3 a-9, c=2 a$.\n\n$f^{\\prime}(x)=a x^{2... | [
"$[-27-18 \\sqrt{2}, -27+18 \\sqrt{2}]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
588 | 请写出所有三个数 (正数) 均为质数, 且公差为 8 的等差数列. | [
"不妨设这三个数位分别是 $a, a+8, a+16$\n\n由 $8=3 \\times 2+2,16=3 \\times 5+1$.\n\n若 $a=3 n+1\\left(n \\in \\mathbf{N}^{*}\\right)$, 则 $a+8=3 n+1+3$ $\\times 2+2=3 \\times(n+3)$ 不是质数,\n\n若 $a=3 n+2\\left(n \\in \\mathbf{N}^{*}\\right)$, 则 $a+16=3 n+2+3$ $\\times 5+1=3(n+6)$ 不是质数,\n\n所以 $a$ 只能是 $3 n$, 而 $a$ 是质数,故 $a=3$, 此时的三个数... | [
"$3,11,19$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
589 | 已知 $|P M|-|P N|=2 \sqrt{2}$, $M(-2,0), N(2,0)$,
求点 $P$ 的轨迹 $W$; | [
"解 由双曲线的定义可知, 曲线 $W$ 是以 $M(-2,0), N(2,0)$ 为焦点的双曲线的右支, 且 $c=$ $2, a=\\sqrt{2}$,易知 $b=\\sqrt{2}$.\n\n故曲线 $W$ 的方程为 $x^{2}-y^{2}=2(x \\geqslant \\sqrt{2})$."
] | [
"$x^{2}-y^{2}=2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Equation | null | Open-ended | Plane Geometry | Math | Chinese |
590 | 已知 $|P M|-|P N|=2 \sqrt{2}$, $M(-2,0), N(2,0)$,
直线 $y=k(x-2)$ 与 $W$ 交于点 $A 、 B$, 求 $S_{\triangle A O B}$ ( $O$ 为原点). | [
"设 $A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right)$, 易得 $k>$ 1 或 $k<-1$.\n\n由题意建立方程组 $\\left\\{\\begin{array}{l}x^{2}-y^{2}=2 \\\\ y=k(x-2)\\end{array}\\right.$ 消去 $y$ 得\n\n$$\n\\left(1-k^{2}\\right) x^{2}+4 k^{2} x-4 k^{2}-2=0,\n$$\n\n由韦达定理得\n\n$$\n\\begin{aligned}\n& x_{1}+x_{2}=\\frac{4 k^{2}}{k^{2}-... | [
"$\\frac{2 \\sqrt{2} \\sqrt{k^{2}+k^{4}}}{k^{2}-1}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Plane Geometry | Math | Chinese |
591 | 12 名职员 (其中 3 名为男性) 被平均分配到 3 个部门.
求此 3 名男性被分别分到不同部门的概率; | [
"12 名职员被平均分配到 3 个部门的方法种数为: $\\mathrm{C}_{14}^{2} \\mathrm{C}_{8}^{4} \\mathrm{C}_{4}^{4}$.\n\n3 名男性被分别分到不同部门的方法种数为: $\\mathrm{C}_{3}^{1} \\mathrm{C}_{9}^{3} \\mathrm{C}_{2}^{1} \\mathrm{C}_{6}^{3}$. 所以 3 名男性被分别分到不同部门的概率是 $P_{1}=\\frac{\\mathrm{C}_{3}^{1} \\mathrm{C}_{9}^{3} \\mathrm{C}_{2}^{1} \\mathrm{C}_{6}^{3}}{... | [
"$\\frac{16}{55}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
592 | 12 名职员 (其中 3 名为男性) 被平均分配到 3 个部门.
求此 3 名男性被分到同一部门的概率; | [
"12 名职员被平均分配到 3 个部门的方法种数为: $\\mathrm{C}_{14}^{2} \\mathrm{C}_{8}^{4} \\mathrm{C}_{4}^{4}$.\n\n3 名男性被分到同一部门的方法种数为: $\\mathrm{C}_{3}^{1} \\mathrm{C}_{9}^{1} \\mathrm{C}_{8}^{4} \\mathrm{C}_{4}^{4}$, 所以 3 名男性被分到同一部门的概率是 $P_{2}=\\frac{\\mathrm{C}_{3}^{1} \\mathrm{C}_{9}^{1} \\mathrm{C}_{8}^{4} \\mathrm{C}_{4}^{4}}{\\ma... | [
"$\\frac{3}{55}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
593 | 12 名职员 (其中 3 名为男性) 被平均分配到 3 个部门.
若有一男性被分到指定部门,求其他 2 人被分到其他不同部门的概率. | [
"12 名职员被平均分配到 3 个部门的方法种数为: $\\mathrm{C}_{14}^{2} \\mathrm{C}_{8}^{4} \\mathrm{C}_{4}^{4}$.\n\n有一男性被分到指定部门，其他 2 人被分到其他不同部门的方法种数为: $\\mathrm{C}_{9}^{3} \\mathrm{C}_{2}^{1} \\mathrm{C}_{6}^{3}$,\n\n所以有一男性被分到指定部门，其他 2 人被分到其他不同部门的概率为 $\\mathrm{P}_{3}=\\frac{C_{9}^{3} C_{2}^{1} C_{6}^{3}}{\\mathrm{C}_{14}^{2} \\mathrm{C}... | [
"$\\frac{16}{165}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
594 | 已知 $f(x)=a x^{2}+8 x+b, g(x)=b x^{2}+8 x+a$, 且 $f(x)_{\min }+g(x)_{\min }=0$, 求 $\mathrm{f}(\mathrm{x})_{\min } , \mathrm{~g}(\mathrm{x})_{\min }$ . | [
"根据条件知 $\\mathrm{f}(\\mathrm{x}) 、 \\mathrm{~g}(\\mathrm{x})$ 存在最小值, 所以 $\\mathrm{a}>0, \\mathrm{~b}>0$ .\n\n又因为 $f(x)_{\\min }=\\frac{4 a b-64}{4 a}=\\frac{a b-16}{a}, g(x)_{\\min }=\\frac{4 a b-64}{4 b}=\\frac{a b-16}{b}$, 所以 $\\frac{a b-16}{a}+\\frac{a b-16}{b}=$ $\\frac{(\\mathrm{ab}-16)(\\mathrm{a}+\\mathrm{b}... | [
"$0, 0$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Elementary Functions | Math | Chinese |
599 | 求 $\sin \frac{\pi}{n} \sin \frac{2 \pi}{n} \cdots \sin \frac{(n-1) \pi}{n}$ 的值. | [
"设 $\\varepsilon=\\cos \\frac{\\pi}{n}+i \\sin \\frac{\\pi}{n}$ ( $\\mathrm{i}$ 为虚数单位), 则 $1, \\varepsilon, \\varepsilon^{2}, \\cdots \\varepsilon^{2(n-1)}$ 为 $x^{2 n}-1=0$ 的根.\n\n$\\sin \\frac{k \\pi}{n}=\\frac{\\varepsilon^{k}-\\varepsilon^{-k}}{2 i}=\\frac{\\varepsilon^{2 k}-1}{2 i \\varepsilon^{k}}, \\sin \\fra... | [
"$\\frac{n}{2^{n-1}}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Trigonometric Functions | Math | Chinese |
605 | 设 $\varepsilon_{n}=\mathrm{e}^{\frac{2 \pi i}{n}}$, 试求:
$$
\sum_{k=0}^{n-1} \frac{1}{1-\varepsilon_{n}{ }^{k} t}
$$ | [
"$\\because \\varepsilon_{n}=e^{\\frac{2 \\pi}{n}}, \\therefore 1, \\varepsilon_{n}, \\varepsilon_{n}{ }^{2} \\cdots \\varepsilon_{n}{ }^{n-1}$ 为 $x^{n}=1$ 的 $\\mathrm{n}$ 个根.\n\n由根与系数的关系知 $\\sum_{i=1}^{n} \\prod \\varepsilon_{n}{ }^{k_{1}} \\cdots \\varepsilon_{n}{ }^{k_{i}}=0,(1 \\leq i<n)$\n\n$$\n\\begin{aligned... | [
"$\\frac{n}{1-t^{n}}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Elementary Functions | Math | Chinese |
606 | 设 $\varepsilon_{n}=\mathrm{e}^{\frac{2 \pi i}{n}}$, 试求:
$$
\sum_{k=1}^{n-1} \frac{1}{1-\varepsilon_{n}{ }^{k}}
$$ | [
"$\\because \\varepsilon_{n}=e^{\\frac{2 \\pi}{n}}, \\therefore 1, \\varepsilon_{n}, \\varepsilon_{n}{ }^{2} \\cdots \\varepsilon_{n}{ }^{n-1}$ 为 $x^{n}=1$ 的 $\\mathrm{n}$ 个根.\n\n由根与系数的关系知 $\\sum_{i=1}^{n} \\prod \\varepsilon_{n}{ }^{k_{1}} \\cdots \\varepsilon_{n}{ }^{k_{i}}=0,(1 \\leq i<n)$\n\n$$\n\\begin{aligned... | [
"$\\frac{n-1}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Elementary Functions | Math | Chinese |
607 | 设 $\varepsilon_{n}=\mathrm{e}^{\frac{2 \pi i}{n}}$, 试求:
$$
\sum_{k=1}^{n-1} \frac{1}{\left(1-\varepsilon_{n}{ }^{k}\right)\left(1-\varepsilon_{n}{ }^{-k}\right)}
$$ | [
"$$\n\\begin{aligned}\n& \\sum_{k=1}^{n-1} \\frac{1}{\\left(1-\\varepsilon_{n}^{k}\\right)\\left(1-\\varepsilon_{n}^{-k}\\right)}=\\sum_{k=1}^{n-1} \\frac{1}{\\left(1-\\cos \\frac{2 k \\pi}{n}-i \\sin \\frac{2 k \\pi}{n}\\right)\\left(1-\\cos \\frac{2 k \\pi}{n}+i \\sin \\frac{2 k \\pi}{n}\\right)} \\\\\n& =\\sum_{... | [
"$\\frac{1}{12}\\left(n^{2}-1\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Elementary Functions | Math | Chinese |
608 | 求所有的函数 $f:(0,+\infty) \rightarrow(0,+\infty)$, 使对任意的正数 $x$, 都有此式成立: $f(f(x))=12 x-f(x)$ | [
"记 $f$ 的 $n$ 次迭代为 $f^{(n)}(x)$ ，下以归纳法证明，对任意 $n \\geq 2$ ，有\n\n$f^{(n)}(x)=\\frac{4 \\times 3^{n}+3 \\times(-4)^{n}}{7} x+\\frac{3^{n}-(-4)^{n}}{7} f(x)$\n\n当 $n=2$ 时，命题亦即 $f^{(2)}(x)=\\frac{4 \\times 9+3 \\times 16}{7} x+\\frac{9-16}{7} f(x)$\n\n即 $f\\left(f(x)\\right)=12 x-f(x)$, 已知其成立\n\n假设对于 $n=k \\geq 2$, 有 $f^... | [
"$f(x) = 3x$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Elementary Functions | Math | Chinese |
614 |
定义映射 $f:\{a, b, c\} \rightarrow\{a, b, c\}, t \mapsto t^{2}-2$, 求 $f(a), f(b), f(c)$ | [
"$p(-2)=-1, p(0)=1, p(1)=-1, p(2)=3$, 而 $p(x)$ 为多项式是连续函数,由零点存在定理知 $p(x)$ 在 $(-2,0),(0,1),(1,2)$ 内分别有一个实根, 而 $p(x)$ 为 3 次多项式至多有 3 个根, 故这三个根为 $p(x)$ 全部的根\n\n若 $x=t$ 为 $p$ 的一个根, 则 $t^{3}-3 t+1=0$, 则可得 $t^{2}-2=1-\\frac{1}{t}$\n\n则 $p\\left(t^{2}-2\\right)=\\left(1-\\frac{1}{t}\\right)^{3}-3\\left(1-\\frac{1}{t}\\right... | [
"$c, a, b$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Algebra | Math | Chinese |
616 | 设数列 $\left\{a_{n}\right\}$ 如下定义 $(n \geq 2): a_{n}=\sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{\cdots \sqrt{1+n}}}}}$
求 $\lim _{n \rightarrow \infty} a_{n}$. | [
"记 $a_{m, n}=\\sqrt{1+m \\sqrt{1+(m+1) \\sqrt{\\cdots \\sqrt{1+n}}}}(m \\leq n)$, 那么显然有 $a_{m, n}=\\sqrt{1+m \\sqrt{1+(m+1) \\sqrt{\\cdots \\sqrt{1+n}}}}>1$, 从而 $m<n$ 时有 $\\left|m+1-a_{m, n}\\right|=|m+1-\\sqrt{1+m \\sqrt{1+(m+1) \\sqrt{\\cdots \\sqrt{1+n}}}}|$\n\n$=\\frac{\\left|(m+1)^{2}-(1+m \\sqrt{1+(m+1) \\sqr... | [
"$3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
618 | 在单位正方形 ${A B C D}$ 内 (包括边界) 自由选取若干个节点 (数目任意), 并与 ${A}, {B}, {C}, {D}$四点用直线段连成一个连通网络 (连通图), 求这样的网络总长度的最小值. | [
"(1)当网络没有增加节点（即只选取正方形的顶点或边上的点作节点）时, 网络长度最小值相当于正方形的三边长即为 ${3}$;\n\n(2) 当网络增加一个正方形内部的节点 ${E}$ 时, 如下图, 易知, 此时网络总长度最小为 ${E} {A}+{E} {B}+{E} {C}+{E} {D} \\geq {A} {C}+{B} {D}$, 即当 ${E}$ 为正方形中心 ${O}$ 时, 网络总长度最小, 最小值为 $2 \\sqrt{2}$;\n\n<img_4205>\n\n(3) 当网络增加正方形内部的两个节点 $M, N$ 时, 我们用局部调整法来确定 $M, N$ 的位置,使得 $(A M+B M)+M N+(C... | [
"$1+\\sqrt{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
621 | 一天 ${n}(\geq 2)$ 个人组成了一个新的班集体, 每个人向校长交一元的注册费, 接下来每一天都发生如下事情: 每一个前一天组建的班集体都分裂成两个团体, 其中含至少两人的团体组建成一个新的班集体, 每个新班集体的成员向校长交一元的注册费, 只有一名成员的团体不再组建新的班集体, 上述过程直到不能继续进行下去为止, 求校长一共可能收到的注册费的最大值和最小值. | [
"记校长一共可能收到的注册费的最大值是 ${M}({n})$, 最小值是 ${m}({n})$, 某种组合方式校长收到的注册费为 $d(n)$, 知 $M(n) \\geq d(n) \\geq m(n)$\n\n(1) 先证明 $M(n)=\\frac{n(n+1)}{2}-1$, 我们用归纳法证明, 当 $n=2$ 时显然成立,假设 $n \\leq k-1(k \\geq 3, k \\in N)$ 时成立, 则当 $n=k$ 时, 第二天分为有 $p$ 和 $k-p$ 人的两个班,则 $d(k)=k+d(p)+d(k-p) \\leq k+M(p)+M(k-p)$ ，\n\n由归纳假设知 $M(p)=\\frac{p... | [
"$\\frac{n(n+1)}{2}-1, n\\left[\\log _{2} n\\right]+2\\left(n-2^{\\left[\\log _{2} n\\right]}\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Algebra | Math | Chinese |
622 | 求方程 $x^{5}+10 x^{3}+20 x-4=0$ 的所有根. | [
"设 $x=z-\\frac{2}{z}$, 则方程 $x^{5}+10 x^{3}+20 x-4=0$ 可化为 $z^{5}-\\frac{32}{z^{5}}-4=0$\n\n于是 $z^{5}=-4$ 或 $z^{5}=8$, 故 $z=\\sqrt[5]{8} e^{\\frac{2 k \\pi}{5} i}$ 或 $z=-\\sqrt[5]{4} e^{\\frac{2 k \\pi}{5} i}$ (其中 $k=0,1,2,3,4$ )\n\n于是相应的 $x=\\sqrt[5]{8} e^{\\frac{2 k \\pi}{5} i}-\\sqrt[5]{4} e^{-\\frac{2 k \\pi}{5} ... | [
"$x=\\sqrt[5]{8} e^{\\frac{2 k \\pi}{5} i}-\\sqrt[5]{4} e^{-\\frac{2 k \\pi}{5} i}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Elementary Functions | Math | Chinese |
623 | 设函数 $f(x)=\sin x+\sqrt{3} \cos x+1$,
求函数 $f(x)$ 在 $\left[0, \frac{\pi}{2}\right]$ 上的最大值与最小值; | [
"由条件知 $f(x)=2 \\sin \\left(x+\\frac{\\pi}{3}\\right)+1$,\n\n由 $0 \\leq x \\leq \\frac{\\pi}{2}$ 知, $\\frac{\\pi}{3} \\leq x+\\frac{\\pi}{3} \\leq \\frac{5 \\pi}{6}$, 于是 $\\frac{1}{2} \\leq \\sin \\left(x+\\frac{\\pi}{3}\\right) \\leq 1$\n\n所以 $x=\\frac{\\pi}{2}$ 时, $f(x)$ 有最小值 $2 \\times \\frac{1}{2}+1=2$ ；\n\n当 $x... | [
"$2，3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
624 | 设函数 $f(x)=\sin x+\sqrt{3} \cos x+1$,
若实数 $a, b, c$ 使得 $a f(x)+b f(x-c)=1$ 对任意 $x \in R$ 恒成立, 求 $\frac{b \cos c}{a}$ 的值. | [
"由条件可知\n\n$2 a \\sin \\left(x+\\frac{\\pi}{3}\\right)+2 b \\sin \\left(x+\\frac{\\pi}{3}-c\\right)+a+b=1$ 对任意的 $x \\in R$ 恒成立,\n\n$\\therefore 2 a \\sin \\left(x+\\frac{\\pi}{3}\\right)+2 b \\sin \\left(x+\\frac{\\pi}{3}\\right) \\cdot \\cos c-2 b \\cos \\left(x+\\frac{\\pi}{3}\\right) \\cdot \\sin c+(a+b-1)=0$\n\n... | [
"$-1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
625 | 已知 $a, b, c \in R^{+}$, 满足 $a b c(a+b+c)=1$,
求 $S=(a+c)(b+c)$ 的最小值; | [
"因为 $(a+c)(b+c)=a b+a c+b c+c^{2}=a b+(a+b+c) c=a b+\\frac{1}{a b}$ $\\geq 2 \\sqrt{a b \\cdot \\frac{1}{a b}}=2$, 等号成立的条件是 $a b=1$,当 $a=b=1, c=\\sqrt{2}-1$ 时, $S$ 可取最小值 2 ."
] | [
"$2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
626 | 已知 $a, b, c \in R^{+}$, 满足 $a b c(a+b+c)=1$,
当 $S$ 取最小值时, 求 $c$ 的最大值. | [
"因为 $(a+c)(b+c)=a b+a c+b c+c^{2}=a b+(a+b+c) c=a b+\\frac{1}{a b}$ $\\geq 2 \\sqrt{a b \\cdot \\frac{1}{a b}}=2$, 等号成立的条件是 $a b=1$,当 $a=b=1, c=\\sqrt{2}-1$ 时, $S$ 可取最小值 2 .\n\n当 $S$ 取最小值时, $a b=1$, 从而 $c(a+b+c)=1$,\n\n即 $c^{2}+(a+b) c-1=0$, 令 $t=a+b$, 则 $t \\geq 2 \\sqrt{a b}=2$\n\n从而 $c=\\frac{-t+\\sqrt{t^{2}+4}}... | [
"$\\sqrt{2}-1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
627 | 直线 $y=k x+1$ 与双曲线 $x^{2}-y^{2}=1$ 的左支交于 $A 、 B$ 两点, 直线 $l$ 经过点 $(-2,0)$ 和 $A B$的中点, 求直线 $l$ 在 $y$ 轴的截距 $b$ 的取值范围. | [
"将直线 $y=k x+1$ 与双曲线 $x^{2}-y^{2}=1$ 方程联立得 $\\left\\{\\begin{array}{l}y=k x+1 \\\\ x^{2}-y^{2}=1\\end{array}\\right.$\n\n化简得 $\\left(k^{2}-1\\right) x^{2}+2 k x+2=0$ (1)\n\n由题设知方程(1)有两负根, 因此 $\\left\\{\\begin{array}{l}\\Delta=4 k^{2}-8\\left(k^{2}-1\\right)>0 \\\\ x_{1}+x_{2}=-\\frac{2 k}{k^{2}-1}<0 \\\\ x_{1} \\cdo... | [
"$(-\\infty,-2-\\sqrt{2}) \\cup(2,+\\infty)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Geometry | Math | Chinese |
631 | 设等比数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 且 $S_{n}=2^{n}+r$ ( $r$ 为常数),
记 $b_{n}=2\left(1+\log _{2} a_{n}\right) \quad\left(n \in \mathbf{N}^{*}\right)$.
求数列 $\left\{a_{n} b_{n}\right\}$ 的前 $n$ 项和 $T_{n}$; | [
"由条件易知 $a_{1}=2+r, a_{2}=S_{2}-S_{1}=2, a_{3}=S_{3}-S_{2}=4$,\n\n又由 $a_{2}^{2}=a_{1} a_{3}$ 得 $r=-1$.\n\n于是 $S_{n}=2^{n}-1$. 故 $a_{n}=2^{n-1}, b_{n}=2\\left(1+\\log _{2} a_{n}\\right)=2 n, a_{n} b_{n}=n \\cdot 2^{n}$.\n\n因此 $T_{n}=1 \\times 2^{1}+2 \\times 2^{2}+\\cdots+(n-1) \\times 2^{n-1}+n \\times 2^{n} \\quad ... | [
"$(n-1) \\cdot 2^{n+1}+2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
632 | 设等比数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 且 $S_{n}=2^{n}+r$ ( $r$ 为常数),
记 $b_{n}=2\left(1+\log _{2} a_{n}\right) \quad\left(n \in \mathbf{N}^{*}\right)$.
若对于任意的正整数 $n$, 都有 $\frac{1+b_{1}}{b_{1}} \cdot \frac{1+b_{2}}{b_{2}} \cdot \cdots \cdot \frac{1+b_{n}}{b_{n}} \geq k \sqrt{n+1}$ 成立,求实数 $k$ 的最大值. | [
"因为 $k \\leq \\frac{1}{\\sqrt{n+1}} \\cdot \\frac{1+b_{1}}{b_{1}} \\cdot \\frac{1+b_{2}}{b_{2}} \\cdot \\cdots \\cdot \\frac{1+b_{n}}{b_{n}}=\\frac{1}{\\sqrt{n+1}} \\cdot \\frac{1+2}{2} \\cdot \\frac{1+4}{4} \\cdot \\cdots \\cdot \\frac{1+2 n}{2 n}$,\n\n构造 $f(n)=\\frac{1}{\\sqrt{n+1}} \\cdot \\frac{1+2}{2} \\cdot \... | [
"$\\frac{3}{4} \\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
634 | 已知抛物线 $y^{2}=2 p x$ 过定点 $C(1,2)$, 在抛物线上任取不同于点 $C$ 的一点 $A$, 直线 $A C$ 与直线 $y=x+3$交于点 $P$, 过点 $P$ 作 $x$ 轴的平行线交抛物线于点 $B$.
直线 $A B$ 过定点，求定点的坐标; | [
"由抛物线 $y^{2}=2 p x$ 过定点 $C(1,2)$,\n\n可得抛物线方程为 $y^{2}=4 x$.\n\n设点 $A$ 坐标为 $\\left(\\frac{y_{0}^{2}}{4}, y_{0}\\right)\\left(y_{0} \\neq 2\\right)$,\n\n则直线 $A C$ 的方程为 $y-2=\\frac{y_{0}-2}{\\frac{y_{0}^{2}}{4}-1}(x-1)$,\n\n即 $y-2=\\frac{4}{y_{0}+2}(x-1) ，$\n\n与 $y=x+3$ 联立解得 $P$ 点坐标为 $\\left(\\frac{-y_{0}-6}{y_{0}-2}, ... | [
"$(3,2)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Tuple | null | Open-ended | Geometry | Math | Chinese |
635 | 已知抛物线 $y^{2}=2 p x$ 过定点 $C(1,2)$, 在抛物线上任取不同于点 $C$ 的一点 $A$, 直线 $A C$ 与直线 $y=x+3$交于点 $P$, 过点 $P$ 作 $x$ 轴的平行线交抛物线于点 $B$.
求 $\triangle A B C$ 面积的最小值. | [
"由抛物线 $y^{2}=2 p x$ 过定点 $C(1,2)$, 可得抛物线方程为 $y^{2}=4 x$.\n\n设直线 $A B$ 的方程为 $x=m y+a$, 与抛物线方程联立得, $y^{2}-4 m y-4 a=0$.\n\n设 $A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right)$, 则 $y_{1}+y_{2}=4 m, y_{1} y_{2}=-4 a$,\n\n$P$ 点坐标为 $B\\left(y_{2}-3, y_{2}\\right)$, 因为 $A P$ 过定点 $C$,\n\n所以 $\\frac{y_{2}-2}{y_{2}-... | [
"$4 \\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
637 | 已知数列 $\left\{a_{n}\right\}$ 满足: $a_{1}=a, a_{n+1}=\frac{5 a_{n}-8}{a_{n}-1}\left(n \in N^{*}\right)$.
若 $a=3$, 数列 $\left\{\frac{a_{n}-2}{a_{n}-4}\right\}$ 成等比数列, 求出数列 $\left\{a_{n}\right\}$ 的通项公式; | [
"因为 $\\frac{a_{n+1}-2}{a_{n+1}-4}=\\frac{\\frac{5 a_{n}-8}{a_{n}-1}-2}{\\frac{5 a_{n}-8}{a_{n}-1}-4}=3 \\cdot \\frac{a_{n}-2}{a_{n}-4}$,\n\n所以, 数列 $\\left\\{\\frac{a_{n}-2}{a_{n}-4}\\right\\}$ 成等比数列.\n\n于是 $\\frac{a_{n}-2}{a_{n}-4}=\\frac{a_{1}-2}{a_{1}-4} \\cdot 3^{n-1}$, 解得 $a_{n}=\\frac{4 \\cdot 3^{n-1}+2}{3^{n-... | [
"$\\frac{4 \\cdot 3^{n-1}+2}{3^{n-1}+1}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
638 | 已知数列 $\left\{a_{n}\right\}$ 满足: $a_{1}=a, a_{n+1}=\frac{5 a_{n}-8}{a_{n}-1}\left(n \in N^{*}\right)$.
若对任意的正整数 $n$, 都有 $a_{n}>3$, 求实数 $a$ 的取值范围. | [
"因为 $a_{n}>3$ 对任意的正整数 $n$ 都成立, 故 $a=a_{1}>3$.\n\n因为 $\\frac{a_{n+1}-2}{a_{n+1}-4}=\\frac{\\frac{5 a_{n}-8}{a_{n}-1}-2}{\\frac{5 a_{n}-8}{a_{n}-1}-4}=3 \\cdot \\frac{a_{n}-2}{a_{n}-4}$,\n\n所以, 数列 $\\left\\{\\frac{a_{n}-2}{a_{n}-4}\\right\\}$ 成等比数列.\n\n于是 $\\frac{a_{n}-2}{a_{n}-4}=\\frac{a_{1}-2}{a_{1}-4} \\cdot 3^{n... | [
"$(3,+\\infty)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | Chinese |
639 | 1993 年, 美国数学家 F.Smarandache 提出许多数论问题, 引起国内外相关学者的关注, 其中之一便是著名的 Smarandache 函数. 正整数 $n$ 的 Smarandache 函数定义为:
$$
S(n)=\min \left\{m\left|m \in N^{*}, n\right| m !\right\} .
$$
比如: $S(2)=2, S(3)=3, S(6)=3$.
求 $S(16)$ 和 $S(2016)$ 的值; | [
"因为 $16=2^{4}$, 故 $S(16)=6$.\n\n由 $2016=2^{5} \\times 3^{2} \\times 7$, 知 $S(2016)=\\max \\left\\{S\\left(2^{5}\\right), S\\left(3^{2}\\right), S(7)\\right\\}$.\n\n而 $S(7)=7, S\\left(3^{2}\\right)=6, S\\left(2^{5}\\right)=8$, 故 $S(2016)=8$."
] | [
"$6,8$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
640 | 1993 年, 美国数学家 F.Smarandache 提出许多数论问题, 引起国内外相关学者的关注, 其中之一便是著名的 Smarandache 函数. 正整数 $n$ 的 Smarandache 函数定义为:
$$
S(n)=\min \left\{m\left|m \in N^{*}, n\right| m !\right\} .
$$
比如: $S(2)=2, S(3)=3, S(6)=3$.
若 $S(n)=7$, 求正整数 $n$ 的最大值; | [
"由 $S(n)=7$, 知 $n \\mid 7$ !, 从而 $n \\leq 7$ !.\n\n又因为 $S(7 !)=7$, 所以正整数 $n$ 的最大值是 $7 !=5040$."
] | [
"$5040$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
644 | 设 $\alpha, \beta$ 为实数, 若对任意实数 $x, y, z$. 有 $a(x y+y z+z x) \leq M \leq \beta\left(x^{2}+y^{2}+z^{2}\right)$ 恒成立, 其中 $M=\sqrt{x^{2}+x y+y^{2}} \cdot \sqrt{y^{2}+y z+z^{2}}+\sqrt{y^{2}+y z+z^{2}} \cdot \sqrt{z^{2}+z x+x^{2}}+\sqrt{z^{2}+z x+x^{2}} \cdot \sqrt{x^{2}+x y+y^{2}}$.求 $\alpha$ 的最大值和 $\beta$ 的最小值. | [
"取 $x=y=z=1$, 有 $3 \\alpha \\leq 9 \\leq 3 \\beta$, 则 $\\alpha \\leq 3, \\beta \\geq 3$.\n\n先证: $M \\geq 3(x y+y z+z x)$ 对任意的实数 $x, y, z$ 成立.\n\n因为 $\\sqrt{x^{2}+x y+y^{2}} \\cdot \\sqrt{y^{2}+y z+z^{2}}=\\sqrt{\\left[\\left(x+\\frac{y}{2}\\right)^{2}+\\frac{3}{4} y^{2}\\right] \\cdot\\left[\\left(z+\\frac{y}{2}\\r... | [
"$3,3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
646 | 已知 $a$ 为常数, 函数
$f(x)=\ln \frac{1-x}{1+x}-a x$.
若 $a=-\frac{8}{3}$, 求 $f(x)$ 的极大值和极小值. | [
"注意到, $a=-\\frac{8}{3}<-2$.\n\n由 $f^{\\prime}(x)=0$, 知驻点为 $x=-\\frac{1}{2}$ 或 $\\frac{1}{2}$.\n\n当 $-1<x<-\\frac{1}{2}$ 时, $f^{\\prime}(x)<0$ ；\n\n当 $-\\frac{1}{2}<x<\\frac{1}{2}$ 时, $f^{\\prime}(x)>0$;\n\n当 $\\frac{1}{2}<x<1$ 时, $f^{\\prime}(x)<0$.\n\n故 $f(x)$ 的极小值为 $f\\left(-\\frac{1}{2}\\right)=-\\frac{4}{3}+\\l... | [
"$\\frac{4}{3}-\\ln 3, -\\frac{4}{3}+\\ln 3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
647 | 已知对一切的 $x \in \mathbf{R}$ 恒有 $3 \sin ^{2} x-\cos ^{2} x+4 a \cos x+a^{2} \leqslant 31$.
求实数 $a$ 的取值范围. | [
"设 $f(x)=3 \\sin ^{2} x-\\cos ^{2} x+4 a \\cos x+a^{2}-31$.\n\n则 $f(x)=-4 \\cos ^{2} x+4 a \\cos x+a^{2}-28$.\n\n令 $t=\\cos x$. 则 $t \\in[-1,1]$.\n\n故当 $t \\in[-1,1]$ 时, 恒有\n\n$g(t)=-4 t^{2}+4 a t+a^{2}-28 \\leqslant 0$.\n\n注意到, 二次函数 $g(t)$ 的对称轴为 $t=\\frac{a}{2}$.\n\n(1) 当 $\\frac{a}{2}<-1$, 即 $a<-2$ 时, $g(t)$ 在区间 ... | [
"$[-4,4]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | Chinese |
648 | 已知 $k$ 为给定正整数, 数列 $\left\{a_{n}\right\}$ 满足
$a_{1}=3, a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3\left(n \in \mathbf{Z}_{+}\right)$,其中, $S_{n}$ 是数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和.
令 $b_{n}=\frac{1}{n} \log _{3} a_{1} a_{2} \cdots a_{n}\left(n \in \mathbf{Z}_{+}\right)$, 记 $T_{k}=\sum_{i=1}^{2 k}\left|b_{i}-\frac{3}{2}\right|$.
若 $T_{k} \in \mathbf{Z}_{+}$, 求 $k$ 的所有可能值. | [
"由题意知\n\n$a_{2}=\\left(3^{\\frac{2}{2 k-1}}-1\\right) \\times 3+3=3^{\\frac{2 k+1}{2 k-1}}$.\n\n又 $a_{n+1}=\\left(3^{\\frac{2}{2 k-1}}-1\\right) S_{n}+3$,\n\n$a_{n}=\\left(3^{\\frac{2}{2 k-1}}-1\\right) S_{n-1}+3(n \\geqslant 2)$,\n\n故 $a_{n+1}-a_{n}=\\left(3^{\\frac{2}{2 k-1}}-1\\right) a_{n}$\n\n$\\Rightarrow a_{... | [
"$1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
649 | 过椭圆 $\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$ 的右焦点 $F$ 作两条垂直的弦 $A B 、 C D$. 设 $A B 、 C D$ 的中点分别为 $M 、 N$.
直线 $M N$ 必过定点, 求此定点; | [
"由题意知 $F(1,0)$.\n\n(i) 当弦 $A B 、 C D$ 的斜率均存在时, 设 $A B$ 的斜率为 $k$. 则 $C D$ 的斜率为 $-\\frac{1}{k}$.\n\n设 $l_{A B}: y=k(x-1)$.\n\n代人椭圆方程 $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, 得\n\n$\\left(3 k^{2}+2\\right) x^{2}-6 k^{2} x+\\left(3 k^{2}-6\\right)=0$.\n\n故 $x_{M}=\\frac{x_{A}+x_{B}}{2}=\\frac{3 k^{2}}{3 k^{2}+2}$,\n\n$y_... | [
"$\\left(\\frac{3}{5}, 0\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Tuple | null | Open-ended | Geometry | Math | Chinese |
650 | 过椭圆 $\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$ 的右焦点 $F$ 作两条垂直的弦 $A B 、 C D$. 设 $A B 、 C D$ 的中点分别为 $M 、 N$.
若弦 $A B 、 C D$ 的斜率均存在, 求 $\triangle F M N$面积 $S$ 的最大值. | [
"由题意知 $F(1,0)$.\n\n(i) 当弦 $A B 、 C D$ 的斜率均存在时, 设 $A B$ 的斜率为 $k$. 则 $C D$ 的斜率为 $-\\frac{1}{k}$.\n\n设 $l_{A B}: y=k(x-1)$.\n\n代人椭圆方程 $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, 得\n\n$\\left(3 k^{2}+2\\right) x^{2}-6 k^{2} x+\\left(3 k^{2}-6\\right)=0$.\n\n故 $x_{M}=\\frac{x_{A}+x_{B}}{2}=\\frac{3 k^{2}}{3 k^{2}+2}$,\n\n$y_... | [
"$\\frac{4}{25}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
651 | 设实数 $\omega>0$, 已知函数 $f(x)=\sin ^{2} \omega x+\sqrt{3} \sin \omega x \cdot \sin \left(\omega x+\frac{\pi}{2}\right) \quad$ 的最小正周期是 $\frac{\pi}{2}$.求 $f(x)$ 在 $\left[\frac{\pi}{8}, \frac{\pi}{4}\right]$ 上的最大值与最小值. | [
"$f(x)=\\frac{1-\\cos 2 \\omega x}{2}+\\frac{\\sqrt{3}}{2} \\sin 2 \\omega x=\\frac{\\sqrt{3}}{2} \\sin 2 \\omega x-\\frac{1}{2} \\cos 2 \\omega x+\\frac{1}{2}$\n\n$$\n=\\sin \\left(2 \\omega x-\\frac{\\pi}{6}\\right)+\\frac{1}{2}\n$$\n\n由条件知 $T=\\frac{2 \\pi}{2 \\omega}=\\frac{\\pi}{2}$, 则 $\\omega=2$.\n\n于是 $f(x)... | [
"$\\frac{3}{2} , 1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
652 | 已知函数 $f(x)=\frac{x^{3}+3 x}{3 x^{2}+1}$, 数列 $\left\{x_{n}\right\}$ 满足: $x_{1}=2, x_{n+1}=f\left(x_{n}\right)\left(n \in N^{*}\right)$,
记 $b_{n}=\log _{3}\left(\frac{x_{n+1}-1}{x_{n+1}+1}\right) \quad\left(n \in N^{*}\right)$.
求数列 $\left\{b_{n}\right\}$ 的通项公式; | [
"$\\frac{x_{n+1}-1}{x_{n+1}+1}=\\frac{f\\left(x_{n}\\right)-1}{f\\left(x_{n}\\right)+1}=\\frac{\\frac{x_{n}^{3}+3 x_{n}}{3 x_{n}^{2}+1}-1}{\\frac{x_{n}^{3}+3 x_{n}}{3 x_{n}^{2}+1}+1}=\\frac{x_{n}^{3}-3 x_{n}^{2}+3 x_{n}-1}{x_{n}^{3}+3 x_{n}^{2}+3 x_{n}+1}=\\left(\\frac{x_{n}-1}{x_{n}+1}\\right)^{3}$\n\n于是 $\\log _{... | [
"$b_{n}=-3^{n-1}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
653 | 已知函数 $f(x)=\frac{x^{3}+3 x}{3 x^{2}+1}$, 数列 $\left\{x_{n}\right\}$ 满足: $x_{1}=2, x_{n+1}=f\left(x_{n}\right)\left(n \in N^{*}\right)$,
记 $b_{n}=\log _{3}\left(\frac{x_{n+1}-1}{x_{n+1}+1}\right) \quad\left(n \in N^{*}\right)$.
记 $c_{n}=-n b_{n}\left(n \in N^{*}\right)$, 求数列 $\left\{c_{n}\right\}$ 的前 $n$ 项和公式 $T_{n}$. | [
"$\\frac{x_{n+1}-1}{x_{n+1}+1}=\\frac{f\\left(x_{n}\\right)-1}{f\\left(x_{n}\\right)+1}=\\frac{\\frac{x_{n}^{3}+3 x_{n}}{3 x_{n}^{2}+1}-1}{\\frac{x_{n}^{3}+3 x_{n}}{3 x_{n}^{2}+1}+1}=\\frac{x_{n}^{3}-3 x_{n}^{2}+3 x_{n}-1}{x_{n}^{3}+3 x_{n}^{2}+3 x_{n}+1}=\\left(\\frac{x_{n}-1}{x_{n}+1}\\right)^{3}$\n\n于是 $\\log _{... | [
"$T_{n}=\\frac{(2 n-1) \\cdot 3^{n}+1}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
654 | 已知点 $B(0,1), P 、 Q$ 为椭圆 $\frac{x^{2}}{4}+y^{2}=1$ 上异于点 $B$ 的任意两点, 且 $B P \perp B Q$.
若点 $B$ 在线段 $P Q$ 上的射影为 $M$, 求 $M$ 的轨迹方程; | [
"设 $P\\left(x_{1}, y_{1}\\right) 、 Q\\left(x_{2}, y_{2}\\right), P Q$ 的方程为 $y=k x+m$,\n\n与椭圆方程联立消去 $y$ 得: $\\left(1+4 k^{2}\\right) x^{2}+8 k m x+4 m^{2}-4=0$,\n\n所以 $x_{1}+x_{2}=\\frac{-8 k m}{4 k^{2}+1}, x_{1} x_{2}=\\frac{4 m^{2}-4}{4 k^{2}+1}$,\n\n由 $B P \\perp B Q$ 得 $\\frac{y_{1}-1}{x_{1}} \\cdot \\frac{y_{2}... | [
"$x^{2}+\\left(y-\\frac{1}{5}\\right)^{2}=\\left(\\frac{4}{5}\\right)^{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Equation | null | Open-ended | Geometry | Math | Chinese |
655 | 已知点 $B(0,1), P 、 Q$ 为椭圆 $\frac{x^{2}}{4}+y^{2}=1$ 上异于点 $B$ 的任意两点, 且 $B P \perp B Q$.
求线段 $P Q$ 的中垂线 $l$ 在 $x$ 轴上的截距的取值范围. | [
"$P Q$ 方程为 $y=k x-\\frac{3}{5}$, 所以 $\\frac{x_{1}+x_{2}}{2}=\\frac{12 k}{5\\left(4 k^{2}+1\\right)}, \\frac{y_{1}+y_{2}}{2}=\\frac{-3}{5\\left(4 k^{2}+1\\right)}$\n\n所以 $P Q$ 中垂线方程为 $y+\\frac{3}{5\\left(4 k^{2}+1\\right)}=-\\frac{1}{k}\\left(x-\\frac{12 k}{5\\left(4 k^{2}+1\\right)}\\right)$,\n\n其在 $x$ 轴上的截距为 $b=\\... | [
"$\\left[-\\frac{9}{20}, \\frac{9}{20}\\right]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Geometry | Math | Chinese |
657 | 若实数 $x_{0}$ 满足 $f\left(x_{0}\right)=x_{0}$, 则称 $x=x_{0}$ 为 $f(x)$ 的不动点.
已知函数 $f(x)=x^{3}+a x^{2}+b x+3$, 其中 $a, b$ 为常数.
若 $a=0$ 时, 存在一个实数 $x_{0}$, 使得 $x=x_{0}$ 既是 $f(x)$ 的不动点, 又是 $f(x)$ 的极值点. 求实数 $b$ 的值; | [
"由条件知 $\\left\\{\\begin{array}{l}3 x_{0}^{2}+b=0 \\\\ x_{0}^{3}+b x_{0}+3=x_{0}\\end{array}\\right.$,\n\n于是 $2 x_{0}^{3}+x_{0}-3=0$, 即 $\\left(x_{0}-1\\right)\\left(2 x_{0}^{2}+2 x_{0}+3\\right)=0$, 解得 $x_{0}=1$\n\n从而 $b=-3$."
] | [
"$-3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
659 | 已知双曲线 $\frac{x^{2}}{4}-\frac{y^{2}}{3}=1$, 设其实轴端点为 $A_{1} 、 A_{2}$, 点 $P$ 是双曲线上不同于 $A_{1}$ 、 $A_{2}$ 的一个动点, 直线 $P A_{1} 、 P A_{2}$ 分别与直线 $x=1$ 交于 $M_{1} 、 M_{2}$ 两点. 以线段 $M_{1} M_{2}$ 为直径的圆必经过定点，求定点坐标. | [
"由已知可设 $A_{1}(-2,0), A_{2}(2,0)$, 双曲线上动点 $P$ 的坐标为 $\\left(x_{0}, y_{0}\\right)$ 且 $y_{0} \\neq 0$,则 $\\frac{x_{0}^{2}}{4}-\\frac{y_{0}^{2}}{3}=1$.\n\n因为直线 $P A_{1}$ 的方程为 $y=\\frac{y_{0}}{x_{0}+2}(x+2)$, 直线 $P A_{2}$ 的方程为 $y=\\frac{y_{0}}{x_{0}-2}(x-2)$,\n\n所以 $\\left.M_{1}\\left(1, \\frac{3 y_{0}}{x_{0}+2}\\right),... | [
"$\\left(-\\frac{1}{2}, 0\\right),\\left(\\frac{5}{2}, 0\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Geometry | Math | Chinese |
660 | 设 $x, y, z$ 为正实数, 求 $\left(x+\frac{1}{y}+\sqrt{2}\right)\left(y+\frac{1}{z}+\sqrt{2}\right)\left(z+\frac{1}{x}+\sqrt{2}\right)$ 的最小值. | [
"记 $T=\\left(x+\\frac{1}{y}+\\sqrt{2}\\right)\\left(y+\\frac{1}{z}+\\sqrt{2}\\right)\\left(z+\\frac{1}{x}+\\sqrt{2}\\right)$,\n\n当 $x=y=z=1$ 时, $T$ 有最小值 $(2+\\sqrt{2})^{3}=20+14 \\sqrt{2}$.\n\n下证: $\\quad T \\geq 20+14 \\sqrt{2}$.\n\n $T=\\left(x y z+\\frac{1}{x y z}\\right)+\\sqrt{2}\\left(x y+y z+z x+\\frac{1}{x ... | [
"$20+14 \\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
661 | 已知数列 $\left\{a_{n}\right\}$ 满足: $a_{1}=1, a_{n+1}=\frac{1}{8} a_{n}^{2}+m\left(n \in \mathbf{N}^{*}\right)$, 若对任意正整数 $n$, 都有 $a_{n}<4$, 求实数 $m$ 的最大值. | [
"因为 $a_{n+1}-a_{n}=\\frac{1}{8} a_{n}^{2}-a_{n}+m=\\frac{1}{8}\\left(a_{n}-4\\right)^{2}+m-2 \\geq m-2$,\n\n故 $a_{n}=a_{1}+\\sum_{k=1}^{n-1}\\left(a_{k+1}-a_{k}\\right) \\geq 1+(m-2)(n-1)$.\n\n若 $m>2$, 注意到 $n \\rightarrow+\\infty$ 时, $(m-2)(n-1) \\rightarrow+\\infty$,\n\n因此, 存在充分大的 $n$, 使得 $1+(m-2)(n-1)>4$, 即 $a_{n... | [
"$2$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
662 | 设函数 $f(x)=p x-\frac{p}{x}-2 \ln x$.
若 $f(x)$ 在其定义域内为单调递增函数, 求实数 $p$ 的取值范围; | [
"函数 $f(x)$ 的定义域为 $(x \\mid x>0\\}$.\n\n由 $f(x)=p x-\\frac{p}{x}-2 \\ln x$ 知 $f^{\\prime}(x)=p+\\frac{p}{x^{2}}-\\frac{2}{x}$,\n\n要使 $f(x)$ 在其定义域 $(0,+\\infty)$ 内为单调递增函数, 只须 $f^{\\prime}(x) \\geq 0$,\n\n即 $p x^{2}-2 x+p \\geq 0$ 在 $(0,+\\infty)$ 内恒成立.\n\n于是 $p \\geq \\frac{2 x}{x^{2}+1}$, 注意到: $\\frac{2 x}{x^{2}+1} ... | [
"$[1, +\\infty)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | Chinese |
663 | 设函数 $f(x)=p x-\frac{p}{x}-2 \ln x$.
设 $g(x)=\frac{2 e}{x}$, 且 $p>0$, 若在 $[1, e]$ 上至少存在一点 $x_{0}$, 使得 $f\left(x_{0}\right)>g\left(x_{0}\right)$ 成立,求实数 $p$ 的取值范围; | [
"注意到 $g(x)=\\frac{2 e}{x}$ 在 $[1, e]$ 上是减函数,\n\n所以 $g(x)_{\\text {min }}=g(e)=2, g(x)_{\\text {max }}=g(1)=2 e$, 即 $g(x) \\in[2,2 e]$.\n\n当 $0<p<1$ 时, 由 $x \\in[1, e]$, 得 $x-\\frac{1}{x} \\geqslant 0$,\n\n故 $f(x)=p\\left(x-\\frac{1}{x}\\right)-2 \\ln x<x-\\frac{1}{x}-2 \\ln x<2$, 不合题意.\n\n当 $p \\geqslant 1$ 时, 由 (1... | [
"$\\left(\\frac{4 e}{e^{2}-1},+\\infty\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | Chinese |
670 | 已知对任意 $x 、 y 、 z \geqslant 0$ 有 $x^{3}+y^{3}+z^{3}-3 x y z \geqslant c|(x-y)(y-z)(z-x)|$.
求 $c$ 的最大值. | [
"不等式左边\n\n$$\n\\begin{aligned}\n& =(x+y+z)\\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\\right) \\\\\n& =\\frac{(x+y+z)\\left[(x-z)^{2}+(y-z)^{2}+(z-x)^{2}\\right]}{2} .\n\\end{aligned}\n$$\n\n不妨设 $x \\geqslant y \\geqslant z \\geqslant 0$.\n\n若 $x=y$ 或 $y=z$, 对于任意实数 $c$, 不等式均成立.\n\n若 $x>y>z \\geqslant 0$, 固定 $x-y 、 y-z$, 则... | [
"$\\left(\\frac{\\sqrt{6}+3 \\sqrt{2}}{2}\\right)^{\\sqrt[4]{3}}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
671 | 用 $r(n)$ 表示 $n$ 被 $1,2, \cdots, n$ 除所得余数的和. 试求所有正整数 $m(1<m \leqslant 2014)$,使得
$r(m)=r(m-1)$. | [
"记 $n$ 被 $k(1 \\leqslant k \\leqslant n)$ 除所得的余数为 $r_{k}(n)$.\n\n于是, $r_{k}(n)=n-k\\left[\\frac{n}{k}\\right]$, 其中, $[x]$ 表示不超过实数 $x$ 的最大整数.\n\n则 $r(m)=\\sum_{k=1}^{m} r_{k}(m)$\n\n$=\\sum_{k=1}^{m}\\left(m-k\\left[\\frac{m}{k}\\right]\\right)=m^{2}-\\sum_{k=1}^{m} k\\left[\\frac{m}{k}\\right]$.\n\n故 $r(m)=r(m-1)$\... | [
"$2^{s}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Number Theory | Math | Chinese |
674 | 已知 $a 、 b 、 c$ 是三个互异实数.若在二次方程
$$
\begin{aligned}
& x^{2}+a x+b=0, \quad \quad (1) \\
& x^{2}+b x+c=0, \quad \quad (2)\\
& x^{2}+c x+a=0 \quad \quad (3)
\end{aligned}
$$
中任意两个均恰有一.个公共根, 求 $a^{2}+b^{2}+$ $c^{2}$ 的值. | [
"由方程(1)、(2)知其公共根为 $p=\\frac{b-c}{b-a}$.\n\n同理,方程(2)、(3),方程(1)、(3)的公共根分别为 $q=\\frac{c-a}{c-b}, r=\\frac{a-b}{a-c}$.\n\n于是, $p q r=-1$.\n\n若 $p 、 q 、 r$ 中有两个相等, 不妨设 $p=q$, 则三个方程有公共根 $p$. 于是,\n\n$$\np=r \\Rightarrow p=r=q=-1 \\Rightarrow a=b=c,\n$$\n\n矛盾.\n\n下设 $p 、 q 、 r$ 互异. 则三个方程有形式\n\n$$\n\\begin{aligned}\n& (x-p)... | [
"$6$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
680 | 在 $4 n \times 4 n\left(n \in \mathbf{N}_{+}\right)$实数表中, 每个数的绝对值均不超过 1 , 所有数的和为 0 . 已知数表中的每行各数之和的绝对值、每列各数之和的绝对值均不小于 $c$. 求 $c$ 的最大可能值. | [
"如下: 取数表中第 $i$ 行、第 $j$ 列处的数为\n\n$$\na_{i j}=\\left\\{\\begin{array}{cl}\n0, & 1 \\leqslant i 、 j \\leqslant 2 n \\text { 或 } 2 n+1 \\leqslant i 、 j \\leqslant 4 n ； \\\\\n1, & 1 \\leqslant i \\leqslant 2 n, 2 n+1 \\leqslant j \\leqslant 4 n ; \\\\\n-1, & 2 n+1 \\leqslant i \\leqslant 4 n, 1 \\leqslant j \\leqslant ... | [
"$2 n$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Number Theory | Math | Chinese |
687 | 若 $n$ 为完全平方数或 $n$ 到距离其最近的完全平方数的距离为完全立方数, 则称 $n$为 “好数”. 例如, $\left|45^{2}-2017\right|=2^{3}$, 即 2017 为好数. 对于正整数 $N$, 令 $S(N)$ 为集合 $\{1$, $2, \cdots, N\}$ 中好数的个数.
求 $S(2018)$; | [
"对于正整数 $n, n^{2}$ 左侧的完全平方数为 $n^{2}-2 n+1, n^{2}$ 右侧的完全平方数为 $n^{2}+2 n+1$.\n\n于是,距离区间 $\\left[n^{2}-n+1, n^{2}+n\\right]$ 内整数最近的完全平方数均为 $n^{2}$.\n\n记 $[x]$ 表示不超过实数 $x$ 的最大整数.\n\n设 $s=[\\sqrt[3]{n-1}], t=[\\sqrt[3]{n}]$.\n\n则区间 $\\left[n^{2}-n+1, n^{2}+n\\right]$ 内的好数依次为\n\n$n^{2}, n^{2}-1^{3}, n^{2}-2^{3}, \\cdots, ... | [
"$241$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
692 | 用不交于内点的对角线将多边形分为一系列三角形,必要时可以在形内取若干个点作为所分出的三角形的顶点, 这些顶点称为 “辅助顶点”. 辅助顶点之间的连线、辅助顶点与顶点之间的连线均称为对角线. 任何两条对角线均不得交于内点. 这种将多边形划分为一系列三角形的方法称为多边形的 “三角形剖分”.
设 $n$ 为不小于 5 的正整数. 为了将一个正 $n$ 边形剖分为一系列的钝角三角形,试求所分成的三角形个数 $m$ 的最小可能值. | [
"若不添加辅助顶点, 则所有分出的三角形的所有顶点均在正 $n$ 边形的 $n$ 个顶点上.故正 $n$ 边形的外接圆为所有这些三角形的公共的外接圆. 于是, 它们共有一个外心, 必落在某个所分出的三角形的内部或边上. 从而,该三角形为锐角三角形或直角三角形.\n\n这表明,若想所分出的三角形均为钝角三角形,则必须添加辅助顶点.\n\n当所添加的辅助顶点的数目为 $k$ 时, 所分成的三角形的所有内角和为\n\n$$\n(n-2) \\pi+2 k \\pi=(n+2 k-2) \\pi,\n$$\n\n于是, 所分出的三角形共有 $n+2 k-2$ 个. 这表明,欲分成的三角形均为钝角三角形,则所分出三角形个数 $m \\... | [
"$n$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | Chinese |
700 | 求最小的质数 $p$, 满足 $(p, N)=1$,其中, $N$ 为所有满足以下条件的 $\left(a_{0}, a_{1}, \cdots\right.$, $a_{2012}$ ) 的个数:
(1) $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ 是 $0,1, \cdots, 2012$的一个排列;
(2) 对 2013 的任意一个正约数 $m$ 及所有的 $n(0 \leqslant n<n+m \leqslant 2012)$ 有
$m \mid\left(a_{n+m}-a_{n}\right)$. | [
"由(2) 知对任意 $i 、 j(0 \\leqslant i<j \\leqslant 2012)$及 $m=3^{\\alpha} \\times 11^{\\beta} \\times 61^{\\gamma}(\\alpha 、 \\beta 、 \\gamma \\in\\{0,1\\})$, 有\n\n$a_{i} \\equiv a_{j}(\\bmod m) \\Leftrightarrow i \\equiv j(\\bmod m) \\quad \\quad ①$.\n\n令\n\n$A_{i}=\\{x \\in \\mathrm{N} \\mid 0 \\leqslant x \\leqslant ... | [
"$67$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
704 | 设数列 $\left\{x_{n}\right\}$ 满足
$x_{1}=1, x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right]$.
求 $x_{2012}$ 的个位数字. | [
"显然, $x_{2}=7$, 且对任意正整数 $n, x_{n}$ 为正整数.\n\n由高斯函数的性质得\n\n$4 x_{n}+\\sqrt{11} x_{n}>x_{n+1}=4 x_{n}+\\left[\\sqrt{11} x_{n}\\right]>4 x_{n}+\\sqrt{11} x_{n}-1$.\n\n上式两边同乘以 $4-\\sqrt{11}$ 得\n\n$5 x_{n}>(4-\\sqrt{11}) x_{n+1}>5 x_{n}-(4-\\sqrt{11})$\n\n$\\Rightarrow 4 x_{n+1}-5 x_{n}<\\sqrt{11} x_{n+1}$\n\n$<4 x_{n+1}... | [
"$3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
705 | 对 $1,2, \cdots, n\left(n \in \mathbf{N}_{+}, n>2\right.$ 012)的任意一个排列 $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, 定义:
$$
L=\sum_{i=1}^{n}\left|x_{i}-\sqrt{3} x_{i+1}\right|
$$
其中, $x_{n+1}=x_{1}$. 试求 $L_{\text {max }} 、 L_{\text {min }}$ 及取得最大值时的所有排列 $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ 的个数. | [
"显然, $L$ 为 $i 、 \\sqrt{3} i(i=1,2, \\cdots, n)$ 中 $n$个取正号、 $n$ 个取负号的数之和.\n\n为使 $L$ 取最大值, 则上述 $2 n$ 个数中较大的 $n$ 个数取正号, 较小的 $n$ 个数取负号.\n\n设 $\\sqrt{3} i(i=1,2, \\cdots, a)$ 及 $j(j=1,2, \\cdots, b)$\n\n取负号,其余取正号. 则\n\n$\\left\\{\\begin{array}{l}a+b=n, \\\\ \\sqrt{3}(a+1)>b, \\\\ b+1>\\sqrt{3} a\\end{array} \\Rightarrow... | [
"$L_{\\max }=\\sqrt{3}\\left[\\frac{n(n+1)}{2}-a(a+1)\\right]+\\left[\\frac{n(n+1)}{2}-b(b+1)\\right], L_{\\min }=(\\sqrt{3}-1) \\frac{n(n+1)}{2}+4-2 \\sqrt{3}, n(n-a-1) ! \\cdot a !$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Combinatorics | Math | Chinese |
706 | 若正实数 $x 、 y$ 满足 $x^{3}+y^{3}+3 x y=1$, 求 $\left(x+\frac{1}{x}\right)^{3}+\left(y+\frac{1}{y}\right)^{3}$ 的最小值。 | [
"由 $0=x^{3}+y^{3}-1+3 x y=(x+y-1)\\left((x-y)^{2}+x y+x+y+1\\right)$,\n\n$(x-y)^{2}+x y+x+y+1>0$,\n\n则 $x+y-1=0 \\Rightarrow x+y=1$.\n\n设 $f(t)=\\left(t+\\frac{1}{t}\\right)^{3}(t>0)$. 则\n\n$f^{\\prime}(t)=3\\left(t+\\frac{1}{t}\\right)^{2}\\left(1-\\frac{1}{t^{2}}\\right)$,\n\n$f^{\\prime \\prime}(t)=3\\left(2\\le... | [
"$\\frac{125}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
707 | 已知单位正方体 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 的棱 $A B 、 A_{1} D_{1} 、 A_{1} B_{1} 、 B C$ 的中点分别为 $L 、 M 、 N$ 、 $K$. 求四面体 $L M N K$ 的内切球半径。 | [
"设四面体 $L M N K$ 的内切球半径为 $r$.\n\n因为 $\\angle N L K=\\angle L N M=90^{\\circ}$, 所以,\n\n$S_{\\triangle L M N}=S_{\\triangle L N K}=\\frac{1}{2} \\times 1 \\times \\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{2}}{4}$.\n\n又 $\\angle M L K=\\angle M N K=90^{\\circ}$, 则\n\n$S_{\\triangle L M K}=S_{\\triangle M N K}=\\frac{1}{2} \\ti... | [
"$\\frac{\\sqrt{3}-\\sqrt{2}}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
708 | 已知 $P$ 为双曲线 $\Gamma: \frac{x^{2}}{463^{2}}-\frac{y^{2}}{389^{2}}=1$ 上一点, 过点 $P$ 作直线 $l$, 与双曲线 $\Gamma$ 的两条渐近线分别交于点 $A 、 B$. 若 $P$ 为线段 $A B$ 的中点, $O$ 为坐标原点, 求$S_{\triangle O A B}$。 | [
"设双曲线 $\\Gamma$ 的两条渐近线为\n\n$l_{1}: y=\\frac{389}{463} x, l_{2}: y=-\\frac{389}{463} x ;$\n\n$l$ 与 $l_{1} 、 l_{2}$ 分别交于点 $A 、 B$.\n\n设 $P\\left(x_{0}, y_{0}\\right), A\\left(x_{1}, \\frac{389}{463} x_{1}\\right)$,\n\n$B\\left(x_{2},-\\frac{389}{463} x_{2}\\right)$.\n\n则 $x_{0}=\\frac{x_{1}+x_{2}}{2}, y_{0}=\\frac{38... | [
"$180107$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
709 | 对于任意正实数 $x 、 y$,求
$$
\frac{(x y+x+y)(x+y+1)}{(x+y)(x+1)(y+1)}
$$
的取值范围。 | [
"设 $f(x, y)=\\frac{(x y+x+y)(x+y+1)}{(x+y)(x+1)(y+1)}$. 则\n\n$f(x, y)=1+\\frac{x y}{(x+y)(x+1)(y+1)}>1$.\n\n当固定 $y$,且 $x \\rightarrow 0$ 时, $f(x, y) \\rightarrow 1$.\n\n由均值不等式得\n\n$x+y \\geqslant 2 \\sqrt{x y}, x+1 \\geqslant 2 \\sqrt{x}, y+1 \\geqslant 2 \\sqrt{y}$.\n\n则 $f(x, y) \\leqslant 1+\\frac{x y}{2 \\sqrt{... | [
"$\\left(1, \\frac{9}{8}\\right]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | Chinese |
710 | 在平面直角坐标系中,已知 $A(13,43)$, $B(-36,19 \sqrt{2}), C(18,-11 \sqrt{14})$. 求 $\triangle A B C$ 的垂心 $H$ 的坐标。 | [
"设 $A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right), C\\left(x_{3}, y_{3}\\right)$.\n\n因为 $x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}=x_{3}^{2}+y_{3}^{2}=2018$, 所以, $\\triangle A B C$ 的外心为坐标原点 $O$.\n\n因此, $H\\left(x_{1}+x_{2}+x_{3}, y_{1}+y_{2}+y_{3}\\right)$.\n\n故 $H(-5,43+19 \\sqrt{2}-11 \\sqrt{14})$."
] | [
"$(-5,43+19 \\sqrt{2}-11 \\sqrt{14})$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Tuple | null | Open-ended | Geometry | Math | Chinese |
711 | 设函数
$f(x)=\frac{x(1-x)}{x^{3}-x+1}(0<x<1)$.
若 $f(x)$ 的最大值为 $f\left(x_{0}\right)$, 求 $x_{0}$的大小。 | [
"令 $f^{\\prime}(x)=\\frac{x^{4}-2 x^{3}+x^{2}-2 x+1}{\\left(x^{3}-x+1\\right)^{2}}=\\frac{\\left(x^{2}-(\\sqrt{2}+1) x+1\\right)\\left(x^{2}+(\\sqrt{2}-1) x+1\\right)}{\\left(x^{3}-x+1\\right)^{2}}=0$.\n\n则 $\\left(x^{2}-(\\sqrt{2}+1) x+1\\right)\\left(x^{2}+(\\sqrt{2}-1) x+1\\right)=0$.\n\n由于 $x^{2}+(\\sqrt{2}-1) ... | [
"$\\frac{\\sqrt{2}+1-\\sqrt{2 \\sqrt{2}-1}}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
712 | 若对于任意正整数 $n \geqslant 3$, 均有
$\sum_{i=1}^{n} \frac{1}{n+i}+\frac{5}{12} \log _{a}(a-1)>\frac{1}{5}$,
求实数 $a$ 的取值范围。 | [
"设 $f(n)=\\sum_{i=1}^{n} \\frac{1}{n+i}(n \\in \\mathbf{Z}, n \\geqslant 3)$. 则\n\n$f(n+1)-f(n)$\n\n$=\\sum_{i=1}^{n+1} \\frac{1}{n+1+i}-\\sum_{i=1}^{n} \\frac{1}{n+i}$\n\n$=\\frac{1}{2 n+1}+\\frac{1}{2 n+2}-\\frac{1}{n+1}$\n\n$=\\frac{1}{2 n+1}-\\frac{1}{2 n+2}>0$,\n\n即 $f(n)$ 严格递增.\n\n于是, 对任意的正整数 $n \\geqslant 3$... | [
"$\\left(\\frac{1+\\sqrt{5}}{2},+\\infty\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | Chinese |
713 | 已知正整数 $a_{1}, a_{2}, \cdots, a_{2018}$ 满足
$a_{1}<a_{2}<\cdots<a_{2018}$.
对于 $i=1,2, \cdots, 2018, b_{i}$ 为 $a_{1}, a_{2}, \cdots$, $a_{2018}$ 中不超过 $i$ 的正整数个数. 求
$\frac{\sum_{k=1}^{2018} a_{k}+\sum_{k=1}^{a_{2018}} b_{k}}{a_{2018}+1}$的值。 | [
"当 $k<a_{1}$ 时, $b_{k}=0$;\n\n当 $a_{j} \\leqslant k<a_{j+1}(j=1,2, \\cdots, 2017)$ 时,\n\n$b_{k}=j$;\n\n当 $k=a_{2018}$ 时, $b_{k}=2018$.\n\n故 $\\sum_{k=1}^{a_{2018}} b_{k}=\\sum_{j=1}^{2017} j\\left(a_{j+1}-a_{j}\\right)+2018$\n\n$=-\\sum_{k=1}^{2018} a_{k}+2018\\left(a_{2018}+1\\right)$.\n\n因此, $\\frac{\\sum_{k=1}^{... | [
"$2018$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
714 | 已知
椭圆 $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$
的长轴的左、右端点分别为 $A 、 B$, 左、右焦点分别为 $F_{1} 、 F_{2}$. 若存在椭圆 $\Gamma$ 上的点 $P$, 使得 $P F_{1} 、 P F_{2}$ 三等分 $\angle A P B$, 在所有满足上述条件的椭圆 $\Gamma$ 中, 试求不同的离心率 $e$ 的个数. | [
"由 $S_{\\triangle P A F_{1}}=S_{\\triangle P B F_{2}}$, 得\n\n$P A \\cdot P F_{1}=P B \\cdot P F_{2}$.\n\n故 $P$ 必为短轴的端点, 不妨设点 $P(0, b)$.\n\n由直线 $P A 、 P F_{1}$ 的斜率分别为 $\\frac{b}{a} 、 \\frac{b}{c}$, 知\n\n$\\tan \\angle A P F_{1}=\\frac{\\frac{b}{c}-\\frac{b}{a}}{1+\\frac{b}{c} \\cdot \\frac{b}{a}}=\\frac{b(a-c)}{a c+... | [
"$1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
715 | 对于任意实数 $a_{1}, a_{2}, \cdots, a_{5}$ $\in[0,1]$, 求 $\prod_{1 \leqslant i<j \leqslant 5}\left|a_{i}-a_{j}\right|$ 的最大值. | [
"设 $f\\left(a_{1}, a_{2}, \\cdots, a_{5}\\right)=\\prod_{1 \\leqslant i<j \\leqslant 5}\\left|a_{i}-a_{j}\\right|$.\n\n由于 $f\\left(a_{1}, a_{2}, \\cdots, a_{5}\\right)$ 的最大值必在 $a_{1}$, $a_{2}, \\cdots, a_{5}$ 互不相等时取到，不妨设\n\n$a_{1}>a_{2}>\\cdots>a_{5}$.\n\n要使 $f\\left(a_{1}, a_{2}, \\cdots, a_{5}\\right)$ 取到最大值,必有 $... | [
"$\\frac{3 \\sqrt{21}}{38416}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
717 | 已知双曲线 $C: \frac{x^{2}}{4}-y^{2}=1$, 过点 $M(1,-1)$的直线 $l$ 与双曲线 $C$ 的右支交于 $A 、 B$ 两点,与 $x$ 轴交于点 $N$. 设
$\overrightarrow{M A}=\lambda_{1} \overrightarrow{A N}, \overrightarrow{M B}=\lambda_{2} \overrightarrow{B N}\left(\lambda_{1}, \lambda_{2} \in \mathbf{R}\right)$.
求 $\frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}$ 的取值范围. | [
"易知, 直线的斜率存在, 设为 $k$.\n\n则直线 $l: y+1=k(x-1), N\\left(\\frac{1}{k}+1,0\\right)$.\n\n设 $A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right)$.\n\n将直线 $l$ 代人双曲线 $C$ 的方程得\n\n$\\left(1-4 k^{2}\\right) x^{2}+8 k(k+1) x-4(k+1)^{2}-4=0$.\n\n由方程(1)有两个不同的正根 $x_{1} 、 x_{2}$, 知\n\n$\\left\\{\\begin{array}{l}\\Delta>0, \\... | [
"$(-\\infty,-\\frac{74}{35}) \\cup(2,+\\infty)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Geometry | Math | Chinese |
718 | 定义在区间 $[1,2017]$ 上的函数 $f(x)$满足 $f(1)=f(2017)$, 且对于任意的 $x 、 y \in$ $[1,2017]$, 均有 $|f(x)-f(y)| \leqslant 2|x-y|$. 若实数 $m$ 满足对于任意的 $x 、 y \in[1,2017]$,均有 $|f(x)-f(y)| \leqslant m$, 求 $m$ 的最小值. | [
"先证明: 对于任意的 $x 、 y \\in[1,2017]$,均有 $|f(x)-f(y)| \\leqslant 2016$.\n\n事实上,若 $|x-y| \\leqslant 1008$, 则\n\n$|f(x)-f(y)| \\leqslant 2|x-y| \\leqslant 2016 ;$\n\n若 $|x-y|>1008$, 不妨假设 $1 \\leqslant x<y \\leqslant$ 2017 , 则\n\n$$\n\\begin{aligned}\n& |f(x)-f(y)| \\\\\n& =|f(x)-f(1)+f(2017)-f(y)| \\\\\n& \\leqslant|f(x)-... | [
"$2016$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
719 | 在平面直角坐标系中, 已知抛物线 $\Gamma: y=a x^{2}$ ( $a$ 为给定的正实数). 对于给定的正实数 $b$, 若以 $P(0, b)$ 为圆心、变量 $r(r>0)$为半径的圆与抛物线 $\Gamma$ 有四个交点 $A 、 B 、 C$ 、 $D$, 求以 $A 、 B 、C 、 D$ 为顶点的四边形面积 $S$ 的最大值. | [
"设以 $P(0, b)$ 为圆心、变量 $r$ 为半径的圆是 $\\Gamma_{r}$, 不妨假设点 $A 、 B 、 C 、 D$ 按顺时针的次序排列, 且 $A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right)$ 在第一象限. 则 $x_{1}>x_{2}>0, y_{1}>y_{2}>0$.\n\n由于圆 $\\Gamma_{r}$ 的方程为 $x^{2}+(y-b)^{2}=r^{2}$,与 $y=a x^{2}$ 联立得\n\n$$\ny^{2}+\\left(\\frac{1}{a}-2 b\\right) y+b^{2}-r^{2}=0 \\t... | [
"$\\frac{4 \\sqrt{6}(2 a b-1)^{\\frac{3}{2}}}{9 a^{2}}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Geometry | Math | Chinese |
721 | 数列 $\left\{x_{n}\right\}$ 满足
$x_{0}=1, x_{1}=6$,
$x_{2}=x_{1}+\sin x_{1}=5.72 \cdots$,
$x_{3}=x_{2}+\cos x_{2}=6.56 \cdots$.
一般地，
若 $x_{n} \geqslant x_{n-1}$, 则 $x_{n+1}=x_{n}+\sin x_{n}$;
若 $x_{n}<x_{n-1}$, 则 $x_{n+1}=x_{n}+\cos x_{n}$.
对任意的 $n \in \mathbf{Z}_{+}$, 求最小常数 $c$ 满足$x_{n}<c$. | [
"首先指出: 若 $x<3 \\pi$, 则\n\n$x+\\sin x=x+\\sin (3 \\pi-x)<x+(3 \\pi-x)=3 \\pi$.\n\n下面证明: $x_{n}<3 \\pi\\left(n \\in \\mathbf{Z}_{+}\\right)$.\n\n若不然,存在 $k \\in \\mathbf{Z}_{+}$, 使得\n\n$x_{k} \\geqslant 3 \\pi, x_{k-1}<3 \\pi$,\n\n则 $x_{k}=x_{k-1}+\\cos x_{k-1}$.\n\n故 $x_{k-1}=x_{k}-\\cos x_{k-1} \\geqslant 3 \\pi-1>\... | [
"$3 \\pi$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | Chinese |
722 | 某国足协规定: 在联赛中比赛双方胜一场得 $a$ 分, 平一场得 $b$ 分,负一场得 0 分, 其中, 实数 $a>b>0$. 若一支球队经过 $n$场比赛后总得分恰有 2015 种可能,求 $n$ 的最小值. | [
"假设某支球队在这 $n$ 场比赛中胜 $x$ 场、平 $y$ 场、负 $z$ 场, 其中, $x 、 y 、 z \\in \\mathbf{N}$, 且 $x+y+z=n$. 则该队在经过 $n$ 场比赛后总得分为 $a x+b y(0 \\leqslant x+y \\leqslant n)$.\n\n故要考虑该球队经过 $n$ 场比赛后总得分有多少种可能, 只需考虑 $a x+b y$ 的取值有多少种可能.\n\n(1) 若 $\\frac{a}{b} \\in \\mathbf{R} \\backslash \\mathbf{Q}$, 则\n\n$$\na x+b y=a x^{\\prime}+b y^{\\pr... | [
"$62$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
728 | 已知集合
$$
\begin{aligned}
& M=\left\{(x, y) \mid y \geqslant \frac{1}{4} x^{2}\right\}, \\
& N=\left\{(x, y) \mid y \leqslant-\frac{1}{4} x^{2}+x+7\right\}, \\
& D_{r}\left(x_{0}, y_{0}\right)=\left\{(x, y) \mid\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2} \leqslant r^{2}\right\} .
\end{aligned}
$$
试求最大的 $r$, 使得 $D_{r}\left(x_{0}, y_{0}\right) \subset M \cap N$. | [
"设 $f(x)=\\frac{1}{4} x^{2}$ ，\n\n$g(x)=-\\frac{1}{4} x^{2}+x+7=-\\frac{1}{4}(x-2)^{2}+8$.\n\n则 $M \\cap N \\neq \\varnothing \\Leftrightarrow f(x) \\leqslant g(x)$ 有实解\n\n$\\Leftrightarrow x^{2}-2 x-14 \\leqslant 0$ 有实解.\n\n故 $M \\cap N \\neq \\varnothing$.\n\n在平面直角坐标系中, $y=f(x)$ 与 $y=$ $g(x)$ 的顶点分别为 $O(0,0) 、 A(2... | [
"$\\sqrt{\\frac{25-5 \\sqrt{5}}{2}}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | Chinese |
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