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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
357 | 已知动圆 $P$ 在圆 $E:(x+1)^{2}+y^{2}=\frac{1}{4}$ 外部且与圆 $E$ 相切, 同时还在圆 $F:(x-1)^{2}+y^{2}=\frac{49}{4}$ 内部与圆 $F$ 相切,
求动圆圆心 $P$ 的轨迹 $C$ 的方程; | [
"设 $P(x, y)$, 动圆 $P$ 半径为 $r$ 由题意 $\\left\\{\\begin{array}{l}|P E|=r+\\frac{1}{2} \\\\ |P F|=\\frac{7}{2}-r\\end{array}\\right.$\n\n$$\n\\therefore|P E|+|P F|=4>|E F|=2\n$$\n\n$\\therefore$ 轨迹 $C$ 是以 $E, F$ 为焦点的椭圆, 设其方程为: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ 则 $\\left\\{\\begin{array}{l}a=2 \\\\ b=\\... | [
"$\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=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 | Plane Geometry | Math | Chinese |
358 | 已知动圆 $P$ 在圆 $E:(x+1)^{2}+y^{2}=\frac{1}{4}$ 外部且与圆 $E$ 相切, 同时还在圆 $F:(x-1)^{2}+y^{2}=\frac{49}{4}$ 内部与圆 $F$ 相切,
设 $A(2,0)$, 过点 $A$ 的直线 $l$ 与轨迹 $C$ 交于点 $B$ (点 $B$ 不在 $X$ 轴上),垂直于 $l$ 的直线与 $l$ 交于点 $M$, 与 $y$ 轴交于点 $H$ .若 $B F \perp H F$, 且 $\angle M O A \leq \angle M A O$, 求直线 $l$ 斜率的取值范围. | [
"设 $P(x, y)$, 动圆 $P$ 半径为 $r$ 由题意 $\\left\\{\\begin{array}{l}|P E|=r+\\frac{1}{2} \\\\ |P F|=\\frac{7}{2}-r\\end{array}\\right.$\n\n$$\n\\therefore|P E|+|P F|=4>|E F|=2\n$$\n\n$\\therefore$ 轨迹 $C$ 是以 $E, F$ 为焦点的椭圆, 设其方程为: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ 则 $\\left\\{\\begin{array}{l}a=2 \\\\ b=\\... | [
"$\\left(-\\infty,-\\frac{\\sqrt{6}}{4}\\right] \\cup\\left[\\frac{\\sqrt{6}}{4},+\\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 | Plane Geometry | Math | Chinese |
359 | 设函数 $f(x)=a \ln x+\frac{2 a}{x}-\frac{e^{x}}{x^{2}}$.
若 $a \leq e$, 求 $f(x)$ 的最大值. | [
"$$\nf^{\\prime}(x)=\\frac{a}{x}-\\frac{2 a}{x^{2}}-\\frac{x e^{x}-2 e^{x}}{x^{3}}=\\frac{a x(x-2)-e^{x}(x-2)}{x^{3}}=\\frac{(x-2)\\left(a x-e^{x}\\right)}{x^{3}} \\quad(x>0)\n$$\n\n(1)当 $a \\leq 0$ 时 $a x-e^{x}<0$ 令 $f^{\\prime}(x)>0$ 则 $0<x<2$; 令 $f^{\\prime}(x)<0$ 则 $x>2$;\n\n$\\therefore f(x)$ 在 $(0,2)$ 单调递增, $... | [
"$a \\ln 2+a-\\frac{e^{2}}{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 | Elementary Functions | Math | Chinese |
361 | 在直角坐标系 $x O y$ 中, 以坐标原点为极点, $x$ 轴的正半轴为极轴建立极坐标系, 半圆 $C$ 的极坐标方程为: $\rho=2 \sin \vartheta, \vartheta \in\left(\frac{\pi}{4}, \frac{3}{4} \pi\right)$.
直线 $l$ 与两坐标轴的交点分别为 $A, B$, 其中 $A(0,-2)$, 点 $D$ 在半圆 $C$上, 且直线 $C D$ 的倾斜角是直线 $l$ 倾斜角 2 倍.若 $\triangle A B D$ 的面积为 $1+\sqrt{3}$,求点 $D$ 的直角坐标. | [
"设 $l$ 的倾斜角为 $\\alpha$, 则 $C D$ 的倾斜角为 $2 \\alpha\\left(0<\\alpha<\\frac{\\pi}{2}\\right)$\n\n$\\therefore D(\\cos 2 \\alpha, 1+\\sin 2 \\alpha)$\n\n$\\because l$ 的方程为 $y=x \\tan \\alpha-2$ 即 $x \\tan \\alpha-y-2=0$\n\n$\\therefore$ 令 $y=0$, 则 $x=\\frac{2}{\\tan \\alpha} \\therefore B\\left(\\frac{2}{\\tan \\alpha},... | [
"$\\left(-\\frac{1}{2}, 1+\\frac{\\sqrt{3}}{2}\\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 | Polar Coordinates and Parametric Equations | Math | Chinese |
362 | 已知函数 $f(x)=2|x-1|-a, g(x)=-|2 x+m| \quad(a \in R, m \in Z)$.若关于 $x$ 不等式 $g(x) \geq-1$ 的整数解有且仅有一个值为$-4$.
求 $m$ 的值. | [
"$\\because g(x) \\geq-1 \\therefore-|2 x+m| \\geq-1 \\therefore|2 x+m| \\leq 1$\n\n$\\therefore \\frac{-m-1}{2} \\leq x \\leq \\frac{-m+1}{2}$\n\n$\\because$ 不等式 $g(x) \\geq-1$ 的整数解有且仅有一个值为 $-4$.\n\n$\\therefore-5<\\frac{-m-1}{2} \\leq-4 \\leq \\frac{-m+1}{2}<-3$\n\n$\\therefore 7<m<9 \\quad \\because m \\in Z \\q... | [
"$8$"
] | 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 | Inequality | Math | Chinese |
363 | 已知函数 $f(x)=2|x-1|-a, g(x)=-|2 x+m| \quad(a \in R, m \in Z)$.若关于 $x$ 不等式 $g(x) \geq-1$ 的整数解有且仅有一个值为$-4$.
若函数 $y=f(x)$ 的图象恒在函数 $y=\frac{1}{2} g(x)$ 的上方, 求 $a$ 的取值范围. | [
"$\\because$ 函数 $y=f(x)$ 的图象恒在函数 $y=\\frac{1}{2} g(x)$ 的上方\n\n$\\therefore f(x)-\\frac{1}{2} g(x)>0$\n\n$\\therefore a<2|x-1|+|x+4|$ 对任意 $x \\in R$ 恒成立\n\n$\\because 2|x-1|+|x+4|=|x-1|+|x-1|+|x+4| \\geq|x-1|+5 \\geq 5$, 当且仅当 $x=1$ 时, “=” 成立\n\n$\\therefore a<5$."
] | [
"$(-\\infty, 5)$"
] | 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 | Interval | null | Open-ended | Inequality | Math | Chinese |
364 | 已知数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 且 $S_{n}=n^{2}+n, n \in N^{*}$, 数列 $\left\{b_{n}\right\}$ 满足 $b_{n}=2^{a}, n \in N^{*}$.
求 $a_{n}, b_{n}$. | [
"当 $n=1$ 时, $a_{1}=S_{1}=2$\n\n当 $n \\geq 2$ 时, $a_{n}=S_{n}-S_{n-1}=\\left(n^{2}+n\\right)-\\left[(n-1)^{2}+(n-1)\\right]=2 n$\n\n$\\therefore a_{n}=2 n(n \\geq 1), b_{n}=2^{2 n}=4^{n}$."
] | [
"$2 n, 4^{n}$"
] | 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 | true | null | Expression | null | Open-ended | Sequence | Math | Chinese |
365 | 已知数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 且 $S_{n}=n^{2}+n, n \in N^{*}$, 数列 $\left\{b_{n}\right\}$ 满足 $b_{n}=2^{a}, n \in N^{*}$.
求数列 $\left\{a_{n} \cdot b_{n}\right\}$ 的前 $n$ 项和 $T_{n}$. | [
"当 $n=1$ 时, $a_{1}=S_{1}=2$\n\n当 $n \\geq 2$ 时, $a_{n}=S_{n}-S_{n-1}=\\left(n^{2}+n\\right)-\\left[(n-1)^{2}+(n-1)\\right]=2 n$\n\n$\\therefore a_{n}=2 n(n \\geq 1), b_{n}=2^{2 n}=4^{n}$.\n\n$a_{n} \\cdot b_{n}=2 n \\cdot 4^{n}$\n\n设 $\\left\\{n \\cdot 4^{n}\\right\\}$ 的前 $n$ 项和为 $H_{n}$\n\n$H_{n}=1 \\times 4^{1}+2... | [
"$\\frac{8}{9}+\\frac{2(3 n-1)}{9} \\cdot 4^{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 |
366 | 根据某省的高考改革方案, 考生应在 3 门理科学科 (物理、化学、生物) 和 3 门文科学科 (历史、政治、地理) 6 门学科中选择 3 门学科数加考试.
假设考生甲理科成绩较好，决定至少选择两门理科学科，那么该同学的选科方案有几种？ | [
"选科方案中两门理科的情况有: 物化历, 物化政, 物化地, 物生历, 物生政, 物生地,化生历, 化生政, 化生地;\n\n选科方案中三门理科的情况有: 物化生;\n\n因此至少选择两门理科学科的选科方案有 10 种."
] | [
"$10$"
] | 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 |
367 | 根据某省的高考改革方案, 考生应在 3 门理科学科 (物理、化学、生物) 和 3 门文科学科 (历史、政治、地理) 6 门学科中选择 3 门学科数加考试.
假设每门学科被选中的概率是相同的，求选科方案中有包括物理但不包括历史的概率. | [
"选科方案的情况有: 物化生, 物化历, 物化政, 物化地, 物生历, 物生政, 物生地,化生历、化生政, 化生地, 历政地, 历政物, 历政化, 历政生, 历地物, 历地化, 历地生,政地物, 政地化, 政地生, 共 20 种.\n\n选科方案中有包括物理但不包括历史的情况有: 物化生, 物化政, 物化地, 物生政, 物生地,物政地, 共 6 种.\n\n所以概率 $P=\\frac{6}{20}=0.3$."
] | [
"$0.3$"
] | 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 |
370 | 已知抛物线 $C: y^{2}=2 p x$ 经过点 $A(1,2)$, 经过点 $(1,1)$ 且斜率为 $k$ 的直线与抛物线 $C$ 交于不同的两点 $P$, $Q$ (均异于点 $A$ ), 设直线 $A P$ 的斜率为 $k_{1}$, 直线 $A Q$ 的斜率为 $k_{2}$.
当 $k=1$ 时, 求 $k_{1}+k_{2}$ 的值. | [
"由题意知, $4=2 p \\times 1, \\therefore p=2$, 抛物线 $C: y^{2}=4 x$\n\n当 $k=1$ 时, 直线 $l: y=x$, 与抛物线联立后得 $x^{2}=4 x, x_{1}=0, x_{2}=4$, 不妨设\n\n$P(0,0), Q(4,4), \\quad k_{1}=2, \\quad k_{2}=\\frac{2}{3}, \\quad \\therefore k_{1}+k_{2}=\\frac{8}{3}$\n."
] | [
"$\\frac{8}{3}$"
] | 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 |
371 | 已知抛物线 $C: y^{2}=2 p x$ 经过点 $A(1,2)$, 经过点 $(1,1)$ 且斜率为 $k$ 的直线与抛物线 $C$ 交于不同的两点 $P$, $Q$ (均异于点 $A$ ), 设直线 $A P$ 的斜率为 $k_{1}$, 直线 $A Q$ 的斜率为 $k_{2}$.
求 $k_{1}^{2}+k_{2}^{2}$ 的取值范围. | [
"由题意得：\n\n$4=2 p \\times 1, \\therefore p=2$, 抛物线 $C: y^{2}=4 x$\n\n直线 $l: y=k x+1-k$, 与抛物线联立 $\\left\\{\\begin{array}{c}y=k x+1-k \\\\ y^{2}=4 x\\end{array}\\right.$, 得\n\n$k y^{2}-4 y+4-4 k=0$.\n\n设 $P\\left(x_{1}, y_{1}\\right), Q\\left(x_{2}, y_{2}\\right)$\n\n$y_{1}+y_{2}=\\frac{4}{k}, \\quad y_{1} \\cdot y_{2... | [
"$\\left[\\frac{5}{3},+\\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 | Plane Geometry | Math | Chinese |
373 | 设函数 $f(x)=e^{x}-x^{2} e^{x}-a x-1$.
若对于任意的 $x_{1}, x_{2} \in(1,+\infty), x_{1} \neq x_{2}$, 都有 $\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}<x_{1}+x_{2}$ 恒成立, 求实数 $a$ 的取值范围. | [
"不妨设 $x_{1}>x_{2}>1$, 则 $f\\left(x_{1}\\right)-f\\left(x_{2}\\right)<x_{1}^{2}-x_{2}^{2}$, 即 $f\\left(x_{1}\\right)-x_{1}^{2}<f\\left(x_{2}\\right)-x_{2}^{2}, $\n\n设 $g(x)=e^{x}-x^{2} e^{x}-a x-1-x^{2}, g^{\\prime}(x)=e^{x}-\\left(2 x e^{x}+x^{2} e^{x}\\right)-a-2 x \\leq 0$ 在 $x \\in(1,+\\infty)$ 上恒成立.\n\n设 $h(x)=... | [
"$[-2 e-2, +\\infty)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
374 | 在直角坐标系$xOy$中，直线$l$的参数方程为$\left\{\begin{array}{l}x=2 + t \\ y=4 - 2t\end{array}\right.$,（$t$为参数） 曲线$C$的参数方程为$\left\{\begin{array}{l}x=3 \cos \theta \\ y=2 \sin \theta\end{array}\right.$（$\theta$为参数）.
写出直线 $l$ 与曲线 $C$ 的普通方程; | [
"直线 $l: 2 x+y-8=0 ;$\n\n由线 $C: \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$."
] | [
"$2 x+y-8=0 , \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$"
] | 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 | true | null | Equation | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
375 | 在直角坐标系$xOy$中，直线$l$的参数方程为$\left\{\begin{array}{l}x=2 + t \\ y=4 - 2t\end{array}\right.$,（$t$为参数） 曲线$C$的参数方程为$\left\{\begin{array}{l}x=3 \cos \theta \\ y=2 \sin \theta\end{array}\right.$（$\theta$为参数）.
过曲线 $C$ 上任意一点 $A$ 作与 $l$ 夹角为 $\alpha$ 的直线, 交直线 $l$ 于点 $B$, 若 $\tan \alpha=\frac{1}{2}$, 求 $|A B|$ 的最大值和最小值. | [
"直线 $l: 2 x+y-8=0 ;$\n\n由线 $C: \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$.\n\n椭圆 $C$ 上的点到直线 $l$ 的距离 $d=\\frac{|6 \\cos \\theta+2 \\sin \\theta-8|}{\\sqrt{2^{2}+1^{2}}}=\\frac{|\\sqrt{40} \\sin (\\theta+\\varphi)-8|}{\\sqrt{5}}$, 其中 $\\varphi$ 是第一象限角, 且 $\\tan \\varphi=3$,\n\n$\\therefore d_{\\min }=\\frac{|2 \\sqrt{10}-8... | [
"$8+2 \\sqrt{10}, 8-2 \\sqrt{10}$"
] | 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 | Polar Coordinates and Parametric Equations | Math | Chinese |
376 | 已知函数 $f(x)=|2 x+1|-|3 x-a|(a>0)$
当 $a=3$ 时, 求不等式 $f(x)>2$ 的解集. | [
"当 $a=3$ 时, $f(x)=|2 x+1|-|3 x-3|=\\left\\{\\begin{array}{l}-x+4, x \\geq 1 \\\\ 5 x-2,-\\frac{1}{2}<x<1 \\\\ x-4, x \\leq-\\frac{1}{2}\\end{array}\\right.$\n\n当 $x \\geq 1$ 时， $4-x>2, \\therefore x \\in[1,2)$\n\n当 $-\\frac{1}{2}<x<1$ 时, $5 x-2>2, \\therefore x \\in\\left(\\frac{4}{5}, 1\\right)$\n\n当 $x \\leq-\\fr... | [
"$\\left(\\frac{4}{5}, 2\\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 | Inequality | Math | Chinese |
377 | 已知函数 $f(x)=|2 x+1|-|3 x-a|(a>0)$
若 $f(x)$ 的图像与 $x$ 轴围成的三角形的面积大于 $\frac{5}{3}$, 求 $a$ 的取值范围. | [
"$\\because a>0$, 则\n\n$f(x)=|2 x+1|-|3 x-a|=\\left\\{\\begin{array}{l}-x+1+a, x \\geq \\frac{a}{3} \\\\ 5 x+1-a,-\\frac{1}{2}<x<\\frac{a}{3} \\\\ x-a-1, x \\leq-\\frac{1}{2}\\end{array}\\right.$\n\n今 $5 x_{1}+1-a=0 . \\therefore x_{1}=\\frac{a-1}{5}$\n\n今 $-x_{2}+1+a=0, \\therefore x_{2}=1+a$\n\n$x_{2}-x_{1}=\\fra... | [
"$(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 |
378 | 在 $\triangle \mathrm{ABC}$ 中, 角 $\mathrm{A}, \mathrm{B}, \mathrm{C}$ 对应的边分别为 $\mathrm{a}, \mathrm{b}, \mathrm{c}$, 且满足 $b \cos A-a \sin B=0$.
求角 $\mathrm{A}$ 的大小. | [
"$\\because b \\cos A-a \\sin B=0$\n\n由正弦定理得： $\\sin B \\cos A-\\sin A \\sin B=0$\n\n又 $\\because 0<\\mathrm{B}<\\pi \\quad \\therefore \\sin B \\neq 0, \\cos A-\\sin A=0$\n\n$\\therefore \\tan A=1$\n\n$\\because 0<\\mathrm{A}<\\pi$\n\n$\\therefore A=\\frac{\\pi}{4}$."
] | [
"$\\frac{\\pi}{4}$"
] | 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 | Trigonometric Functions | Math | Chinese |
379 | 在 $\triangle \mathrm{ABC}$ 中, 角 $\mathrm{A}, \mathrm{B}, \mathrm{C}$ 对应的边分别为 $\mathrm{a}, \mathrm{b}, \mathrm{c}$, 且满足 $b \cos A-a \sin B=0$.
已知 $b=\sqrt{2}, A=\frac{\pi}{4}, \triangle \mathrm{ABC}$ 的面积为 1 , 求边 $\mathrm{a}$. | [
"$\\because b=\\sqrt{2}, A=\\frac{\\pi}{4}, S_{\\triangle A B C}=1$\n\n$\\therefore \\frac{1}{2} b c \\sin A=1 \\quad$，得 $c=2$\n\n由余弦定理得, $a^{2}=b^{2}+c^{2}-2 b c \\cos A=2$\n\n得 $a=\\sqrt{2}$."
] | [
"$\\sqrt{2}$"
] | 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 | Trigonometric Functions | Math | Chinese |
382 | 双十一之后, 网购粉丝们期待的双十二已然到来, 为了解双十二消费者购物情况和电商的营业情况，做如下数据分析.据悉 12 月 12 日.有 2000 名网购者在某购物网站进行网购消费（消费金额不超过 1000 元），其中有女士 1100 名，男士:900 名，该购物网站为优化营销策略, 根据性别采用分层抽样的方法从这 2000 名网购者中抽取 200 名进行分析, 如表.（消费金额单位：元）
女士消费情况:
| 消费金额 | $(0,200)$ | $[200,400)$ | $[400,600)$ | $[600,800)$ | $[800,1000]$ |
| :------- | :---------- | :------------ | :------------ | :------------ | :------------- |
| 人数 | 10 | 25 | 35 | 30 | $x$ |
男士消费情况:
| 消费金额 | $(0,200)$ | $[200,400)$ | $[400,600)$ | $[600,800)$ | $[800,1000]$ |
| :------- | :---------- | :------------ | :------------ | :------------ | :------------- |
| 人数 | 15 | 30 | 25 | $y$ | 5 |
附:
$$
K^{2}=\frac{n(a d-b c)^{2}}{(a+\mathrm{b})(\mathrm{c}+\mathrm{d})(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{d})}, \quad n=a+b+c+d
$$
| $P\left(K^{2} \geq k_{0}\right)$ | 0.10 | 0.05 | 0.025 | 0.010 | 0.005 |
| :--------------------------------- | :---- | :---- | :---- | :---- | :---- |
| $k_{0}$ | 2.706 | 3.841 | 5.024 | 6.635 | 7.879 |
计算 $\mathrm{x}, \mathrm{y}$ 的值. | [
"依原意,女士应抽取 110 名,男士应抽取 90 名,故 $x=10, y=15$."
] | [
"$10, 15$"
] | 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 | true | null | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
383 | 双十一之后, 网购粉丝们期待的双十二已然到来, 为了解双十二消费者购物情况和电商的营业情况，做如下数据分析.据悉 12 月 12 日.有 2000 名网购者在某购物网站进行网购消费（消费金额不超过 1000 元），其中有女士 1100 名，男士:900 名，该购物网站为优化营销策略, 根据性别采用分层抽样的方法从这 2000 名网购者中抽取 200 名进行分析, 如表.（消费金额单位：元）
女士消费情况:
| 消费金额 | $(0,200)$ | $[200,400)$ | $[400,600)$ | $[600,800)$ | $[800,1000]$ |
| :------- | :---------- | :------------ | :------------ | :------------ | :------------- |
| 人数 | 10 | 25 | 35 | 30 | $x$ |
男士消费情况:
| 消费金额 | $(0,200)$ | $[200,400)$ | $[400,600)$ | $[600,800)$ | $[800,1000]$ |
| :------- | :---------- | :------------ | :------------ | :------------ | :------------- |
| 人数 | 15 | 30 | 25 | $y$ | 5 |
附:
$$
K^{2}=\frac{n(a d-b c)^{2}}{(a+\mathrm{b})(\mathrm{c}+\mathrm{d})(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{d})}, \quad n=a+b+c+d
$$
| $P\left(K^{2} \geq k_{0}\right)$ | 0.10 | 0.05 | 0.025 | 0.010 | 0.005 |
| :--------------------------------- | :---- | :---- | :---- | :---- | :---- |
| $k_{0}$ | 2.706 | 3.841 | 5.024 | 6.635 | 7.879 |
在抽出的 200 名且消费金额在 [800, 1000]（单位: 元）网购者中随机选出 2 名发放网购红包, 求选出的 2 名网购者都是男士的概率; | [
"消费金额在 $[800,1000]$ (单位:元)的网影者共有 15 名, 从中选出 2 名共有 105 种选法,若 2 名网则者都是男士,共有 10 种选法,所以选出的 2 名网则者都是男士的概率为 $\\frac{10}{105}=\\frac{2}{21} $."
] | [
"$\\frac{2}{21}$"
] | 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 |
385 | 已知椭圆 $\mathrm{M}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的两个顶点分别为 $\mathrm{A}(-\mathrm{a}, 0), \mathrm{B}(\mathrm{a}, 0)$,点 $\mathrm{P}$ 为椭圆上异于 $\mathrm{A}, \mathrm{B}$ 的点, 设直线 $\mathrm{PA}$ 的斜率为 $k_{1}$, 直线 $\mathrm{PB}$ 的斜率为 $k_{2}, k_{1} k_{2}=-\frac{1}{2}$ .
求椭圆 $C$ 的离心率; | [
"设 $P\\left(x_{0}, y_{0}\\right)$, 代入椭圆的方程有: $\\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1$,\n\n整理得: $y_{0}^{2}=-\\frac{b^{2}}{a^{2}}\\left(x_{0}^{2}-a^{2}\\right)$,\n\n又 $k_{1}=\\frac{y_{0}}{x_{0}+a}, k_{2}=\\frac{y_{0}}{x_{0}-a}$, 所以 $k_{1} k_{2}=\\frac{y_{0}^{2}}{x_{0}^{2}-a^{2}}=-\\frac{1}{2}$,\n\n联立两个方程... | [
"$\\frac{\\sqrt{2}}{2}$"
] | 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 | Plane Geometry | Math | Chinese |
386 | 已知椭圆 $\mathrm{M}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的两个顶点分别为 $\mathrm{A}(-\mathrm{a}, 0), \mathrm{B}(\mathrm{a}, 0)$,点 $\mathrm{P}$ 为椭圆上异于 $\mathrm{A}, \mathrm{B}$ 的点, 设直线 $\mathrm{PA}$ 的斜率为 $k_{1}$, 直线 $\mathrm{PB}$ 的斜率为 $k_{2}, k_{1} k_{2}=-\frac{1}{2}$ .
若 $\mathrm{b}=1$ ， 设直线 $l$ 与 $x$ 轴交于 $\mathrm{D}\left(-1, 0\right)$, 与椭圆交于 $M, N$ 两点, 求 $\triangle O M N$ 的面积的最大值. | [
"设 $P\\left(x_{0}, y_{0}\\right)$, 代入椭圆的方程有: $\\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1$,\n\n整理得: $y_{0}^{2}=-\\frac{b^{2}}{a^{2}}\\left(x_{0}^{2}-a^{2}\\right)$,\n\n又 $k_{1}=\\frac{y_{0}}{x_{0}+a}, k_{2}=\\frac{y_{0}}{x_{0}-a}$, 所以 $k_{1} k_{2}=\\frac{y_{0}^{2}}{x_{0}^{2}-a^{2}}=-\\frac{1}{2}$,\n\n联立两个方程... | [
"$\\frac{\\sqrt{2}}{2}$"
] | 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 | Plane Geometry | Math | Chinese |
388 | 设函数 $f(x)=\ln x+\frac{a-1}{x}, g(x)=a x-3$
当 $\mathrm{a}=1$ 时, 记 $h(x)=f(x) \cdot g(x)$, 存在整数 $\lambda$, 使得关于 $x$ 的不等式 $2 \lambda \geq h(x)$ 有解，请求出 $\lambda$ 的最小值. | [
"当 $a=1$ 时, $f(x)=\\ln x, g(x)=x-3, h(x)=(x-3) \\ln x$\n\n所以 $h^{\\prime}(x)=\\ln x+1-\\frac{3}{x}$ 单调递增,\n\n$h^{\\prime}\\left(\\frac{3}{2}\\right)=\\ln \\frac{3}{2}+1-2<0, h^{\\prime}(2)=\\ln 2+1-\\frac{3}{2}>0$.\n\n所以存在唯一 $x_{0} \\in\\left(\\frac{3}{2}, 2\\right)$, 使得 $h^{\\prime}\\left(x_{0}\\right)=0$, 即 $\\ln... | [
"$0$"
] | 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 | false | null | Numerical | null | Open-ended | Elementary Functions | Math | Chinese |
389 | 在直角坐标系 $x o y$ 中, 圆 $\mathrm{C}$ 的参数方程为 $\left\{\begin{array}{l}x=\cos \varphi \\ y=1+\sin \varphi\end{array}\right.$ ( $\varphi$ 为参数), 以 $O$ 为极点, $x$轴的非负半轴为极轴建立极坐标系.
半圆 $C$ (圆心为点 $C$) 的参数方程为 $\left\{\begin{array}{l}x=\cos \varphi \\ y=1+\sin \varphi\end{array} \varphi\right.$ 为参数, $\{\varphi \in(0, \pi)\}$.
求圆 $C$ 的普通方程. | [
"由圆 $C$ 的参数方程 $\\left\\{\\begin{array}{c}x=2 \\cos \\varphi \\\\ y=2+2 \\sin \\varphi\\end{array}\\right.$ ( $\\varphi$ 为参数) 知, 圆 $C$ 的圆心为 $(0,2)$,半径为 2 , 圆 $C$ 的普通方程为 $x^{2}+(y-2)^{2}=4$."
] | [
"$x^{2}+(y-2)^{2}=4$"
] | 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 | Equation | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
390 | 在直角坐标系 $x o y$ 中, 圆 $\mathrm{C}$ 的参数方程为 $\left\{\begin{array}{l}x=\cos \varphi \\ y=1+\sin \varphi\end{array}\right.$ ( $\varphi$ 为参数), 以 $O$ 为极点, $x$轴的非负半轴为极轴建立极坐标系.
半圆 $C$ (圆心为点 $C$) 的参数方程为 $\left\{\begin{array}{l}x=\cos \varphi \\ y=1+\sin \varphi\end{array} \varphi\right.$ 为参数, $\{\varphi \in(0, \pi)\}$.
直线 $l$ 的极坐标方程是 $2 \rho \sin \left(\theta+\frac{\pi}{6}\right)=5 \sqrt{3}$, 射线 $O\mathrm{M}: \theta=\frac{\pi}{6}$ 与圆 $C$ 的交点为 $O 、 \mathrm{P}$,与直线 $l$ 的交点为 $\mathrm{Q}$, 求线段 $\mathrm{PQ}$ 的长. | [
"由圆 $C$ 的参数方程 $\\left\\{\\begin{array}{c}x=2 \\cos \\varphi \\\\ y=2+2 \\sin \\varphi\\end{array}\\right.$ ( $\\varphi$ 为参数) 知, 圆 $C$ 的圆心为 $(0,2)$,半径为 2 , 圆 $C$ 的普通方程为 $x^{2}+(y-2)^{2}=4$.\n\n将 $x=\\rho \\cos \\theta, y=\\rho \\sin \\theta$ 代入 $x^{2}+(y-2)^{2}=4$. 得圆 $C$ 的极坐标方程为 $\\rho=4 \\sin \\theta$.\n\n设 $P\\le... | [
"$3$"
] | 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 | Polar Coordinates and Parametric Equations | Math | Chinese |
391 | 已知函数 $f(x)=|x-1|+|x-2|$.
求不等式 $f(x) \geq 3$ 的解集. | [
"$f(x)=|x-1|+|x-2|=\\begin{cases}-2 x+3 & (x \\leq 1) \\\\ 1 & (1<x<2) \\\\ 2 x-3 & (x \\geq 2)\\end{cases}$\n\n当 $x \\leq 1$ 时, 由 $-2 x+3 \\geq 3$, 得 $x \\leq 0$\n\n当 $1<x<2$ 时，由 $1 \\geq 3$, 得 $x \\in \\varnothing$\n\n当 $x \\geq 2$ 时, 由 $2 x-3 \\geq 3$, 得 $x \\geq 3$\n\n所以不等式 $f(x) \\geq 3$ 的解集为 $\\{x \\mid x \\l... | [
"$(-\\infty, 0] \\cup [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 | Inequality | Math | Chinese |
392 | 已知函数 $f(x)=|x-1|+|x-2|$.
若存在实数 $x$ 满足 $f(x) \leq-a^{2}+a+5$, 求实数 $a$ 的最大值. | [
"$\\because|x-1|+|x-2| \\geq|(x-1)-(x-2)|=1$.\n\n$\\therefore$ 依题意有 $-a^{2}+a+7 \\geq 1$, 即 $a^{2}-a-6 \\leq 0$\n\n解得 $-2 \\leq a \\leq 3$\n\n故 $a$ 的最大值为 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 | Inequality | Math | Chinese |
393 | 已知等比数列 $\left\{a_{n}\right\}$ 的各项均为正数, 并且 $2 a_{1}+3 a_{2}=33$, $a_{2} a_{4}=27 a_{3}$. 数列 $\left\{b_{n}\right\}$ 满足 $b_{n}=\log _{3} a_{n+1}, n \in N^{+}$.
求数列 $\left\{b_{n}\right\}$ 的通项公式. | [
"由 $2 a_{1}+3 a_{2}=33, a_{2} a_{4}=27 a_{3}$ 得 $a_{1}=3, q=3 \\therefore a_{n}=3^{n}$\n\n$b_{n}=\\log _{3} a_{n+1}$, 则 $b_{n}=n+1$."
] | [
"$b_{n}=n+1$"
] | 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 | Expression | null | Open-ended | Sequence | Math | Chinese |
394 | 已知等比数列 $\left\{a_{n}\right\}$ 的各项均为正数, 并且 $2 a_{1}+3 a_{2}=33$, $a_{2} a_{4}=27 a_{3}$. 数列 $\left\{b_{n}\right\}$ 满足 $b_{n}=\log _{3} a_{n+1}, n \in N^{+}$.
若 $c_{n}=a_{n} \cdot b_{n}$, 求数列 $\left\{c_{n}\right\}$ 的前 $n$ 项和 $S_{n}$. | [
"由 $2 a_{1}+3 a_{2}=33, a_{2} a_{4}=27 a_{3}$ 得 $a_{1}=3, q=3 \\therefore a_{n}=3^{n}$\n\n$b_{n}=\\log _{3} a_{n+1}$, 则 $b_{n}=n+1$.\n\n$c_{n}=(n+1) 3^{n}$\n\n记 $S_{n}=2 \\times 3+3 \\times 3^{2}+4 \\times 3^{3}+\\cdots+(n+1) 3^{n}$\n\n$3 S_{n}=2 \\times 3^{2}+3 \\times 3^{3}+4\\times3^{4} + \\cdots+(n+1)3^{n+1}$\n... | [
"$S_{n}=\\frac{(2 n+1) 3^{n+1}}{4}-\\frac{3}{4}$"
] | 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 | Sequence | Math | Chinese |
398 | 19. (12 分) 为了对中考成绩进行分析, 某中学从分数在 70 分(满分 100 分)以上的全体同学中随机抽出 8 位, 他们的数学、物理分数对应如下表:
| 数学$X$ | 75 | 76 | 76 | 77 | 83 | 84 | 84 | 85 |
| :-------- | :- | :- | :- | :- | :- | :- | :- | :- |
| 物理$Y$ | 77 | 78 | 81 | 81 | 81 | 82 | 83 | 85 |
附:
| $P\left(K^{2} \geq k_{0}\right)$ | 0.10 | 0.05 | 0.010 | 0.005 |
| :--------------------------------: | :---: | :---: | :---: | :---: |
| $k_{0}$ | 2.706 | 3.841 | 6.635 | 7.879 |
$K^{2}=\frac{n(a d-b c)^{2}}{(a+b)(c+d)(a+c)(b+d)}$
从物理或数学分数在 80 分以上的同学中任意挑选 2 名, 求这 2 名同学的数学与物理分数恰好都在 80 分以上的概率. | [
"由已知数表可以看出,物理或数学分数在 80 分以上的同学共 6 人,其中 4 人的物理与数学分数都在 80 分以上, 设这 4 人分别为 $A_{1}, A_{2}, A_{3}, A_{4}$, 另外 2 人为 $B_{1}, B_{2}$, 则从中任选 2 人的所有基本事件为\n\n$A_{1} A_{2}, A_{1} A_{3}, A_{1} A_{4}, A_{1} B_{1}, A_{1} B_{2}$,\n\n$A_{2} A_{3}, A_{2} A_{4}, A_{2} B_{1}, A_{2} B_{2}$\n\n$A_{3} A_{4}, A_{3} B_{1}, A_{3} B_{2}$,\n\n$... | [
"$\\frac{2}{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 | Probability and Statistics | Math | Chinese |
401 | 已知函数 $f(x)=x \ln x, g(x)=-x^{2}+a x-3$.
求函数 $f(x)$ 在 $[t, t+1](t>0)$ 上的最小值. | [
"$f^{\\prime}(x)=\\ln x+1$\n\n$\\therefore f(x)$ 在 $\\left(0, \\frac{1}{e}\\right)$ 为减函数, 在 $\\left(\\frac{1}{e},+\\infty\\right)$ 为增函数\n\n(1) 当 $0<t<\\frac{1}{e}$ 时, $f(x)$ 在 $\\left[t, \\frac{1}{e}\\right)$ 为减函数, 在 ${\\left[\\frac{1}{e}, t+1\\right]}$ 为增函数,\n\n$\\therefore f(x)_{\\min }=f\\left(\\frac{1}{e}\\righ... | [
"$-\\frac{1}{e} , t\\ln t$"
] | 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 | Elementary Functions | Math | Chinese |
402 | 已知函数 $f(x)=x \ln x, g(x)=-x^{2}+a x-3$.
若对任意的 $x \in\left[\frac{1}{e^{2}}, e^{2}\right](e$ 是自然对数的底数, $e=2.71828 \cdots)$, 不等式 $f(x) \geq \frac{1}{2} g(x)$ 恒成立, 求实数 $a$ 的取值范围. | [
"由题意可知, $2 x \\ln x+x^{2}-a x+3 \\geq 0$ 在 $\\left[\\frac{1}{e^{2}}, e^{2}\\right]$ 上恒成立, 即\n\n$a \\leq \\frac{2 x \\ln x+x^{2}+3}{x}=2 \\ln x+x+\\frac{3}{x}$ 在 $\\left[\\frac{1}{e^{2}}, e^{2}\\right]$ 上恒成立,\n\n令 $h(x)=2 \\ln x+x+\\frac{3}{x}$, 即 $a \\leq h(x)_{\\text {min }}$\n\n$\\because h^{\\prime}(x)=\\frac{2}... | [
"$(-\\infty, 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 | Elementary Functions | Math | Chinese |
403 | 在直角坐标系 $x o y$ 中, 已知圆雉曲线 $C:\left\{\begin{array}{l}x=\sqrt{3} \cos \theta, \\ y=\sqrt{2} \sin \theta,\end{array}\right.$ ( $\theta$ 为参数) 和直线 $l:\left\{\begin{array}{l}x=-1+t \cos \alpha, \\ y=t \sin \alpha\end{array}\right.$ ( $t$ 为参数, $\alpha$ 为 $l$ 的倾斜角, 且 $\left.0<\alpha<\pi\right)$ 相交于不同两点 $A, B$.
当 $\alpha=30^{\circ}$ 时, 求交点 $A 、 B$ 的坐标. | [
"$C: \\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, 轨迹为椭圆,\n\n其焦点为 $F_{1}(-1,0)$,\n\n$\\therefore$ 直线 $A B$ 的方程为: $y=\\frac{\\sqrt{3}}{3}(x+1),$\n\n联立可得, $2 x^{2}+3 \\cdot \\frac{1}{3}(x+1)^{2}=6$. , 化简得, $3 x^{2}+2 x-5=0$\n\n可得交点坐标为 $\\left(1, \\frac{2 \\sqrt{3}}{3}\\right),\\left(-\\frac{5}{3},-\\frac{2 \\sqrt{3}}{9}\\rig... | [
"$\\left(1, \\frac{2 \\sqrt{3}}{3}\\right),\\left(-\\frac{5}{3},-\\frac{2 \\sqrt{3}}{9}\\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 | Polar Coordinates and Parametric Equations | Math | Chinese |
404 | 在直角坐标系 $x o y$ 中, 已知圆雉曲线 $C:\left\{\begin{array}{l}x=\sqrt{3} \cos \theta, \\ y=\sqrt{2} \sin \theta,\end{array}\right.$ ( $\theta$ 为参数) 和直线 $l:\left\{\begin{array}{l}x=-1+t \cos \alpha, \\ y=t \sin \alpha\end{array}\right.$ ( $t$ 为参数, $\alpha$ 为 $l$ 的倾斜角, 且 $\left.0<\alpha<\pi\right)$ 相交于不同两点 $A, B$.
若 $|P A| \cdot|P B|=\frac{8}{5}$, 其中 $P(-1,0)$, 求直线 $l$ 的斜率. | [
"直线 AB 的参数方程为 $\\left\\{\\begin{array}{l}x=-1+t \\cos \\alpha, \\\\ y=t \\sin \\alpha\\end{array}\\right.$ ( $t$ 为参数),\n\n上式代入椭圆 $C$ 的方程式中得: $\\left(2 \\cos ^{2} \\alpha+3 \\sin ^{2} \\alpha\\right) t^{2}-4 \\cos \\alpha t-4=0,$\n\n$\\therefore|P A| \\cdot|P B|=\\left|t_{1} t_{2}\\right|=\\frac{4}{2 \\cos ^{2} \\al... | [
"$+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 | true | null | Numerical | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
405 | 已知函数 $f(x)=|2 x-a|+|x+3|, g(x)=2 x+4, a$ 是实数.
当 $a=1$ 时, 求不等式 $f(x) \geq g(x)$ 的解集; | [
"当 $a=1, f(x)=|2 x-1|+|x+3|$\n\n当 $x \\leq-3$ 时，原不等式化为 $-3 x-2 \\geq 2 x+4$, 得 $x \\leq-3 .$\n\n当 $-3<x \\leq \\frac{1}{2}$ 时, 原不等式化为 $4-x \\geq 2 x+4$, 得 $-3<x \\leq 0 .$\n\n当 $x>\\frac{1}{2}$ 时, 原不等式化为 $3 x+2 \\geq 2 x+4$, 得 $x \\geq 2$\n\n综上, 不等式 $f(x) \\geq g(x)$ 的解集为 $\\{x \\mid x \\leq 0$ 或 $x \\geq 2\\}$."
] | [
"$(-\\infty, 0] \\cup [2, +\\infty)$"
] | 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 | Inequality | Math | Chinese |
406 | 已知函数 $f(x)=|2 x-a|+|x+3|, g(x)=2 x+4, a$ 是实数.
设 $a>-6$, 当 $x \in\left[-3, \frac{a}{2}\right)$ 时, $f(x) \geq g(x)$, 求 $a$ 的取值范围. | [
"当 $x \\in\\left[-3, \\frac{a}{2}\\right)$ 时, $f(x)=-x+a+3$\n\n不等式 $f(x) \\geq g(x)$ 化为 $-x+a+3 \\geq 2 x+4$.\n\n所以 $3 x \\leq a-1$ 对 $x \\in\\left[-3, \\frac{a}{2}\\right)$ 都成立.\n\n故 $\\frac{3 a}{2} \\leq a-1$, 即 $a \\leq-2$. 综上, $a$ 的取值范围为 $-6<a \\leq-2 .$"
] | [
"$(-6, -2]$"
] | 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 | Inequality | Math | Chinese |
407 | 已知圆 $O: x^{2}+y^{2}=4$ 与曲线 $C: y=3|x-t|, A(m, n), B(s, p)\left(m, n, s, p \in N^{*}\right)$ 为曲线 $C$ 上的两点, 使得圆 $O$ 上的任意一点到点 $A$ 的距离与到点 $B$ 的距离之比为定值 $k(k>1)$, 求 $t$ 的值. | [
"取圆上的点 $C(2,0), D(-2,0), E(0,2), F(0,-2)$, 依题意有:\n\n$$\n\\left\\{\\begin{array}{l}\n\\frac{(2-m)^{2}+n^{2}}{(2-s)^{2}+p^{2}}=\\frac{(2+m)^{2}+n^{2}}{(2+s)^{2}+p^{2}}=\\frac{m}{s} \\\\\n\\frac{m^{2}+(2-n)^{2}}{s^{2}+(p-2)^{2}}=\\frac{m^{2}+(2+n)^{2}}{s^{2}+(2+p)^{2}}=\\frac{n}{p}\n\\end{array}\\right.\n$$\n\n于是: $\\... | [
"$\\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 | Conic Sections | Math | Chinese |
409 | 实数 $a 、 b 、 c$ 满足 $a^{2}+b^{2}+c^{2}=\lambda(\lambda>0)$, 试求
$$
f=\min \left\{(a-b)^{2}, \quad(b-c)^{2}, \quad(c-a)^{2}\right\}
$$
的最大值. | [
"由 $\\mathrm{i}$ 对称性, 不妨设 $a \\geq b \\geq c$,\n\n从而:\n\n$$\na-b>a-c>0\n$$\n\n于是有:\n\n$$\n\\begin{aligned}\nf & =\\min \\left\\{(a-b)^{2}, \\quad(b-c)^{2}, \\quad(c-a)^{2}\\right\\}=\\min \\left\\{(a-b)^{2}, \\quad(b-c)^{2}\\right\\} \\\\\n& \\leq(a-b)(b-c) \\\\\n& \\leq\\left[\\frac{(a-b)+(b-c)}{2}\\right]^{2} \\\... | [
"$\\frac{\\lambda}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Elementary Functions | Math | Chinese |
411 | 已知抛物线 $C_{1}$ 的顶点 $(\sqrt{2}-1,1)$, 焦点 $\left(\sqrt{2}-\frac{3}{4}, 1\right)$, 另一抛物线 $C_{2}$ 的方程 $y^{2}-a y+x+2 b=0, C_{1}$与 $C_{2}$ 在一个交点处它们的切线互相垂直. 其中 $C_{2}$ 必过定点, 求该点的坐标. | [
"$C_{1}$ 的 $p=\\frac{1}{2}$, 方程 $(y-1)^{2}=x-(\\sqrt{2}-1)$, 即 $y^{2}-2 y-x+\\sqrt{2}=0$.\n\n设交点为 $\\left(x_{0}, y_{0}\\right)$, 则 $C_{1}$ 的切线方程为 $y_{0} y-\\left(y+y_{0}\\right)-\\frac{1}{2}\\left(x+x_{0}\\right)+\\sqrt{2}=0$, 即\n\n$$\n2\\left(y_{0}-1\\right) y-x-2 y_{0}-x_{0}+2 \\sqrt{2}=0\n$$\n\n同理可得, $C_{2}$ 的切线... | [
"$\\left(\\sqrt{2}-\\frac{1}{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 | Tuple | null | Open-ended | Conic Sections | Math | Chinese |
413 | 已知二次函数 $f(x)=x^{2}-16 x+p+3$.
若函数在区间 $[-1,1]$ 上存在零点, 求实数 $p$ 的取值范围; | [
"因为二次函数 $f(x)=x^{2}-16 x+p+3$ 的对称轴是 $x=8$, 所以函数 $f(x)$ 在区间 $[-1,1]$ 上单调递减, 则函数 $f(x)$ 在区间 $[-1,1]$ 上存在零点须满足 $f(-1) \\cdot f(1) \\leq 0$, 即 $(1+16+p+3)(1-16+p+3) \\leq 0$,解得 $-20 \\leq p \\leq 12$."
] | [
"$[-20,12]$"
] | 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 |
414 | 已知二次函数 $f(x)=x^{2}-16 x+p+3$.
求常数 $q(q \geq 0)$, 当 $x \in[q, 10]$ 时, $f(x)$ 的值域为区间 $D$, 且 $D$ 的长度为 $12-q$.（记 区间 $[a, b](a<b)$ 的长度为 $b-a$.） | [
"假设存在常数 $q$ 满足题意, 分三种情况求解:\n\n(1) 当 $\\left\\{\\begin{array}{l}q<8 \\\\ 8-q \\geq 10-8 \\\\ q \\geq 0\\end{array}\\right.$ 时, 即 $0 \\leq q \\leq 6$ 时.\n\n当 $x=8$ 时, 取到最小值 $f(8)$; 当 $x=q$ 时, 取到最大值 $f(q)$, 所以 $f(x)$ 的值域为: $[f(8), f(q)]$, 即 $\\left[p-61, q^{2}-16 q+p+3\\right]$, 所以区间长度为 $q^{2}-16 q+p+3-(p-61)=q^{2}-16... | [
"$8,9,\\frac{15-\\sqrt{17}}{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 | Elementary Functions | Math | Chinese |
416 | 已知数列 $\left\{a_{n}\right\}$ 的奇数项是首项为 1 的等差数列, 偶数项是首项为 2 的等比数列, 数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 且满足 $S_{5}=2 a_{4}+a_{5}, a_{9}=a_{3}+a_{4}$.
若 $a_{m} a_{m+1}=a_{m+2}$, 求正整数 $m$ 的值; | [
"设等差数列的公差为 $d$, 等比数列的公比为 $q$, 则 $a_{1}=1, a_{2}=2, a_{3}=1+d, a_{4}=2 q, a_{9}=1+4 d$.\n\n因为 $S_{5}=2 a_{4}+a_{5}$, 所以 $a_{1}+a_{2}+a_{3}=a_{4}$, 即 $4 d=2 q$, 又 $a_{9}=a_{3}+a_{4}$, 所以 $1+4 d=1+d+2 q$, 解得: $d=2, q=3$, 所以对于 $k \\in \\mathbf{N}^{*}$, 有 $a_{2 k-1}=1+(k-1) \\cdot 2=2 k-1, a_{2 k}=2 \\cdot 3^{k-1}$, 故\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 | Sequence | Math | Chinese |
417 | 已知数列 $\left\{a_{n}\right\}$ 的奇数项是首项为 1 的等差数列, 偶数项是首项为 2 的等比数列, 数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 且满足 $S_{5}=2 a_{4}+a_{5}, a_{9}=a_{3}+a_{4}$.
存在正整数 $m$, 使得 $\frac{S_{2 m}}{S_{2 m-1}}$ 恰好为数列 $\left\{a_{n}\right\}$ 中的一项。 求出所有满足条件的 $m$ 值; | [
"设等差数列的公差为 $d$, 等比数列的公比为 $q$, 则 $a_{1}=1, a_{2}=2, a_{3}=1+d, a_{4}=2 q, a_{9}=1+4 d$.\n\n因为 $S_{5}=2 a_{4}+a_{5}$, 所以 $a_{1}+a_{2}+a_{3}=a_{4}$, 即 $4 d=2 q$, 又 $a_{9}=a_{3}+a_{4}$, 所以 $1+4 d=1+d+2 q$, 解得: $d=2, q=3$, 所以对于 $k \\in \\mathbf{N}^{*}$, 有 $a_{2 k-1}=1+(k-1) \\cdot 2=2 k-1, a_{2 k}=2 \\cdot 3^{k-1}$, 故\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 | true | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
418 | 设 $a_{n}=2^{n}, n \in \mathbf{N}^{+}$, 数列 $\left\{b_{n}\right\}$ 满足 $b_{1} a_{n}+b_{2} a_{n-1}+\cdots+b_{n} a_{1}=2^{n}-\frac{n}{2}-1$, 求数列 $\left\{a_{n} \cdot b_{n}\right\}$ 的前 $n$ 项和. | [
"由 $a_{1}=2$ 及 $b_{1} a_{1}=2-\\frac{1}{2}-1$, 可得 $b_{1}=\\frac{1}{4}$.\n\n当 $n \\geq 2$ 时, 由已知条件有 $\\left\\{\\begin{array}{l}b_{1} \\cdot 2^{n-1}+b_{2} \\cdot 2^{n-2}+\\cdots+b_{n-1} \\cdot 2=2^{n-1}-\\frac{n-1}{2}-1 \\cdots (1)\\\\ b_{1} \\cdot 2^{n}+b_{2} \\cdot 2^{n-1}+\\cdots+b_{n-1} \\cdot 2^{2}+b_{n} \\cdot ... | [
"$\\frac{(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 | Sequence | Math | Chinese |
419 | 已知椭圆的中心在原点 $O$, 焦点在 $x$ 轴上, 离心率为 $\frac{\sqrt{3}}{2}$, 且过点 $\left(\sqrt{2}, \frac{\sqrt{2}}{2}\right)$. 设不过原点 $O$ 的直线 $l$ 与该椭圆交于点 $P$ 和 $Q$, 且直线 $O P, P Q, O Q$ 的斜率构成等比数列, 求 $\triangle O P Q$ 面积的取值范围. | [
"设椭圆方程为 $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, 则 $\\left\\{\\begin{array}{l}\\frac{c}{a}=\\frac{\\sqrt{3}}{2}, \\\\ \\frac{2}{a^{2}}+\\frac{1}{2 b^{2}}=1\\end{array}\\right.$, 解之得 $\\left\\{\\begin{array}{l}a=2 \\\\ b=1\\end{array}\\right.$, 故椭圆方程为\n\n$$\n\\frac{x^{2}}{4}+y^{2}=1\n$$\n\n由题设可知直线 $l$ 的... | [
"$(0,1)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Conic Sections | Math | Chinese |
422 | 已知 $\cos (\alpha+\beta)=\cos \alpha+\cos \beta$, 试求 $\cos \alpha$ 的最大值. | [
"由题意得: $\\cos \\alpha \\cos \\beta-\\sin \\alpha \\sin \\beta=\\cos \\alpha+\\cos \\beta$, 则\n\n$$\n(\\cos \\alpha-1) \\cos \\beta-\\sin \\alpha \\sin \\beta-\\cos \\alpha=0\n$$\n\n记点 $P(\\cos \\beta, \\sin \\beta)$, 直线 $l:(\\cos \\alpha-1) x-\\sin \\alpha y-\\cos \\alpha=0$, 则点 $P$ 的轨迹方程为单位圆:\n\n$$\nx^{2}+y^{2}=1 ... | [
"$\\sqrt{3}-1$"
] | 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 | Trigonometric Functions | Math | Chinese |
423 | 已知方程 $17 x^{2}-16 x y+4 y^{2}-34 x+16 y+13=0$ 在 $x O y$ 平面上表示一椭圆, 试求它的对称中心及对称轴. | [
"易知点 $A(1,1), B(1,-1)$ 均在该椭圆上.\n\n设椭圆的对称中心为 $(a, b)$, 则由对称性可知点 $A^{\\prime}(2 a-1,2 b-1), B^{\\prime}(2 a-1,2 b+1)$ 均在该椭圆上, 代入方程得:\n\n$$\n\\begin{aligned}\n& 17(2 a-1)^{2}-16(2 a-1)(2 b-1)+4(2 b-1)^{2}-34(2 a-1)+16(2 b-1)+13=0 \\cdots (1)\\\\\n& 17(2 a-1)^{2}-16(2 a-1)(2 b+1)+4(2 b+1)^{2}-34(2 a-1)+16(2 b+1)+13=0\\... | [
"$(1,0), y=\\frac{13 \\pm 5 \\sqrt{17}}{16}(x-1)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple,Expression | null | Open-ended | Conic Sections | Math | Chinese |
424 | 在数列 $\left\{a_{n}\right\}$ 中, $a_{1}, a_{2}$ 是给定的非零整数, $a_{n+2}=\left|a_{n+1}-a_{n}\right|$.
若 $a_{16}=4, a_{17}=1$, 求 $a_{2018}$. | [
"因为 $a_{16}=4, a_{17}=1, a_{18}=3, a_{19}=2, a_{20}=1, a_{21}=1, a_{22}=0, a_{23}=1, a_{24}=1, a_{25}=0, \\cdots$ ，所以自第 20 项起, 每三个相邻的项周期地取值 $1,1,0$, 又 $2018=19+666 \\times 3+1$, 故 $a_{2018}=1 .$"
] | [
"$1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
426 | 已知 $\triangle A B C$ 的三边长分别为 $a, b, c$, 且满足 $a b c=2(a-1)(b-1)(c-1)$, 其中存在边长均为整数的 $\triangle A B C$, 求出三边长(a,b,c)的所有可能组合,边长按照从小到大排列; | [
"不妨设 $a \\geq b \\geq c$, 显然 $c \\geq 2$.\n\n若 $c \\geq 5$, 此时有\n\n$$\n\\frac{1}{a} \\leq \\frac{1}{b} \\leq \\frac{1}{c} \\leq \\frac{1}{5}\n$$\n\n由 $a b c=2(a-1)(b-1)(c-1)$ 可得\n\n$$\n\\frac{1}{2}=\\left(1-\\frac{1}{a}\\right)\\left(1-\\frac{1}{b}\\right)\\left(1-\\frac{1}{c}\\right) \\geq\\left(\\frac{4}{5}\\righ... | [
"$(3,7,8) , (4,5,6)$"
] | 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 | true | null | Tuple | null | Open-ended | Plane Geometry | Math | Chinese |
427 | $\triangle A B C$ 中, $A, B, C$ 所对的边分别为 $a, b, c, \tan C=\frac{\sin A+\sin B}{\cos A+\cos B}$, $\sin (B-A)=\cos C$.
求 $A, C$. | [
"因为 $\\tan C=\\frac{\\sin A+\\sin B}{\\cos A+\\cos B}$, 即 $\\frac{\\sin C}{\\cos C}=\\frac{\\sin A+\\sin B}{\\cos A+\\cos B}$,\n\n所以 $\\sin C \\cos A+\\sin C \\cos B=\\cos C \\sin A+\\cos C \\sin B$,\n\n即 $\\sin C \\cos A-\\cos C \\sin A=\\cos C \\sin B-\\sin C \\cos B$,\n\n得 $\\sin (C-A)=\\sin (B-C)$.\n\n所以 $C-A=B... | [
"$\\frac{\\pi}{4}, \\frac{\\pi}{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 |
428 | $\triangle A B C$ 中, $A, B, C$ 所对的边分别为 $a, b, c, \tan C=\frac{\sin A+\sin B}{\cos A+\cos B}$, $\sin (B-A)=\cos C$.
若 $S_{\triangle A B C}=3+\sqrt{3}$, 求 $a, c$. | [
"因为 $\\tan C=\\frac{\\sin A+\\sin B}{\\cos A+\\cos B}$, 即 $\\frac{\\sin C}{\\cos C}=\\frac{\\sin A+\\sin B}{\\cos A+\\cos B}$,\n\n所以 $\\sin C \\cos A+\\sin C \\cos B=\\cos C \\sin A+\\cos C \\sin B$,\n\n即 $\\sin C \\cos A-\\cos C \\sin A=\\cos C \\sin B-\\sin C \\cos B$,\n\n得 $\\sin (C-A)=\\sin (B-C)$.\n\n所以 $C-A=B... | [
"$2 \\sqrt{2}, 2 \\sqrt{3}$"
] | 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 | true | null | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
429 | 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=1, a_{n+1}=2 a_{n}+1\left(n \in N^{*}\right)$.
求数列 $\left\{a_{n}\right\}$ 的通项公式. | [
"$\\because a_{n-1}=2 a_{n}+1\\left(n \\in N^{*}\\right)$.\n\n$\\therefore a_{n+1}+1=2\\left(a_{n}+1\\right), \\therefore\\left\\{a_{n}+1\\right\\}$ 是以 $a_{1}+1=2$ 为首项, 2 为公比的等比数列.\n\n$\\therefore a_{n}+1=2^{n}$, 即 $a_{n}=2^{n}-1$"
] | [
"$a_{n}=2^{n}-1$"
] | 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 | Sequence | Math | Chinese |
438 | 已知函数 $f(x)=\frac{m x-n}{x}-\ln x, m, n \in R$.
若函数 $f(x)$ 在 $(2, f(2))$ 处的切线与直线 $x-y=0$ 平行, 求实数 $n$ 的值; | [
"由 $f^{\\prime}(x)=\\frac{n-x}{x^{2}}, f^{\\prime}(2)=\\frac{n-2}{4}$, 由于函数 $f(x)$ 在 $(2, f(2))$ 处的切线与直线 $x-y=0$ 平行,\n\n故 $\\frac{n-2}{4}=1$, 解得 $n=6$."
] | [
"$6$"
] | 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 | Elementary Functions | Math | Chinese |
439 | 已知函数 $f(x)=\frac{m x-n}{x}-\ln x, m, n \in R$.
试讨论当$n \leq 1 $时，函数 $f(x)$ 在区间 $[1,+\infty)$ 上最大值. | [
"$f^{\\prime}(x)=\\frac{n-x}{x^{2}}(x>0)$, 由 $f^{\\prime}(x)<0$ 时, $x>n ; f^{\\prime}(x)>0$ 时, $x<n$,\n\n所以(1)当 $n \\leq 1$ 时, $f(x)$ 在 $[1,+\\infty)$ 上单调递减,\n\n故 $f(x)$ 在 $[1,+\\infty)$ 上的最大值为 $f(1)=m-n$;\n\n(2)当 $n>1, f(x)$ 在 $[1, n)$ 上单调递增, 在 $(n,+\\infty)$ 上单调递减,\n\n故 $f(x)$ 在 $[1,+\\infty)$ 上的最大值为 $f(n)=m-1-\\... | [
"$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 | Elementary Functions | Math | Chinese |
441 | 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=3, a_{n+1}=\frac{3 a_{n}-4}{9 a_{n}+15}\left(n \in N^{*}\right)$, 求数列 $\left\{a_{n}\right\}$ 通项 $a_{n}$. | [
"考虑不动点 $x, \\because a_{n+1}=\\frac{3 a_{n}-4}{9 a_{n}+15}, \\therefore x=\\frac{3 x-4}{9 x+15}$ ，\n\n即 $9 x^{2}+12 x+4=0, \\therefore x_{1,2}=-\\frac{2}{3}, \\therefore a_{n+1}+\\frac{2}{3}=\\frac{3 a_{n}-4}{9 a_{n}+15}+\\frac{2}{3}=\\frac{3 a_{n}+2}{3 a_{n}+5}$,\n\n即 $\\frac{1}{a_{n+1}+\\frac{2}{3}}=\\frac{3\\lef... | [
"$a_{n}=\\frac{49-22 n}{3(11 n-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 | Expression | null | Open-ended | Sequence | Math | Chinese |
442 | 过直线 $l: x-2 y-20=0$ 上的点 $P$ 作椭圆 $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ 的切线 $P M, P N$, 切点分别为 $M, N$ 连接 $M N$.
当点 $P$ 在直线 $l$ 上运动时,直线 $M N$ 恒过一定点 $Q$,求点 $Q$ 的坐标. | [
"设 $P\\left(2 y_{0}+20, y_{0}\\right)$, 则直线 $M N$ 的方程为 $\\frac{x\\left(2 y_{0}+20\\right)}{16}+\\frac{y y_{0}}{9}=1$\n\n整理得 $9\\left(y_{0}+10\\right)\\left(x-\\frac{4}{5}\\right)+8 y_{0}\\left(y+\\frac{9}{10}\\right)=0$ (2)\n\n$\\therefore$ 直线 $M N$ 恒过一定点 $Q\\left(\\frac{4}{5},-\\frac{9}{10}\\right)$"
] | [
"$\\left(\\frac{4}{5},-\\frac{9}{10}\\right)$"
] | 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 | Plane Geometry | Math | Chinese |
447 | 求不定方程 $x+y+z+w=25$ 的满足 $x<y$ 的正整数解 $(x, y, z, w)$ 的组数. | [
"令 $Y-X=\\alpha, \\alpha \\in \\mathbf{N}^{*}$, 则 $2 x+\\alpha+z+w=25$, 当 $x$ 取遍 $1 \\sim 11$ 时, $\\alpha+z+w=25-2 x$ 的正整数解组数为 $C_{24-2}^{2}$, 所以总共 $C_{2}^{2}+C_{4}^{2}+C_{6}^{2}+\\cdots+C_{2} 2^{2}=946, C_{2 n}^{2}=n(2 n-1)$, 所以\n\n$$\n\\sum C_{2 n}^{2}=2 \\sum n^{2}-\\sum n=\\frac{n(n+1)(2 n+1)}{3}-\\frac{n(n+1)}... | [
"$\\frac{n(n+1)(4 n-1)}{6}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | Chinese |
448 | 设 $a, b, c, d$ 是实数, 求
$$
F(a, b, c, d)=a^{2}+b^{2}+c^{2}+d^{2}+a b+a c+a d+b c+b d+c d+a+b+c+d
$$
的最小值. | [
"$$\n\\left\\{\\begin{array}{l}\n\\frac{\\partial F}{\\partial a}=2 a+b+c+d+1=0 \\\\\n\\frac{\\partial F}{\\partial b}=2 b+a+c+d+1=0 \\\\\n\\frac{\\partial F}{\\partial c}=2 c+a+b+d+1=0 \\\\\n\\frac{\\partial F}{\\partial d}=2 d+a+b+c+1=0\n\\end{array} \\quad \\text {, 所以当 } a=b=c=d=-\\frac{1}{5} \\text { 时, } F(a,... | [
"$-\\frac{2}{5}$"
] | 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 | Derivative | Math | Chinese |
449 | 设 $n$ 为给定的正整数, 考虑平面直角坐标系 $x O y$ 中的点集
$$
T=\{(x, y)|| x|=| y \mid \leq n, x, y \in \mathbf{Z}\}
$$
对 $T$ 中的两点 $P, Q$, 当且仅当 $|P Q|=\sqrt{2}$ 或 $P Q$ 与两条坐标轴之一平行时, 称 $P, Q$ 是 “相邻的”, 将 $T$ 中的每个点染上红、蓝、绿三种颜色之一，要求任意两个相邻点被染不同的颜色，求染色方式的数目. | [
"从 $(0,0)$ 点向外一共有 $n$ 层正方形, 染色要求: 正方形相邻顶点颜色不同, 与上一层相邻点也不同.记 $(0,0)$ 点染了 3 号色, 第 1 个正方形四个顶点 $\\begin{array}{ll}1 & 2 \\\\ 2 & 1\\end{array}$ 染色:\n\n$A$ 类: $\\begin{array}{ll}1 & 2 \\\\ 2 & 1\\end{array}$ (一共用了 2 色) 上层 $A$ 类, 之后一层的染色情况:\n\n$$\n\\left\\{\\begin{array}{lllllllllll}\nA \\text { 三种 } & 2 & 1 & 3 & 1 & 2 ... | [
"$\\frac{6}{\\sqrt{33}}\\left[\\left(\\frac{7+\\sqrt{33}}{2}\\right)^{n}-\\left(\\frac{7-\\sqrt{33}}{2}\\right)^{n}\\right]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | Chinese |
450 | 已知 $O$ 为坐标原点, $N(1,0), M$ 为直线 $x=-1$ 上的动点, $\angle M O N$ 的平分线与直线 $M N$ 交于点 $P$, 记点 $P$ 的轨迹为曲线 $E$.
求曲线 $E$ 的方程. | [
"设 $P(x, y), M(-1, t)$, 易知 $0 \\leq x<1$. 因为 $O P$ 平分 $\\angle M O N$, 所以\n\n$$\n\\overrightarrow{M P}=\\frac{|O M|}{|O N|} \\overrightarrow{P N}=\\sqrt{1+t^{2}} \\overrightarrow{P N}\n$$\n\n所以\n\n$$\n\\begin{aligned}\n& x+1=\\sqrt{1+t^{2}}(1-x) \\cdots (1)\\\\\n& y-t=\\sqrt{1+t^{2}}(0-y)\\cdots (2)\n\\end{aligned}... | [
"$y^{2}-x=0$"
] | 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 | false | null | Equation | null | Open-ended | Conic Sections | Math | Chinese |
451 | 已知 $O$ 为坐标原点, $N(1,0), M$ 为直线 $x=-1$ 上的动点, $\angle M O N$ 的平分线与直线 $M N$ 交于点 $P$, 记点 $P$ 的轨迹为曲线 $E$.
过点 $Q\left(-\frac{1}{2},-\frac{1}{2}\right)$ 作斜率为 $k$ 的直线 $l$, 若直线 $l$ 与曲线 $E$ 恰好有一个公共点, 求 $k$ 的取值范围. | [
"记 $A(1,1), B(1,-1)$, 则 $k_{Q A}=1, k_{Q B}=-\\frac{1}{3}$.\n\n设 $P(x, y), M(-1, t)$, 易知 $0 \\leq x<1$. 因为 $O P$ 平分 $\\angle M O N$, 所以\n\n$$\n\\overrightarrow{M P}=\\frac{|O M|}{|O N|} \\overrightarrow{P N}=\\sqrt{1+t^{2}} \\overrightarrow{P N}\n$$\n\n所以\n\n$$\n\\begin{aligned}\n& x+1=\\sqrt{1+t^{2}}(1-x) \\cdots ... | [
"$\\left(-\\frac{1}{3}, 1\\right] \\cup\\left\\{\\frac{1+\\sqrt{3}}{2}\\right\\}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Conic Sections | Math | Chinese |
452 | 对任意正整数 $m, n$, 定义函数 $f(m, n)$ 如下:
(1) $f(1,1)=1$;
(2) $f(m+1, n)=f(m, n)+2(m+n)$;
(3) $f(m, n+1)=f(m, n)+2(m+n-1)$.
求 $f(m, n)$ 的解析式. | [
"由条件(2)可得:\n\n$$\n\\begin{gathered}\nf(2,1)-f(1,1)=2 \\times(1+1)=2 \\times 2 \\\\\nf(3,1)-f(2,1)=2 \\times(2+1)=2 \\times 3 \\\\\n\\ldots \\\\\n\\cdots \\\\\nf(m, 1)-f(m-1,1)=2 \\times(m-1+1)=2 \\times m\n\\end{gathered}\n$$\n\n将上述 $m-1$ 个等式相加得 $f(m, 1)-f(1,1)=2(2+3+\\cdots+m)$, 而 $f(1,1)=1$, 所以\n\n$$\nf(m, 1)=2(2... | [
"$f(m, n)=m^{2}+2 m n+n^{2}-m-3 n+1$"
] | 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 | Elementary Functions | Math | Chinese |
454 | 已知正数 $a, b$ 满足 $a+b=1$, 求 $M=\sqrt{1+2 a^{2}}+2 \sqrt{\left(\frac{5}{12}\right)^{2}+b^{2}}$ 的最小值. | [
"由柯西不等式可得\n\n$$\n\\left(2 a^{2}+1\\right)\\left(\\frac{1}{2}+\\lambda^{2}\\right) \\geq(a+\\lambda)^{2}, \\quad\\left[b^{2}+\\left(\\frac{5}{12}\\right)^{2}\\right]\\left(1+\\mu^{2}\\right) \\geq\\left(b+\\frac{5}{12} \\mu\\right)^{2}\n$$\n\n所以\n\n$$\nM=\\sqrt{1+2 a^{2}}+2 \\sqrt{\\left(\\frac{5}{12}\\right)^{2}+b^... | [
"$\\frac{5 \\sqrt{34}}{12}$"
] | 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 | Inequality | Math | Chinese |
456 | 设 $O$ 是坐标原点, 双曲线 $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ 上动点 $M$ 处的切线交 $C$ 的两条渐近线于 $A, B$ 两点.
$\triangle A O B$ 的面积 $S$ 是定值，求定值; | [
"$M\\left(x_{0}, y_{0}\\right)$ 处的切线方程为 $\\frac{x_{0} x}{a^{2}}-\\frac{y_{0} y}{b^{2}}=1$.\n\n与渐近线方程联立, 得 $A\\left(x_{1}, y_{1}\\right)=\\left(\\frac{a}{\\frac{x_{0}}{a}+\\frac{y_{0}}{b}}, \\frac{b}{\\frac{x_{0}}{a}+\\frac{y_{0}}{b}}\\right), B\\left(x_{2}, y_{2}\\right)=\\left(\\frac{a}{\\frac{x_{0}}{a}-\\frac{y_{... | [
"$|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 | false | null | Expression | null | Open-ended | Conic Sections | Math | Chinese |
457 | 设 $O$ 是坐标原点, 双曲线 $C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ 上动点 $M$ 处的切线交 $C$ 的两条渐近线于 $A, B$ 两点.
求 $\triangle A O B$ 的外心 $P$ 的轨迹方程. | [
"$M\\left(x_{0}, y_{0}\\right)$ 处的切线方程为 $\\frac{x_{0} x}{a^{2}}-\\frac{y_{0} y}{b^{2}}=1$.\n\n与渐近线方程联立, 得 $A\\left(x_{1}, y_{1}\\right)=\\left(\\frac{a}{\\frac{x_{0}}{a}+\\frac{y_{0}}{b}}, \\frac{b}{\\frac{x_{0}}{a}+\\frac{y_{0}}{b}}\\right), B\\left(x_{2}, y_{2}\\right)=\\left(\\frac{a}{\\frac{x_{0}}{a}-\\frac{y_{... | [
"$a^{2} x^{2}-b^{2} y^{2}=\\frac{1}{4}\\left(a^{2}+b^{2}\\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 | Conic Sections | Math | Chinese |
461 | 已知函数 $f(x)=2 \cos x(\cos x+\sqrt{3} \sin x)-1, x \in \mathbf{R}$.
求函数 $f(x)$ 的单调递增区间; | [
"$f(x)=\\cos 2 x+\\sqrt{3} \\sin 2 x=2 \\sin \\left(2 x+\\frac{\\pi}{6}\\right), f(x)$ 单调递增区间为 $\\left[-\\frac{\\pi}{3}+k \\pi, \\frac{\\pi}{6}+k \\pi\\right], k \\in \\mathbf{Z}$;"
] | [
"$\\left[-\\frac{\\pi}{3}+k \\pi, \\frac{\\pi}{6}+k \\pi\\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 | Trigonometric Functions | Math | Chinese |
462 | 已知函数 $f(x)=2 \cos x(\cos x+\sqrt{3} \sin x)-1, x \in \mathbf{R}$.
设点 $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots, P_{n}\left(x_{n}, y_{n}\right), \cdots$ 都在函数 $y=f(x)$ 的图象上, 且满足
$$
x_{1}=\frac{\pi}{6}, x_{n+1}-x_{n}=\frac{\pi}{2}\left(n \in \mathbf{N}^{*}\right)
$$
求 $y_{1}+y_{2}+\cdots+y_{2018}$ 的值. | [
"设 $t_{n}=2 x_{n}+\\frac{\\pi}{6}, t_{1}=2 x_{1}+\\frac{\\pi}{6}=\\frac{\\pi}{2}, t_{n+1}-t_{n}=\\pi \\Rightarrow t_{n}=\\left(n-\\frac{1}{2}\\right) \\pi$, 所以\n\n$$\ny_{n}=2 \\sin \\left(\\frac{2 n-1}{2} \\pi\\right)=\\left\\{\\begin{array}{ll}\n2 & n=2 k-1 \\\\\n-2 & n=2 k\n\\end{array}(k \\in \\mathbf{Z})\\right... | [
"$0$"
] | 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 | false | null | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
469 | 已知函数 $f(x)=4 \cos x \cdot \sin \left(x+\frac{7 \pi}{6}\right)+a$ 的最大值为 2 .
求 $a$ 的值及 $f(x)$ 的最小正周期 | [
"$$\n\\begin{aligned}\nf(x) & =4 \\cos x \\cdot \\sin \\left(x+\\frac{7 \\pi}{6}\\right)+a=4 \\cos x \\cdot\\left(-\\frac{\\sqrt{3}}{2} \\sin x-\\frac{1}{2} \\cos x\\right)+a \\\\\n& =-2 \\sqrt{3} \\sin x \\cos x-2 \\cos ^{2} x+1-1+a=-\\sqrt{3} \\sin 2 x-\\cos 2 x-1+a \\\\\n& =-2 \\sin \\left(2 x+\\frac{\\pi}{6}\\r... | [
"$1, \\pi$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
470 | 已知函数 $f(x)=4 \cos x \cdot \sin \left(x+\frac{7 \pi}{6}\right)+a$ 的最大值为 2 .
求 $f(x)$ 的单调递减区间 | [
"$$\n\\begin{aligned}\nf(x) & =4 \\cos x \\cdot \\sin \\left(x+\\frac{7 \\pi}{6}\\right)+a=4 \\cos x \\cdot\\left(-\\frac{\\sqrt{3}}{2} \\sin x-\\frac{1}{2} \\cos x\\right)+a \\\\\n& =-2 \\sqrt{3} \\sin x \\cos x-2 \\cos ^{2} x+1-1+a=-\\sqrt{3} \\sin 2 x-\\cos 2 x-1+a \\\\\n& =-2 \\sin \\left(2 x+\\frac{\\pi}{6}\\r... | [
"$\\left[-\\frac{\\pi}{3}+k \\pi, \\frac{\\pi}{6}+k \\pi\\right]$"
] | 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 | Trigonometric Functions | Math | Chinese |
471 | 数列 $\left\{a_{n}\right\}$ 为等差数列, 且满足 $3 a_{5}=8 a_{12}>0$, 数列 $\left\{b_{n}\right\}$ 满足 $b_{n}=a_{n} \cdot a_{n+1} \cdot a_{n+2}\left(n \in \mathbf{N}^{*}\right),\left\{b_{n}\right\}$ 的前 $n$ 项和记为 $S_{n}$. 问: $n$ 为何值时, $S_{n}$ 取得最大值？ | [
"因为 $3 a_{5}=8 a_{12}>0$, 所以 $3 a_{5}=8\\left(a_{5}+7 d\\right)$, 解得 $a_{5}=-\\frac{56}{5} d>0$, 所以 $d<0, a_{1}=-\\frac{76}{5} d$.故 $\\left\\{a_{n}\\right\\}$ 是首项为正数的递减数列.\n\n由\n\n$$\n\\left\\{\\begin{array}{l}\na_{n} \\geq 0 \\\\\na_{n+1} \\leq 0\n\\end{array}\\right.\n$$\n\n即\n\n$$\n\\left\\{\\begin{array}{l}\n-\... | [
"$16$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
473 | 设 $x, y, z \geq 0$, 且至多有一个为 0 , 求
$$
f(x, y, z)=\sqrt{\frac{x^{2}+256 y z}{y^{2}+z^{2}}}+\sqrt{\frac{y^{2}+256 z x}{z^{2}+x^{2}}}+\sqrt{\frac{z^{2}+256 x y}{x^{2}+y^{2}}}
$$
的最小值. | [
"不妨设 $x \\geq y \\geq z$.\n\n情形一 当 $256 y^{3} \\geq x^{2} z$ 时, 因为\n\n$$\n\\begin{aligned}\n& \\frac{x^{2}+256 y z}{y^{2}+z^{2}}-\\frac{x^{2}}{y^{2}}=\\frac{z\\left(256 y^{3}-x^{2} z\\right)}{\\left(y^{2}+z^{2}\\right) y^{2}} \\geq 0 \\\\\n& \\frac{y^{2}+256 z x}{z^{2}+x^{2}}-\\frac{y^{2}}{x^{2}}=\\frac{z\\left(256... | [
"$12$"
] | 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 | Inequality | Math | Chinese |
474 | 已知数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和 $S_{n}$ 满足 $2 S_{n}-n a_{n}=n, n \in \mathbf{N}^{*}$, 且 $a_{2}=3$.
求数列 $\left\{a_{n}\right\}$ 的通项公式; | [
"由 $2 S_{n}-n a_{n}=n$, 得 $2 S_{n+1}-(n+1) a_{n+1}=n+1$, 将上述两式相减, 得\n\n$$\n2 a_{n+1}-(n+1) a_{n+1}+n a_{n}=1 \n$$\n\n所以\n\n$$\nn a_{n}-(n-1) a_{n+1}=1 \\cdots (1)\n$$\n\n所以\n\n$$\n(n+1) a_{n+1}-n a_{n+2}=1\\cdots (2)\n$$\n\n(1)-(2), 得\n\n$$\nn a_{n}-2 n a_{n+1}+n a_{n+2}=0\n$$\n\n所以 $a_{n}+a_{n+2}=2 a_{n+1}$, 所以数... | [
"$a_{n}=2 n-1$"
] | 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 | Sequence | Math | Chinese |
475 | 已知数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和 $S_{n}$ 满足 $2 S_{n}-n a_{n}=n, n \in \mathbf{N}^{*}$, 且 $a_{2}=3$.
设 $b_{n}=\frac{1}{a_{n} \sqrt{a_{n+1}}+a_{n+1} \sqrt{a_{n}}}, T_{n}$ 为数列 $\left\{b_{n}\right\}$ 的前 $n$ 项和, 求使 $T_{n}>\frac{9}{20}$ 成立的最小正整数 $n$ 的值. | [
"由 $2 S_{n}-n a_{n}=n$, 得 $2 S_{n+1}-(n+1) a_{n+1}=n+1$, 将上述两式相减, 得\n\n$$\n2 a_{n+1}-(n+1) a_{n+1}+n a_{n}=1\n$$\n\n所以\n\n$$\nn a_{n}-(n-1) a_{n+1}=1\\cdots (1)\n$$\n\n所以\n\n$$\n(n+1) a_{n+1}-n a_{n+2}=1\\cdots (2)\n$$\n\n(1)-(2), 得\n\n$$\nn a_{n}-2 n a_{n+1}+n a_{n+2}=0\n$$\n\n所以 $a_{n}+a_{n+2}=2 a_{n+1}$, 所以数列 $\... | [
"$50$"
] | 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 |
476 | 已知 $F_{1}, F_{2}$ 分别为椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的左、右焦点, 点 $P\left(\frac{2 \sqrt{6}}{3}, 1\right)$ 在椭圆 $C$ 上, 且 $\triangle F_{1} P F_{2}$ 的垂心为 $H\left(\frac{2 \sqrt{6}}{3},-\frac{5}{3}\right)$.
求椭圆 $C$ 的方程; | [
"设 $F_{1}(-c, 0), F_{2}(c, 0)$, 由 $\\triangle F_{1} P F_{2}$ 的垂心为 $H\\left(\\frac{2 \\sqrt{6}}{3},-\\frac{5}{3}\\right)$, 得 $F_{1} H \\perp P F_{2}$, 所以\n\n$$\nk_{F_{1} H} \\cdot k_{P F_{2}}=\\frac{-\\frac{5}{3}}{\\frac{2 \\sqrt{6}}{3}+c} \\cdot \\frac{1}{\\frac{2 \\sqrt{6}}{3}-c}=-1, \\frac{24}{9}-c^{2}=\\frac{5}{... | [
"$\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Equation | null | Open-ended | Conic Sections | Math | Chinese |
477 | 已知 $F_{1}, F_{2}$ 分别为椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的左、右焦点, 点 $P\left(\frac{2 \sqrt{6}}{3}, 1\right)$ 在椭圆 $C$ 上, 且 $\triangle F_{1} P F_{2}$ 的垂心为 $H\left(\frac{2 \sqrt{6}}{3},-\frac{5}{3}\right)$.
设 $A$ 为椭圆 $C$ 的左顶点, 过点 $F_{2}$ 的直线 $l$ 交椭圆 $C$ 于 $D E$ 两点, 记直线 $A D, A E$ 的斜率分别为 $k_{1}, k_{2}$, 若 $k_{1}+k_{2}=-\frac{1}{2}$, 求直线 $l$ 的方程. | [
"设 $F_{1}(-c, 0), F_{2}(c, 0)$, 由 $\\triangle F_{1} P F_{2}$ 的垂心为 $H\\left(\\frac{2 \\sqrt{6}}{3},-\\frac{5}{3}\\right)$, 得 $F_{1} H \\perp P F_{2}$, 所以\n\n$$\nk_{F_{1} H} \\cdot k_{P F_{2}}=\\frac{-\\frac{5}{3}}{\\frac{2 \\sqrt{6}}{3}+c} \\cdot \\frac{1}{\\frac{2 \\sqrt{6}}{3}-c}=-1, \\frac{24}{9}-c^{2}=\\frac{5}{... | [
"$y=2(x-1)$"
] | 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 | Conic Sections | Math | Chinese |
480 | 已知 $f(x)=e^{x}-m x$.
若 $x>0$ 时, 不等式 $(x-2) f(x)+m x^{2}+2>0$ 恒成立, 求实数 $m$ 的取值范围; | [
"设 $g(x)=(x-2) f(x)+m x^{2}+2$, 则 $g(x)=(x-2) e^{x}+2 m x+2$.\n\n$x>0$ 时, 不等式 $(x-2) f(x)+m x^{2}+2>0$ 恒成立 $\\Leftrightarrow x>0$ 时, $g(x)>0$ 恒成立.\n\n因为\n\n$$\ng^{\\prime}(x)=e^{x}+(x-2) e^{x}+2 m=(x-1) e^{x}+2 m, g^{\\prime \\prime}(x)=e^{x}+(x-1) e^{x}=x e^{x}\n$$\n\n所以 $x>0$ 时, $g^{\\prime \\prime}(x)=x e^{x}>0,... | [
"$\\left[\\frac{1}{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 | Elementary Functions | Math | Chinese |
482 | 设 $M$ 是由有限个正整数构成的集合, 且 $M=A_{1} \cup A_{2} \cup \cdots \cup A_{20}=B_{1} \cup B_{2} \cup \cdots \cup B_{20}$, 这里 $A_{i} \neq \varnothing, B_{i} \neq \varnothing, i=1,2, \cdots, 20$, 并对任意的 $1 \leq i<j \leq 20$, 都有 $A_{i} \cap A_{j}=\varnothing, B_{i} \cap B_{j}=\varnothing$.
已知对任意的 $1 \leq i \leq 20,1 \leq j \leq 20$, 若 $A_{i} \cap B_{j}=\varnothing$, 则 $\left|A_{i} \cup B_{j}\right| \geq 18$, 求集合 $M$ 的元素个数的最小值（这里， $|X|$ 表示集合 $X$ 的元素个数）. | [
"记 $\\min \\left\\{\\left|A_{1}\\right|,\\left|A_{2}\\right|, \\cdots,\\left|A_{20}\\right|,\\left|B_{1}\\right|,\\left|B_{2}\\right|, \\cdots,\\left|B_{20}\\right|\\right\\}=t$.\n\n不妨设 $\\left|A_{1}\\right|=t, A_{1} \\cap B_{i} \\neq \\varnothing, i=1,2, \\cdots, k ; A_{1} \\cap B_{j}=\\varnothing, j=k+1, k+2, \\c... | [
"$180$"
] | 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 |
483 | 已知函数 $y=3 x+\sqrt{x^{2}-2 x}$, 求该函数的值域. | [
"令 $u=x-1$, 则 $y=3 u+3+\\sqrt{u^{2}-1}$, 则 $|u| \\geq 1$.\n\n设 $\\sqrt{u^{2}-1}=|u|-t \\geq 0$, 则 $0<t \\leq|u|_{\\text {min }}=1$, 且 $|u|=\\frac{1}{2}\\left(t+\\frac{1}{t}\\right)$.\n\n当 $u>0$ 时,\n\n$$\ny=\\frac{3}{2}\\left(t+\\frac{1}{t}\\right)+3+\\frac{1}{2}\\left(t+\\frac{1}{t}\\right)-t=t+\\frac{2}{t}+3\n$$\n... | [
"$(-\\infty, 3-2 \\sqrt{2}] \\cup[6,+\\infty)$"
] | 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 | Elementary Functions | Math | Chinese |
484 | 已知椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的离心率 $e=\frac{\sqrt{2}}{2}$, 直线 $y=2 x-1$ 与 $C$ 交于 $A, B$ 两点, 且 $|A B|=\frac{8}{9} \sqrt{5}$.
求椭圆 $C$ 的方程; | [
"由 $e=\\frac{\\sqrt{2}}{2}$ 得 $a=\\sqrt{2} c=\\sqrt{2} b$, 所以椭圆的方程为 $x^{2}+2 y^{2}-2 b^{2}=0$.\n\n由 $\\left\\{\\begin{array}{l}x^{2}+2 y^{2}-2 b^{2}=0 \\\\ y=2 x-1\\end{array}\\right.$ 得 $9 x^{2}-8 x+\\left(2-2 b^{2}\\right)=0$, 所以 $\\Delta=64-36\\left(2-2 b^{2}\\right)$,\n\n由 $|A B|=\\frac{8}{9} \\sqrt{5}$ 得 $\\sq... | [
"$\\frac{x^{2}}{2}+y^{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 | Equation | null | Open-ended | Conic Sections | Math | Chinese |
485 | 已知椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的离心率 $e=\frac{\sqrt{2}}{2}$, 直线 $y=2 x-1$ 与 $C$ 交于 $A, B$ 两点, 且 $|A B|=\frac{8}{9} \sqrt{5}$.
过点 $M(2,0)$ 的直线 $l$ (斜率不为零) 与椭圆 $C$ 交于不同的两点 $E 、 F$($E$ 在点 $F 、 M$ 之间 ).
记 $\lambda=\frac{S_{\triangle O M E}}{S_{\triangle O M F}}$, 求 $\lambda$ 的取值范围. | [
"由 $e=\\frac{\\sqrt{2}}{2}$ 得 $a=\\sqrt{2} c=\\sqrt{2} b$, 所以椭圆 $C$ 的方程为 $x^{2}+2 y^{2}-2 b^{2}=0$.\n\n由 $\\left\\{\\begin{array}{l}x^{2}+2 y^{2}-2 b^{2}=0 \\\\ y=2 x-1\\end{array}\\right.$ 得 $9 x^{2}-8 x+\\left(2-2 b^{2}\\right)=0$, 所以 $\\Delta=64-36\\left(2-2 b^{2}\\right)$,\n\n由 $|A B|=\\frac{8}{9} \\sqrt{5}$ 得 ... | [
"$(0,3+2 \\sqrt{2})$"
] | 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 | Conic Sections | Math | Chinese |
490 | 已知 $O$ 是 $\triangle A B C$ 的外心, 且 $3 \overrightarrow{O A}+4 \overrightarrow{O B}+5 \overrightarrow{O C}=\overrightarrow{0}$, 求 $\cos \angle B A C$ 的值. | [
"设 $\\triangle A B C$ 的外接圆半径 $r=1$, 由已知得 $3 \\overrightarrow{O A}=-4 \\overrightarrow{O B}-5 \\overrightarrow{O C}$, 两边平方得\n\n$$\n\\overrightarrow{O B} \\cdot \\overrightarrow{O C}=-\\frac{4}{5}\n$$\n\n同理可得\n\n$$\n\\overrightarrow{O A} \\cdot \\overrightarrow{O C}=-\\frac{3}{5}, \\quad \\overrightarrow{O A} \\cdot ... | [
"$\\frac{\\sqrt{10}}{10}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
495 | 若函数 $f(x)$ 的定义域为 $(0,+\infty)$, 且满足:
(1) 存在实数 $a \in(1,+\infty)$, 使得 $f(a)=1$;
(2) 当 $m \in R$ 且 $x \in(0,+\infty)$ 时, 有 $f\left(x^{m}\right)-m f(x)=0$ 恒成立.
若当 $t>0$ 时, 不等式 $f\left(t^{2}+4\right)-f(t) \geq 1$ 恒成立, 求实数 $a$ 的取值范围. | [
"因为 $x, y$ 均为正数, 故总存在实数 $m, n$ 使得 $x=a^{m}, y=a^{n}(a>1)$, 所以\n\n$$\nf\\left(\\frac{x}{y}\\right)=f\\left(\\frac{a^{m}}{a^{n}}\\right)=f\\left(a^{m-n}\\right)=(m-n) f(a)=m-n\n$$\n\n又\n\n$$\nf(x)-f(y)=f\\left(a^{m}\\right)-f\\left(a^{n}\\right)=m f(a)-n f(a)=m-n\n$$\n\n所以\n\n$$\nf\\left(\\frac{x}{y}\\right)=f(x)-f(y... | [
"$(1,4]$"
] | 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 | Interval | null | Open-ended | Algebra | Math | Chinese |
496 | 已知数列 $\left\{a_{n}\right\}$ 中 $a_{1}=\frac{1}{2}, a_{n+1}=\frac{1}{2} a_{n}+\frac{2 n+1}{2^{n+1}}\left(n \in \mathbf{N}^{*}\right)$.
求数列 $\left\{a_{n}\right\}$ 的通项公式; | [
"由 $a_{n+1}=\\frac{1}{2} a_{n}+\\frac{2 n+1}{2^{n+1}}\\left(n \\in \\mathbf{N}^{*}\\right)$ 知: $2^{n+1} a_{n+1}=2^{n} a_{n}+2 n+1\\left(n \\in \\mathbf{N}^{*}\\right)$.\n\n令 $b_{n}=2^{n} a_{n}$, 则 $b_{1}=1$ 且 $b_{n+1}=b_{n}+2 n+1\\left(n \\in \\mathbf{N}^{*}\\right)$.\n\n由 $b_{n}=\\left(b_{n}-b_{n-1}\\right)+\\left... | [
"$a_{n}=\\frac{n^{2}}{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 | Algebra | Math | Chinese |
497 | 已知数列 $\left\{a_{n}\right\}$ 中 $a_{1}=\frac{1}{2}, a_{n+1}=\frac{1}{2} a_{n}+\frac{2 n+1}{2^{n+1}}\left(n \in \mathbf{N}^{*}\right)$.
求数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和 $S_{n}$. | [
"由 $a_{n+1}=\\frac{1}{2} a_{n}+\\frac{2 n+1}{2^{n+1}}\\left(n \\in \\mathbf{N}^{*}\\right)$ 知: $2^{n+1} a_{n+1}=2^{n} a_{n}+2 n+1\\left(n \\in \\mathbf{N}^{*}\\right)$.\n\n令 $b_{n}=2^{n} a_{n}$, 则 $b_{1}=1$ 且 $b_{n+1}=b_{n}+2 n+1\\left(n \\in \\mathbf{N}^{*}\\right)$.\n\n由 $b_{n}=\\left(b_{n}-b_{n-1}\\right)+\\left... | [
"$6-\\frac{n^{2}+4 n+6}{2^{n}}$"
] | 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 | Algebra | Math | Chinese |
501 |
已知正数 $x, y$ 满足: $x+y=1$. 求 $\frac{1}{x}+\frac{4}{y}$ 的最小值. | [
"由 $a+b-2 \\sqrt{a b}=(\\sqrt{a}-\\sqrt{b})^{2} \\geq 0$, 故 $a+b \\geq 2 \\sqrt{a b}$.\n\n$\\frac{1}{x}+\\frac{4}{y}=\\left(\\frac{1}{x}+\\frac{4}{y}\\right)(x+y)=1+4+\\frac{y}{x}+\\frac{4 x}{y} \\geq 5+2 \\sqrt{\\frac{y}{x} \\cdot \\frac{4 x}{y}}=9$\n\n等号在 $x=\\frac{1}{3}, y=\\frac{2}{3}$ 处取到, 故最小值为 9."
] | [
"$9$"
] | 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 | Inequality | Math | Chinese |
503 | 已知实数 $x, y$ 满足: $1+\cos ^{2}(x+y-1)=\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}$, 求 $x y$ 的最小值. | [
"$1+\\cos ^{2}(x+y-1)=\\frac{x^{2}+y^{2}+2(x+1)(1-y)}{x-y+1}=\\frac{\\left(x^{2}+y^{2}-2 x y\\right)+2(x-y)+1+1}{x-y+1}$\n\n$$\n=\\frac{(x-y+1)^{2}+1}{x-y+1}=x-y+1+\\frac{1}{x-y+1} .\n$$\n\n由于 $0<1+\\cos ^{2}(x+y-1) \\leq 2$, 故 $x-y+1>0$, 从而 $x-y+1+\\frac{1}{x-y+1} \\geq 2$.\n\n$$\n\\begin{aligned}\n& \\Rightarrow\... | [
"$\\frac{1}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Inequality | Math | Chinese |
506 | 设 $f(m)$ 是正整数 $m$ 的各位数字的乘积, 求方程 $f(m)=m^{2}-10 m-36$ 的正整数解. | [
"设 $m$ 是 $n$ 位正整数, 若 $n \\geq 3$, 则 $m \\geq 10^{n-1} \\geq 100$, 所以\n\n$$\n9^{n} \\geq f(m)=m^{2}-10 m-36=m(m-10)-36 \\geq 90 m-36 \\geq 9 \\cdot 10^{n}-36>10^{n}\n$$\n\n矛盾, 此时无解;\n\n若 $n=1$, 则 $f(m)=m=m^{2}-10 m-36$, 此方程无整数解;\n\n若 $n=2$, 且 $m \\geq 20$, 则 $81 \\geq f(m)=m^{2}-10 m-36 \\geq 400-200-36=164$, 矛盾.\n\... | [
"$13$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Elementary Functions | Math | Chinese |
508 | 设椭圆 $C$ 的左、右顶点为 $A, B(a, 0)$, 过焦点 $F(1,0)$ 作非水平直线 $l$ 与椭圆 $C$ 交于 $P, Q$ 两点, 记直线 $A P, B Q$ 的斜率分别为 $k_{1}, k_{2}$. 其中 $\frac{k_{1}}{k_{2}}$ 为定值, 并求此定值 (用 $a$ 的函数表示). | [
"设 $l: x=t y+1$, 代入椭圆方程 $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-1}=1$ 得\n\n$$\n\\left(\\left(a^{2}-1\\right) t^{2}+a^{2}\\right) y^{2}+2\\left(a^{2}-1\\right) t y-\\left(a^{2}-1\\right)^{2}=0\n$$\n\n设 $P\\left(x_{1}, y_{1}\\right), Q\\left(x_{2}, y_{2}\\right)$, 则\n\n$$\ny_{1}+y_{2}=-\\frac{2\\left(a^{2}-1\\right... | [
"$\\frac{a-1}{a+1}$"
] | 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 | Expression | null | Open-ended | Conic Sections | Math | Chinese |
509 | 设函数 $f(x)=a x^{2}+b x+c(a \neq 0)$ 满足 $|f(0)| \leq 2,|f(2)| \leq 2,|f(-2)| \leq 2$, 求当 $x \in[-2,2]$ 时, $y=|f(x)|$的最大值. | [
"由题意知\n\n$$\n\\left\\{\\begin{array} { l } \n{ c = f ( 0 ) } \\\\\n{ 4 a + 2 b + c = f ( 2 ) } \\\\\n{ 4 a - 2 b + c = f ( - 2 ) }\n\\end{array} \\Rightarrow \\left\\{\\begin{array}{l}\nc=f(0) \\\\\na=\\frac{f(2)+f(-2)-2 f(0)}{8} \\\\\nb=\\frac{f(2)-f(-2)}{4}\n\\end{array}\\right.\\right.\n$$\n\n从而当 $x \\in[-2,2]$ ... | [
"$\\frac{5}{2}$"
] | 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 | Elementary Functions | Math | Chinese |
510 | 已知将函数 $g(x)=\cos x$ 图像上所有点的纵坐标伸长到原来的 2 倍 (横坐标不变), 再将所得图像向右平移 $\frac{\pi}{2}$ 个单位长度得到函数 $y=f(x)$ 的图像, 且关于 $x$ 的方程 $f(x)+g(x)=m$ 在 $[0,2 \pi)$ 内有两个不同的解 $\alpha, \beta$.
求满足题意的实数 $m$ 的取值范围; | [
"将 $g(x)=\\cos x$ 图像上的所有点的纵坐标伸长为原来的 2 倍 (横坐标不变) 得到 $y=2 \\cos x$ 的图像, 再将 $y=2 \\cos x$ 的图像向右平移 $\\frac{\\pi}{2}$ 个单位长度后得到 $y=2 \\cos \\left(x-\\frac{\\pi}{2}\\right)$ 的图像, 故 $f(x)=2 \\sin x$,\n\n$$\nf(x)+g(x)=2 \\sin x+\\cos x=\\sqrt{5} \\sin (x+\\varphi)\n$$\n\n（其中 $\\sin \\varphi=1 / \\sqrt{5}, \\cos \\varphi=2 /... | [
"$(-\\sqrt{5}, \\sqrt{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 | Interval | null | Open-ended | Algebra | Math | Chinese |
511 | 已知将函数 $g(x)=\cos x$ 图像上所有点的纵坐标伸长到原来的 2 倍 (横坐标不变), 再将所得图像向右平移 $\frac{\pi}{2}$ 个单位长度得到函数 $y=f(x)$ 的图像, 且关于 $x$ 的方程 $f(x)+g(x)=m$ 在 $[0,2 \pi)$ 内有两个不同的解 $\alpha, \beta$.
求 $\cos (\alpha-\beta)$ (用含 $m$ 的式子表示). | [
"首先求 $m$的取值范围.\n将 $g(x)=\\cos x$ 图像上的所有点的纵坐标伸长为原来的 2 倍 (横坐标不变) 得到 $y=2 \\cos x$ 的图像, 再将 $y=2 \\cos x$ 的图像向右平移 $\\frac{\\pi}{2}$ 个单位长度后得到 $y=2 \\cos \\left(x-\\frac{\\pi}{2}\\right)$ 的图像, 故 $f(x)=2 \\sin x$,\n\n$$\nf(x)+g(x)=2 \\sin x+\\cos x=\\sqrt{5} \\sin (x+\\varphi)\n$$\n\n（其中 $\\sin \\varphi=1 / \\sqrt{5}, \\c... | [
"$\\frac{2 m^{2}}{5}-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 |
512 | 已知数列 $\left\{a_{n}\right\}$ 满足: $a_{1}=1, a_{n+1}=a_{n}+a_{n}^{2}\left(n \in \mathbf{N}^{*}\right)$. 记
$$
S_{n}=\frac{1}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)}, \quad T_{n}=\sum_{k=1}^{n} \frac{1}{1+a_{k}}
$$
求 $S_{n}+T_{n}$ 的值. | [
"因为 $a_{n+1}=a_{n}+a_{n}^{2}$, 所以 $a_{n+1}=a_{n}\\left(1+a_{n}\\right)$, 所以 $\\frac{1}{1+a_{n}}=\\frac{a_{n}}{a_{n+1}}$, 所以\n\n$$\nS_{n}=\\frac{a_{1}}{a_{2}} \\cdot \\frac{a_{2}}{a_{3}} \\cdot \\frac{a_{3}}{a_{4}} \\cdots \\frac{a_{n}}{a_{n+1}}=\\frac{a_{1}}{a_{n+1}}=\\frac{1}{a_{n+1}}\n$$\n\n因为 $a_{n+1}=a_{n}\\lef... | [
"$1$"
] | 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 |
513 | 如图, 椭圆 $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的左焦点为 $F$, 过点 $F$ 的直线交椭圆于 $A, B$ 两点, 当直线 $A B$ 经过椭圆的一个顶点时, 其倾斜角恰为 $60^{\circ}$.
求该椭圆的离心率; | [
"依题意，当直线 $A B$ 经过椭圆的顶点 $(0, b)$ 时，其倾斜角为 $60^{\\circ}$.\n\n设 $F(-c, 0)$, 则 $\\frac{b}{c}=\\tan 60^{\\circ}=\\sqrt{3}$, 将 $b=\\sqrt{3} c$ 代入 $a^{2}=b^{2}+c^{2}$, 得 $a=2 c$, 所以椭圆的离心率为\n\n$$\ne=\\frac{c}{a}=\\frac{1}{2}\n$$"
] | [
"$\\frac{1}{2}$"
] | 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 | Conic Sections | Math | Chinese |
514 | 如图, 椭圆 $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的左焦点为 $F$, 过点 $F$ 的直线交椭圆于 $A, B$ 两点, 当直线 $A B$ 经过椭圆的一个顶点时, 其倾斜角恰为 $60^{\circ}$.
该线段 $A B$ 的中点为 $G, A B$ 的中垂线与 $x$ 轴, $y$ 轴分别交于 $D ,E $ 两点, 记 $\triangle G D F$ 的面积为 $S_{1}$ , $\triangle O E D$ ( $O$ 为坐标原点) 的面积为 $S_{2}$, 求 $\frac{S_{1}}{S_{2}}$ 的取值范围. | [
"依题意，当直线 $A B$ 经过椭圆的顶点 $(0, b)$ 时，其倾斜角为 $60^{\\circ}$.\n\n设 $F(-c, 0)$, 则 $\\frac{b}{c}=\\tan 60^{\\circ}=\\sqrt{3}$, 将 $b=\\sqrt{3} c$ 代入 $a^{2}=b^{2}+c^{2}$, 得 $a=2 c$, 所以椭圆的离心率为 $e=\\frac{c}{a}=\\frac{1}{2}$\n\n椭圆的方程可设为 $3 x^{2}+4 y^{2}=12 c^{2}$ ,\n\n设 $A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right)... | [
"$(9,+\\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 | Conic Sections | Math | Chinese |
515 | 求曲线 $f(x)=e^{x-1}$ 与曲线 $g(x)=\ln x$ 的公切线的条数. | [
"曲线 $f(x)=e^{x-1}$ 与曲线 $g(x)=\\ln x$ 有两条公切线.\n\n设两曲线的公切线为 $l$, 与曲线 $f(x)=e^{x-1}$ 切于点 $\\left(a, e^{a-1}\\right)$, 与曲线 $g(x)=\\ln x$ 切于点 $(b, \\ln b)$, 则直线 $l$ 的方程既可以写为 $y-e^{a-1}=e^{a-1}(x-a)$, 即: $y=e^{a-1} x+e^{a-1}-a e^{a-1}$;\n\n又可以写为 $y-\\ln b=\\frac{1}{b}(x-b)$, 即: $y=\\frac{1}{b} x+\\ln b-1$.\n\n因为直线 $l$ 为公... | [
"$2$"
] | 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 | Conic Sections | Math | Chinese |
520 | 已知 $\left\{a_{n}\right\}$ 是公差为 $d(d \neq 0)$ 的等差数列, 且 $a_{1}+t^{2}=a_{2}+t^{3}=a_{3}+t$.
求实数 $t, d$ 的值. | [
"由题 $\\left\\{\\begin{array}{l}a_{1}+t^{2}=a_{1}+d+t^{3} \\\\ a_{1}+d+t^{3}=a_{1}+2 d+t\\end{array}\\right.$, 即 $\\left\\{\\begin{array}{l}d=t^{2}-t^{3} \\\\ d=t^{3}-t\\end{array}\\right.$.\n\n因为 $d \\neq 0$, 所以 $t \\neq 0, t \\neq 1$, 所以由 $2 t^{2}-t-1=0$ 得 $t=-\\frac{1}{2}, d=\\frac{3}{8}$."
] | [
"$-\\frac{1}{2}, \\frac{3}{8}$"
] | 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 | Sequence | Math | Chinese |
521 | 已知 $\left\{a_{n}\right\}$ 是公差为 $d(d \neq 0)$ 的等差数列, 且 $a_{1}+t^{2}=a_{2}+t^{3}=a_{3}+t$.
若正整数满足 $m<p<r, a_{m}-2 t^{m}=a_{p}-2 t^{p}=a_{r}-2 t^{r}=0$, 求数组 $(m, p, r)$ 和相应通项公式 $a_{n}$. | [
"由题 $\\left\\{\\begin{array}{l}a_{1}+t^{2}=a_{1}+d+t^{3} \\\\ a_{1}+d+t^{3}=a_{1}+2 d+t\\end{array}\\right.$, 即 $\\left\\{\\begin{array}{l}d=t^{2}-t^{3} \\\\ d=t^{3}-t\\end{array}\\right.$.\n\n因为 $d \\neq 0$, 所以 $t \\neq 0, t \\neq 1$, 所以由 $2 t^{2}-t-1=0$ 得 $t=-\\frac{1}{2}, d=\\frac{3}{8}$;\n\n由\n\n$$\na_{m}-2 t^{... | [
"$(1,3,4), \\frac{3}{8} n-\\frac{11}{8}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple,Expression | null | Open-ended | Sequence | Math | Chinese |
524 | 设函数 $f(x)=e^{x}-1-x$.$f(x)$ 在区间 $\left[0, \frac{1}{n}\right]$ ( $n$ 为正整数) 上的最大值为 $b_{n}$.
求 $b_{n}$. | [
"因为 $f^{\\prime}(x)=e^{x}-1$, 所以, 当 $x \\in\\left[0, \\frac{1}{n}\\right]$ 时, $f^{\\prime}(x) \\geq 0$, 即 $f(x)$ 在 $x \\in\\left[0, \\frac{1}{n}\\right]$ 是增函数, 故 $f(x)$ 在 $x \\in\\left[0, \\frac{1}{n}\\right]$ 上的最大值为 $b_{n}=e^{\\frac{1}{n}}-1-\\frac{1}{n}$."
] | [
"$e^{\\frac{1}{n}}-1-\\frac{1}{n}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Elementary Functions | Math | Chinese |
529 | 已知正整数 $n$ 都可以唯一表示为
$$
n=a_{0}+a_{1} 9+a_{2} 9^{2}+\cdots+a_{m} 9^{m} \cdots (*)
$$
的形式, 其中 $m$ 为非负整数, $a_{j} \in\{0,1, \cdots, 8\}(j=0,1, \cdots, m-1), a_{m} \in\{1, \cdots, 8\}$. 试求 $(*)$ 中的数列 $a_{0}, a_{1}, a_{2}, \cdots, a_{m}$ 严格单调递增或严格单调递减的所有正整数 $n$ 的和. | [
"设 $A$ 和 $B$ 分别表示 (*) 中数列严格递增和递减的所有正整数构成的集合, 符号 $S(M)$ 表示数集 $M$中所有数的和, 并将满足 $(*)$ 式的正整数记为:\n\n$$\nn=\\overline{a_{m} a_{m-1} \\cdots a_{1} a_{0}}\n$$\n\n把集合 $A$ 分成如下两个不交子集 $A_{0}=\\left\\{n \\in A \\mid a_{0}=0\\right\\}$ 和 $A_{1}=\\left\\{n \\in A \\mid a_{0} \\neq 0\\right\\}$, 我们有\n\n$$\nS(A)=S\\left(A_{0}\\righ... | [
"$984374748$"
] | 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 |
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