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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
194 | 已知边长为 4 的正三角形 $A B C . D 、 E 、 F$ 分别是 $B C 、 C A 、 A B$ 上的点,且 $|A E|=|B F|=|C D|=1$, 连结 $A D 、 B E 、 C F$, 交成 $\triangle R Q S$. 点 $P$ 在 $\triangle R Q S$ 内及边上移动, 点 $P$到 $\triangle A B C$ 三边的距离分别记作 $x 、 y 、 z$.
求上述乘积 $x y z$ 的极小值. | [
"利用面积, 易证: (1) 当点 $P$ 在 $\\triangle A B C$ 内部及边上移动时, $x+y+z$为定值 $h=2 \\sqrt{3}$;\n\n(2)过 $P$ 作 $B C$ 的平行线 $l$, 交 $\\triangle A B C$ 的两边于 $G 、 H$. 当点 $P$ 在线段 $G H$上移动时, $y+z$ 为定值, 从而 $x$ 为定值.\n\n(3) 设 $y \\in[\\alpha, \\beta], m$ 为定值. 则函数 $u=y(m-y)$ 在点 $y=\\alpha$ 或 $y=\\beta$ 时取得极小值.\n\n<img_4188>\n\n于是可知, 过 $R$ 作 ... | [
"$\\frac{648}{2197} \\sqrt{3}$"
] | 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 | Geometry | Math | Chinese |
197 | 一张台球桌形状是正六边形 $A B C D E F$, 一个球从 $A B$ 的中点 $P$ 击出, 击中 $B C$ 边上的某点 $Q$, 并且依次碰击 $C D 、 D E 、 E F 、 F A$ 各边, 最后击中 $A B$ 边上的某一点. 设 $\angle B P Q=\vartheta$, 求 $\vartheta$ 的范围.
提示: 利用入射角等于反射角的原理. | [
"只要把这个正六边形经过 5 次对称变换. 则击球时应如图所示, 击球方向在 $\\angle M P N$ 内部时即可.\n\n<img_4211>\n\n\n设 $A B=2$, 以 $P$ 为原点, $P B$ 为 $x$ 轴正方向建立直角坐标系, 点 $M$ 坐标为 $(8,3 \\sqrt{3})$. 点 $N$坐标为 $(10,3 \\sqrt{3})$, 即 $\\vartheta \\in\\left[\\arctan \\frac{3 \\sqrt{3}}{10}, \\arctan \\frac{3 \\sqrt{3}}{8}\\right]$."
] | [
"$\\left[\\arctan \\frac{3 \\sqrt{3}}{10}, \\arctan \\frac{3 \\sqrt{3}}{8}\\right]$"
] | 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 | Geometry | Math | Chinese |
202 | 将编号为 $1,2, \cdots, 9$ 的九个小球随机放置在圆周的九个等分点上, 每个等分点上各有一个小球. 设圆周上所有相邻两球号码之差的绝对值之和为要 $\mathrm{S}$. 求使 $\mathrm{S}$ 达到最小值的放法的概率. (注: 如果某种放法, 经旋转或镜面反射后可与另一种放法重合, 则认为是相同的放法) | [
"九个编号不同的小球放在圆周的九个等分点上, 每点放一个, 相当于九个不同元素在圆周上的一个圆形排列, 故共有 $8 !$ 种放法, 考虑到翻转因素, 则本质不同的放法有 $\\frac{8 !}{2}$ 种。\n\n下求使 $\\mathrm{S}$ 达到最小值的放法数：在圆周上, 从 1 到 9 有优弧与劣弧两条路径, 对其中任一条路径, 设 $x_{1}, x_{2}, \\cdots, x_{k}$ 是依次排列于这段弧上的小球号码, 则\n\n$$\n\\left|1-x_{1}\\right|+\\left|x_{1}-x_{2}\\right|+\\cdots+| x_{k}-9|\\geq|\\left(1-x... | [
"$\\frac{1}{315}$"
] | 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 |
203 | 过抛物线 $y=x^{2}$ 上的一点 $\mathrm{A}(1,1)$ 作抛物线的切线, 分别交 $x$ 轴于 $\mathrm{D}$, 交 $y$ 轴于 B. 点 $\mathrm{C}$ 在抛物线上, 点 $\mathrm{E}$ 在线段 $\mathrm{AC}$ 上, 满足 $\frac{A E}{E C}=\lambda_{1}$; 点 $\mathrm{F}$ 在线段 $\mathrm{BC}$ 上, 满足 $\frac{B F}{F C}=\lambda_{2}$,且 $\lambda_{1}+\lambda_{2}=1$, 线段 $\mathrm{CD}$ 与 $\mathrm{EF}$ 交于点 $\mathrm{P}$. 当点 $\mathrm{C}$ 在抛物线上移动时, 求点 $\mathrm{P}$ 的轨迹方程. | [
"<img_4061>\n\n过抛物线上点$A$的切线斜率为：$y^{\\prime} = 2x|_{x=1}=2$, $\\because$ 切线$AB$的方程为$y=2x-1$. \n\n$\\therefore B、D$ 的坐标为 $B(0,-1), D\\left(\\frac{1}{2}, 0\\right), \\therefore D$ 是线段 $A B$ 的中点.\n\n令 $\\gamma=\\frac{C D}{C P}, t_{1}=\\frac{C A}{C E}=1+\\lambda_{1}, t_{2}=\\frac{C B}{C F}=1+\\lambda_{2}$, 则 $t_{1}+t_{2... | [
"$y=\\frac{1}{3}(3 x-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 | Equation | null | Open-ended | Geometry | Math | Chinese |
204 | 设正数 $a 、 \mathrm{~b} 、 \mathrm{c} 、 x 、 \mathrm{y} 、 \mathrm{z}$ 满足 $c y+b z=a, a z+c x=b ; b x+a y=c$.
求函数 $f(x, y, z)=\frac{x^{2}}{1+x}+\frac{y^{2}}{1+y}+\frac{z^{2}}{1+z}$ 的最小值. | [
"由条件得, $b(a z+c x-b)+c(b x+a y-c)-a(c y+b z-a)=0$,\n\n即 $2 b c x+a^{2}-b^{2}-c^{2}=0 ，$\n\n$\\therefore x=\\frac{b^{2}+c^{2}-a^{2}}{2 b c}$, 同理, 得 $y=\\frac{a^{2}+c^{2}-b^{2}}{2 a c} ; z=\\frac{a^{2}+b^{2}-c^{2}}{2 a b}$.\n\n$\\because a、 b 、 c 、x 、 y 、 z$ 为正数, 据以上三式知,\n\n$b^{2}+c^{2}>a^{2}, a^{2}+c^{2}>b^{2}, a^{2... | [
"$\\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 | Algebra | Math | Chinese |
206 | 某足球邀请赛有十六个城市参加, 每市派出甲、乙两个队, 根据比赛规则, 比赛若干天后进行统计, 发现除 $A$ 市甲队外, 其它各队已比赛过的场数各不相同. 问 $A$ 市乙队已赛过多少场？ | [
"用 32 个点表示这 32 个队, 如果某两队比赛了一场, 则在表示这两个队的点间连一条线. 否则就不连线.\n\n由于, 这些队比赛场次最多 30 场, 最少 0 场, 共有 31 种情况, 现除 $A$ 城甲队外还有 31 个队, 这 31 个队比赛场次互不相同, 故这 31 个队比赛的场次恰好从 0 到 30 都有. 就在表示每个队的点旁注上这队的比赛场次.\n\n考虑比赛场次为 30 的队, 这个队除自己与同城的队外, 与不同城有队都进行了比赛,于是, 它只可能与比赛 0 场的队同城; 再考㱆比赛 29 场的队, 这个队除与同城队及比赛 0 场、 1 场 (只赛 1 场的队已经与比赛 30 场的队赛过 1 场, 故... | [
"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 | Combinatorics | Math | Chinese |
211 | 在 $\triangle A B C$ 中, 内角 $A, B, C$ 的对边分别为 $a, b, c, a=3, \cos C=-\frac{1}{15}, 5 \sin (B+C)=3 \sin (A+C)$.
求边 $c$. | [
"由 $5 \\sin (B+C)=3 \\sin (A+C)$ 得 $5 \\sin A=3 \\sin B$\n\n由正弦定理得 $5 a=3 b$\n\n$\\because a=3, \\therefore b=5$\n\n由余弦定理得 $c^{2}=a^{2}+b^{2}-2 a b \\cos C=36$\n\n$\\therefore c=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 | Trigonometric Functions | Math | Chinese |
212 | 在 $\triangle A B C$ 中, 内角 $A, B, C$ 的对边分别为 $a, b, c, a=3, \cos C=-\frac{1}{15}, 5 \sin (B+C)=3 \sin (A+C)$.
$c=6$，求 $\sin \left(B-\frac{\pi}{3}\right)$ 的值. | [
"由余弦定理求出 $\\cos B=\\frac{5}{9}$\n\n再求出 $\\sin B=\\frac{2 \\sqrt{14}}{9}$\n\n$\\therefore \\sin \\left(B-\\frac{\\pi}{3}\\right)=\\sin B \\cos \\frac{\\pi}{3}-\\cos B \\sin \\frac{\\pi}{3}$\n\n$=\\frac{2 \\sqrt{14}-5 \\sqrt{3}}{18}$."
] | [
"$\\frac{2 \\sqrt{14}-5 \\sqrt{3}}{18}$"
] | 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 |
214 | 有甲乙两个班级进行一门课程的考试, 按照学生考试优秀和不优秀统计成绩后, 得到如下列联表:
班级与成绩列联表
| | 优秀 | 不优秀 | 合计 |
| :--: | :--: | :----: | :--: |
| 甲班 | 20 | | 45 |
| 乙班 | | 40 | |
| 合计 | | | 90 |
参考数据:
| $P\left(K^{2} \geq k_{0}\right)$ | 0.15 | 0.10 | 0.05 | 0.025 | 0.010 | 0.005 | 0.001 |
| :--------------------------------: | :---: | :---: | :---: | :---: | :---: | :---: | :----: |
| $k_{0}$ | 2.072 | 2.706 | 3.841 | 5.024 | 6.635 | 7.879 | 10.828 |
参考公式: $K^{2}=\frac{n(a d-b c)^{2}}{(a+b)(a+c)(c+d)(b+d)}$
针对调查的 90 名同学, 各班都想办法要提高班级同学的优秀率, 甲班决定从调查的 45 名同学中按分层抽样的方法随机抽取 9 名同学组成学习互助小组, 每单元学习结束后在这 9 人中随机抽取 2 人负责制作本单元思维导图, 设这 2 人中优秀人数为 $X$, 求 $X$ 的数学期望. | [
"甲班 45 人, 按照分层抽样的方法随机抽取 9 人、其中优秀 4 人, 不优秀 5 人.故 $X$ 的应有可能取值为 $0,1,2$\n\n$$\nP(X=0)=\\frac{C_{4}^{0} C_{5}^{2}}{C_{9}^{2}}=\\frac{5}{18}, P(X=1)=\\frac{C_{4}^{1} C_{5}^{1}}{C_{9}^{2}}=\\frac{5}{9}, \\quad P(X=2)=\\frac{C_{4}^{2} C_{5}^{0}}{C_{9}^{2}}=\\frac{1}{6}\n$$\n\n故 $X$ 的分布列为:\n\n| $X$ | 0 | ... | [
"$\\frac{8}{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 | Probability and Statistics | Math | Chinese |
218 | 已知椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ 过 $\left(1, \frac{\sqrt{3}}{2}\right)$ 和 $\left(-\sqrt{2}, \frac{\sqrt{2}}{2}\right)$ 两点.
求椭圆的标准方程. | [
"$\\frac{x^{2}}{4}+y^{2}=1$."
] | [
"$\\frac{x^{2}}{4}+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 | Plane Geometry | Math | Chinese |
219 | 已知椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ 过 $\left(1, \frac{\sqrt{3}}{2}\right)$ 和 $\left(-\sqrt{2}, \frac{\sqrt{2}}{2}\right)$ 两点.
已知 $P$ 为椭圆 $C$ 上不同于顶点的任意一点, $A, B$ 为椭圆的左. 右顶点, 直线 $A P, B P$ 分别与定直线 $l: x=-6$相交于 $M, N$ 两点, 设线段 $M N$ 中点为 $Q$, 若 $\overrightarrow{A Q} \cdot \overrightarrow{B Q}$ 的值晨小, 求此时 $Q$ 的坐标. | [
"$\\frac{x^{2}}{4}+y^{2}=1$;\n\n令 $A(-2,0), B(2,0)$, 设 $P(x, y)(x y \\neq 0)$,\n\n$k_{A P} \\cdot k_{B P}=\\frac{y}{x+2} \\cdot \\frac{y}{x-2}=\\frac{y^{2}}{x^{2}-4}=\\frac{1-\\frac{x^{2}}{4}}{x^{2}-4}=-\\frac{1}{4}$\n\n$\\therefore$ 令 $k_{A P}=k, k_{B P}=-\\frac{1}{4 k}(k \\neq 0)$\n\n$l_{A P}: y=k(x+2), \\therefo... | [
"$(-6,0)$"
] | 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 | Tuple | null | Open-ended | Plane Geometry | Math | Chinese |
221 | 在平面直角坐标系中, 以坐标原点为极点, $x$ 轴的正半轴为极轴建立极坐标系, 已知直线 $l$ 的极坐标方程为
$\rho \sin \left(\theta-\frac{\pi}{6}\right)+3=0$, 椭圆 $\mathrm{C}$ 的参数方程为 $\left\{\begin{array}{l}x=2 \cos \alpha \\ y=\sin \alpha\end{array}\right.$ ( $\alpha$ 为参数).
求直线 $l$ 的直角坐标方程; | [
"$\\rho\\left(\\sin \\theta \\cos \\frac{\\pi}{6}-\\cos \\theta \\sin \\frac{\\pi}{6}\\right)+3=0$\n\n$\\frac{\\sqrt{3}}{2} \\rho \\sin \\theta-\\frac{1}{2} \\rho \\cos \\theta+3=0$\n\n$\\therefore$ 直线 $l: x-\\sqrt{3} y-6=0$."
] | [
"$x-\\sqrt{3} y-6=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 | Polar Coordinates and Parametric Equations | Math | Chinese |
222 | 在平面直角坐标系中, 以坐标原点为极点, $x$ 轴的正半轴为极轴建立极坐标系, 已知直线 $l$ 的极坐标方程为
$\rho \sin \left(\theta-\frac{\pi}{6}\right)+3=0$, 椭圆 $\mathrm{C}$ 的参数方程为 $\left\{\begin{array}{l}x=2 \cos \alpha \\ y=\sin \alpha\end{array}\right.$ ( $\alpha$ 为参数).
椭圆 $C$ 上一点到直线 $l$ 的最小距离是多少? | [
"$\\rho\\left(\\sin \\theta \\cos \\frac{\\pi}{6}-\\cos \\theta \\sin \\frac{\\pi}{6}\\right)+3=0$\n\n$\\frac{\\sqrt{3}}{2} \\rho \\sin \\theta-\\frac{1}{2} \\rho \\cos \\theta+3=0$\n\n$\\therefore$ 直线 $l: x-\\sqrt{3} y-6=0$.\n\n设 $P(2 \\cos \\alpha, \\sin \\alpha)$\n\n点 $P$ 到$l$的距离\n\n$d=\\frac{|2 \\cos \\alpha-\\... | [
"$\\frac{6-\\sqrt{7}}{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 | Polar Coordinates and Parametric Equations | Math | Chinese |
223 | 已知函数 $f(x)=|x+1|-|x-2|$.
求不等式 $f(x)<1$ 的解集. | [
"$f(x)=\\left\\{\\begin{array}{l}3, x \\geq 2, \\\\ 2 x-1,-1<x<2, \\\\ -3, x \\leq-1 .\\end{array}\\right.$\n\n$\\therefore x \\in(-\\infty, 1)$\n\n$\\therefore$ 不等式的解集为 $(-\\infty, 1)$."
] | [
"$(-\\infty, 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 | Inequality | Math | Chinese |
224 | 已知函数 $f(x)=|x+1|-|x-2|$.
若 $f(x)+2|x-2| \geq m$ 恒成立, 求实数 $m$ 的取值范围. | [
"设 $g(x)=f(x)+2|x-2|$, 则 $m \\leq g(x)_{\\text {min }}=3$."
] | [
"$(-\\infty, 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 | Interval | null | Open-ended | Inequality | Math | Chinese |
225 | 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 (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)$.
求半圆 C 的极坐标方程. | [
"半圆 C 的直角坐标方程为 $x^{2}+(y-1)^{2}=1(y>1)$,\n\n它的极坐标方程是 $\\rho=2 \\sin \\theta, \\theta \\in\\left(\\frac{\\pi}{4}, \\frac{3 \\pi}{4}\\right)$."
] | [
"$\\rho=2 \\sin \\theta$"
] | 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 |
226 | 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 (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)$.
若一直线与量坐标轴的交点分别为 $\mathrm{A}, \mathrm{B}$, 其中 $\mathrm{A}(0,-2)$, 点 $\mathrm{D}$ 在半圆 $\mathrm{C}$ 上, 且真线 $\mathrm{CD}$ 的倾斜角是直线 $A B$ 倾斜角的 2 倍, 若 $\triangle A B D$ 的面积为4,点 $D$ 的直角坐标. | [
"设直线 $l$ 的倾斜角为 $\\alpha$, 则直线 $l$ 的方程为 $y=x \\tan \\alpha-2$,\n\n$D(\\cos 2 \\alpha, 1+\\sin 2 \\alpha), 2 \\alpha \\in(0, \\pi),|A B|=\\frac{2}{\\sin \\alpha}$\n\n点 $D$ 到直线 $l$ 的距离为 $|\\sin \\alpha \\cos 2 \\alpha-\\cos \\alpha \\sin 2 \\alpha-3 \\cos \\alpha|=3 \\cos \\alpha+\\sin \\alpha$\n\n由 $\\triangle A B D$... | [
"$(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 | Tuple | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
227 | 已知函数 $f(x)=|x+2|-m|x-1|$.
若 $\mathrm{m}=-2$ 时, 解不等式 $f(x) \geq 5$; | [
"当 $m=-2$ 时, $f(x)=\\left\\{\\begin{array}{l}-3 x, x<-2 \\\\ 4-x,-2 \\leq x \\leq 1 \\\\ 3 x, x>1\\end{array}\\right.$\n\n由 $f(x)$ 的单调性及 $f(-1)=f\\left(\\frac{5}{3}\\right)=5$,\n\n得 $f(x) \\geq 5$ 的解集为 $\\left\\{x \\mid x \\leq-1\\right.$ 或 $\\left.x \\geq \\frac{5}{3}\\right\\}$."
] | [
"$(-\\infty, -1] \\cup [\\frac{5}{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 |
228 | 已知函数 $f(x)=|x+2|-m|x-1|$.
若 $f(x) \leq m(x+5)$, 求 $\mathrm{m}$ 的最小值. | [
"由 $f(x) \\leq m|x+5|$ 得 $\\mathbb{m} \\geq \\frac{|x+2|}{|x-1|+|x+5|}$,\n\n由 $|x-1|+|x+5| \\geq 2|x+2|$ 得 $\\frac{|x+2|}{|x-1|+|x+5|} \\leq \\frac{1}{2}$, 得 $m \\geq \\frac{1}{2}$.\n\n（当且仅当 $x \\geqslant 4$ 或 $x \\leqslant-6$ 时等号成立）\n\n故 $m$ 的最小值为 $\\frac{1}{2}$."
] | [
"$\\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 | Inequality | Math | Chinese |
229 | 在 $\triangle A B C$ 中, $A B=3, A C=1, \angle A=60^{\circ}$.
求 $\sin \angle A C B$. | [
"$\\because A B=3, A C=1, \\angle A=60^{\\circ}$, 所以由余弦定理可知,\n\n$B C^{2}=3^{2}+1^{2}-2 \\times 3 \\times 1 \\times \\cos 60^{\\circ}, \\quad \\therefore B C=\\sqrt{7}$.\n\n根据正弦定理, $\\frac{3}{\\sin \\angle A C B}=\\frac{\\sqrt{7}}{\\frac{\\sqrt{3}}{2}}, \\therefore \\sin \\angle A C B=\\frac{3 \\sqrt{21}}{14}$."
] | [
"$\\frac{3 \\sqrt{21}}{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 | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
230 | 在 $\triangle A B C$ 中, $A B=3, A C=1, \angle A=60^{\circ}$.
若 $D$ 为 $B C$ 的中点, 求 $A D$ 的长度. | [
"$\\because A B=3, A C=1, \\angle A=60^{\\circ}$, 所以由余弦定理可知,\n\n$B C^{2}=3^{2}+1^{2}-2 \\times 3 \\times 1 \\times \\cos 60^{\\circ}, \\quad \\therefore B C=\\sqrt{7}$.\n\n根据正弦定理, $\\frac{3}{\\sin \\angle A C B}=\\frac{\\sqrt{7}}{\\frac{\\sqrt{3}}{2}}, \\therefore \\sin \\angle A C B=\\frac{3 \\sqrt{21}}{14}$.\n\n$... | [
"$\\frac{\\sqrt{13}}{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 |
234 | 设盒子中装有 6 个红球, 4 个白球, 2 个黑球, 且规定：取出一个红球得 $a$ 分, 取出一个白球得 $b$ 分,取出一个黑球得 $c$ 分, 其中 $a, b, c$ 都为正整数.
当 $a=1$ 时, 从该盒子中任取（每球取到的机会均等） 1 个球, 记随机变量 $\eta$ 为取出此球所得分数. 若 $E \eta=\frac{5}{3}, D \eta=\frac{5}{9}$, 求 $b$ 和 $c$. | [
"由题意知 $\\eta$ 的分布列为\n\n| $\\eta$ | 1 | $b$ | $c$ |\n| :------- | :-------------- | :-------------- | :-------------- |\n| $P$ | $\\frac{1}{2}$ | $\\frac{1}{3}$ | $\\frac{1}{6}$ |\n\n$E \\eta=\\frac{1}{2}+\\frac{1}{3} b+\\frac{1}{6} c=\\frac{5}{3}, \\quad D(\\eta)=\\left(1-\\frac... | [
"$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 | Probability and Statistics | Math | Chinese |
235 | 设椭圆 $C: \frac{x^{2}}{4}+y^{2}=1$ 的右焦点为 $F$, 过点 $(m, 0)(|m| \geq 1)$ 作直线 $l$ 与椭圆 $C$ 交于 $A, B$ 两点, 且坐标原点 $O(0,0)$ 到直线 $l$ 的距离为 1 .
当 $m=1$ 时, 求直线 $A F$ 的方程. | [
"焦点 $F(\\sqrt{3}, 0)$, 当 $m=1$ 时, 直线 $l: x=1$, 点 $A\\left(1, \\pm \\frac{\\sqrt{3}}{2}\\right), k_{A F}=\\frac{\\frac{\\sqrt{3}}{2}-0}{1-\\sqrt{3}}$ 或 $\\frac{-\\frac{\\sqrt{3}}{2}-0}{1-\\sqrt{3}}$\n\n$\\therefore$ 直线 $A F$ 的方程为: $y=-\\frac{\\sqrt{3}+3}{4}(x-\\sqrt{3})$ 或 $y=\\frac{\\sqrt{3}+3}{4}(x-\\sqrt{3})$."
] | [
"$y=-\\frac{\\sqrt{3}+3}{4}(x-\\sqrt{3}) , y=\\frac{\\sqrt{3}+3}{4}(x-\\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 | Expression | null | Open-ended | Plane Geometry | Math | Chinese |
236 | 设椭圆 $C: \frac{x^{2}}{4}+y^{2}=1$ 的右焦点为 $F$, 过点 $(m, 0)(|m| \geq 1)$ 作直线 $l$ 与椭圆 $C$ 交于 $A, B$ 两点, 且坐标原点 $O(0,0)$ 到直线 $l$ 的距离为 1 .
求 $\triangle A B F$ 面积的最大值. | [
"当直线 $l$ 的斜率不存在时, $m= \\pm 1, S_{\\triangle A B F}=\\frac{3-\\sqrt{3}}{2}$ 或 $\\frac{3+\\sqrt{3}}{2}$ .\n\n当直线 $l$ 的斜率存在时, 设直线 $l: y=k(x-m)$, 联立方程 $\\left\\{\\begin{array}{l}\\frac{x^{2}}{4}+y^{2}=1 \\\\ y=k(x-m)\\end{array}\\right.$, 得 $\\left(1+4 k^{2}\\right) x^{2}-8 m k^{2} x+4 m^{2} k^{2}-4=0$. 设 $A\\left(x_{1... | [
"$\\frac{3+\\sqrt{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 | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
239 | 在直角坐标系 $x O y$ 中, 直线 $l$ 的参数方程为 $\left\{\begin{array}{l}x=-2+\frac{\sqrt{2}}{2} t \\ y=-4+\frac{\sqrt{2}}{2} t\end{array}\right.$ ( $t$ 为参数 $)$, 点 $M(-2,-4)$. 以坐标原点为极点, $x$ 轴正半轴为极轴建立极坐标系, 曲线 $C$ 的极坐标方程为 $\rho \sin ^{2} \theta-2 a \cos \theta=0(a>0)$.
当 $a=1$ 时，求曲线 $C$ 的直角坐标方程. | [
"曲线 $C: y^{2}=2 x$."
] | [
"$y^{2}=2 x$"
] | 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 |
240 | 在直角坐标系 $x O y$ 中, 直线 $l$ 的参数方程为 $\left\{\begin{array}{l}x=-2+\frac{\sqrt{2}}{2} t \\ y=-4+\frac{\sqrt{2}}{2} t\end{array}\right.$ ( $t$ 为参数 $)$, 点 $M(-2,-4)$. 以坐标原点为极点, $x$ 轴正半轴为极轴建立极坐标系, 曲线 $C$ 的极坐标方程为 $\rho \sin ^{2} \theta-2 a \cos \theta=0(a>0)$.
设曲线 $C$ 与直线 $l$ 交于点 $A, B$, 若 $|A B|^{2}=|M A| \cdot|M B|$, 求 $a$ 的值. | [
"曲线 $C: y^{2}=2 a x$, 直线 $l$ 的参数方程代入曲线 $C$ 的直角坐标方程得,\n\n$t^{2}-(8 \\sqrt{2}+2 \\sqrt{2} a) t+8 a+32=0$,\n\n设交点 $A, B$ 所对参数分别为 $t_{1}, t_{2}$,\n\n则 $t_{1}+t_{2}=8 \\sqrt{2}+2 \\sqrt{2} a, t_{1} \\cdot t_{2}=8 a+32$,\n\n$\\therefore|A B|=\\left|t_{1}-t_{2}\\right|=\\sqrt{\\left(t_{1}+t_{2}\\right)^{2}-4 t_{1} t_{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 | Polar Coordinates and Parametric Equations | Math | Chinese |
241 | 已知 $f(x)=|x+2|-|a x-3|$.
当 $a=2$ 时, 求不等式 $f(x)>2$ 的解集. | [
"$f(x)=|x+2|-|2 x-3|= \\begin{cases}5-x & x \\geq \\frac{3}{2} \\\\ 3 x-1 & -2<x<\\frac{3}{2} \\\\ x-5 & x \\leq-2\\end{cases}$\n\n当 $x \\geq \\frac{3}{2}$ 时, $5-x>2, \\therefore x \\in\\left[\\frac{3}{2}, 3\\right)$\n\n当 $-2<x<\\frac{3}{2}$ 时, $3 x-1>2, \\therefore x \\in\\left(1, \\frac{3}{2}\\right)$\n\n当 $x \\l... | [
"$(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 | Interval | null | Open-ended | Inequality | Math | Chinese |
243 | 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=1, a_{n+1}=2 a_{n}+3 \times 2^{n-1}$, 设 $b_{n}=\frac{a_{n}}{2^{n}}$.
求 $b_{1}, b_{2}, b_{3}$. | [
"$a_{2}=5, a_{3}=16, b_{1}=\\frac{1}{2}, b_{2}=\\frac{5}{4}, b_{3}=2 ;$"
] | [
"$\\frac{1}{2},\\frac{5}{4},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 |
245 | 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=1, a_{n+1}=2 a_{n}+3 \times 2^{n-1}$, 设 $b_{n}=\frac{a_{n}}{2^{n}}$.
求数列 $\left\{a_{n}\right\}$ 的通项公式. | [
"$b_{1}=\\frac{1}{2}$,\n\n$\\because \\frac{a_{n+1}}{2^{n+1}}=\\frac{a_{n}}{2^{n}}+\\frac{3}{4}, \\therefore b_{n+1}-b_{n}=\\frac{3}{4}, \\therefore\\left\\{b_{n}\\right\\}$ 是等差数列;\n\n$\\therefore b_{n}=\\frac{1}{2}+(n-1) \\times \\frac{3}{4}=\\frac{1}{4}(3 n-1) ， \\therefore a_{n}=(3 n-1) \\cdot 2^{n-2}$."
] | [
"$(3 n-1) \\cdot 2^{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 | Expression | null | Open-ended | Sequence | Math | Chinese |
246 | 某集团公司计划从甲分公司中的 3 位员工 $A_{1} 、 A_{2} 、 A_{3}$ 和乙分公司中的 3 位员工 $B_{1} 、 B_{2} 、 B_{3}$ 选择 2 位员工去国外工作.
若从这 6 名员工中任选 2 名, 求这 2 名员工都是甲分公司的概率. | [
"由题意得, 从 6 名员工中任选 2 名, 其一切可能的结果组成的基本事件有:\n\n$\\left\\{A_{1}, A_{2}\\right\\},\\left\\{A_{1}, A_{3}\\right\\},\\left\\{A_{2}, A_{3}\\right\\},\\left\\{A_{1}, B_{1}\\right\\},\\left\\{A_{1}, B_{2}\\right\\},\\left\\{A_{1}, B_{3}\\right\\}$,\n\n$\\left\\{A_{2}, B_{1}\\right\\},\\left\\{A_{2}, B_{2}\\right\\},\\left\\{... | [
"$\\frac{1}{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 |
247 | 某集团公司计划从甲分公司中的 3 位员工 $A_{1} 、 A_{2} 、 A_{3}$ 和乙分公司中的 3 位员工 $B_{1} 、 B_{2} 、 B_{3}$ 选择 2 位员工去国外工作.
若从甲分公司和乙分公司中各任选 1 名员工，求这 2 名员工包括 $A_{1}$ 但不包括 $B_{1}$ 的概率. | [
"从甲分公司和乙分公司各任选 1 名员工, 其一切可能的结果组成的基本事件有: $\\left\\{A_{1}, B_{1}\\right\\},\\left\\{A_{1}, B_{2}\\right\\},\\left\\{A_{1}, B_{3}\\right\\},\\left\\{A_{2}, B_{1}\\right\\},\\left\\{A_{2}, B_{2}\\right\\},\\left\\{A_{2}, B_{3}\\right\\}\\left\\{A_{3}, B_{1}\\right\\},\\left\\{A_{3}, B_{2}\\right\\},\\left\\{A_{3}, B_{3... | [
"$\\frac{2}{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 | Probability and Statistics | Math | Chinese |
250 | 已知 $O$ 为坐标原点, 抛物线 $C: y^{2}=4 x$, 点 $A(-2,0)$, 设直线 $l$ 与 $C$ 交于不同的两点 $P 、 Q$.
若直线 $l \perp x$ 轴, 求直线 $P A$ 的斜率的取值范围. | [
"当点 $P$ 在第一象限时, 设 $P(t, 2 \\sqrt{t}), k_{P A}=\\frac{2 \\sqrt{t}-0}{t+2}=\\frac{2}{\\sqrt{t}+\\frac{2}{\\sqrt{t}}} \\leq \\frac{2}{2 \\sqrt{2}}=\\frac{\\sqrt{2}}{2}$,\n\n$\\therefore k_{P A} \\in\\left(0, \\frac{\\sqrt{2}}{2}\\right]$, 同理, 当点 $P$ 在第四象限时, $\\therefore k_{P A} \\in\\left[-\\frac{\\sqrt{2}}{2}, 0\\rig... | [
"$\\left[-\\frac{\\sqrt{2}}{2}, 0\\right) \\cup\\left(0, \\frac{\\sqrt{2}}{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 | Plane Geometry | Math | Chinese |
253 | 已知函数 $f(x)=e^{2 x}-3 a e^{x}-2 a^{2} x$.
若对于任意的 $x_{1}, x_{2} \in R, x_{1} \neq x_{2}$, 都有 $\frac{f\left(x_{1}\right)-f\left(x_{2}\right)}{x_{1}-x_{2}}>-2 a^{2}-1$ 恒成立, 求实数 $a$ 的取值范围. | [
"$\\because \\frac{f\\left(x_{1}\\right)-f\\left(x_{2}\\right)}{x_{1}-x_{2}}>-2 a^{2}-1, \\therefore$ 不妨设 $x_{1}>x_{2}$,\n\n则 $f\\left(x_{1}\\right)+\\left(2 a^{2}+1\\right) x_{1}>f\\left(x_{2}\\right)+\\left(2 a^{2}+1\\right) x_{2}$,\n\n$\\therefore f(x)+\\left(2 a^{2}+1\\right) x$ 在 $R$ 上单调递增:\n\n记 $g(x)=f(x)+\\l... | [
"$(-\\infty, \\frac{2 \\sqrt{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 | false | null | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
254 | 在直角坐标系 $x O y$ 中, 直线 $l$ 的参数方程为 $\left\{\begin{array}{l}x=1+t \cos \alpha \\ y=1+t \sin \alpha\end{array}\right.$, ( $t$ 为参数). 以坐标原点为极点, $x$ 轴正半轴为极轴建立极坐标系, 曲线 $C$ 的极坐标方程为 $\rho^{2}-4 \rho \cos \theta-1=0$.
求曲线 $C$ 的直角坐标方程; | [
"$x^{2}+y^{2}-4 x-1=0$."
] | [
"$x^{2}+y^{2}-4 x-1=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 | Polar Coordinates and Parametric Equations | Math | Chinese |
255 | 在直角坐标系 $x O y$ 中, 直线 $l$ 的参数方程为 $\left\{\begin{array}{l}x=1+t \cos \alpha \\ y=1+t \sin \alpha\end{array}\right.$, ( $t$ 为参数). 以坐标原点为极点, $x$ 轴正半轴为极轴建立极坐标系, 曲线 $C$ 的极坐标方程为 $\rho^{2}-4 \rho \cos \theta-1=0$.
设曲线 $C$ 与直线 $l$ 交于点 $A, B$, 求 $|A B|$ 的最小值. | [
"当 $\\alpha=n \\pi+\\frac{\\pi}{2}(n \\in Z)$ 时, 直线 $l: x=1$, 此时 $|A B|=4 $;\n\n当 $\\alpha \\neq n \\pi+\\frac{\\pi}{2}(n \\in Z)$ 时, 设直线 $l: y-1=k(x-1)$, 圆心 $(2,0)$ 到直线 $l$ 的距离最大值为 $d_{\\text {max }}=\\sqrt{2}$, 此时 $|A B|=2 \\sqrt{5-2}=2 \\sqrt{3}, \\because 4>2 \\sqrt{3}, \\therefore|A B|_{\\text {min }}=2 \\sqrt... | [
"$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 | false | null | Numerical | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
256 | 已知函数 $f(x)=|x-2|+|x+1|, x \in R$, 其最小值为 $t$.
求 $t$ 的值. | [
"$f(x)=\\left\\{\\begin{array}{c}2 x-1, x>2 \\\\ 3,-1 \\leq x \\leq 2 \\\\ -2 x+1, x<-1\\end{array}, f(x)_{\\text {min }}=3\\right.$, 即 $t=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 |
258 | 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=2, a_{n+1}=3 a_{n}+2$, 设 $b_{n}=a_{n}+1$.
求 $b_{1}, b_{2}, b_{3}$. | [
"$b_{1}=3, b_{2}=9, b_{3}=27$."
] | [
"$3, 9, 27$"
] | 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 |
260 | 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=2, a_{n+1}=3 a_{n}+2$, 设 $b_{n}=a_{n}+1$.
求数列 $\left\{a_{n}\right\}$ 的通项公式. | [
"$\\because \\frac{b_{n+1}}{b_{n}}=\\frac{a_{n+1}+1}{a_{n}+1}=\\frac{3 a_{n}+3}{a_{n}+1}=3, \\therefore\\left\\{b_{n}\\right\\}$ 是等比数列.\n\n可得 $b_{n}=3^{n}, a_{n}=3^{n}-1$."
] | [
"$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 |
266 | 在直角坐标系 $x O y$ 中, 直线 $l$ 的参数方程为 $\left\{\begin{array}{l}x=2+t \cos \alpha \\ y=1+t \sin \alpha\end{array}\right.$ ( $t$ 为参数). 以坐标原点为极点, $x$ 轴正半轴为极轴建立极坐标系, 曲线 $C$ 的极坐标方程为 $\rho^{2}-6 \rho \sin \theta-7=0$.
求曲线 $C$ 的直角坐标方程. | [
"曲线 $C: x^{2}+y^{2}-6 y-7=0$."
] | [
"$x^{2}+y^{2}-6 y-7=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 | Polar Coordinates and Parametric Equations | Math | Chinese |
267 | 在直角坐标系 $x O y$ 中, 直线 $l$ 的参数方程为 $\left\{\begin{array}{l}x=2+t \cos \alpha \\ y=1+t \sin \alpha\end{array}\right.$ ( $t$ 为参数). 以坐标原点为极点, $x$ 轴正半轴为极轴建立极坐标系, 曲线 $C$ 的极坐标方程为 $\rho^{2}-6 \rho \sin \theta-7=0$.
设曲线 $C$ 与直线 $l$ 交于点 $A, B$, 若点 $Q$ 的坐标为 $(2,1)$, 求 $|Q A|+|Q B|$ 的最小值. | [
"曲线 $C: x^{2}+y^{2}-6 y-7=0$.\n\n直线 $l$ 的参数方程代入曲线 $C$ 的直角坐标方程得,\n\n$t^{2}+4(\\cos \\alpha-\\sin \\alpha) t-8=0$,\n\n设交点 $A, B$ 所对参数分别为 $t_{1}, t_{2}$,\n\n则 $t_{1}+t_{2}=-4(\\cos \\alpha-\\sin \\alpha), t_{1} \\cdot t_{2}=-8$,\n\n$\\therefore|Q A|+|Q B|=\\left|t_{1}\\right|+\\left|t_{2}\\right|=\\left|t_{1}-t_{2}\\r... | [
"$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 | Polar Coordinates and Parametric Equations | Math | Chinese |
268 | 设函数 $f(x)=|x+3|+|x-a|-10$.
当 $a=1$ 时, 求不等式 $f(x)>0$ 的解集. | [
"当 $a=1$ 时, $f(x)=|x+3|+|x-1|-10=\\left\\{\\begin{array}{c}2 x-8, x \\geq 1 \\\\ -6,-3<x<1 \\\\ -2 x-12, x \\leq-3\\end{array}\\right.$, 所以 $f(x)>0$ 的解集为 $(-\\infty,-6) \\cup(4,+\\infty)$."
] | [
"$(-\\infty,-6) \\cup(4,+\\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 |
269 | 设函数 $f(x)=|x+3|+|x-a|-10$.
如果对任意的 $x$, 不等式 $f(x)>0$ 恒成立, 求 $a$ 的取值范围. | [
"$\\because|x+3|+|x-a|-10 \\geq \\mid 3+a \\mid-10$\n\n$\\therefore|3+a|-10>0, \\quad \\therefore a \\in(-\\infty,-13) \\cup(7,+\\infty)$."
] | [
"$(-\\infty,-13) \\cup(7,+\\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 |
270 | 已知 $\triangle A B C$ 的面积为 $S$, 其外接圆半径为 $R$,三个内角 $A, B, C$ 所对的边分别为 $a, b, c, 2 R\left(\sin ^{2} A-\sin ^{2} C\right)=$ $(\sqrt{3} a-b) \sin B$.
求角 $C$. | [
"根据题意, 有\n\n$$\n2 R \\cdot \\sin (A+C) \\cdot \\sin (A-C)=(\\sqrt{3} a-b) \\cdot \\sin B\n$$\n\n即\n\n$$\n\\sin (A-C)=\\sqrt{3} \\sin A-\\sin B,\n$$\n\n也即\n\n$$\n\\sqrt{3} \\sin A=\\sin (A-C)+\\sin (A+C),\n$$\n\n于是\n\n$$\n\\sqrt{3} \\sin A=2 \\sin A \\cdot \\cos C\n$$\n\n解得 $C=\\frac{\\pi}{6}$."
] | [
"$\\frac{\\pi}{6}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
271 | 已知 $\triangle A B C$ 的面积为 $S$, 其外接圆半径为 $R$,三个内角 $A, B, C$ 所对的边分别为 $a, b, c, 2 R\left(\sin ^{2} A-\sin ^{2} C\right)=$ $(\sqrt{3} a-b) \sin B$.
若 $\left(\frac{\sqrt{S}}{2 R}\right)^{2}=\sin ^{2} A-(\sin B-\sin C)^{2}, a=4$, 求 $c$ 及 $\triangle A B C$ 的面积. | [
"根据题意, 有\n\n$$\nS=a^{2}-(b-c)^{2},\n$$\n\n也即\n\n$$\n\\sqrt{p(p-a)(p-b)(p-c)}=4(p-b)(p-c),\n$$\n\n其中 $p$ 为 $\\triangle A B C$ 的半周长. 因此\n\n$$\n\\sqrt{\\frac{(p-b)(p-c)}{p(p-a)}}=\\frac{1}{4}\n$$\n\n也即\n\n$$\n\\tan \\frac{A}{2}=\\frac{1}{4}\n$$\n\n从而\n\n$$\n\\sin A=\\frac{8}{17}\n$$\n\n由正弦定理, 可得 $c=\\frac{17}{4}$. 又\n... | [
"$\\frac{17}{4}, \\frac{15+8 \\sqrt{3}}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Trigonometric Functions | Math | Chinese |
274 | 已知圆 $B:(x+\sqrt{2})^{2}+y^{2}=16$, 定点 $A(\sqrt{2}, 0), P$ 是圆周上任一点, 线段 $A P$ 的垂直平分线与 $B P$交于点 $Q$.
求点 $Q$ 的轨迹 $C$ 的方程; | [
"根据题意, 有\n\n$$\nQ A+Q B=Q P+Q B=P B=4\n$$\n\n因此 $Q$ 的轨迹是以 $A, B$ 为焦点, 4 为长轴长的椭圆, 其方程为\n\n$$\n\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1.\n$$"
] | [
"$\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Equation | null | Open-ended | Plane Geometry | Math | Chinese |
275 | 已知圆 $B:(x+\sqrt{2})^{2}+y^{2}=16$, 定点 $A(\sqrt{2}, 0), P$ 是圆周上任一点, 线段 $A P$ 的垂直平分线与 $B P$交于点 $Q$.
直线 $l$ 过点 $A$ 且与 $x$ 轴不重合, 直线 $l$ 交曲线 $C$ 于 $M, N$ 两点, 过 $A$ 且与 $l$ 垂直的直线与圆 $B$ 交于 $D, E$ 两点, 求四边形 $M D N E$ 面积的取值范围. | [
"根据题意, 设直线 $l$ 的倾斜角为 $\\theta$, 则 $\\theta \\neq 0$, 且\n\n$$\nl: x \\sin \\theta-y \\cos \\theta-\\sqrt{2} \\cdot \\sin \\theta=0\n$$\n\n于是直线 $M N$ 的方程可以表示为\n\n$$\nx \\cos \\theta+y \\sin \\theta-\\sqrt{2} \\cos \\theta=0,\n$$\n\n因此点 $B$ 到直线 $M N$ 的距离\n\n$$\nd=2 \\sqrt{2} \\cdot|\\cos \\theta|,\n$$\n\n进而\n\n$$\nD E... | [
"$[8,8 \\sqrt{2})$"
] | 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 | Plane Geometry | Math | Chinese |
278 | 在直角坐标系中, 直线 $l$ 的参数方程为 $\left\{\begin{array}{l}x=-\sqrt{3} t, \\ y=3+t,\end{array}\right.$ 其中 $t$ 为参数, 以坐标原点为极点, $x$ 轴正半轴为极轴建立极坐标系, 曲线 $C$ 的极坐标方程为 $\rho^{2} \cos ^{2} \theta+3 \rho^{2} \sin ^{2} \theta=3$.
过曲线 $C$ 上任意一点 $P$ 作与 $l$ 夹角为 $45^{\circ}$ 的直线, 交 $l$ 于点 $A$, 求 $|P A|$ 的最大值与最小值. | [
"直线 $l: x+\\sqrt{3} y-3 \\sqrt{3}=0$, 过 $P(\\sqrt{3} \\cos \\theta, \\sin \\theta)$ 作直线 $l$ 的垂线, 垂足为 $H$, 则\n\n$$\n\\begin{aligned}\nP A & =\\sqrt{2} \\cdot P H \\\\\n& =\\sqrt{2} \\cdot \\frac{3 \\sqrt{3}-\\sqrt{3}(\\sin \\theta+\\cos \\theta)}{2} \\\\\n& =\\frac{3 \\sqrt{6}-2 \\sqrt{3} \\sin \\left(\\theta+\\frac... | [
"$\\frac{3 \\sqrt{6}+2 \\sqrt{3}}{2}, \\frac{3 \\sqrt{6}-2 \\sqrt{3}}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
279 | 已知函数 $f(x)=|2 x-1|-|x+2|$.
存在 $x_{0} \in \mathbb{R}$, 使得 $f\left(x_{0}\right)+2 a^{2} \leqslant 4 a$, 求实数 $a$ 的取值范围. | [
"根据题意, 有\n\n$$\n\\exists x \\in \\mathbb{R}, f(x) \\leqslant-2 a^{2}+4 a\n$$\n\n也即\n\n$$\n\\min _{x \\in \\mathbb{R}}\\{f(x)\\} \\leqslant-2 a^{2}+4 a,\n$$\n\n也即\n\n$$\n2 a^{2}-4 a-\\frac{5}{2} \\leqslant 0,\n$$\n\n解得实数 $a$ 的取值范围是 $\\left[-\\frac{1}{2}, \\frac{5}{2}\\right]$."
] | [
"$\\left[-\\frac{1}{2}, \\frac{5}{2}\\right]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Inequality | Math | Chinese |
280 | 已知函数 $f(x)=|2 x-1|-|x+2|$.
存在 $x_{0} \in \mathbb{R}$, 使得 $f\left(x_{0}\right)+2 a^{2} \leqslant 4 a$.设实数$a$的取值范围中的最大数为 $a_{0}$, 若 $\frac{1}{a^{2}}+\frac{4}{b^{2}}+\frac{9}{c^{2}}=a_{0}$。存在 $a, b, c$, 使得$a^{2}+4 b^{2}+9 c^{2}$ 有最小值, 求出该最小值. | [
"根据柯西不等式有\n\n$$\n\\frac{5}{2}=\\frac{1}{a^{2}}+\\frac{4}{b^{2}}+\\frac{9}{c^{2}} \\geqslant \\frac{(1+4+9)^{2}}{a^{2}+4 b^{2}+9 c^{2}},\n$$\n\n因此\n\n$$\na^{2}+4 b^{2}+9 c^{2} \\geqslant \\frac{392}{5}\n$$\n\n等号当\n\n$$\na^{2}=b^{2}=c^{2}=\\frac{28}{5}\n$$\n\n时取得, 进而可得所求代数式的最小值为 $\\frac{392}{5}$."
] | [
"$\\frac{392}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Inequality | Math | Chinese |
281 | 已知向量 $\vec{m}=(2 \cos x, \sin x), \vec{n}=(\cos x, 2 \sqrt{3} \cos x) \quad(x \in \mathbf{R})$,设函数 $f(x)=\vec{m} \cdot \vec{n}-1$.
求函数 $f(x)$ 的单调减区间. | [
"$f(x)=\\vec{m} \\cdot \\vec{n}-1=2 \\cos ^{2} x+2 \\sqrt{3} \\sin x \\cos x-1=\\cos 2 x+\\sqrt{3} \\sin 2 x$.\n\n$$\n=2 \\sin \\left(2 x+\\frac{\\pi}{6}\\right)\n$$\n\n$\\because x \\in \\mathrm{R}$, 由 $2 k \\pi+\\frac{\\pi}{2} \\leq 2 x+\\frac{\\pi}{6} \\leq 2 k \\pi+\\frac{3 \\pi}{2}$ 得\n\n$$\nk \\pi+\\frac{\\pi... | [
"$\\left[k \\pi+\\frac{\\pi}{6}, k \\pi+\\frac{2 \\pi}{3}\\right]$"
] | 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 |
282 | 已知向量 $\vec{m}=(2 \cos x, \sin x), \vec{n}=(\cos x, 2 \sqrt{3} \cos x) \quad(x \in \mathbf{R})$,设函数 $f(x)=\vec{m} \cdot \vec{n}-1$.
已知 $\triangle A B C$ 的三个内角分别为 $A, B, C$, 若 $f(A)=2, B=\frac{\pi}{4}$, 边 $A B=\sqrt{6}+\sqrt{2}$, 求边 $B C$. | [
"$\\because f(A)=2$, 即 $2 \\sin \\left(2 A+\\frac{\\pi}{6}\\right)=2, \\because$ 角 $A$ 为锐角, 得 $A=\\frac{\\pi}{6}$,\n\n$$\n\\text { 又 } B=\\frac{\\pi}{4}, \\therefore C=\\frac{7}{12} \\pi, \\therefore \\sin C=\\sin \\frac{7 \\pi}{12}=\\sin \\left(\\frac{\\pi}{4}+\\frac{\\pi}{3}\\right)=\\frac{\\sqrt{6}+\\sqrt{2}}{4}... | [
"$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 |
285 | 2018 年元旦期间, 某品牌轿车销售商为了促销，给出了两种优惠方案, 顾客只能选择其中的一种, 方案一: 每满 6 万元, 可减 6 千元. 方案二: 金额超过 6 万元 (含 6 万元), 可摇号三次, 其规则是依次装有 2 个幸运号、 2 个吉祥号的一号摇号机, 装有 2 个幸运号、 2 个吉祥号的二号摇号机, 装有 1 个幸运号、 3 个吉祥号的三号摇号机各摇号一次, 其优惠情况为:若摇出 3 个幸运号则打 6 折; 若摇出 2 个幸运号则打 7 折; 若摇出 1 个幸运号则打 8 折；若没有摇出幸运号则不打折.
若某型号的车正好 6 万元, 两个顾客都选中第二种方案, 求至少有一名顾客比选择方案一更优惠的概率; | [
"要使选择方案二比选择方案一更优惠, 则需要至少摸出 1 个幸运球,设顾客不打折即三次没摸出幸运球为事件 $A$, 则\n\n$P(A)=\\frac{2 \\times 2 \\times 3}{4 \\times 4 \\times 4}=\\frac{3}{16}$,\n\n故所求概率 $P=1-P(A) P(A)=1-\\left(\\frac{3}{16}\\right)^{2}=\\frac{247}{256}$."
] | [
"$\\frac{247}{256}$"
] | 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 |
287 | 已知动点 $D$ 与两定点 $A(-2,0) 、 B(2,0)$ 连线的斜率的乘积为 $-\frac{1}{2}$,圆 $C$ 的圆心在原点, 半径为 $\sqrt{2}$.
若动点 $D$ 的轨迹为曲线 $E$, 求曲线 $E$ 方程; | [
"设点 $D(x, y)$, 则 $k_{D A} \\cdot k_{D B}=\\frac{y}{x+2} \\cdot \\frac{y}{x-2}=-\\frac{1}{2}$,\n\n即 $2 y^{2}+x^{2}=4$, 整理得 $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, 其中 $x \\neq \\pm 2$."
] | [
"$\\frac{x^{2}}{4}+\\frac{y^{2}}{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 | Plane Geometry | Math | Chinese |
288 | 已知动点 $D$ 与两定点 $A(-2,0) 、 B(2,0)$ 连线的斜率的乘积为 $-\frac{1}{2}$,圆 $C$ 的圆心在原点, 半径为 $\sqrt{2}$.
若动点 $P$ 在曲线 $E$ 上, 点 $Q$ 在圆 $C$ 上, 且 $P, Q$ 在 $y$ 轴两侧, 线段 $P Q$ 平行于 $x$ 轴. 直线 $A P$ 交 $y$ 轴于点 $M$, 直线 $B P$ 交 $y$ 轴于点 $N$, 求 $\angle M Q N$. | [
"设点 $D(x, y)$, 则 $k_{D A} \\cdot k_{D B}=\\frac{y}{x+2} \\cdot \\frac{y}{x-2}=-\\frac{1}{2}$,\n\n即 $2 y^{2}+x^{2}=4$, 整理得 $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, 其中 $x \\neq \\pm 2$.\n\n设直线 $A P$ 为 $y=k(x+2)$, 联立椭圆 $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ 得\n\n$$\n\\left(1+2 k^{2}\\right) x^{2}+8 k^{2} x+8 k^{2}-4=0\n$... | [
"$\\frac{\\pi}{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 | Plane Geometry | Math | Chinese |
289 | 已知函数 $f(x)=\frac{m \ln x}{x}+n, g(x)=x^{2}\left(f(x)-\frac{1}{x}-\frac{a}{2}\right)$ $(m, n, a \in R)$, 且曲线 $y=f(x)$ 在点 $(1, f(1))$ 处的切线方程为 $y=x-1$.
求实数 $m, n$ 的值. | [
"函数 $f(x)$ 的定义域为 $(0,+\\infty), f^{\\prime}(x)=\\frac{m(1-\\ln x)}{x^{2}},$\n\n因为 $f(x)$ 的图象在点 $(1, f(1))$ 处的切线方程为 $y=x-1$,\n\n所以 $\\left\\{\\begin{array}{l}f^{\\prime}(1)=m=1 \\\\ f(1)=\\frac{m \\ln 1}{1}+n=0\\end{array}\\right.$. 解得 $m=1, n=0$."
] | [
"$1, 0$"
] | 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 |
290 | 已知函数 $f(x)=\frac{m \ln x}{x}+n, g(x)=x^{2}\left(f(x)-\frac{1}{x}-\frac{a}{2}\right)$ $(m, n, a \in R)$, 且曲线 $y=f(x)$ 在点 $(1, f(1))$ 处的切线方程为 $y=x-1$.
函数 $f(x)$ 的最大值. | [
"函数 $f(x)$ 的定义域为 $(0,+\\infty), f^{\\prime}(x)=\\frac{m(1-\\ln x)}{x^{2}},$\n\n因为 $f(x)$ 的图象在点 $(1, f(1))$ 处的切线方程为 $y=x-1$,\n\n所以 $\\left\\{\\begin{array}{l}f^{\\prime}(1)=m=1 \\\\ f(1)=\\frac{m \\ln 1}{1}+n=0\\end{array}\\right.$. 解得 $m=1, n=0$.\n\n所以 $f(x)=\\frac{\\ln x}{x}$. 故 $f^{\\prime}(x)=\\frac{1-\\ln x}{x^... | [
"$\\frac{1}{e}$"
] | 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 |
291 | 已知函数 $f(x)=\frac{m \ln x}{x}+n, g(x)=x^{2}\left(f(x)-\frac{1}{x}-\frac{a}{2}\right)$ $(m, n, a \in R)$, 且曲线 $y=f(x)$ 在点 $(1, f(1))$ 处的切线方程为 $y=x-1$.
当 $a \in\left(-e, \frac{1}{e}\right)$ 时, 记函数 $g(x)$ 的最小值为 $b$, 求实数 $b$ 的取值范围. | [
"函数 $f(x)$ 的定义域为 $(0,+\\infty), f^{\\prime}(x)=\\frac{m(1-\\ln x)}{x^{2}},$\n\n因为 $f(x)$ 的图象在点 $(1, f(1))$ 处的切线方程为 $y=x-1$,\n\n所以 $\\left\\{\\begin{array}{l}f^{\\prime}(1)=m=1 \\\\ f(1)=\\frac{m \\ln 1}{1}+n=0\\end{array}\\right.$. 解得 $m=1, n=0$.\n\n所以 $f(x)=\\frac{\\ln x}{x}$. 故 $f^{\\prime}(x)=\\frac{1-\\ln x}{x^... | [
"$\\left(-\\frac{e}{2},-\\frac{3}{2 e}\\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 |
292 | 在直角坐标系 $x O y$ 中, 曲线 $C_{1}$ 的参数方程为 $\left\{\begin{array}{l}x=t \\ y=m+t\end{array}\right.$ ( $t$ 为参数, $m \in R$ ), 以原点 $O$ 为极点, $x$ 轴的非负半轴为极轴建立极坐标系, 曲线 $C_{2}$的极坐标方程为 $\rho^{2}=\frac{3}{3-2 \cos ^{2} \theta}(0 \leq \theta \leq \pi)$.
写出曲线 $C_{1}$ 的普通方程 | [
"由曲线 $C_{1}$ 的参数方程, 消去参数 $t$, 可得 $C_{1}$ 的普通方程为: $x-y+m=0$."
] | [
"$x-y+m=0$"
] | 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 | Polar Coordinates and Parametric Equations | Math | Chinese |
293 | 在直角坐标系 $x O y$ 中, 曲线 $C_{1}$ 的参数方程为 $\left\{\begin{array}{l}x=t \\ y=m+t\end{array}\right.$ ( $t$ 为参数, $m \in R$ ), 以原点 $O$ 为极点, $x$ 轴的非负半轴为极轴建立极坐标系, 曲线 $C_{2}$的极坐标方程为 $\rho^{2}=\frac{3}{3-2 \cos ^{2} \theta}(0 \leq \theta \leq \pi)$.
写出曲线 $C_{2}$ 的直角坐标方程. | [
"由曲线 $C_{2}$ 的极坐标方程得 $3 \\rho^{2}-2 \\rho^{2} \\cos ^{2} \\theta=3, \\theta \\in[0, \\pi]$,\n\n$\\therefore$ 曲线 $C_{2}$ 的直角坐标方程为 $\\frac{x^{2}}{3}+y^{2}=1(0 \\leq y \\leq 1)$."
] | [
"$\\frac{x^{2}}{3}+y^{2}=1$"
] | 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 | Polar Coordinates and Parametric Equations | Math | Chinese |
294 | 在直角坐标系 $x O y$ 中, 曲线 $C_{1}$ 的参数方程为 $\left\{\begin{array}{l}x=t \\ y=m+t\end{array}\right.$ ( $t$ 为参数, $m \in R$ ), 以原点 $O$ 为极点, $x$ 轴的非负半轴为极轴建立极坐标系, 曲线 $C_{2}$的极坐标方程为 $\rho^{2}=\frac{3}{3-2 \cos ^{2} \theta}(0 \leq \theta \leq \pi)$.
已知点 $P$ 是曲线 $C_{2}$ 上一点, 若点 $P$ 到曲线 $C_{1}$ 的最小距离为 $2 \sqrt{2}$, 求 $m$的值. | [
"由曲线 $C_{1}$ 的参数方程, 消去参数 $t$, 可得 $C_{1}$ 的普通方程为: $x-y+m=0$.\n\n由曲线 $C_{2}$ 的极坐标方程得 $3 \\rho^{2}-2 \\rho^{2} \\cos ^{2} \\theta=3, \\theta \\in[0, \\pi]$,\n\n$\\therefore$ 曲线 $C_{2}$ 的直角坐标方程为 $\\frac{x^{2}}{3}+y^{2}=1(0 \\leq y \\leq 1)$\n\n设曲线 $C_{2}$ 上任意一点 $P$ 为 $(\\sqrt{3} \\cos \\alpha, \\sin \\alpha), \\alpha \... | [
"$-4-\\sqrt{3},6$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
295 | 已知函数$f(x)=\frac{1}{3}|x-a|(a \in R)$.
当 $a=2$ 时, 解不等式 $\left|x-\frac{1}{3}\right|+f(x) \geq 1$ ； | [
"当 $a=2$ 时, 原不等式可化为 $|3 x-1|+|x-2| \\geq 3$.\n\n(1) 当 $x \\leq \\frac{1}{3}$ 时, 原不等式可化为 $-3 x+1+2-x \\geq 3$, 解得 $x \\leq 0$, 所以 $x \\leq 0 ;$\n\n(2)当 $\\frac{1}{3}<x<2$ 时, 原不等式可化为 $3 x-1+2-x \\geq 3$, 解得 $x \\geq 1$, 所以 $1 \\leq x<2 ;$\n\n(3)当 $x \\geq 2$ 时, 原不等式可化为 $3 x-1-2+x \\geq 3$, 解得 $x \\geq \\frac{3}{2}$, ... | [
"$(-\\infty, 0] \\cup [1, +\\infty)$"
] | 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 |
296 | 已知函数$f(x)=\frac{1}{3}|x-a|(a \in R)$.
设不等式 $\left|x-\frac{1}{3}\right|+f(x) \leq x$ 的解集为 $M$, 若 $\left[\frac{1}{3}, \frac{1}{2}\right] \subseteq M$, 求实数 $a$ 的取值范围. | [
"不等式 $\\left|x-\\frac{1}{3}\\right|+f(x) \\leq x$ 可化为 $|3 x-1|+|x-a| \\leq 3 x ，$\n\n依题意不等式 $|3 x-1|+|x-a| \\leq 3 x$ 在 $\\left[\\frac{1}{3}, \\frac{1}{2}\\right]$ 恒成立,\n\n所以 $3 x-1+|x-a| \\leq 3 x$, 即 $|x-a| \\leq 1,$\n\n即 $a-1 \\leq x \\leq a+1$, 所以 $\\left\\{\\begin{array}{l}a-1 \\leq \\frac{1}{3} \\\\ a+1 \\geq... | [
"$\\left[-\\frac{1}{2}, \\frac{4}{3}\\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 |
297 | 已知函数 $f(x)=4 \sin x \sin \left(x+\frac{\pi}{6}\right)-\sqrt{3}$.
求 $f(x)$ 的最小正周期: | [
"$f(x)=4 \\sin x \\sin \\left(x+\\frac{\\pi}{6}\\right)-\\sqrt{3}=4 \\sin x\\left(\\frac{\\sqrt{3}}{2} \\sin x+\\frac{1}{2} \\cos x\\right)-\\sqrt{3}$\n\n$$\n\\begin{aligned}\n& =2 \\sqrt{3} \\sin ^{2} x+2 \\sin x \\cos x-\\sqrt{3}=\\sqrt{3}(1-\\cos 2 x)+\\sin 2 x-\\sqrt{3} \\\\\n& =\\sin 2 x-\\sqrt{3} \\cos 2 x=2 ... | [
"$\\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 | Elementary Functions | Math | Chinese |
298 | 已知函数 $f(x)=4 \sin x \sin \left(x+\frac{\pi}{6}\right)-\sqrt{3}$.
在锐角 $\triangle A B C$ 中, $a, b, c$ 分别为角 $A, B, C$ 的对边, 且满足 $a \sin B+b \cos 2 A=b \cos A$, 求 $f(B)$ 的取值范围. | [
"依题意, 由正弦定理, $\\sin A \\sin B+\\sin B \\cos 2 A=\\sin B \\cos A$, .\n\n因为在三角形中 $\\sin B \\neq 0$, 所以 $\\cos 2 A=\\cos A-\\sin A$.\n\n即 $(\\cos A-\\sin A)(\\cos A+\\sin A-1)=0$,\n\n当 $\\cos A-\\sin A=0$ 时, $\\tan A=1, A=\\frac{\\pi}{4}$ ：\n\n当 $\\cos A+\\sin A=1$ 时, 两边平方得 $1+2 \\sin A \\cos A=1$, 故 $\\sin A \\cos A=... | [
"$(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 | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
304 | 已知 $A, B$ 是 $x$ 轴正半轴上两点 ( $A$ 在 $B$ 的左侧), 且 $|A B| = a(a>0)$, 过 $A, B$ 作 $x$ 轴的垂线, 与抛物线 $y^{2}=2 p x(p>0)$ 在第一象限分别交于 $D, C$ 两点.
若 $a=p$, 点 $A$ 与抛物线 $y^{2}=2 p x$ 的焦点重合, 求直线 $C D$ 的斜率; | [
"由 $A\\left(\\frac{p}{2}, 0\\right)$, 则 $B\\left(\\frac{p}{2}+a, 0\\right), D\\left(\\frac{p}{2}, p\\right)$, 则 $C\\left(\\frac{p}{2}+a, \\sqrt{p^{2}+2 p a}\\right)$,\n\n又 $a=p$, 所以 $k_{C D}=\\frac{\\sqrt{3} p-p}{\\frac{3 p}{2}-\\frac{p}{2}}=\\sqrt{3}-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 | Plane Geometry | Math | Chinese |
305 | 已知 $A, B$ 是 $x$ 轴正半轴上两点 ( $A$ 在 $B$ 的左侧), 且 $|A B| = a(a>0)$, 过 $A, B$ 作 $x$ 轴的垂线, 与抛物线 $y^{2}=2 p x(p>0)$ 在第一象限分别交于 $D, C$ 两点.
若 $O$ 为坐标原点, 记 $\triangle O C D$ 的面积为 $S_{1}$, 梯形 $A B C D$ 的面积为 $S_{2}$, 求 $\frac{S_{1}}{S_{2}}$ 的取值范围. | [
"设直线 $C D$ 的方程为: $y=k x+b(k \\neq 0)$, 设 $C\\left(x_{1}, y_{1}\\right), D\\left(x_{2}, y_{2}\\right)$,\n\n由 $\\left\\{\\begin{array}{l}y=k x+b \\\\ y^{2}=2 p x\\end{array}\\right.$, 得 $k y^{2}-2 p y+2 p b=0$,\n\n所以 $\\Delta=4 p^{2}-8 p k b>0$, 得 $k b<\\frac{p}{2}$,\n\n又 $y_{1}+y_{2}=\\frac{2 p}{k}, y_{1} y_{2}=\\fr... | [
"$(0,\\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 | Interval | null | Open-ended | Plane Geometry | Math | Chinese |
308 | 已知曲线 $C$ 的参数方程为 $\left\{\begin{array}{l}x=\frac{\sqrt{10}}{2} \cos \theta \\ y=\sin \theta\end{array}\right.$ ( $\theta$ 为参数), 以平面直角坐标系的原点为极点, $x$ 的非负半轴为极轴建立极坐标系.
求曲线 $C$ 的极坐标方程. | [
"由 $\\left\\{\\begin{array}{l}x=\\frac{\\sqrt{10}}{2} \\cos \\theta \\\\ y=\\sin \\theta\\end{array}\\right.$, 得到曲线 $C$ 的普通方程是: $\\frac{2 x^{2}}{5}+y^{2}=1, $\n\n又 $x=\\rho \\cos \\theta, y=\\rho \\sin \\theta$, 代入得, $5 \\rho^{2} \\sin ^{2} \\theta+2 \\rho^{2} \\cos ^{2} \\theta=5$, 即 $\\rho^{2}=\\frac{5}{5 \\sin ^... | [
"$\\rho^{2}=\\frac{5}{5 \\sin ^{2} \\theta+2 \\cos^{2} \\theta}$"
] | 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 |
309 | 已知曲线 $C$ 的参数方程为 $\left\{\begin{array}{l}x=\frac{\sqrt{10}}{2} \cos \theta \\ y=\sin \theta\end{array}\right.$ ( $\theta$ 为参数), 以平面直角坐标系的原点为极点, $x$ 的非负半轴为极轴建立极坐标系.
$P 、 Q$ 为曲线 $C$ 上两点, 若 $\overrightarrow{O P} \cdot \overrightarrow{O Q}=0$, 求 $\frac{|\overrightarrow{O P}|^{2} \cdot |\overrightarrow{O Q}|^{2}}{|\overrightarrow{O P}|^{2}+|\overrightarrow{O Q}|^{2}}$ 的值. | [
"由 $\\left\\{\\begin{array}{l}x=\\frac{\\sqrt{10}}{2} \\cos \\theta \\\\ y=\\sin \\theta\\end{array}\\right.$, 得到曲线 $C$ 的普通方程是: $\\frac{2 x^{2}}{5}+y^{2}=1, $\n\n又 $x=\\rho \\cos \\theta, y=\\rho \\sin \\theta$, 代入得, $5 \\rho^{2} \\sin ^{2} \\theta+2 \\rho^{2} \\cos ^{2} \\theta=5$, 即 $\\rho^{2}=\\frac{5}{5 \\sin ^... | [
"$\\frac{5}{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 | Polar Coordinates and Parametric Equations | Math | Chinese |
310 | 已知函数 $f(x)=|x+2|-a|2 x-1|, a \in \mathbf{R}$.
当 $a=1$ 时, 求不等式 $f(x) \geq 0$ 的解集. | [
"当 $a=1$ 时, 由 $f(x) \\geq 0$, 即 $|x+2| \\geq|2 x-1|$, 两边平方, 得:\n\n$x^{2}+4 x+4 \\geq 4 x^{2}-4 x+1$,\n\n即 $3 x^{2}-8 x-3 \\leq 0$,\n\n解得: $-\\frac{1}{3} \\leq x \\leq 3$,\n\n所以不等式 $f(x) \\geq 0$ 的解集为: $\\left\\{x \\mid-\\frac{1}{3} \\leq x \\leq 3\\right\\}$."
] | [
"$[-\\frac{1}{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 | Interval | null | Open-ended | Inequality | Math | Chinese |
311 | 已知函数 $f(x)=|x+2|-a|2 x-1|, a \in \mathbf{R}$.
若存在 $x \in \mathbf{R}$, 使得不等式 $f(x)>a$ 成立, 求 $a$ 的取值范围. | [
"若存在 $x \\in \\mathbf{R}$, 使得不等式 $f(x)>a$ 成立, 即 $|x+2|-a|2 x-1|>a$ 成立,\n\n所以存在 $x \\in \\mathbf{R}$, 使得 $a<\\frac{|x+2|}{|2 x-1|+1}$ 成立, 令 $g(x)=\\frac{|x+2|}{|2 x-1|+1}$, 只需 $a<g_{\\text {max }}(x)$ 即可.\n\n又函数 $g(x)=\\frac{|x+2|}{|2 x-1|+1}=\\left\\{\\begin{array}{l}\\frac{1}{2}+\\frac{3}{2(x-1)}, x<-2 \\\\ -\\fra... | [
"$(-\\infty, \\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 | Interval | null | Open-ended | Inequality | Math | Chinese |
312 | 已知 $S_{2}=3, a_{n+1}=S_{n}+1$
求通项公式 $a_{n}$. | [
"$\\because a_{n+1}=S_{n}+1$, 当 $n=1$ 时, $a_{2}=a_{1}+1$,\n\n$$\n\\text { 又 } \\because \\mathrm{a}_{1}+a_{2}=3, \\therefore a_{1}=1, a_{2}=2\n$$\n\n当$n>1$时，$a_{n}=S_{n-1} + 1, a_{n+1}-a_{n}=a_{n},\\therefore a_{n+1}=2a_{n}, \\therefore \\left \\{a_{n}\\right \\}$为等比数列.\n\n且公比 $\\mathrm{q}=2$,\n\n$a_{n}=2^{n-1} ;$"... | [
"$2^{n-1}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Sequence | Math | Chinese |
313 | 已知 $S_{2}=3, a_{n+1}=S_{n}+1$
若 $b_{n}=a_{n}\left(\log _{2} a_{n}+1\right)\left(n \in N_{+}\right)$, 求数列 $\left\{b_{n}\right\}$ 的前 $\mathrm{n}$ 项和为 $T_{n}$ | [
"$\\because a_{n+1}=S_{n}+1$, 当 $n=1$ 时, $a_{2}=a_{1}+1$,\n\n$$\n\\text { 又 } \\because \\mathrm{a}_{1}+a_{2}=3, \\therefore a_{1}=1, a_{2}=2\n$$\n\n当$n>1$时，$a_{n}=S_{n-1} + 1, a_{n+1}-a_{n}=a_{n},\\therefore a_{n+1}=2a_{n}, \\therefore \\left \\{a_{n}\\right \\}$为等比数列.\n\n且公比 $\\mathrm{q}=2$,\n\n$a_{n}=2^{n-1} ;$\... | [
"$(n-1) \\cdot 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 |
317 | “不怕同桌是学霸, 就怕学霸过暑假.”说的是暑假学习的重要性.现对某中学高中学生暑假学习时间进行调查, 随机调查男生、女生各 50 名, 其中每学习时间超过 6 小时的学生为 “甲组”, 否则为 “乙组”, 调查结果如下:
| | 甲组 | 乙组 | 合计 |
| :--: | :--: | :--: | :--: |
| 男生 | 28 | 22 | 50 |
| 女生 | 30 | 20 | 50 |
| 合计 | 58 | 42 | 100 |
参考公式: $K^{2}=\frac{n(a d-b c)^{2}}{(a+b)(c+d)(a+c)(b+d)}$, 其中 $n=a+b+c+d$ 为样本容量.
参考数据:
| $P\left(K^{2} \geq k_{0}\right)$ | 0.50 | 0.40 | 0.25 | 0.05 | 0.025 | 0.010 |
| :--------------------------------: | :---: | :---: | :---: | :---: | :---: | :---: |
| $k_{0}$ | 0.455 | 0.708 | 1.323 | 3.841 | 5.024 | 6.635 |
现从调查的女生中按分层抽样的方法选出 5 人赠送学习软件各 1 份,求所抽取 5 人中 “甲组”有3人，“乙组” 有2人.接着从抽取的 5 人中再随机抽取 2 人赠送 200 元的学习资料, 求 “这 2 人中都在甲组或都在乙组” 的概率. | [
"设: 甲组的三人为 A, B, C, 乙组的二人为 $a, b$, 则从 5 人中选取 2 人的基本事件有: $(A, B)(A, C)(A, a)(A, b)(B, C)(B, a)(B, b)(C, a)(C, b)(a, b)$ 共 10 种.\n\n其中都在甲组或都在乙组的有（A, B）(A, C ）(B, C) ( $a, b)$ ，共 4 种.\n\n所以 “所选取 2 人中都在甲组或都在乙组” 的概率 $\\mathrm{P}=\\frac{4}{10}=\\frac{2}{5}$."
] | [
"$\\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 |
318 | 已知中心在原点, 焦点在 $x$ 轴上的椭圆 C 过点 $\left(1, \frac{\sqrt{3}}{2}\right)$, 离心率为 $\frac{\sqrt{3}}{2}, A_{1}$, $A_{2}$ 是椭圆 $C$ 的长轴的两个端点 ( $A_{2}$ 位于 $A_{1}$ 右侧), $B$ 是椭圆在 $\mathrm{y}$ 轴正半轴的交点.
求椭圆 $C$ 的标准方程; | [
"设椭圆的方程为 $\\frac{x^{2}}{\\mathrm{a}^{2}}+\\frac{y^{2}}{b^{2}}=1$\n\n$$\n\\left\\{\\begin{array}{l}\n\\frac{1}{a^{2}}+\\frac{3}{4} \\cdot \\frac{1}{b^{2}}=1 \\\\\n\\frac{c}{a}=\\frac{\\sqrt{3}}{2}, \\\\\na^{2}=b^{2}+c^{2}\n\\end{array}\\right.\n$$\n\n$\\therefore a=2, b=1,$所以椭圆的方程为$\\frac{\\mathrm{x}^{2}}{4}+y^{2}=1... | [
"$\\frac{x^{2}}{4}+y^{2}=1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Equation | null | Open-ended | Plane Geometry | Math | Chinese |
320 | 已知函数 $f(x)=2 \ln x-a x^{2}+3 x, a \in R$.
若 $f(1)=2$, 求函数 $f(x)$ 的最大值. | [
"因为 $f(1)=2$, 所以 $-a+3=2$, 所以 $a=1$,\n\n$$\n\\begin{aligned}\n\\therefore & f(x)=2 \\ln x-x^{2}+3 x \\\\\n\\therefore & f^{\\prime}(x)=\\frac{2}{x}-2 x+3=-\\frac{(2 x+1)(x-2)}{x}\n\\end{aligned}\n$$\n\n由 $f^{\\prime}(x)>0$ 得, $0<x<2$, 有 $f^{\\prime}(x)<0$ 得, $x>2$,\n\n$\\therefore f(x)$ 在 $(0,2)$ 为增函数, 在 $(2,+\\inf... | [
"$2 \\ln 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 | Elementary Functions | Math | Chinese |
322 | 在直角坐标系 $x 0 y$ 中, 以坐标原点为极点, $\mathrm{x}$ 轴正半轴为极轴建立极坐标系,曲线 $\mathrm{C}_{1}$ 的极坐标方程为 $\rho \cos \theta=6(\rho>0)$.
$M$ 为曲线 $C_{1}$ 的动点, 点 $P$ 在线段 $O M$ 上, 且满足 $|O M| \cdot|O P|=36$, 求点 $P$ 的轨迹 $C_{2}$ 的直角坐标方程. | [
"设 $\\mathrm{P}$ 的极坐标为 $(\\rho, \\theta)(\\rho>0), \\mathrm{M}$ 的极坐标为 $\\left(\\rho_{1}, \\theta\\right)\\left(\\rho_{1}>0\\right)$\n\n由题设知 $|\\mathrm{OP}|=\\rho,|\\mathrm{OM}|=\\rho_{1}=\\frac{6}{\\cos \\theta}$,\n\n由 $|\\mathrm{OM}||\\mathrm{OP}|=36$ 得 $\\mathrm{C}_{2}$ 的极坐标方程 $\\rho=6 \\cos \\theta(\\rho>0)$\n\n... | [
"$(x-3)^{2}+y^{2}=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 | Equation | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
323 | 在直角坐标系 $x 0 y$ 中, 以坐标原点为极点, $\mathrm{x}$ 轴正半轴为极轴建立极坐标系,曲线 $\mathrm{C}_{1}$ 的极坐标方程为 $\rho \cos \theta=6(\rho>0)$.
设点 $\mathrm{A}$ 的极坐标为 $\left(4, \frac{\pi}{3}\right)$, 点 $\mathrm{B}$ 在曲线 $\mathrm{C}_{2}$ 上, 求 $\triangle O A B$ 面积的最大值 | [
"设点 B 的极坐标为 $\\left(\\rho_{B}, \\alpha\\right)\\left(\\rho_{B}>0\\right)$. 由题设知 $|0 \\mathrm{~A}|=4, \\rho_{B}=6 \\cos \\alpha,$\n\n于是 $\\triangle \\mathrm{OAB}$ 面积 $S=\\frac{1}{2}|\\mathrm{OA}| \\cdot \\rho_{\\mathrm{B}} \\sin \\angle A O B=12 \\cos \\alpha\\left|\\sin \\left(\\alpha-\\frac{\\pi}{3}\\right)\\right... | [
"$6+3 \\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 | Polar Coordinates and Parametric Equations | Math | Chinese |
324 | 已知函数 $f(x)=|x+1|-|x-2|$
求不等式 $f(x) \geqslant 2$ 的解集. | [
"$f(x)=|x+1|-|x-2| \\geqslant 2$\n\n当 $x<-1$ 时, $f(x)=-x-1+x-2=-3 \\geqslant 2$ 不成立\n\n当 $-1 \\leq x \\leq 2$ 时, $f(x)=x+1+x-2=2 x-1 \\geqslant 2, \\therefore x \\geqslant \\frac{3}{2} \\therefore \\frac{3}{2} \\leqslant x \\leqslant 2$\n\n当 $x>2$ 时, $f(x)=x+1-x+2=3 \\geqslant 2$ 恒成立, $\\therefore x>2$\n\n综上所述: $f(... | [
"$\\left[\\frac{3}{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 | Inequality | Math | Chinese |
325 | 已知函数 $f(x)=|x+1|-|x-2|$
若不等式 $f(x) \geqslant a^{2}+4 a$ 恒成立, 求 $a$ 的取值范围. | [
"$\\because|f(x)| \\leqslant|x+1-x+2|=3, \\therefore-3 \\leqslant f(x) \\leqslant 3$\n\n$$\n\\therefore f(x)_{\\text {min }}=-3 \\text {, }\n$$\n\n$\\because f(x) \\geqslant a^{2}+4 a$ 恒成立, $\\therefore a^{2}+4 a \\leqslant-3$\n\n所以 $a$ 的范围是 $[-3,-1]$."
] | [
"$[-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 | Interval | null | Open-ended | Inequality | Math | Chinese |
326 | 在 $\triangle A B C$ 中, 内角 $A, B, C$ 的对边分别为 $a, b, c$, 已知 $\frac{b^{2}+c^{2}-a^{2}}{b c}=\frac{2 \sin B-\sin A}{\sin C}$.
求角 $C$ 的值. | [
"由条件和正弦定理可得 $\\frac{b^{2}+c^{2}-a^{2}}{b}=2 b-a$\n\n整理得 $b^{2}+a^{2}-c^{2}=a b$ 从而由余弦定理得 $\\cos C=\\frac{1}{2}$.\n\n又 $\\because C$ 是三角形的内角\n\n$\\therefore C=\\frac{\\pi}{3}$."
] | [
"$\\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 | false | null | Expression | null | Open-ended | Trigonometric Functions | Math | Chinese |
327 | 在 $\triangle A B C$ 中, 内角 $A, B, C$ 的对边分别为 $a, b, c$, 已知 $\frac{b^{2}+c^{2}-a^{2}}{b c}=\frac{2 \sin B-\sin A}{\sin C}$.
若 $a+b=4$, 当边 $c$ 取最小值时，求 $\triangle A B C$ 的面积. | [
"由条件和正弦定理可得 $\\frac{b^{2}+c^{2}-a^{2}}{b}=2 b-a$\n\n整理得 $b^{2}+a^{2}-c^{2}=a b$ 从而由余弦定理得 $\\cos C=\\frac{1}{2}$.\n\n又 $\\because C$ 是三角形的内角\n\n$\\therefore C=\\frac{\\pi}{3}$.\n\n由余弦定理得 $c^{2}=a^{2}+b^{2}-2 a b \\cos C=a^{2}+b^{2}-a b$,\n\n$\\because a+b=4, \\quad \\therefore c^{2}=a^{2}+b^{2}-a b=(a+b)^{2}-3 a b=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 | Trigonometric Functions | Math | Chinese |
328 | 网约车的兴起,丰富了民众出行的选择,为民众出行提供便利的同时也解决了很多劳动力的就业问题,据某著名网约车公司 “滴*打车” 官网显示, 截止目前, 该公司已经累计解决退伍军人转业为兼职或专职司机三百多万人次.梁某即为此类网约车司机，据梁某自己统计某一天出车一次的总路程数可能的取值是 $20 、 22 、 24 、 26 、$ 28、30(km)， 它们出现的概率依次是 $0.1 、 0.2 、 0.3 、 0.1 、 t 、 2 t$.
求这一天中梁某一次行驶路程 $\mathrm{X}$ 的均值和方差. | [
"由概率分布的性质有 $0.1+0.2+0.3+0.1+t+2 t=1$.\n\n所以 $t=0.1,$\n\n$\\therefore X$ 的分布列为:\n\n| $\\mathrm{X}$ | 20 | 22 | 24 | 26 | 28 | 30 |\n| :------------: | :-: | :-: | :-: | :-: | :-: | :-: |\n| $\\mathrm{P}$ | 0.1 | 0.2 | 0.3 | 0.1 | 0.1 | 0.2 |\n\n$\\therefore E(X)=20 \\times 0.1+22 \\times 0.2+24 \\times 0.3+26 \\time... | [
"$25 , 10.6$"
] | 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 | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
329 | 网约车的兴起,丰富了民众出行的选择,为民众出行提供便利的同时也解决了很多劳动力的就业问题,据某著名网约车公司 “滴*打车” 官网显示, 截止目前, 该公司已经累计解决退伍军人转业为兼职或专职司机三百多万人次.梁某即为此类网约车司机，据梁某自己统计某一天出车一次的总路程数可能的取值是 $20 、 22 、 24 、 26 、$ 28、30(km)， 它们出现的概率依次是 $0.1 、 0.2 、 0.3 、 0.1 、 t 、 2 t$.
网约车计费细则如下：起步价为 5 元，行驶路程不超过 $3 \mathrm{~km}$ 时，租车费为 5 元，若行驶路程超过 $3 \mathrm{~km}$ ，则按每超出 $1 \mathrm{~km}$ （不足 $1 \mathrm{~km}$ 也按 $1 \mathrm{~km}$ 计程）收费 3 元计费. 依据以上条件, 计算梁某一天中出车一次收入的均值和方差. | [
"由已知设梁某一天出车一次的收入为 $Y$ 元,\n\n则 $Y=3(X-3)+5=3 X-4,(X>3, X \\in N),$\n\n$\\therefore E(Y)=E(3 X-4)=3 E(X)-4=3 \\times 25-4=71$ (元),\n\n$D(Y)=D(3 X-4)=32 D(X)=95.4$."
] | [
"$71 , 95.4$"
] | 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 | true | null | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
334 | 在平面直角坐标系 $x O y$ 中, 直线 $l$ 的参数方程为 $\left\{\begin{array}{l}x=1+t \\ y=t\end{array}\right.$ ( $t$ 为参数), 以坐标原点为极点, $x$ 轴正半轴为极轴的极坐标系中, 圆 $C_{1}$ 的极坐标方程为 $\rho^{2}-2 a \rho \cos \theta+a^{2}-4=0(a>0)$.
若直线 $l$ 与圆 $C_{1}$ 相切, 求 $a$ 的值. | [
"直线 $l$ 的普通方程为 $y=x-1$.\n\n圆 $C_{1}$ 的直角坐标方程为 $(x-a)^{2}+y^{2}=4$.\n\n因直线 $l$ 与圆 $C_{1}$ 相切,所以 $\\frac{|a-1|}{\\sqrt{2}}=2$, 由于 $a>0$ 解得 $a=2 \\sqrt{2}+1$."
] | [
"$2 \\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 | Polar Coordinates and Parametric Equations | Math | Chinese |
335 | 在平面直角坐标系 $x O y$ 中, 直线 $l$ 的参数方程为 $\left\{\begin{array}{l}x=1+t \\ y=t\end{array}\right.$ ( $t$ 为参数), 以坐标原点为极点, $x$ 轴正半轴为极轴的极坐标系中, 圆 $C_{1}$ 的极坐标方程为 $\rho^{2}-2 a \rho \cos \theta+a^{2}-4=0(a>0)$.
若直线 $l$ 与曲线 $C_{2}:\left\{\begin{array}{l}x=2 \cos \theta \\ y=\sqrt{3} \sin \theta\end{array}\right.$ ( $\theta$ 为参数), 交于 $A, B$ 两点, 点 $C(2,1)$, 求 $|A C|+|B C|$. | [
"曲线 $C_{2}$ 的普通方程为 $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, 点 $C(2,1)$ 在直线 $y=x-1$ 上,\n\n所以直线 $l$ 的参数方程可以写为 $\\left\\{\\begin{array}{l}x=2+\\frac{\\sqrt{2}}{2} t \\\\ y=1+\\frac{\\sqrt{2}}{2} t\\end{array}\\right.$ ( $t$ 为参数),\n\n将上式代入 $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ 得 $\\frac{7}{2} t^{2}+10 \\sqrt{2} t+4=0$.\n... | [
"$\\frac{20 \\sqrt{2}}{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 | Polar Coordinates and Parametric Equations | Math | Chinese |
336 | 已知函数 $f(x)=|x+1|+|3 x+a|$, 若 $f(x)$ 的最小值为 1 .
求实数 $a$ 的值. | [
"$f(x)=|x+1|+|3 x+a|$\n\n(1) 当$a>3$时，即$-1>-\\frac{a}{3}$ , $f(x)=\\left\\{\\begin{array}{l}-4x-1-a, x \\leq -\\frac{a}{3}\\\\ 2x+a-1, -\\frac{a}{3}<x<-1 \\\\ 4x+a+1, x \\geq -1\\end{array}\\right.$,\n\n$\\because f(-1)-f\\left(-\\frac{a}{3}\\right)=(-3+a)-\\left(\\frac{a}{3}-1\\right)=\\frac{2(a-3)}{3}>0$\n\n$\\the... | [
"$0 , 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 | Numerical | null | Open-ended | Inequality | Math | Chinese |
337 | 已知函数 $f(x)=|x+1|+|3 x+a|$, 若 $f(x)$ 的最小值为 1 .
若 $a>0$, 且 $m, n$ 均为正实数, 且满足 $m+n=\frac{a}{2}$, 求 $m^{2}+n^{2}$ 的最小值. | [
"由题意知, $m+n=3, \\because m>0, n>0$,\n\n$\\therefore(m+n)^{2}=m^{2}+n^{2}+2 m n \\leq\\left(m^{2}+n^{2}\\right)+\\left(m^{2}+n^{2}\\right)=2\\left(m^{2}+n^{2}\\right)$\n\n即 $m^{2}+n^{2} \\geq \\frac{1}{2}(m+n)^{2},$\n\n当且仅当 $m=n=\\frac{3}{2}$ 时取 “ $=”$\n\n$\\therefore m^{2}+n^{2} \\geq \\frac{9}{2} ， \\therefore m^{... | [
"$\\frac{9}{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 | Inequality | Math | Chinese |
338 | 已知等比数列 $\left\{a_{n}\right\}$ 各项都是正数, $S_{n}$ 为其前 $n$ 项和, $a_{3}=8, S_{3}=14$.
求数列 $\left\{a_{n}\right\}$ 的通项公式. | [
"等比数列 $\\left\\{a_{n}\\right\\}$ 中, $a_{3}=8, S_{3}=14$, 可列方程组 $\\left\\{\\begin{array}{l}a_{1} q^{2}=8 \\\\ a_{1}+a_{1} q=6\\end{array} \\right.$\n\n由于 $\\left\\{a_{n}\\right\\}$ 各项都是正数, $\\therefore q>0$, 可得 $\\left\\{\\begin{array}{l}a_{1}=2 \\\\ q=2\\end{array}\\right.$\n\n$\\therefore a_{n}=2^{n}$."
] | [
"$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 | Sequence | Math | Chinese |
339 | 已知等比数列 $\left\{a_{n}\right\}$ 各项都是正数, $S_{n}$ 为其前 $n$ 项和, $a_{3}=8, S_{3}=14$.
设 $\left\{a_{n}-b_{n}\right\}$ 是首项为 1 , 公差为 3 的等差数列, 求数列 $\left\{b_{n}\right\}$ 的通项公式及其前 $n$ 项和 $T_{n}$. | [
"等比数列 $\\left\\{a_{n}\\right\\}$ 中, $a_{3}=8, S_{3}=14$, 可列方程组 $\\left\\{\\begin{array}{l}a_{1} q^{2}=8 \\\\ a_{1}+a_{1} q=6\\end{array} \\right.$\n\n由于 $\\left\\{a_{n}\\right\\}$ 各项都是正数, $\\therefore q>0$, 可得 $\\left\\{\\begin{array}{l}a_{1}=2 \\\\ q=2\\end{array}\\right.$\n\n$\\therefore a_{n}=2^{n}$.\n\n$\\becau... | [
"$2^{n}-3 n+2, 2^{n+1}-\\frac{3}{2} n^{2}+\\frac{n}{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 | true | null | Expression | null | Open-ended | Sequence | Math | Chinese |
340 | 某公司生产甲、乙两种不同规格的产品, 并且根据质量的测试指标分数进行划分, 其中分数不小于 70 的为合格品, 否则为次品, 现随机抽取两种产品各 100 件进行检测, 其结果如下:
| 测试指标分数 | $[50,60)$ | $[60,70)$ | $[70,80)$ | $[80,90)$ | $[90,100]$ |
| :----------- | :---------- | :---------- | :---------- | :---------- | :----------- |
| 产品甲 | 7 | 13 | 40 | 32 | 8 |
| 产品乙 | 9 | 21 | 40 | 24 | 6 |
根据表中数据，估计甲、乙两种产品的不合格率. | [
"甲产品的不合格率为 $P_{1}=\\frac{7+13}{100}=20 \\%$, 乙产品的不合格率为 $P_{2}=\\frac{9+21}{100}=30 \\%$."
] | [
"$20，30$"
] | 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 | %,% | Numerical | null | Open-ended | Probability and Statistics | Math | Chinese |
341 | 某公司生产甲、乙两种不同规格的产品, 并且根据质量的测试指标分数进行划分, 其中分数不小于 70 的为合格品, 否则为次品, 现随机抽取两种产品各 100 件进行检测, 其结果如下:
| 测试指标分数 | $[50,60)$ | $[60,70)$ | $[70,80)$ | $[80,90)$ | $[90,100]$ |
| :----------- | :---------- | :---------- | :---------- | :---------- | :----------- |
| 产品甲 | 7 | 13 | 40 | 32 | 8 |
| 产品乙 | 9 | 21 | 40 | 24 | 6 |
若按合格与不合格的比例抽取 5 件甲产品, 再从这 5 件甲产品中随机抽取 2 件, 求这 2 件产品全是合格品的概率. | [
"由题意, 若按合格与不合格的比例抽取 5 件甲产品, 则其中恰有 1 件次品, 4 件合格品, 因而可设这 5 件甲产品分别为 a,b,c,d,E, 其中小写字母代表合格品, E 代表次品, 从中随机抽取 2 件，则所有可能的情况为 $\\mathrm{ab}, \\mathrm{ac}, \\mathrm{ad}, \\mathrm{aE}, \\mathrm{bc}, \\mathrm{bd}, \\mathrm{bE}, \\mathrm{cd}, \\mathrm{cE}, \\mathrm{dE}$ ，共 10种, 设 “这 2 件产品全是合格品” 为事件 M, 则事件 M 所包含的情况为 ab,ac,ad,bc... | [
"$\\frac{3}{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 |
343 | 设抛物线 $C: y^{2}=2 p x(p>0)$, 过焦点 $F$ 的直线 $l$ 与 $C$ 交于 $A$,$B$ 两点.
当 $|A B|$ 的最小值为 4 时, 求抛物线 $C$ 的方程. | [
"当直线 $l$ 的斜率不存在时, $A\\left(\\frac{p}{2}, p\\right), B\\left(\\frac{p}{2},-p\\right)$, 此时 $|A B|=2 p$.\n\n当直线 $l$ 的斜率存在时, 设为 $k$, 此时 $l: y=k\\left(x-\\frac{p}{2}\\right)$, 与抛物线方程联立:\n\n$\\left\\{\\begin{array}{l}y=k\\left(x-\\frac{p}{2}\\right) \\\\ y^{2}=2 p x\\end{array}\\right.$, 消去 $y$, 可得: $k^{2} x^{2}-\\left(k... | [
"$y^{2}=4 x$"
] | 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 |
346 | 已知函数 $f(x)=x^{3}+a x-2 \ln x$.
若 $f(x) \geq 0$ 在定义域内恒成立, 求实数 $a$ 的取值范围. | [
"$\\because f(x)=x^{3}+a x-2 \\ln x \\geq 0$ 在 $(0,+\\infty)$ 上恒成立, $\\therefore$ 当 $x \\in(0,+\\infty)$ 时,\n\n$g(x)=x^{2}+a-\\frac{2 \\ln x}{x} \\geq 0$ 恒成立\n\n$g^{\\prime}(x)=2 x-2 \\frac{(\\ln x)^{\\prime} \\cdot x-\\ln x \\cdot x^{\\prime}}{x^{2}}=2 \\frac{x^{3}+\\ln x-1}{x}$ .\n\n令 $h(x)=x^{3}+\\ln x-1$, 可得 $h... | [
"$[-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 | Elementary Functions | Math | Chinese |
347 | 已知在平面直角坐标系 $x O y$ 中, 以 $O$ 为极点, $x$ 轴正半轴为极轴建立极坐标系, $P$ 点的极坐标为 $\left(3 \sqrt{2},-\frac{\pi}{4}\right)$, 曲线 $C$ 的极坐标方程为 $\rho^{2}-4 \rho \cos \theta-2 \rho \sin \theta+1=0$.
写出点 $P$ 的直角坐标及曲线 $C$ 的直角坐标方程; | [
"$P(3,-3), \\mathrm{C}:(x-2)^{2}+(y-1)^{2}=4$."
] | [
"$(3,-3), (x-2)^{2}+(y-1)^{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 | true | null | Tuple,Equation | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
348 | 已知在平面直角坐标系 $x O y$ 中, 以 $O$ 为极点, $x$ 轴正半轴为极轴建立极坐标系, $P$ 点的极坐标为 $\left(3 \sqrt{2},-\frac{\pi}{4}\right)$, 曲线 $C$ 的极坐标方程为 $\rho^{2}-4 \rho \cos \theta-2 \rho \sin \theta+1=0$.
若 $Q$ 为 $C$ 上的动点, 求 $P Q$ 的中点 $M$ 到直线 $l:\left\{\begin{array}{l}x=2-2 t \\ y=4+2 t\end{array}\right.$ ( $t$ 为参数) 的距离的最大值. | [
"$P(3,-3), \\mathrm{C}:(x-2)^{2}+(y-1)^{2}=4$.\n\n设 $\\mathrm{Q}(2 \\cos \\theta+2,2 \\sin \\theta+1)$, 则 $\\mathrm{PQ}$ 的中点 $\\mathrm{M}\\left(\\cos \\theta+\\frac{5}{2}, \\sin \\theta-1\\right)$.\n\n直线 $l: x+y-6=0$.\n\n则点 $\\mathrm{M}$ 到直线 $l$ 的距离为 $\\mathrm{d}=\\frac{\\left|\\cos \\theta+\\sin \\theta-\\frac{9}{... | [
"$1+\\frac{9 \\sqrt{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 | Numerical | null | Open-ended | Polar Coordinates and Parametric Equations | Math | Chinese |
349 | 已知不等式 $|2 x+4|+|x-1| \geq m$ 的解集为 $R$.
求实数 $m$ 的取值范围. | [
"$f(x)=|2 x+4|+|x-1|=\\left\\{\\begin{array}{l}3 x+3, x \\geq 1 \\\\ x+5,-2<x<1, \\\\ -3 x-3, x \\leq-2\\end{array}\\right.$ 所以 $f(x)$ 的值域为 $[3,+\\infty)$,由不等式 $|2 x+4|+|x-1| \\geq m$ 的解集为 $\\mathrm{R}$ 可知, $m \\leq 3 $."
] | [
"$(-\\infty, 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 | Interval | null | Open-ended | Inequality | Math | Chinese |
350 | 已知不等式 $|2 x+4|+|x-1| \geq m$ 的解集为 $R$.
若 $m$ 的最大值为 $n$, 当正数 $a, b$ 满足 $\frac{2}{2 a+b}+\frac{1}{a+3 b}=n$ 时, 求 $17 a+11 b$ 的最小值. | [
"$n=3, \\frac{2}{2 a+b}+\\frac{1}{a+3 b}=3$\n\n当 $a, b>0$ 时, $17 a+11 b=\\frac{1}{3}\\left(\\frac{2}{2 a+b}+\\frac{1}{a+3 b}\\right)[8(2 a+b)+(a+3 b)]$ $=\\frac{1}{3}\\left[16+\\frac{2(a+3 b)}{2 a+b}+\\frac{8(2 a+b)}{a+3 b}+1\\right] \\geq \\frac{1}{3}(17+2 \\sqrt{2 \\times 8})=\\frac{25}{3}$.\n\n当且仅当 $\\left\\{\\b... | [
"$\\frac{25}{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 |
351 | 已知数列 $\left\{a_{n}\right\}$ 中, $a_{1}=2, a_{2}=6$, 且数列 $\left\{a_{n+1}-a_{n}\right\}$ 是公差为 2 的等差数列.
求 $a_{n}$. | [
"$\\because$ 数列 $\\left\\{a_{n+1}-a_{n}\\right\\}$ 是首项为 $a_{2}-a_{1}=4$, 公差为 2 的等差数列\n\n$$\n\\begin{aligned}\n& \\therefore a_{n+1}-a_{n}=2 n+2 \\\\\n& \\therefore a_{2}-a_{1}=2 \\times 1+2 \\\\\n& a_{3}-a_{2}=2 \\times 2+2 \\\\\n& \\ldots \\ldots . \\\\\n& a_{n}-a_{n-1}=2 \\times(n-1)+2(n \\geq 2)\n\\end{aligned}\... | [
"$a_{n}=n^{2}+n$"
] | 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 | Sequence | Math | Chinese |
352 | 已知数列 $\left\{a_{n}\right\}$ 中, $a_{1}=2, a_{2}=6$, 且数列 $\left\{a_{n+1}-a_{n}\right\}$ 是公差为 2 的等差数列.
$a_{n}=n^{2}+n(n \geq 1)$, 记数列 $\left\{\frac{1}{a_{n}}\right\}$ 的前 $n$ 项和为 $S_{n}$, 求满足不等式 $S_{n}>\frac{2016}{2017}$ 的 $n$ 的最小值. | [
"$\\because \\frac{1}{a_{n}}=\\frac{1}{n(n+1)}=\\frac{1}{n}-\\frac{1}{n+1}$\n\n$\\therefore S_{n}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\ldots+\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right)=1-\\frac{1}{n+1}$\n\n令 $S_{n}>\\frac{2016}{2017}$ 即 $1-\\frac{1}{n+1}>\\frac{2016}{2017} \\there... | [
"$2017$"
] | 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 |
356 | 身体质量指数 BMI（简称体质指数）是目前国际上常用的衡量人体胖瘦程度以及是否健康的一个标准.现为分析胖瘦程度对空腹血糖的影响, 从志愿者中随机抽取 12 名志愿者测定 BMI 值及空腹血糖 GLU 指标值 (单位: mmol/L) 如下表所示:
| 人员编号 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| :----------------------: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: | :-: |
| BMI 值$\mathrm{x}$ | 14 | 17 | 18 | 19 | 20 | 22 | 23 | 26 | 27 | 29 | 30 | 31 |
| GLU 指标值$\mathrm{y}$ | 2.5 | 4.5 | 4.8 | 4.9 | 5.5 | 5.6 | 5.8 | 6.1 | 6.4 | 6.9 | 7.1 | 9.5 |
参考公式: $\hat{b}=\frac{\sum_{i=1}^{n} x_{i} y_{i}-n \bar{x} \bar{y}}{\sum_{i=1}^{n} x_{i}^{2}-n \bar{x}^{2}}=\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)}{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}, \quad \hat{a}=\bar{y}-\hat{b} \bar{x}$;
相关系数 $r=\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)}{\sqrt{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}}}$;
参考数据: $\sum_{i=1}^{12}\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)=95.8 ，$
$$
\sum_{i=1}^{12}\left(x_{i}-\bar{x}\right)^{2}=342, \quad \sum_{i=1}^{12}\left(y_{i}-\bar{y}\right)^{2}=\sum_{i=1}^{12} y_{i}^{2}-12 y^{2}=31.56
$$
$s=\sqrt{\frac{1}{12}\left[\sum_{i=1}^{12}\left(y_{i}-\bar{y}\right)^{2}\right]} \approx 1.6, \sqrt{38} \approx 6.2, \sqrt{31.56} \approx 5.6, \sqrt{0.6964} \approx 0.83$
求 $y$ 与 $x$ 的线性回归方程, 并预测 GLU 指标值 $y=9.8$ 时 BMI 值 $x$ 是多少？ | [
"$\\because \\hat{b}=\\frac{\\sum_{i=1}^{12}\\left(x_{i}-\\bar{x}\\right)\\left(y_{i}-\\bar{y}\\right)}{\\sum_{i=1}^{12}\\left(x_{i}-\\bar{x}\\right)^{2}}=\\frac{95.8}{342} \\approx 0.28$\n\n又 $\\because \\bar{x}=23, \\bar{y}=5.8$\n\n$\\therefore \\hat{a}=5.8-0.28 \\times 23=-0.64 \\therefore$ 线性回归方程为: $\\hat{y}=0.... | [
"$\\hat{y}=0.28 x-0.64, 37.29$"
] | 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,Numerical | ,1e-2 | Open-ended | Probability and Statistics | Math | Chinese |
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