录入2023届全国高考甲乙文科卷

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wangweiye7840 2023-06-12 16:15:26 +08:00
parent 9dcab5874a
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#修改起始id,出处,文件名
starting_id = 18104
raworigin = ""
filename = r"C:\Users\weiye\Documents\wwy sync\临时工作区\自拟题目14.tex"
editor = "20230611\t王伟叶"
starting_id = 18170
raworigin = "2023届全国高考"
filename = r"C:\Users\wangweiye\Documents\wwy sync\临时工作区\23届高考文科.tex"
editor = "20230612\t王伟叶"
indexed = True
IndexDescription = "试题"

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@ -465404,6 +465404,926 @@
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"content": "设全集$U=\\{1,2,3,4,5\\}$, 集合$M=\\{1,4\\}$, $N=\\{2,5\\}$, 则$N \\cup \\overline{M}=$\\bracket{20}.\n\\fourch{$\\{2,3,5\\}$}{$\\{1,3,4\\}$}{$\\{1,2,4,5\\}$}{$\\{2,3,4,5\\}$}",
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"content": "$\\dfrac{5(1+\\mathrm{i}^3)}{(2+\\mathrm{i})(2-\\mathrm{i})}=$\\bracket{20}.\n\\fourch{$-1$}{$1$}{$1-\\mathrm{i}$}{$1+\\mathrm{i}$}",
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"018172": {
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"content": "已知向量$\\overrightarrow{a}=(3,1)$, $\\overrightarrow{b}=(2,2)$, 则$\\cos \\langle\\overrightarrow{a}+\\overrightarrow{b}, \\overrightarrow{a}-\\overrightarrow{b}\\rangle=$\\bracket{20}.\n\\fourch{$\\dfrac{1}{17}$}{$\\dfrac{\\sqrt{17}}{17}$}{$\\dfrac{\\sqrt{15}}{15}$}{$\\dfrac{2 \\sqrt{5}}{5}$}",
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"018173": {
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"content": "某校文艺部有$4$名学生, 其中高一、高二年级各$2$名. 从这$4$名学生中随机选$2$名组织校文艺汇演, 则这$2$名学生来自不同年级的概率为\\bracket{20}.\n\\fourch{$\\dfrac{1}{6}$}{$\\dfrac{1}{3}$}{$\\dfrac{1}{2}$}{$\\dfrac{2}{3}$}",
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"018174": {
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"content": "已知向量$S_n$为等差数列$\\{a_n\\}$的前$n$项和若$a_2+a_6=10$, $a_4 a_8=45$, 则$S_5=$\\bracket{20}.\n\\fourch{$25$}{$22$}{$20$}{$15$}",
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"content": "执行如图的程序框图, 则输出的$B=$\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, node distance = 10pt]\n\\node [draw, rounded corners] (start) {开始};\n\\node [draw, trapezium, trapezium left angle = 60, trapezium right angle = 120, below = of start] (input) {输入$n=3$, $A=1$, $B=2$, $k=1$};\n\\node [draw, diamond, aspect = 2, below = of input] (switch) {$k\\le n$};\n\\node [draw, below = of switch] (step1) {$A=A+B$};\n\\node [draw, below = of step1] (step2) {$B=A+B$};\n\\node [draw, below = of step2] (step3) {$k=k+1$};\n\\node [draw, trapezium, trapezium left angle = 60, trapezium right angle = 120, below = of step3] (output) {输出$B$};\n\\node [draw, rounded corners, below = of output] (end) {结束};\n\\foreach \\i/\\j in {start/input,input/switch,step1/step2,step2/step3,output/end}\n{\\draw [->] (\\i) -- (\\j);};\n\\draw [->] (switch) -- (step1) node [midway, right] {是};\n\\coordinate [right = 15pt of switch] (stepx);\n\\coordinate [left = 15pt of step3] (stepy);\n\\draw [->] (switch) -- (stepx) node [midway, above] {否} -- (stepx|-output) -- (output);\n\\draw [->] (step3) -- (stepy) -- (stepy|-switch) -- (switch);\n\\end{tikzpicture}\n\\end{center}\n\\fourch{$21$}{$34$}{$55$}{$89$}",
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"content": "设$F_1, F_2$为椭圆$C: \\dfrac{x^2}{5}+y^2=1$的两个焦点, 点$P$在$C$上, 若$\\overrightarrow{PF_1} \\cdot \\overrightarrow{PF_2}=0$, 则$|PF_1| \\cdot|PF_2|=$\\bracket{20}.\n\\fourch{$1$}{$2$}{$4$}{$5$}",
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"content": "曲线$y=\\dfrac{\\mathrm{e}^x}{x+1}$在点$(1, \\dfrac{\\mathrm{e}}{2})$处的切线方程为\\bracket{20}.\n\\fourch{$y=\\dfrac{\\mathrm{e}}{4} x$}{$y=\\dfrac{\\mathrm{e}}{2} x$}{$y=\\dfrac{\\mathrm{e}}{4} x+\\dfrac{\\mathrm{e}}{4}$}{$y=\\dfrac{\\mathrm{e}}{2} x+\\dfrac{3 \\mathrm{e}}{4}$}",
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"018178": {
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"content": "已知双曲线$C: \\dfrac{x^2}{a^2}-\\dfrac{y^2}{b^2}=1$($a>0$, $b>0$)的离心率为$\\sqrt{5}$, $C$的一条渐近线与圆$(x-2)^2+(y-3)^2=1$交于$A, B$两点, 则$|AB|=$\\bracket{20}.\n\\fourch{$\\dfrac{\\sqrt{5}}{5}$}{$\\dfrac{2 \\sqrt{5}}{5}$}{$\\dfrac{3 \\sqrt{5}}{5}$}{$\\dfrac{4 \\sqrt{5}}{5}$}",
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"018179": {
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"content": "在三棱锥$P-ABC$中, $\\triangle ABC$是边长为$2$的等边三角形, $PA=PB=2$, $PC=\\sqrt{6}$, 则该棱锥的体积为\\bracket{20}.\n\\fourch{$1$}{$\\sqrt{3}$}{$2$}{$3$}",
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"018180": {
"id": "018180",
"content": "已知函数$f(x)=\\mathrm{e}^{-(x-1)^2}$. 记$a=f(\\dfrac{\\sqrt{2}}{2})$, $b=f(\\dfrac{\\sqrt{3}}{2})$, $c=f(\\dfrac{\\sqrt{6}}{2})$, 则\\bracket{20}.\n\\fourch{$b>c>a$}{$b>a>c$}{$c>b>a$}{$c>a>b$}",
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"content": "函数$y=f(x)$的图像由$y=\\cos (2 x+\\dfrac{\\pi}{6})$的图像向左平移$\\dfrac{\\pi}{6}$个单位长度得到, 则$y=f(x)$的图像与直线$y=\\dfrac{1}{2} x-\\dfrac{1}{2}$的交点个数为\\bracket{20}.\n\\fourch{$1$}{$2$}{$3$}{$4$}",
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"content": "记$S_n$为等比数列$\\{a_n\\}$的前$n$项和. 若$8S_6=7S_3$, 则$\\{a_n\\}$的公比为\\blank{50}.",
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"018183": {
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"content": "若$f(x)=(x-1)^2+a x+\\sin (x+\\dfrac{\\pi}{2})$为偶函数, 则$a=$\\blank{50}.",
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"content": "若$x, y$满足约束条件$\\begin{cases}3 x-2 y \\leq 3, \\\\ -2 x+3 y \\leq 3,\\\\ x+y \\geq 1,\\end{cases}$ 则$z=3 x+2 y$的最大值为\\blank{50}.",
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"content": "在正方体$ABCD-A_1B_1C_1D_1$中, $AB=4$, $O$为$AC_1$的中点, 若该正方体的棱与球$O$的球面有公共点, 则球$O$的半径的取值范围是\\blank{50}.",
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"content": "记$\\triangle ABC$的内角$A, B, C$的对边分别为$a, b, c$, 已知$\\dfrac{b^2+c^2-a^2}{\\cos A}=2$.\\\\\n(1) 求$b c$;\\\\\n(2) 若$\\dfrac{a \\cos B-b \\cos A}{a \\cos B+b \\cos A}-\\dfrac{b}{c}=1$, 求$\\triangle ABC$的面积.",
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"content": "如图, 在三棱柱$ABC-A_1B_1C_1$中, $A_1C \\perp$平面$ABC$, $\\angle ACB=90^{\\circ}$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0,0) node [left] {$C$} coordinate (C);\n\\draw (0,0,{sqrt(2)}) node [left] {$A$} coordinate (A);\n\\draw (2,0,0) node [right] {$B$} coordinate (B);\n\\draw (0,{sqrt(2)},0) node [above] {$A_1$} coordinate (A_1);\n\\draw ($(A_1)+(C)-(A)$) node [above] {$C_1$} coordinate (C_1);\n\\draw ($(A_1)+(B)-(A)$) node [right] {$B_1$} coordinate (B_1);\n\\draw (A)--(B)--(B_1)--(A_1)--cycle(B)--(A_1)--(C_1)--(B_1);\n\\draw [dashed] (A)--(C)--(B)(C)--(A_1)(C)--(C_1);\n\\end{tikzpicture}\n\\end{center}\n(1) 证明: 平面$ACC_1A_1 \\perp$平面$BB_1C_1C$;\\\\\n(2) 设$AB=A_1B$, $AA_1=2$, 求四棱锥$A_1-BB_1C_1C$的高.",
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"content": "一项试验旨在研究臭氧效应, 试验方案如下: 选$40$只小白鼠, 随机地将其中$20$只分配到试验组, 另外$20$只分配到对照组, 试验组的小白鼠饲养在高浓度臭氧环境, 对照组的小白鼠饲养在正常环境, 一段时间后统计每只小白鼠体重的增加量(单位: $\\text{g}$). 试验结果如下\\\\\n对照组的小白鼠体重的增加量从小到大排序为:\n\\begin{center}\n\\begin{tabular}{cccccccccc}\n15.2 & 18.8 & 20.2 & 21.3 & 22.5 & 23.2 & 25.8 & 26.5 & 27.5 & 30.1 \\\\\n32.6 & 34.3 & 34.8 & 35.6 & 35.6 & 35.8 & 36.2 & 37.3 & 40.5 & 43.2\n\\end{tabular}\n\\end{center}\n试验组的小白鼠体重的增加量从小到大排序为:\n\\begin{center}\n\\begin{tabular}{cccccccccc}\n7.8 & 9.2 & 11.4 & 12.4 & 13.2 & 15.5 & 16.5 & 18.0 & 18.8 & 19.2 \\\\ 19.8 & 20.2 & 21.6 & 22.8 & 23.6 & 23.9 & 25.1 & 28.2 & 32.3 & 36.5\n\\end{tabular}\n\\end{center}\n(1) 计算试验组的样本平均数;\\\\\n(2) (I) 求$40$只小白鼠体重的增加量的中位数$m$, 再分别统计两样本中小于$m$与不小于$m$的数据的个数, 完成如下列联表:\n\\begin{center}\n\\begin{tabular}{|c|c|c|}\n\\hline &$<m$&$\\geq m$\\\\\n\\hline 对照组 & & \\\\\n\\hline 试验组 & & \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n(II) 根据(I)中的列联表, 能否有$95 \\%$的把握认为小白鼠在高浓度臭氧环境中与在正常环境中体重的增加量有差异?、、\n附: $\\chi^2=\\dfrac{n(a d-b c)^2}{(a+b)(b+c)(a+c)(b+d)}$.\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline$k_0$& 0.10 & 0.05 & 0.010 & 0.005 & 0.001 \\\\\n\\hline$p(\\chi^2 \\geq k_0)$& 2.706 & 3.841 & 6.635 & 7.879 & 10.828 \\\\\n\\hline\n\\end{tabular}\n\\end{center}",
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"content": "已知函数$f(x)=a x-\\dfrac{\\sin x}{\\cos ^2 x}, x \\in(0, \\dfrac{\\pi}{2})$.\\\\\n(1) 当$a=1$时, 讨论$f(x)$的单调性;\\\\\n(2) 若$f(x)+\\sin x<0$, 求$a$的取值范围.",
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"content": "已知直线$x-2 y+1=0$与抛物线$C: y^2=2 p x$($p>0$) 交于$A, B$两点, $|AB|=4 \\sqrt{15}$.\\\\\n(1) 求$p$;\\\\\n(2) 设$F$为$C$的焦点, $M, N$为$C$上两点, 且$\\overrightarrow{FM} \\cdot \\overrightarrow{FN}=0$, 求$\\triangle MFN$面积的最小值.",
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"content": "已知$P(2,1)$, 直线$l: \\begin{cases}x=2+t \\cos \\alpha, \\\\ y=1+t \\sin \\alpha\\end{cases}$($t$为参数), $l$与$x$轴, $y$轴正半轴交于$A, B$两点, $|PA| \\cdot|PB|=4$.\\\\\n(1) 求$\\alpha$的值;\\\\\n(2) 以原点为极点, $x$轴的正半轴为极轴建立极坐标系, 求$l$的极坐标方程.",
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"content": "已知$a>0$, $f(x)=2|x-a|-a$.\\\\\n(1) 求不等式$f(x)<x$的解集;\\\\\n(2) 若函数$y=f(x)$的图像与$x$轴围城的三角形面积为$2$, 求$a$.",
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"018193": {
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"content": "$|2+\\mathrm{i}^2+2 \\mathrm{i}^3|=()$\\bracket{20}.\n\\fourch{$1$}{$2$}{$\\sqrt{5}$}{$5$}",
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"content": "设全集$U=\\{0,1,2,4,6,8\\}$, 集合$M=\\{0,4,6\\}$, $N=\\{0,1,6\\}$, 则$M \\cup \\overline{N}=$\\bracket{20}.\n\\fourch{$\\{0,2,4,6,8\\}$}{$\\{0,1,4,6,8\\}$}{$\\{1,2,4,6,8\\}$}{$U$}",
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"018195": {
"id": "018195",
"content": "如图, 网格纸上绘制的是个零件的三视图, 网格小正方形的边长为$1$, 则该零件的表面积为\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\foreach \\i in {0,1,...,6}\n{\\draw [gray] (\\i,0) --++ (0,5);\n\\draw [gray] ({\\i+7},0) --++ (0,5);\n\\draw [gray] (\\i,-6) --++ (0,5);};\n\\foreach \\i in {0,1,...,5}\n{\\draw [gray] (0,\\i) --++ (6,0);\n\\draw [gray] (7,\\i) --++ (6,0);\n\\draw [gray] (0,{\\i-6}) --++ (6,0);};\n\\draw [ultra thick] (2,1) rectangle (4,4) (2,3) -- (4,3);\n\\draw [ultra thick] (9,1) --++ (2,0) --++ (0,2) --++ (-1,0) --++ (0,1) --++ (-1,0) --cycle;\n\\draw [ultra thick] (2,-2) rectangle (4,-4) (2,-3) --++ (2,0);\n\\end{tikzpicture}\n\\end{center}\n\\fourch{$24$}{$26$}{$28$}{$30$}",
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"018196": {
"id": "018196",
"content": "在$\\triangle ABC$中, 内角$A, B, C$的对边分别是$a, b, c$, 若$a \\cos B-b \\cos A=c$, 且$C=\\dfrac{\\pi}{5}$, 则$\\angle B=$\\bracket{20}.\n\\fourch{$\\dfrac{\\pi}{10}$}{$\\cdot \\dfrac{\\pi}{5}$}{$\\dfrac{3 \\pi}{10}$}{$\\dfrac{2 \\pi}{5}$}",
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"018197": {
"id": "018197",
"content": "已知函数$f(x)=\\dfrac{x \\mathrm{e}^x}{\\mathrm{e}^{a x}-1}$是偶函数, 则实数$a=$\\bracket{20}.\n\\fourch{$-2$}{$-1$}{$1$}{$2$}",
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"018198": {
"id": "018198",
"content": "正方形$ABCD$的边长是$2$, $E$是$AB$的中点, 则$\\overrightarrow{EC} \\cdot \\overrightarrow{ED}=$\\bracket{20}.\n\\fourch{$\\sqrt{5}$}{$3$}{$2 \\sqrt{5}$}{$5$}",
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"genre": "选择题",
"ans": "",
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"018199": {
"id": "018199",
"content": "已知$O$是平面直角坐标系的原点, 在区域$\\{(x, y) | 1 \\leq x^2+y^2 \\leq 4\\}$内随机取一点$A$, 则直线$OA$的倾斜角不大于$\\dfrac{\\pi}{4}$的概率为\\bracket{20}.\n\\fourch{$\\dfrac{1}{8}$}{$\\dfrac{1}{6}$}{$\\dfrac{1}{4}$}{$\\dfrac{1}{2}$}",
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"genre": "选择题",
"ans": "",
"solution": "",
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},
"018200": {
"id": "018200",
"content": "函数$f(x)=x^3+a x+2$存在$3$个零点, 则$a$的取值范围是\\bracket{20}.\n\\fourch{$(-\\infty,-2)$}{$(-\\infty,-3)$}{$(-4,-1)$}{$(-3,0)$}",
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"genre": "选择题",
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"solution": "",
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"018201": {
"id": "018201",
"content": "某学校举办作文比赛, 共$6$个主题, 每位参赛同学从中随机抽取一个主题准备作文, 则甲、乙两位参赛同学抽到不同主题概率为\\bracket{20}.\n\\fourch{$\\dfrac{5}{6}$}{$\\dfrac{2}{3}$}{$\\dfrac{1}{2}$}{$\\dfrac{1}{3}$}",
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"tags": [],
"genre": "选择题",
"ans": "",
"solution": "",
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"remark": "",
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},
"018202": {
"id": "018202",
"content": "函数$f(x)=\\sin (\\omega x+\\varphi)$在区间$(\\dfrac{\\pi}{6}, \\dfrac{2 \\pi}{3})$上单调递增, 直线$x=\\dfrac{\\pi}{6}$和$x=\\dfrac{2 \\pi}{3}$是函数$y=f(x)$图像的两条对称轴, 则$f(-\\dfrac{5 \\pi}{12})=$\\bracket{20}.\n\\fourch{$-\\dfrac{\\sqrt{3}}{2}$}{$-\\dfrac{1}{2}$}{$\\dfrac{1}{2}$}{$\\dfrac{\\sqrt{3}}{2}$}",
"objs": [],
"tags": [],
"genre": "选择题",
"ans": "",
"solution": "",
"duration": -1,
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},
"018203": {
"id": "018203",
"content": "已知实数$x, y$满足$x^2+y^2-4 x-2 y-4=0$, 则$x-y$的最大值是\\bracket{20}.\n\\fourch{$1+\\dfrac{3 \\sqrt{2}}{2}$}{4}{$1+3 \\sqrt{2}$}{7}",
"objs": [],
"tags": [],
"genre": "选择题",
"ans": "",
"solution": "",
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},
"018204": {
"id": "018204",
"content": "设$A, B$为双曲线$x^2-\\dfrac{y^2}{9}=1$上两点, 下列四个点中, 可以为线段$AB$中点的是\\bracket{20}.\n\\fourch{$(1,1)$}{$(-1,2)$}{$(1,3)$}{$(-1,-4)$}",
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"solution": "",
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},
"018205": {
"id": "018205",
"content": "已知点$A(1, \\sqrt{5})$在抛物线$C: y^2=2 p x$上, 则$A$到$C$的准线的距离为\\blank{50}.",
"objs": [],
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"genre": "填空题",
"ans": "",
"solution": "",
"duration": -1,
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},
"018206": {
"id": "018206",
"content": "若$\\theta \\in(0, \\dfrac{\\pi}{2}), \\tan \\theta=\\dfrac{1}{2}$, 则$\\sin \\theta-\\cos \\theta=$\\blank{50}.",
"objs": [],
"tags": [],
"genre": "填空题",
"ans": "",
"solution": "",
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"018207": {
"id": "018207",
"content": "若$x, y$满足约束条件$\\begin{cases}x-3 y \\leq-1 \\\\x+2 y \\leq 9, \\\\\n3 x+y \\geq 7,\\end{cases}$则$z=2 x-y$的最大值为\\blank{50}.",
"objs": [],
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"genre": "填空题",
"ans": "",
"solution": "",
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"origin": "2023届全国高考乙卷文科试题15",
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},
"018208": {
"id": "018208",
"content": "已知点$S, A, B, C$均在半径为$2$的球面上, $\\triangle ABC$是边长为$3$的等边三角形, $SA \\perp$平面$ABC$, 则$SA=$\\blank{50}.",
"objs": [],
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"genre": "填空题",
"ans": "",
"solution": "",
"duration": -1,
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"origin": "2023届全国高考乙卷文科试题16",
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"018209": {
"id": "018209",
"content": "某厂为比较甲乙两种工艺对橡胶产品伸缩率的处理效应, 进行$10$次配对试验, 每次配对试验选用材质相同的两个橡胶产品, 随机地选其中一个用甲工艺处理, 另一个用乙工艺处理, 测量处理后的橡胶产品的伸缩率. 甲、乙两种工艺处理后的橡胶产品的伸缩率分别记为$x_i$, $y_i$($i=1,2, \\cdots, 10$). 试验结果如下:\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}\n\\hline 试验序号$i$& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\\\n\\hline 伸缩率$x_i$& 545 & 533 & 551 & 522 & 575 & 544 & 541 & 568 & 596 & 548 \\\\\n\\hline 伸缩率$y_i$& 536 & 527 & 543 & 530 & 560 & 533 & 522 & 550 & 576 & 536 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n记$z_i=x_i-y_i$($i=1,2, \\cdots, 10$), 记$z_1, z_2, \\cdots, z_{10}$的样本平均数为$\\overline {z}$, 样本方差为$s^2$.\\\\\n(1) 求$\\overline {z}$, $s^2$;\\\\\n(2) 判断甲工艺处理后的橡胶产品的伸缩率较乙工艺处理后的橡胶产品的伸缩率是否有显著提高(如果$\\overline {z} \\geq 2 \\sqrt{\\dfrac{s^2}{10}}$, 则认为甲工艺处理后的橡胶产品的伸缩率较乙工艺处理后的橡胶产品的伸缩率有显著提高, 否则不认为有显著提高).",
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"genre": "解答题",
"ans": "",
"solution": "",
"duration": -1,
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"origin": "2023届全国高考乙卷文科试题17",
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"018210": {
"id": "018210",
"content": "记$S_n$为等差数列$\\{a_n\\}$的前$n$项和, 已知$a_2=11$, $S_{10}=40$.\n(1) 求$\\{a_n\\}$的通项公式;\\\\\n(2) 求数列$\\{|a_n|\\}$的前$n$项和$T_n$.",
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"genre": "解答题",
"ans": "",
"solution": "",
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"origin": "2023届全国高考乙卷文科试题18",
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"018211": {
"id": "018211",
"content": "如图, 在三棱锥$P-ABC$中, $AB \\perp BC$, $AB=2$, $BC=2 \\sqrt{2}$, $PB=PC=\\sqrt{6}$, $BP$、$AP$、$BC$的中点分别为$D$、$E$、$O$, 点$F$在$AC$上, $BF \\perp AO$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, z = {(-45:0.5cm)}, scale = 1.5]\n\\draw (0,0,0) node [left] {$B$} coordinate (B);\n\\draw (0,0,2) node [below] {$A$} coordinate (A);\n\\draw ({2*sqrt(2)},0,0) node [right] {$C$} coordinate (C);\n\\draw ({sqrt(2)},{sqrt(3)},-1) node [above] {$P$} coordinate (P);\n\\draw (A)--(C)--(P)--(B)--cycle;\n\\draw [dashed] (B)--(C);\n\\draw (A)--(P);\n\\draw ($(B)!0.5!(C)$) node [above left] {$O$} coordinate (O);\n\\draw ($(A)!0.5!(C)$) node [below right] {$F$} coordinate (F);\n\\draw ($(P)!0.5!(B)$) node [left] {$D$} coordinate (D);\n\\draw ($(P)!0.5!(A)$) node [right] {$E$} coordinate (E);\n\\draw (A)--(D) (E)--(F)(B)--(E);\n\\draw [dashed] (A)--(O)(B)--(F)--(O)--(D)(O)--(P);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: $EF\\parallel$平面$ADO$;\\\\\n(2) 若$\\angle POF=120^{\\circ}$, 求三棱锥$P-ABC$的体积.",
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"ans": "",
"solution": "",
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"origin": "2023届全国高考乙卷文科试题19",
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"remark": "",
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"018212": {
"id": "018212",
"content": "已知函数$f(x)=(\\dfrac{1}{x}+a) \\ln (x+1)$.\\\\\n(1) 当$a=-1$时, 求曲线$y=f(x)$在$(1, f(1))$处的切线方程;\\\\\n(2) 若函数$f(x)$在$(0,+\\infty)$单调递增, 求$a$的取值范围.",
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"ans": "",
"solution": "",
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"origin": "2023届全国高考乙卷文科试题20",
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"remark": "",
"space": "4em",
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"018213": {
"id": "018213",
"content": "已知椭圆$C: \\dfrac{y^2}{a^2}+\\dfrac{x^2}{b^2}=1$($a>b>0$)的离心率是$\\dfrac{\\sqrt{5}}{3}$, 点$A(-2,0)$在$C$上.\\\\\n(1) 求$C$的方程;\\\\\n(2) 过点$(2,3)$的直线交$C$于$P, Q$两点, 直线$AP, AQ$与$y$轴的交点分别为$M, N$, 证明: 线段$MN$的中点为定点.",
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"genre": "解答题",
"ans": "",
"solution": "",
"duration": -1,
"usages": [],
"origin": "2023届全国高考乙卷文科试题21",
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"remark": "",
"space": "4em",
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"018214": {
"id": "018214",
"content": "在直角坐标系$x O y$中, 以坐标原点$O$为极点, $x$轴正半轴为极轴建立极坐标系, 曲线$C_1$的极坐标方程为$\\rho=2 \\sin \\theta$($\\dfrac{\\pi}{4} \\leq \\theta \\leq \\dfrac{\\pi}{2}$), $C_2: \\begin{cases}x=2 \\cos \\alpha, \\\\ y=2 \\sin \\alpha\\end{cases}$($\\alpha$为参数, $\\dfrac{\\pi}{2}<\\alpha<\\pi)$.\\\\\n(1) 写出$C_1$的直角坐标方程;\\\\\n(2) 若直线$y=x+m$既与$C_1$没有公共点, 也与$C_2$没有公共点, 求$m$的取值范围.",
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"ans": "",
"solution": "",
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"space": "4em",
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"018215": {
"id": "018215",
"content": "已知$f(x)=2|x|+|x-2|$.\\\\\n(1) 求不等式$f(x) \\leq 6-x$的解集;\\\\\n(2) 在直角坐标系$x O y$中, 求不等式组$\\begin{cases}f(x) \\leq y, \\\\ x+y-6 \\leq 0\\end{cases}$所确定的平面区域的面积.",
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"genre": "解答题",
"ans": "",
"solution": "",
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"020001": {
"id": "020001",
"content": "判断下列各组对象能否组成集合, 若能组成集合, 指出是有限集还是无限集.\\\\\n(1) 上海市控江中学$2022$年入学的全体高一年级新生;\\\\\n(2) 中国现有各省的名称;\\\\\n(3) 太阳、$2$、上海市;\\\\\n(4) 大于$10$且小于$15$的有理数;\\\\\n(5) 末位是$3$的自然数;\\\\\n(6) 影响力比较大的中国数学家;\\\\\n(7) 方程$x^2+x-3=0$的所有实数解;\\\\ \n(8) 函数$y=\\dfrac 1x$图像上所有的点;\\\\ \n(9) 在平面直角坐标系中, 到定点$(0, 0)$的距离等于$1$的所有点;\\\\\n(10) 不等式$3x-10<0$的所有正整数解;\\\\\n(11) 所有的平面四边形.",