收录2023届黄浦高三二模试题

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WangWeiye 2023-04-13 18:49:44 +08:00
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filename = r"C:\Users\weiye\Documents\wwy sync\临时工作区\自拟题目11.tex"
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"content": "设集合$A=\\{1,3,5,7,9\\}$, $B=\\{x | 2 \\leq x \\leq 5\\}$, 则$A \\cap B=$\\blank{50}.",
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"content": "函数$y=4 \\cos 2 x+3$的最小正周期为\\blank{50}.",
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"content": "若函数$y=x^a$的图像经过点$(2,16)$与$(3, m)$, 则$m$的值为\\blank{50}.",
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"content": "已知复数$z_1, z_2$在复平面内的对应点关于虚轴对称, 且$z_1=2+\\mathrm{i}$(i 为虚数单位), 则$z_1 z_2=$\\blank{50}.",
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"content": "以抛物线$y^2=4 x$的焦点为圆心、且与该抛物线的准线相切的圆的方程为\\blank{50}.",
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"content": "已知$m$是$m-2$与$4$的等差中项, 且$(m+x)^5=a_0+a_1 x+a_2 x^2+a_3 x^3+a_4 x^4+a_5 x^5$, 则$a_3$的值为\\blank{50}.",
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"content": "已知函数$y=f(x)$是定义在$\\mathbf{R}$上的奇函数, 且当$x<0$时, $f(x)=\\mathrm{e}^{a x}$. 若$f(\\ln 2)=-4$, 则实数$a$的值为\\blank{50}.",
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"content": "如图, 某学具可看成将一个底面半径与高都为$10 \\text{cm}$的圆柱挖去一个圆锥(此圆锥的顶点是圆柱的下底面圆心, 底面是圆柱的上底面) 所得到的几何体, 则该学具的表面积为$\\text{cm}^2$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\fill [pattern = north east lines] (-1,1) arc (180:360:1 and 0.25) -- (0,0) -- cycle;\n\\fill [pattern = north west lines] (0,1) ellipse (1 and 0.25);\n\\draw (0,1) ellipse (1 and 0.25);\n\\draw (-1,0) arc (180:360:1 and 0.25);\n\\draw [dashed] (-1,0) arc (180:0:1 and 0.25);\n\\draw (-1,0) --++ (0,1) (1,0) --++ (0,1);\n\\draw [dashed] (-1,1) -- (0,0) -- (1,1);\n\\end{tikzpicture}\n\\end{center}",
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"content": "若函数$y=f(x)$的图像可由函数$y=3 \\sin 2 x-\\sqrt{3} \\cos 2 x$的图像向右平移$\\varphi$($0<\\varphi<\\pi$)个单位所得到, 且函数$y=f(x)$在区间$[0, \\dfrac{\\pi}{2}]$上是严格减函数, 则$\\varphi=$\\blank{50}.",
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"content": "若每经过一天某种物品的价格变为原来的$1.02$倍的概率为$0.5$, 变为原来的$0.98$倍的概率也为$0.5$, 则经过$6$天该物品的价格较原来价格增加的概率为\\blank{50}.",
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"content": "如图, 在直角梯形$ABCD$中, $AD\\parallel BC$, $\\angle ABC=90^{\\circ}$, $AD=2$, $BC=1$, 点$P$是腰$AB$上的动点, 则$|2 \\overrightarrow{PC}+\\overrightarrow{PD}|$的最小值为\\blank{50}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0) node [below] {$A$} coordinate (A);\n\\draw (2.5,0) node [below] {$B$} coordinate (B);\n\\draw (0,2) node [left] {$D$} coordinate (D);\n\\draw (2.5,1) node [right] {$C$} coordinate (C);\n\\draw ($(A)!0.6!(B)$) node [below] {$P$} coordinate (P);\n\\draw (D)--(A)--(B)--(C)--cycle;\n\\draw [->] (P) -- (C);\n\\draw [->] (P) -- (D);\n\\end{tikzpicture}\n\\end{center}",
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"content": "已知实数$a, b, c$满足: $a+b+c=0$与$a^2-b c=3$, 则$a b c$的取值范围为\\blank{50}.",
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"content": "若直线$(a-1) x+y-1=0$与直线$3 x-a y+2=0$垂直, 则实数$a$的值为\\bracket{20}.\n\\fourch{$\\dfrac{1}{2}$}{$\\dfrac{3}{2}$}{$\\dfrac{1}{4}$}{$\\dfrac{3}{4}$}",
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"content": "从装有两个红球和两个白球的口袋内任取两个球, 那么互斥而不对立的事件是\\bracket{20}.\n\\twoch{``恰好有一个白球''与``都是红球''}{``至多有一个白球''与``都是红球''}{``至多有一个白球''与``都是白球''}{``至多有一个白球''与``至多有一个红球''}",
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"content": "如图, $\\triangle ABD$与$\\triangle BCD$都是等腰直角三角形, 其底边分别为$BD$与$BC$, 点$E$、$F$分别为线段$BD$、$AC$的中点, 设二面角$A-BD-C$的大小为$\\alpha$, 当$\\alpha$在区间$(0, \\pi)$内变化时, 下列结论正确的是\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 1.5]\n\\draw (-1,0,0) node [left] {$B$} coordinate (B);\n\\draw (1,0,0) node [right] {$D$} coordinate (D);\n\\draw (0,0,1) node [below] {$C$} coordinate (C);\n\\draw (0,1,0) node [above] {$A$} coordinate (A);\n\\draw ($(B)!0.5!(D)$) node [above] {$E$} coordinate (E);\n\\draw ($(A)!0.5!(C)$) node [left] {$F$} coordinate (F);\n\\draw (A)--(B)--(C)--(D)--cycle (A)--(C);\n\\draw [dashed] (B)--(D)(E)--(F);\n\\end{tikzpicture}\n\\end{center}\n\\twoch{存在某一$\\alpha$值, 使得$AC \\perp BD$}{存在某一$\\alpha$值, 使得$EF \\perp BD$}{存在某一$\\alpha$值, 使得$EF \\perp CD$}{存在某一$\\alpha$值, 使得$AB \\perp CD$}",
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"content": "设数列$\\{a_n\\}$的前$n$项的和为$S_n$, 若对任意的$n \\in \\mathbf{N}$, $n\\ge 1$, 都有$S_n<a_{n+1}$, 则称数列$\\{a_n\\}$为``$K$数列''. 关于命题: \\textcircled{1} 存在等差数列$\\{a_n\\}$, 使得它是``$K$数列''; \\textcircled{2} 若$\\{a_n\\}$是首项为正数、公比为$q$的等比数列, 则$q \\in[2,+\\infty)$是$\\{a_n\\}$为``$K$数列''的充要条件. 下列判断正确的是\\bracket{20}.\n\\twoch{\\textcircled{1}和\\textcircled{2}都为真命题}{\\textcircled{1}为真命题, \\textcircled{2}为假命题}{\\textcircled{1}为假命题, \\textcircled{2}为真命题}{\\textcircled{1}和\\textcircled{2}都为假命题}",
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"content": "在$\\triangle ABC$中, $\\cos A=-\\dfrac{5}{13}$, $\\cos B=\\dfrac{3}{5}$.\\\\\n(1) 求$\\sin C$的值;\\\\\n(2) 若$AB=4$, 求$\\triangle ABC$的周长和面积.",
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"content": "如图, 多面体$A_1C_1D_1ABCD$是由棱长为$3$的正方体$ABCD-A_1B_1C_1D_1$沿平面$A_1BC_1$截去一角所得到. 在棱$A_1C_1$上取一点$E$, 过点$D_1, C, E$的平面交棱$BC_1$于点$F$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\def\\l{2}\n\\draw (0,0,0) node [below left] {$A$} coordinate (A);\n\\draw (A) ++ (\\l,0,0) node [below right] {$B$} coordinate (B);\n\\draw (A) ++ (\\l,0,-\\l) node [right] {$C$} coordinate (C);\n\\draw (A) ++ (0,0,-\\l) node [left] {$D$} coordinate (D);\n\\draw (A) -- (B) -- (C);\n\\draw [dashed] (A) -- (D) -- (C);\n\\draw (A) ++ (0,\\l,0) node [left] {$A_1$} coordinate (A1);\n\\draw (C) ++ (0,\\l,0) node [above right] {$C_1$} coordinate (C1);\n\\draw (D) ++ (0,\\l,0) node [above left] {$D_1$} coordinate (D1);\n\\draw (A1)--(C1) -- (D1) -- cycle;\n\\draw (A) -- (A1) (A1)--(B)--(C1) (C) -- (C1);\n\\draw [dashed] (D) -- (D1);\n\\draw ($(A1)!{1/3}!(C1)$) node [below] {$E$} coordinate (E);\n\\draw ($(B)!{1/3}!(C1)$) node [left] {$F$} coordinate (F);\n\\draw (D1)--(E)--(F)--(C);\n\\draw [dashed] (C)--(D1);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: $EF\\parallel A_1B$;\\\\\n(2) 若$C_1E=2EA_1$, 求点$E$到平面$A_1D_1CB$的距离以及$ED_1$与平面$A_1D_1CB$所成角的大小.",
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"content": "将某工厂的工人按年龄分成两组: ``$35$周岁及以上''、``$35$周岁以下'', 从每组中随机抽取$80$人, 将他们的绩效分数分成$5$组: $[50,60),[60,70),[70,80),[80,90),[90,100]$, 分别加以统计, 得到下列频率分布直方图. 该工厂规定绩效分数不少于$80$者为生产标兵.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, xscale = 0.05, yscale = 60]\n\\draw [->] (40,0) -- (42,0) -- (44,-0.003) -- (46,0.003) -- (48,0)-- (120,0) node [below] {绩效分数};\n\\draw [->] (40,0) -- (40,0.05) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw (40,0) node [below left] {$O$};\n\\foreach \\i/\\j in {50/0.005,60/0.035,70/0.035,80/0.020,90/0.005}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (10,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {70/0.035,80/0.020,90/0.005}\n{\\draw [dashed] (\\i,\\j) -- (40,\\j) node [left] {$\\k$};};\n\\draw (100,0) node [below] {$100$};\n\\draw (75,-0.01) node {$35$周岁及以上组};\n\\end{tikzpicture}\n\\begin{tikzpicture}[>=latex, xscale = 0.05, yscale = 60]\n\\draw [->] (40,0) -- (42,0) -- (44,-0.003) -- (46,0.003) -- (48,0)-- (120,0) node [below] {绩效分数};\n\\draw [->] (40,0) -- (40,0.05) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw (40,0) node [below left] {$O$};\n\\foreach \\i/\\j in {50/0.005,60/0.025,70/0.0325,80/0.0325,90/0.005}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (10,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {60/0.025,80/0.0325,90/0.005}\n{\\draw [dashed] (\\i,\\j) -- (40,\\j) node [left] {$\\k$};};\n\\draw (100,0) node [below] {$100$};\n\\draw (75,-0.01) node {$35$周岁以下组};\n\\end{tikzpicture}\n\\end{center}\n(1) 请列出$2 \\times 2$列联表, 并判断能否有$95 \\%$的把握认为是否为生产标兵与工人所在的年龄组有关;\\\\\n(2) 若已知该工厂工人中生产标兵的占比为$30 \\%$, 试估计该厂$35$周岁以下的工人所占的百分比以及生产标兵中$35$周岁以下的工人所占的百分比.\\\\\n附: $\\chi^2=\\dfrac{n(a d-b c)^2}{(a+b)(c+d)(a+c)(b+d)}$.\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline$P(x^2 \\geq k)$& 0.100 & 0.050 & 0.010 & 0.001 \\\\\n\\hline$k$& 2.706 & 3.841 & 6.635 & 10.828 \\\\\n\\hline\n\\end{tabular}",
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"content": "已知双曲线$C$的中心在坐标原点, 左焦点$F_1$与右焦点$F_2$都在$x$轴上, 离心率为$3$, 过点$F_2$的动直线$l$与双曲线$C$交于点$A$、$B$, 设$\\dfrac{|AF_2| \\cdot|BF_2|}{|AB|^2}=\\lambda$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.2]\n\\draw [->] (-6,0) -- (6,0) node [below] {$x$};\n\\draw [->] (0,-12) -- (0,12) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\draw [domain = {-2*sqrt(30)}:{2*sqrt(30)}, samples = 100] plot ({sqrt(1+\\x*\\x/8)},\\x);\n\\draw [domain = {-2*sqrt(30)}:{2*sqrt(30)}, samples = 100] plot ({-sqrt(1+\\x*\\x/8)},\\x);\n\\filldraw (3,0) circle (0.15) node [below] {$F_2$} coordinate (F_2);\n\\filldraw (-3,0) circle (0.15) node [below] {$F_2$} coordinate (F_2);\n\\end{tikzpicture}\n\\end{center}\n(1) 求双曲线$C$的渐近线方程;\\\\\n(2) 若点$A$、$B$都在双曲线$C$的右支上, 求$\\lambda$的最大值以及$\\lambda$取最大值时$\\angle AF_1B$的正切值; (关于求$\\lambda$的最值, 某学习小组提出了如下的思路可供参考: \\textcircled{1} 利用基本不等式求最值; \\textcircled{2} 设$\\dfrac{|AF_2|}{|AB|}$为$\\mu$, 建立相应数量关系并利用它求最值; \\textcircled{3} 设直线$l$的斜率为$k$, 建立相应数量关系并利用它求最值)\\\\\n(3) 若点$A$在双曲线$C$的左支上(点$A$不是该双曲线的顶点), 且$\\lambda=1$, 求证: $\\triangle AF_1B$是等腰三角形, 且$AB$边的长等于双曲线$C$的实轴长的$2$倍.",
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"015079": {
"id": "015079",
"content": "三个互不相同的函数$y=f(x), y=g(x)$与$y=h(x)$在区间$D$上恒有$f(x) \\geq h(x) \\geq g(x)$或恒有$f(x) \\leq h(x) \\leq g(x)$, 则称$y=h(x)$为$y=f(x)$与$y=g(x)$在区间$D$上的``分割函数''.\\\\\n(1) 设$h_1(x)=4 x$, $h_2(x)=x+1$, 试分别判断$y=h_1(x)$、$y=h_2(x)$是否是$y=2 x^2+2$与$y=-x^2+4 x$在区间$(-\\infty,+\\infty)$上的``分割函数'', 请说明理由;\\\\\n(2) 求所有的二次函数, 使得该函数是$y=2 x^2+2$与$y=4 x$在区间$(-\\infty,+\\infty)$上的``分割函数'';\\\\\n(3) 若$[m, n] \\subseteq[-2,2]$, 且存在实数$k, b$, 使得$y=k x+b$为$y=x^4-4 x^2$与$y=4 x^2-16$在区间$[m, n]$\n上的``分割函数'', 求$n-m$的最大值.",
"objs": [],
"tags": [],
"genre": "解答题",
"ans": "",
"solution": "",
"duration": -1,
"usages": [],
"origin": "2023届黄浦区高三二模试题21",
"edit": [
"202304012\t王伟叶"
],
"same": [],
"related": [],
"remark": "",
"space": "12ex"
},
"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.",