收录2023届青浦高三二模试题

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WangWeiye 2023-04-13 19:02:20 +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"
editor = "202304012\t王伟叶"

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"content": "若空间中两条直线$a$、$b$确定一个平面, 则$a$、$b$的位置关系为\\blank{50}.",
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"content": "已知复数$z$满足$\\overline {z} \\cdot \\mathrm{i}=4+3 \\mathrm{i}$, 则$|z|=$\\blank{50}.",
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"content": "已知向量$\\overrightarrow {a}=(1,0)$和$\\overrightarrow {b}=(\\sqrt{3}, 1)$, 则$\\overrightarrow {b}$在$\\overrightarrow {a}$方向上的投影是\\blank{50}.",
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"content": "过点$P(-1,3)$, 与直线$x+\\sqrt{3} y+1=0$垂直的直线方程为\\blank{50}.",
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"content": "已知集合$A=\\{x | y=\\ln (3-x)\\}$, $B=\\{x | x>a\\}$, 若$A \\cap B=\\varnothing$, 则实数$a$的取值范围为\\blank{50}.",
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"content": "已知圆柱的底面直径和高都等于球的直径, 圆柱的体积为$16 \\pi$, 则球的表面积为\\blank{50}.",
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"content": "已知函数$y=a x^2+b x+c$的图像如图所示, 则不等式$(a x+b)(b x+c)(c x+a)<0$的解集是\\blank{50}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\draw [->] (-1,0) -- (4,0) node [below] {$x$};\n\\draw [->] (0,-1) -- (0,4) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\draw [domain = -0.5:3.5] plot (\\x,{(\\x-1)*(\\x-2)});\n\\draw (1,0) node [below] {$1$} (2,0) node [below] {$2$};\n\\end{tikzpicture}\n\\end{center}",
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"content": "已知函数$y=f(x)$是定义在$\\mathbf{R}$上的奇函数, 且满足$f(2+x)=-f(2-x)$, $f(1)=1$, 则$f(1)+f(2)+\\cdots+f(2023)=$\\blank{50}.",
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"content": "如图所示, 要在两山顶$M$、$N$间建一索道, 需测量两山顶$M$、$N$间的距离. 已知两山的海拔高度分别是$MC=100 \\sqrt{3}$米和$NB=50 \\sqrt{2}$米, 现选择海平面上一点$A$为观测点, 从$A$点测得$M$点的仰角$\\angle MAC=60^{\\circ}$, $N$点的仰角$\\angle NAB=30^{\\circ}$以及$\\angle MAN=45^{\\circ}$, 则$MN$等于\\blank{50}米.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.6]\n\\draw (0,0) node [left] {$C$} coordinate (C);\n\\draw (3,0) node [right] {$B$} coordinate (B);\n\\draw (1.7,-0.5) node [below] {$A$} coordinate (A);\n\\draw (C) ++ (0,3) node [above] {$M$} coordinate (M);\n\\draw (B) ++ (0,1.5) node [above] {$N$} coordinate (N);\n\\draw (C)--(A)--(B)--(N)--(M)--cycle(M)--(A)--(N);\n\\draw [dashed] (C)--(B);\n\\end{tikzpicture}\n\\end{center}",
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"content": "已知数列$\\{a_n\\}$满足$a_n=a n^2+n$, 若满足$a_1<a_2<a_3<a_4<a_5<a_6$且对任意$n \\in[9,+\\infty)$, 都有$a_n>a_{n+1}$, 则实数$a$的取值范围是\\blank{50}.",
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"content": "如图, 已知$F_1, F_2$分别是椭圆$C: \\dfrac{x^2}{a^2}+\\dfrac{y^2}{b^2}=1$($a>b>0$)的左、右焦点, $M, N$为椭圆上两点, 满足$F_1M\\parallel F_2N$, 且$|F_2N|: |F_2M|: |F_1M|=1: 2: 3$, 则椭圆$C$的离心率为\\blank{50}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.6]\n\\draw [->] (-2.5,0) -- (2.5,0) node [below] {$x$};\n\\draw [->] (0,-2) -- (0,2) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\draw (0,0) ellipse ({sqrt(5)} and {sqrt(3)});\n\\draw ({-sqrt(2)},0) node [below] {$F_1$} coordinate (F_1);\n\\draw ({sqrt(2)},0) node [below] {$F_2$} coordinate (F_2);\n\\draw ({sqrt(2)/2},{3*sqrt(3)/sqrt(10)}) node [above] {$M$} coordinate (M);\n\\draw ($(F_2)+1/3*(M)-1/3*(F_1)$) node [above] {$N$} coordinate (N);\n\\draw (F_1)--(M)--(F_2)--(N);\n\\end{tikzpicture}\n\\end{center}",
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"content": "已知函数$y=\\sqrt{1-x^2}$, $-\\dfrac{1}{2} \\leq x \\leq \\dfrac{1}{2}$的图像绕着原点按逆时针方向旋转$\\theta$($0 \\leq \\theta \\leq \\pi$)弧度, 若得到的图像仍是函数图像, 则$\\theta$可取值的集合为\\blank{50}.",
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"content": "设$\\overrightarrow{e_1}$、$\\overrightarrow{e_2}$是两个不平行的向量, 则下列四组向量中, 不能组成平面向量的一个基的是\\bracket{20}.\n\\fourch{$\\overrightarrow{e_1}+\\overrightarrow{e_2}$和$\\overrightarrow{e_1}-\\overrightarrow{e_2}$}{$\\overrightarrow{e_1}+2 \\overrightarrow{e_2}$和$\\overrightarrow{e_2}+2 \\overrightarrow{e_1}$}{$3 \\overrightarrow{e_1}-2 \\overrightarrow{e_2}$和$4 \\overrightarrow{e_2}-6 \\overrightarrow{e_1}$}{$\\overrightarrow{e_2}$和$\\overrightarrow{e_2}+\\overrightarrow{e_1}$}",
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"content": "已知$n$为正整数, 则``$n$是$3$的倍数''是``$(x^4-\\dfrac{2}{x^2})^n$的二项展开式中存在常数项''的 \\bracket{20}条件.\n\\fourch{充分非必要}{必要非充分}{充要}{既不充分也不必要}",
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"content": "某产品的广告费$x$(单位: 万元) 与销售额$y$(单位: 万元) 的统计数据如下表:\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline 广告费$x$(万元) & 2 & 3 & 4 & 5 \\\\\n\\hline 销售额$y$(万元) & 26 & 39 & 49 & 54 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n根据上表可得回归方程$y=\\hat{a} x+\\hat{b}$中$\\hat{a}=9.4$, 据此模型可预测当广告费为$6$万元时, 销售额约为\\bracket{20}.\n\\fourch{$63.6$万元}{$65.5$万元}{$67.7$万元}{$72.0$万元}",
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"content": "已知数列$\\{a_n\\}$满足$a_1=1$, $a_{n+1}-a_n=(-\\dfrac{1}{2})^n$, 存在正偶数$n$使得$(a_n-\\lambda)(a_{n+1}+\\lambda)>0$, 且对任意正奇数$n$有$(a_n-\\lambda)(a_{n+1}+\\lambda)<0$, 则实数$\\lambda$的取值范围是\\bracket{20}.\n\\fourch{$(-\\dfrac{2}{3}, 1]$}{$(-\\infty,-\\dfrac{2}{3}] \\cup(1,+\\infty)$}{$(-\\dfrac{3}{4}, \\dfrac{2}{3})$}{$(-\\dfrac{3}{4},-\\dfrac{2}{3}]$}",
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"content": "已知函数$y=f(x)$的表达式为$f(x)=\\sqrt{3} \\sin (x+\\dfrac{\\pi}{6}) \\cos (x+\\dfrac{\\pi}{6})+\\cos ^2(x-\\dfrac{\\pi}{3})$.\\\\\n(1) 求函数$y=f(x)$的最小正周期及图像的对称轴的方程;\\\\\n(2) 求函数$y=f(x)$在$(0, \\dfrac{\\pi}{2})$上的值域.",
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"content": "如图, 在直三棱柱$ABC-A_1B_1C_1$中, 底面$\\triangle ABC$是等腰直角三角形, $AC=BC=AA_1=2$, $D$为侧棱$AA_1$的中点.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0,0) node [left] {$C$} coordinate (C);\n\\draw (0,0,2) node [left] {$A$} coordinate (A);\n\\draw (2,0,0) node [right] {$B$} coordinate (B);\n\\draw (0,2,0) node [above] {$C_1$} coordinate (C_1);\n\\draw (2,2,0) node [right] {$B_1$} coordinate (B_1);\n\\draw (0,2,2) node [left] {$A_1$} coordinate (A_1);\n\\draw ($(A)!0.5!(A_1)$) node [left] {$D$} coordinate (D);\n\\draw (A)--(B)--(B_1)--(C_1)--(A_1)--cycle(A_1)--(B_1)(D)--(B_1);\n\\draw [dashed] (A)--(C)--(B)(C)--(C_1)(C)--(B_1)(C)--(D)--(C_1);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: $BC \\perp$平面$ACC_1A_1$;\\\\\n(2) 求二面角$B_1-CD-C_1$的正弦值.",
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"content": "在全民抗击新冠疫情期间, 某校开展了``停课不停学''活动, 一个星期后, 某校随机抽取了$100$名居家学习的高二学生进行问卷调查, 得到学生每天学习时间 (单位: $\\text{h}$) 的频率分布直方图如下, 若被抽取的这$100$名学生中, 每天学习时间不低于$8$小时有$30$人.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, xscale = 1, yscale = 5]\n\\draw [->] (5,0) -- (5.2,0) -- (5.3,0.04) -- (5.5,-0.04) -- (5.6,0) -- (10,0) node [below right] {每天学习时间(h)};\n\\draw [->] (5,0) -- (5,0.75) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw (5,0) node [below left] {$O$};\n\\foreach \\i/\\j in {6/0.14,6.5/0.26,7/0.42,7.5/0.58,8/0.38,8.5/0.22}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (0.5,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {6/0.14,6.5/0.26/a,7/0.42,7.5/0.58,8/0.38/b,8.5/0.22}\n{\\draw [dashed] (\\i,\\j) -- (5,\\j) node [left] {$\\k$};};\n\\draw (9,0) node [below] {$9$};\n\\end{tikzpicture}\n\\end{center}\n(1) 求频率分布直方图中实数$a, b$的值;\\\\\n(2) 每天学习时间在$[6.0,6.5)$的$7$名学生中, 有$4$名男生, $3$名女生, 现从中抽$2$人进行电话访谈, 已知抽取的学生有男生, 求抽取的$2$人恰好为一男一女的概率;\\\\\n(3) 依据所抽取的样本, 从每天学习时间在$[6.0,6.5)$和$[7.0,7.5)$的学生中按比例分层抽样抽取$8$人, 再从这$8$人中选$3$人进行电话访谈, 求抽取的$3$人中每天学习时间在$[6.0,6.5)$的人数$X$的分布和数学期望.",
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"content": "如图, 已知$A$、$B$、$C$是抛物线$\\Gamma_1: x^2=y$上的三个点, 且直线$CB$、$CA$分别与抛物线$\\Gamma_2: y^2=4 x$相切, $F$为抛物线$\\Gamma_1$的焦点.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\draw [->] (-3,0) -- (3,0) node [below] {$x$};\n\\draw [->] (0,-4) -- (0,4) node [left] {$y$};\n\\draw (0,0) node [below right] {$O$};\n\\draw [domain = -2:2, samples = 100] plot (\\x,{\\x*\\x});\n\\draw [domain = {-sqrt(12)}:{sqrt(12)}, samples = 100] plot ({\\x*\\x/4},\\x);\n\\filldraw (0,0.25) node [right] {$F$} coordinate (F) circle (0.03);\n\\draw (-1,1) node [below] {$C$} coordinate (C);\n\\draw ({(1+sqrt(5))/2},{(3+sqrt(5))/2}) node [above] {$B$} coordinate (B);\n\\draw ({(1-sqrt(5))/2},{(3-sqrt(5))/2}) node [below] {$A$} coordinate (A);\n\\draw [thick] ($(B)!-0.5!(C)$) -- ($(B)!1.2!(C)$);\n\\draw [thick] ($(A)!-5!(C)$) -- ($(A)!1.5!(C)$);\n\\draw [thick] (A)--(B);\n\\end{tikzpicture}\n\\end{center}\n(1) 若点$C$的横坐标为$x_3$, 用$x_3$表示线段$CF$的长;\\\\\n(2) 若$CA \\perp CB$, 求点$C$的坐标;\\\\\n(3) 证明: 直线$AB$与抛物线$\\Gamma_2$相切.",
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"015100": {
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"content": "设$y=f(x)$、$y=g(x)$是定义域为$\\mathbf{R}$的函数, 当$g(x_1) \\neq g(x_2)$时, 记$\\delta(x_1, x_2)=\\dfrac{f(x_1)-f(x_2)}{g(x_1)-g(x_2)}$.\\\\\n(1) 已知$y=g(x)$在区间$I$上严格增, 且对任意$x_1, x_2 \\in I$, $x_1 \\neq x_2$, 有$\\delta(x_1, x_2)>0$, 证明: 函数$y=f(x)$在区间$I$上严格增;\\\\\n(2) 已知$g(x)=\\dfrac{1}{3} x^3+a x^2-3 x$, 且对任意$x_1, x_2 \\in \\mathbf{R}$, 当$g(x_1) \\neq g(x_2)$时, 有$\\delta(x_1, x_2)>0$, 若当$x=1$时, 函数$y=f(x)$取得极值, 求实数$a$的值;\\\\\n(3) 已知$g(x)=\\sin x$, $f(\\dfrac{\\pi}{2})=1$, $f(-\\dfrac{\\pi}{2})=-1$, 且对任意$x_1, x_2 \\in \\mathbf{R}$, 当$g(x_1) \\neq g(x_2)$时, 有$|\\delta(x_1, x_2)| \\leq 1$, 证明: $f(x)=\\sin x$.",
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"ans": "",
"solution": "",
"duration": -1,
"usages": [],
"origin": "2023届青浦区高三二模试题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.",