收录高三寒假作业试卷04新题
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20240125-151424
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20240125-152122
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024198:024200,000465,024201:024205,019538,024206:024210
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"20221214\t王伟叶"
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"20230209\t王伟叶"
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"20230503\t王伟叶"
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"20230730\t王伟叶"
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"space": "4em",
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"20240105\t杨懿荔"
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@ -650933,6 +650943,274 @@
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"space": "4em",
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"024198": {
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"id": "024198",
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"content": "在长方体 $ABCD-A_1B_1C_1D_1$ 中, $AB=BC=1$, $AA_1=\\sqrt{3}$, 则异面直线 $AD_1$ 与 $DB_1$ 所成角的余弦值为\\blank{50}.",
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"genre": "填空题",
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"ans": "",
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"024199": {
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"id": "024199",
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"content": "平面 $M$ 上有 $4$ 个点, 平面 $N$ 上有 $3$ 个点, 这 $7$ 个点最多可确定\\blank{50}个平面.",
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"024200": {
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"id": "024200",
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"content": "已知 $A(1,1,1)$、$B(-1,0,4)$、$C(2,-2,3)$, 则以 $AB$、$AC$ 为邻边的平行四边形的面积为\\blank{50}.",
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"024201": {
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"id": "024201",
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"content": "若正三棱锥的高为 $1$, 底面边长为 $2 \\sqrt{3}$, 内有一个球与四个面都相切, 则棱锥的内切球的半径为\\blank{50}.",
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"024202": {
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"id": "024202",
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"content": "如图, 在棱长为 $1$ 的正方体 $ABCD-A_1B_1C_1D_1$ 中, $M$、$N$ 分别是 $A_1D_1$、$A_1B_1$ 的中点, 过直线 $BD$ 的平面 $\\alpha \\parallel $ 平面 $AMN$, 则平面 $\\alpha$ 截该正方体所得截面的面积为\\blank{50}.\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 (A_1);\n\\draw (B) ++ (0,\\l,0) node [right] {$B_1$} coordinate (B_1);\n\\draw (C) ++ (0,\\l,0) node [above right] {$C_1$} coordinate (C_1);\n\\draw (D) ++ (0,\\l,0) node [above left] {$D_1$} coordinate (D_1);\n\\draw (A_1) -- (B_1) -- (C_1) -- (D_1) -- cycle;\n\\draw (A) -- (A_1) (B) -- (B_1) (C) -- (C_1);\n\\draw [dashed] (D) -- (D_1);\n\\draw ($(A_1)!0.5!(D_1)$) node [left] {$M$} coordinate (M);\n\\draw ($(A_1)!0.5!(B_1)$) node [below] {$N$} coordinate (N);\n\\draw (M)--(N)--(A);\n\\draw [dashed] (B)--(D)(A)--(M);\n\\end{tikzpicture}\n\\end{center}",
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"genre": "填空题",
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"024203": {
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"id": "024203",
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"content": "我们把平面内与直线垂直的非零向量称为直线的法向量, 在平面直角坐标系中, 利用求动点轨迹方程的方法, 可以求出过点 $A(-3,4)$, 且法向量为 $\\overrightarrow{n}=(1,-2)$ 的直线方程为$1 \\times(x+3)+(-2) \\times(y-4)=0$, 即 $x-2 y+11=0$. 类比以上方法, 在空间直角坐标系中, 经过点 $A(1,2,3)$, 且法向量为 $\\overrightarrow{m}=(-1,-2,1)$ 的平面的方程为\\blank{50}.",
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"genre": "填空题",
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"ans": "",
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"024204": {
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"id": "024204",
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"content": "在正方体 $ABCD-A_1B_1C_1D_1$ 中, $M$、$N$、$Q$ 分别是棱 $D_1C_1$、$A_1D_1$、$BC$ 的中点, 点 $P$ 在 $BD_1$ 上且 $BP=\\dfrac{2}{3}BD_1$, 则下面所有说法中正确的序号是\\blank{50}.\\\\\n\\textcircled{1} $MN \\parallel $ 平面 $APC$;\\\\\n\\textcircled{2} $C_1Q \\parallel $ 平面 $APC$;\\\\\n\\textcircled{3} $A$、$P$、$M$ 三点共线;\\\\\n\\textcircled{4} 平面 $MNQ \\parallel $ 平面 $APC$.",
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"objs": [],
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"genre": "填空题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"024205": {
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"id": "024205",
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"content": "已知圆柱的上底面圆周经过正三棱锥 $P-ABC$ 的三条侧棱的中点, 下底面圆心为此三棱锥底面中心 $O$. 若三棱锥 $P-ABC$ 的高为该圆柱外接球半径的 $2$ 倍, 则该三棱锥的外接球与圆柱外接球的半径之比为\\bracket{20}.\n\\fourch{$2: 1$}{$7: 4$}{$3: 1$}{$5: 3$}",
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"objs": [],
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"tags": [],
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"genre": "选择题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"024206": {
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"id": "024206",
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"content": "如图, 已知圆锥 $SO$ 底面圆的半径 $r=1$, 直径 $AB$ 与直径 $CD$ 垂直, 母线 $SA$ 与底面所成角的大小为 $\\dfrac{\\pi}{3}$.\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.6]\n\\draw (-2,0) node [left] {$A$} coordinate (A);\n\\draw (2,0) node [right] {$B$} coordinate (B);\n\\draw (0,0) node [below] {$O$} coordinate (O);\n\\draw (0,{2*sqrt(3)}) node [above] {$S$} coordinate (S);\n\\draw ($(S)!0.5!(B)$) node [above right] {$E$} coordinate (E);\n\\draw (100:2 and 0.5) node [above] {$D$} coordinate (D);\n\\draw (-80:2 and 0.5) node [below] {$C$} coordinate (C);\n\\draw (A)--(S)--(B)(A) arc (180:360:2 and 0.5);\n\\draw [dashed] (A) arc (180:0:2 and 0.5);\n\\draw [dashed] (A)--(B)(C)--(D)(C)--(E)--(D)(O)--(S);\n\\end{tikzpicture}\n\\end{center}\n(1) 求圆锥 $SO$ 的侧面积;\\\\\n(2) 若 $E$ 为母线 $SA$ 的中点, 求二面角 $E-CD-B$ 的大小.",
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"objs": [],
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"tags": [],
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"genre": "解答题",
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"ans": "",
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"duration": -1,
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"20240125\t毛培菁"
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"space": "4em",
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"024207": {
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"id": "024207",
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"content": "如图, 在四面体 $P-ABC$ 中, $PA=PC=AB=BC=5$, $AC=6$, $PB=4 \\sqrt{2}$, 线段 $AC$、$PA$ 的中点分别为 $O$、$Q$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\draw (0,0,0) node [below] {$O$} coordinate (O);\n\\draw (-3,0,0) node [left] {$A$} coordinate (A);\n\\draw (3,0,0) node [right] {$C$} coordinate (C);\n\\draw (0,4,0) node [above] {$P$} coordinate (P);\n\\draw (0,0,4) node [below] {$B$} coordinate (B);\n\\draw ($(A)!0.5!(P)$) node [above left] {$Q$} coordinate (Q);\n\\draw (A)--(P)--(C)--(B)--cycle(Q)--(B)(P)--(B);\n\\draw [dashed] (A)--(C)(B)--(O)--(P)(O)--(Q);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证:平面 $PAC \\perp$ 平面 $ABC$;\\\\\n(2) 求四面体 $P-OBQ$ 的体积.",
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"024208": {
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"id": "024208",
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"content": "如图, 在三棱锥 $P-ABC$ 中, $PA \\perp$ 底面 $ABC, \\angle BAC=90^{\\circ}$. 点 $D$、$E$、$N$ 分别为棱 $PA$、$PC$、$BC$ 的中点, $M$ 是线段 $AD$ 的中点, $PA=AC=4$, $AB=2$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, x = {(-7:1cm)}, z = {(-135:0.5cm)}, scale = 0.7]\n\\draw (0,0,0) node [below] {$A$} coordinate (A);\n\\draw (0,0,2) node [below] {$B$} coordinate (B);\n\\draw (4,0,0) node [right] {$C$} coordinate (C);\n\\draw (0,4,0) node [above] {$P$} coordinate (P);\n\\draw ($(A)!0.5!(P)$) node [left] {$D$} coordinate (D);\n\\draw ($(C)!0.5!(P)$) node [above right] {$E$} coordinate (E);\n\\draw ($(B)!0.5!(C)$) node [below] {$N$} coordinate (N);\n\\draw ($(A)!0.5!(D)$) node [left] {$M$} coordinate (M);\n\\draw (B)--(C)--(P)--cycle(B)--(E)(N)--(E);\n\\draw [dashed] (A)--(B)(A)--(C)(A)--(P)(B)--(D)--(E)(E)--(M)--(N);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: $MN \\parallel $ 平面 $BDE$;\\\\\n(2) 已知点 $H$ 在棱 $PA$ 上, 且直线 $NH$ 与直线 $BE$ 所成角的余弦值为 $\\dfrac{\\sqrt{7}}{21}$, 求线段 $AH$ 的长.",
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"024209": {
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"id": "024209",
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"content": "已知 $\\overrightarrow{a}=(x_1, y_1, z_1)$, $\\overrightarrow{b}=(x_2, y_2, z_2)$, $\\overrightarrow{c}=(x_3, y_3, z_3)$, 定义一种运算: $(\\overrightarrow{a}\\times \\overrightarrow{b}) \\cdot \\overrightarrow{c}=x_1y_2z_3+x_2y_3z_1+x_3y_1z_2-x_1y_3z_2-x_2y_1z_3-x_3y_2z_1$, 已知四棱锥 $P-ABCD$中, 底面 $ABCD$ 是一个平行四边形, $\\overrightarrow{AB}=(2,-1,4)$, $\\overrightarrow{AD}=(4,2,0)$, $\\overrightarrow{AP}=(-1,2,1)$.\\\\\n(1) 试计算 $(\\overrightarrow{AB}\\times \\overrightarrow{AD}) \\cdot \\overrightarrow{AP}$ 的绝对值的值, 并求证 $PA \\perp$ 面 $ABCD$;\\\\\n(2) 求四棱锥 $P-ABCD$ 的体积, 说明 $(\\overrightarrow{AB}\\times \\overrightarrow{AD}) \\cdot \\overrightarrow{AP}$ 的绝对值的值与四棱锥 $P-ABCD$体积的关系, 并由此猜想向量这一运算 $(\\overrightarrow{AB}\\times \\overrightarrow{AD}) \\cdot \\overrightarrow{AP}$ 的绝对值的几何意义.",
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"objs": [],
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"ans": "",
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"024210": {
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"id": "024210",
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"content": "如图, 在三棱锥 $P-ABC$ 中, $AB=BC=2 \\sqrt{2}$, $PA=PB=PC=AC=4$, $O$ 为 $AC$ 的中点.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.6]\n\\draw (0,0) node [above left] {$O$} coordinate (O);\n\\draw (-2,0) node [left] {$A$} coordinate (A);\n\\draw (2,0) node [right] {$C$} coordinate (C);\n\\draw (0,0,2) node [below] {$B$} coordinate (B);\n\\draw ($(B)!{1/3}!(C)$) node [below right] {$M$} coordinate (M);\n\\draw (0,{2*sqrt(3)},0) node [above] {$P$} coordinate (P);\n\\draw (A)--(B)--(C)--(P)--cycle(P)--(M)(P)--(B);\n\\draw [dashed] (A)--(M)(A)--(C)(P)--(O);\n\\end{tikzpicture}\n\\end{center}\n(1) 证明: $PO \\perp$ 平面 $ABC$;\\\\\n(2) 若点 $M$ 在棱 $BC$ 上, 且二面角 $M-PA-C$ 为 $30^{\\circ}$, 求 $PC$ 与平面 $PAM$ 所成角的正弦值.",
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"objs": [],
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"ans": "",
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"solution": "",
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"duration": -1,
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||||
"origin": "自拟题目",
|
||||
"edit": [
|
||||
"20240125\t毛培菁"
|
||||
],
|
||||
"same": [],
|
||||
"related": [
|
||||
"023091"
|
||||
],
|
||||
"remark": "",
|
||||
"space": "4em",
|
||||
"unrelated": []
|
||||
},
|
||||
"030001": {
|
||||
"id": "030001",
|
||||
"content": "若$x,y,z$都是实数, 则:(填写``\\textcircled{1} 充分非必要、\\textcircled{2} 必要非充分、\\textcircled{3} 充要、\\textcircled{4} 既非充分又非必要''之一)\\\\\n(1) ``$xy=0$''是``$x=0$''的\\blank{50}条件;\\\\\n(2) ``$x\\cdot y=y\\cdot z$''是``$x=z$''的\\blank{50}条件;\\\\\n(3) ``$\\dfrac xy=\\dfrac yz$''是``$xz=y^2$''的\\blank{50}条件;\\\\\n(4) ``$|x |>| y|$''是``$x>y>0$''的\\blank{50}条件;\\\\\n(5) ``$x^2>4$''是``$x>2$'' 的\\blank{50}条件;\\\\\n(6) ``$x=-3$''是``$x^2+x-6=0$'' 的\\blank{50}条件;\\\\\n(7) ``$|x+y|<2$''是``$|x|<1$且$|y|<1$'' 的\\blank{50}条件;\\\\\n(8) ``$|x|<3$''是``$x^2<9$'' 的\\blank{50}条件;\\\\\n(9) ``$x^2+y^2>0$''是``$x\\ne 0$'' 的\\blank{50}条件;\\\\\n(10) ``$\\dfrac{x^2+x+1}{3x+2}<0$''是``$3x+2<0$'' 的\\blank{50}条件;\\\\\n(11) ``$0<x<3$''是``$|x-1|<2$'' 的\\blank{50}条件.",
|
||||
|
|
|
|||
Reference in New Issue