录入2025届作业数据截至20231117

This commit is contained in:
weiye.wang 2023-11-17 21:29:28 +08:00
parent 4d3fbaaa89
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2 changed files with 784 additions and 113 deletions

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ans
usages
22697
$0$.
021041
20231117 2025届高二10班 0.985 0.868
021042
20231117 2025届高二10班 0.868
22698
$x \leq 1$ 且 $y \leq 1$.
021043
20231117 2025届高二10班 0.824
021044
20231117 2025届高二10班 0.824
22699
$(1,2)$.
021045
20231117 2025届高二10班 0.882
021046
20231117 2025届高二10班 0.882
22700
$4$.
021047
20231117 2025届高二10班 0.779
021048
20231117 2025届高二10班 0.868
22701
$14$.
021049
20231117 2025届高二10班 0.824
021041
20231117 2025届高二11班 1.000 1.000
22702
$\dfrac{a+1}{2}$.
021042
20231117 2025届高二11班 0.615
021043
20231117 2025届高二11班 0.654
22703
$\dfrac{1}{8}$.
021044
20231117 2025届高二11班 0.538
021045
20231117 2025届高二11班 0.596
22704
$[0,4)$.
021046
20231117 2025届高二11班 0.462
021047
20231117 2025届高二11班 0.462
22705
$(\dfrac{1}{2}, 1)$.
021048
20231117 2025届高二11班 0.538
021049
20231117 2025届高二11班 0.712
22706
$(-\infty, \dfrac{1}{2}]$.
021041
20231117 2025届高二01班 1.000 0.977
021042
20231117 2025届高二01班 0.861
22707
$20$.
021043
20231117 2025届高二01班 0.814
021044
20231117 2025届高二01班 0.826
22708
$(-\infty, \dfrac{1}{2}]$.
021045
20231117 2025届高二01班 0.802
021046
20231117 2025届高二01班 0.872
22709
D
021047
20231117 2025届高二01班 0.884
021048
20231117 2025届高二01班 0.779
22710
A
021049
20231117 2025届高二01班 0.884
021041
20231117 2025届高二02班 1.000 0.989
22711
C
021042
20231117 2025届高二02班 0.886
021043
20231117 2025届高二02班 0.830
22712
A
021044
20231117 2025届高二02班 0.886
021045
20231117 2025届高二02班 0.852
22713
(1) $(-3,4]$; (2) $(-1,1)$
021046
20231117 2025届高二02班 1.000
021047
20231117 2025届高二02班 0.841
22714
(1) 当 $a=b$ 时, $a^3+b^3=a^2 b+a b^2$; 当 $a \neq b$ 时, $a^3+b^3>a^2 b+a b^2$; (2) 最小值为$1$
021048
20231117 2025届高二02班 0.852
021049
20231117 2025届高二02班 0.909
22715
(1) $200$吨; (2) $100$至$300$吨(含$100$吨与$300$吨)
021041
20231117 2025届高二04班 1.000 0.974
021042
20231117 2025届高二04班 0.782
22716
(1) $\log_2 3$; (2) $[\dfrac{9}{4},+\infty)$; (3) $[8,9)$
021043
20231117 2025届高二04班 0.744
021044
20231117 2025届高二04班 0.654
22717
(1) $\{1,2,3,4\}$ 不是``可分集合'', $\{1,3,5,7,9,11,13\}$ 是``可分集合''; (2) 证明略; (3) 证明略
021045
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021046
20231117 2025届高二04班 0.756
solution
021047
20231117 2025届高二04班 0.628
22713
(1) $A=(-3,3)$, $B=[0,4]$, 故 $A \cup B=(-3,4]$.\\
(2) $A \cap B=B$ 等价于 $B \subseteq A$, 故 $a-2>-3$ 且 $a+2<3$, 故 $a$ 的取值范围是 $(-1,1)$.
021048
20231117 2025届高二04班 0.628
021049
20231117 2025届高二04班 0.705
22714
(1) $a^3+b^3-a^2 b-a b^2=(a-b)^2(a+b)$, 故当 $a=b$ 时, $a^3+b^3=a^2 b+a b^2$; 当 $a \neq b$ 时, $a^3+b^3>a^2 b+a b^2$.\\
(2) 由 (1) 知 $a^3+b^3 \geq a^2 b+a b^2$, 又由于 $a>0$, $b>0$, 故 $\dfrac{a^3+b^3}{a b}\geq \dfrac{a^2 b+a b^2}{a b}$, 即 $\dfrac{a^2}{b}+\dfrac{b^2}{a}\geq a+b=1$,当 $a=b=\dfrac{1}{2}$ 时等号成立. 故 $\dfrac{a^2}{b}+\dfrac{b^2}{a}$ 的最小值是 $1$.
021041
20231117 2025届高二09班 0.947 0.855
021042
20231117 2025届高二09班 0.724
22715
(1) 该小区每月处理 $x$ 吨垃圾, 则每吨垃圾的平均成本: $w=\dfrac{x^2-2 x+40000}{x}=x+\dfrac{40000}{x}-200 \geq 2 \sqrt{40000}-200=200$. 当且仅当 $x=200$ 时等号成立. 故该小区每月分类处理 200 吨垃圾, 才能使得每吨垃圾分类处理的平均成本最低.\\
(2) 题意即: $300 x \geq x^2-200 x+40000$. 解集为 $[100,400]$, 又 $x \leq 300$, 故 $x \in[100,300]$. 所以该小区每月垃圾分类处理量 $x$ 的取值范围是 $[100,300]$.
021043
20231117 2025届高二09班 0.842
021044
20231117 2025届高二09班 0.697
22716
(1) $2^x-\dfrac{3}{2^x}=2 \Leftrightarrow(2^x)^2-2(2^x)-3=0 \Leftrightarrow 2^x=3$ 或 $2^x=-1$ (舍), 故 $x=\log _23$.\\
(2) 题意即: 对任意 $x \in \mathbf{R}$, $2^x+\dfrac{a}{2^x}\geq 3$ 恒成立 $\Leftrightarrow a \geq-(2^x)^2+3(2^x)=-(2^x-\dfrac{3}{2})^2+\dfrac{9}{4}$ 恒成立.
又 $-(2^x-\dfrac{3}{2})^2+\dfrac{9}{4}\leq \dfrac{9}{4}$, 当 $x=\log _2 \dfrac{3}{2}$ 时等号成立.
故 $-(2^x-\dfrac{3}{2})^2+\dfrac{9}{4}$ 的最大值是 $\dfrac{9}{4}$, 故实数 $a$ 的取值范围是 $[\dfrac{9}{4},+\infty)$.\\
(3) 令 $t=2^x \in$($0,+\infty$), 故 $2^x+\dfrac{a}{2^x}=6$ 有两个不同的实数根 $\Leftrightarrow t+\dfrac{a}{t}-6=0$ 有两个不同的正实数根 $\Leftrightarrow t^2-6 t+a=0$ 有两个不同的正实数根.
故 $\begin{cases}\Delta=36-4 a>0,\\6>0,\\a>0 ;\end{cases}$, 解得 $a \in(0,9)$.
又 $|x_1-x_2|=|\log _2 t_1-\log _2 t_2|=|\log _2 \dfrac{t_1}{t_2}| \leq 1$,
故 $\dfrac{1}{2}\leq \dfrac{t_1}{t_2}\leq 2$, 即 $\dfrac{1}{2}\leq \dfrac{3-\sqrt{9-a}}{3+\sqrt{9+a}}\leq 2$, 即 $\sqrt{9-a}\leq 1$, 即 $a \in[8,9]$.
综上, $a$ 的取值范围是 $[8,9)$.
021045
20231117 2025届高二09班 0.658
021046
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22717
(1) 集合 $\{1,2,3,4\}$ 不是``可分集合'', $\{1,3,5,7,9,11,13\}$ 是``可分集合''.\\
(2) 反证法: 假设集合 $A$ 是``可分集合'', 不妨设 $a_1<a_2<a_3<a_4<a_5$.
去掉``$a_1$''后, 剩下的元素可分为两个集合, 可能分别有两个元素, 可能一个集合有 1 个元素, 另一个集合有 3 个元素, 故 $a_2+a_5=a_3+a_4$\textcircled{1} 或 $a_5=a_2+a_3+a_4$\textcircled{2}.\\
去掉``$a_2$''后, 得到 $a_1+a_5=a_3+a_4$\textcircled{3} 或 $a_5=a_1+a_3+a_4$\textcircled{4}. 由\textcircled{1}\textcircled{3}知, $a_1=a_2$, 与集合元素的互异性矛盾, 由\textcircled{1}\textcircled{4}知, $a_1+a_2=0$, 与 $a_1, a_2$ 是正整数矛盾, 由\textcircled{2}\textcircled{3}知, $a_1+a_2=0$, 与 $a_1, a_2$ 是正整数矛盾, 由\textcircled{2}\textcircled{4}知, $a_1=a_2$, 与集合元素的互异性矛盾. 故假设不成立, 所以集合 $A$ 不是``可分集合''.\\
(3) 记 $S=a_1+a_2+\cdots+a_n$, 因集合 $A$ 是``可分集合'', 故 $S-a_1, S-a_2, \cdots, S-a_n$ 是偶数, 故 $a_1, a_2, \cdots, a_n, S$奇偶性相同.\\
情况 1: 当 $a_1$ 是奇数时, 有 $a_1, a_2, \cdots, a_n, S$ 都是奇数, 又 $S=a_1+a_2+\cdots+a_n$ 是 $n$ 个奇数的和, 故 $n$ 是奇数.情况 2: 当 $a_2$ 是偶数时, 因集合 $A$ 是``可分集合'', 故去掉``$a_i$''($i=1,2, \cdots, n$) 后, 剩余的所有元素组成的集合能分为两个交集为空的集合 $B_i, C_i$, 且这两个集合的所有元素之和相等, 记为 $S_{B_i}=S_{C_i}$.\\
下证 $A_1=\{\dfrac{a_1}{2}, \dfrac{a_2}{2}, \cdots, \dfrac{a_n}{2}\}$ 是``可分集合''.
因 $a_1, a_2, \cdots, a_n, S$ 是正偶数, 故 $\dfrac{a_1}{2}, \dfrac{a_2}{2}, \cdots, \dfrac{a_n}{2}$ 均为正整数. 去掉 $\dfrac{a_i}{2}$ 后, 集合 $B_i$ 中的元素分别乘以 $\dfrac{1}{2}$ 后组成集合 $B_i'$, 集合 $C_i$ 中的元素分别乘以 $\dfrac{1}{2}$ 后组成集合 $C_i'$, 因集合 $B_i \cap C_i=\varnothing$, 故 $B_i' \cap C_i'=\varnothing$. 因 $S_{B_i}=S_{C_i}$, 故 $S_{B_i'}=S_{C_i'}=\dfrac{1}{2}S_{B_i}=\dfrac{1}{2}S_{C_i}$.
即 $A_1$ 去掉 $\dfrac{a_i}{2}$ 后的元素组成的集合能可分为两个交集为空的集合, 且这两个集合的所有元素之和相等, 故 $A_1$是``可分集合''.\\
这个步骤可以一直进行下去, 直至 $m$ 次后, $\dfrac{a_1}{2^m}$ 是奇数, 此时因``可分集合''中元素的奇偶性相同, 故由情况 $1$ 可知 $n$ 是奇数.
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"solution": "",
"duration": -1,
"usages": [],
"usages": [
"20231117\t2025届高二01班\t0.854",
"20231117\t2025届高二02班\t0.978",
"20231117\t2025届高二05班\t0.938",
"20231117\t2025届高二07班\t0.822"
],
"origin": "2024届高二校本作业选择性必修第六章",
"edit": [
"20230101\t王伟叶"