From e27267c2d23cc00ca9e2c852869e50881de18282 Mon Sep 17 00:00:00 2001
From: wangweiye7840 <kj_wangweiye@126.com>
Date: Fri, 12 Jan 2024 15:22:32 +0800
Subject: [PATCH] =?UTF-8?q?=E5=BD=95=E5=85=A5=E9=AB=98=E4=B8=80=E6=9C=9F?=
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---
 题库0.3/Problems.json | 52 +++++++++++++++++++++----------------------
 1 file changed, 26 insertions(+), 26 deletions(-)

diff --git a/题库0.3/Problems.json b/题库0.3/Problems.json
index 86c8de94..77072ea8 100644
--- a/题库0.3/Problems.json
+++ b/题库0.3/Problems.json
@@ -632238,7 +632238,7 @@
             "第四单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$5$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632273,7 +632273,7 @@
             "第一单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$a\\neq 0$或$b\\neq 0$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632308,7 +632308,7 @@
             "第四单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$100$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632343,7 +632343,7 @@
             "第二单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$2a$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632378,7 +632378,7 @@
             "第二单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$1$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632413,7 +632413,7 @@
             "第二单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$[\\dfrac{1}{2},1]$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632448,7 +632448,7 @@
             "第四单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$\\dfrac{1}{3}$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632483,7 +632483,7 @@
             "第四单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$\\begin{cases}  2^{n-1},  &n\\geq 2, \\\\ 0, & n=1. \\end{cases}$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632518,7 +632518,7 @@
             "第四单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$40$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632554,7 +632554,7 @@
             "第一单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$[0,\\dfrac{1}{2})$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632589,7 +632589,7 @@
             "第二单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "$(-\\infty,2\\sqrt{2})$",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632624,7 +632624,7 @@
             "第四单元"
         ],
         "genre": "填空题",
-        "ans": "",
+        "ans": "\\textcircled{1}\\textcircled{3}\\textcircled{4}",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632659,7 +632659,7 @@
             "第一单元"
         ],
         "genre": "选择题",
-        "ans": "",
+        "ans": "D",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632694,7 +632694,7 @@
             "第二单元"
         ],
         "genre": "选择题",
-        "ans": "",
+        "ans": "A",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632729,7 +632729,7 @@
             "第四单元"
         ],
         "genre": "选择题",
-        "ans": "",
+        "ans": "C",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632764,7 +632764,7 @@
             "第二单元"
         ],
         "genre": "选择题",
-        "ans": "",
+        "ans": "B",
         "solution": "",
         "duration": -1,
         "usages": [
@@ -632799,8 +632799,8 @@
             "第二单元"
         ],
         "genre": "解答题",
-        "ans": "",
-        "solution": "",
+        "ans": "(1) 不是偶函数, 理由略; (2) 是严格增函数, 理由略",
+        "solution": "(1) 不是偶函数. $f(1)=0$, $f(-1)=2$, $f(-1)\\neq f(1)$.\\\\\n(2) $f(x)$是$(0,+\\infty)$上的严格增函数. 任取$x_1,x_2\\in (0,+\\infty)$, $x_1<x_2$, $f(x_1)-f(x_2)=(x_1^2-\\dfrac{1}{x_1})-(x_2^2-\\dfrac{1}{x_2})=(x_1-x_2)(x_1+x_2+\\dfrac{1}{x_1x_2})$, 因为$x_1-x_2<0$, $x_1+x_2+\\dfrac{1}{x_1x_2}>0$, 故$f(x_1)<f(x_2)$,  $f(x)$是$(0,+\\infty)$上的严格增函数.",
         "duration": -1,
         "usages": [
             "20240111\t2026届高一01班\t0.756\t0.976",
@@ -632834,8 +632834,8 @@
             "第四单元"
         ],
         "genre": "解答题",
-        "ans": "",
-        "solution": "",
+        "ans": "(1) 证明略; (2) $a_n=\\dfrac{3}{n+2}$",
+        "solution": "(1) 由已知可得$\\dfrac{1}{a_{n+1}}=\\dfrac{a_n+3}{3 a_n}=\\dfrac{1}{a_n}+\\dfrac{1}{3}$, 即 $\\dfrac{1}{a_{n+1}}-\\dfrac{1}{a_n}=\\dfrac{1}{3}$是常数, 所以$\\{b_n\\}$ 是等差数列.\\\\\n(2) 由(1) 知 $b_n=b_1+(n-1) \\times \\dfrac{1}{3}=\\dfrac{n+2}{3}$\n所以$a_n=\\dfrac{1}{b_n}=\\dfrac{3}{n+2}$($n\\geq 1$, $n\\in \\mathbf{N}$).",
         "duration": -1,
         "usages": [
             "20240111\t2026届高一01班\t0.967\t0.857",
@@ -632869,8 +632869,8 @@
             "第二单元"
         ],
         "genre": "解答题",
-        "ans": "",
-        "solution": "",
+        "ans": "(1) $5$万度; (2) $8$万度",
+        "solution": "(1) $Z(x)=\\begin{cases}50 x,& 0 \\le x \\le 4, \\\\ -\\dfrac{400}{x^2}-\\dfrac{300}{x}+620, & 4<x \\le 20.\\end{cases}$ 当 $0 \\le x \\le 4$ 时, $Z(x) \\le 200$. 因为增效效益要等于 $544$ 万元, 所以令 $-\\dfrac{400}{x^2}-$ $\\dfrac{300}{x}+620=544$, 即 $19 x^2-75 x-100=0$, 又 $4<x \\le 20$ , 所以 $x=5$. 所以减少用电量 $5$ 万千瓦时, 今年该企业增效效益恰好达到 $544$ 万元.\\\\\n(2) 设企业总效益为 $Q(x)$ 万元, 则 $Q(x)=Z(x)-S(x)+n(x)=\\begin{cases}\n-50 x^2+150 x,& 0 \\le x \\le 4, \\\\\n-\\dfrac{400}{x^2}+\\dfrac{100}{x}+120, & 4<x \\le 20.\n\\end{cases}$\\\\\n当 $0 \\le x \\le 4$ 时, $Q(x)=-50\\left(x-\\dfrac{3}{2}\\right)^2+\\dfrac{225}{2} \\le \\dfrac{225}{2}$; 当 $4<x \\le 20$ 时, $Q(x)=-400\\left(\\dfrac{1}{x}-\\dfrac{1}{8}\\right)^2+$ $\\dfrac{505}{4} \\le Q(8)=\\dfrac{505}{4}$. 因为 $\\dfrac{225}{2}<\\dfrac{505}{4}$, 所以当 $x=8$ 时, $Q(x)$ 取最大值, 即当减少用电量 $8$ 万千瓦时时, 今年该企业总效益最大.",
         "duration": -1,
         "usages": [
             "20240111\t2026届高一01班\t0.976\t0.710",
@@ -632904,8 +632904,8 @@
             "第四单元"
         ],
         "genre": "解答题",
-        "ans": "",
-        "solution": "",
+        "ans": "(1) $a_n=-\\dfrac{3^{n+1}}{4^n}$; $S_n=\\dfrac{3^{n+2}}{4^n}-9$; (2) 证明略; (3) $[-3, 1]$",
+        "solution": "(1) $a_n=-\\dfrac{3^{n+1}}{4^n}$. $S_n=\\dfrac{3^{n+2}}{4^n}-9$.\\\\\n(2) $b_n=-\\dfrac{n-4}{3}a_n=(n-4)(\\dfrac{3}{4})^n$.\\\\\n\\textcircled{1} 当$n=1$时,$T_1=b_1=-\\dfrac{9}{4}=-4\\times(\\dfrac{3}{4})^2$, 所以等式成立.\n\\textcircled{2} 假设$n=k$($k$为正整数)时, 等式成立, 即有$T_k=-4k(\\dfrac{3}{4})^{k+1}$. 那么当$n=k+1$时, 就有$T_{k+1}=T_k+b_{k+1}=T_k+(k-3)(\\dfrac{3}{4})^{k+1}=(-4k+k-3)(\\dfrac{3}{4})^{k+1}=-3(k+1)(\\dfrac{3}{4})^{k+1}=-4(k+1)(\\dfrac{3}{4})^{k+2}$. 等式也成立.\\\\\n根据\\textcircled{1}、\\textcircled{2}, 由数学归纳法可知$T_n=-4n(\\dfrac{3}{4})^{n+1}$($n\\geq 1$, $n\\in \\mathbf{N}$).\\\\\n(3) $T_n \\le m b_n$恒成立 即 $-4 n \\cdot\\left(\\dfrac{3}{4}\\right)^{n+1} \\le m(n-4) \\cdot\\left(\\dfrac{3}{4}\\right)^n$ 恒成立, 即$m(n-4)+3 n \\geq 0$ 恒成立.\\\\\n当 $n=4$ 时, 任意$m\\in \\mathbf{R}$均使不等式恒成立;\\\\\n当 $n\\in \\{1,2,3\\}$ 时, $m \\le-\\dfrac{3 n}{n-4}=-3-\\dfrac{12}{n-4}$, 该不等式恒成立, 即 $m \\le 1$;\\\\\n当 $n\\in \\{5,6,7,\\cdots\\}$ 时, $m \\geq-\\dfrac{3 n}{n-4}=-3-\\dfrac{12}{n-4}$, 得 $m \\ge -3$.\\\\\n因此$m$的取值范围为$-3 \\le m \\le 1$.",
         "duration": -1,
         "usages": [
             "20240111\t2026届高一01班\t0.970\t0.797\t0.427",
@@ -632939,8 +632939,8 @@
             "第二单元"
         ],
         "genre": "解答题",
-        "ans": "",
-        "solution": "",
+        "ans": "(1) 是, 不是; (2) $[1,2]$; (3) 证明略",
+        "solution": "(1) 是, 不是.\\\\\n(2) 记 $I=[0, m], S=\\{f(x) \\mid x \\in I\\}$, 注意到 $f(0)=0 \\in[0, m]$ , 因此, 若 $I$ 为函数 $f(x)$ 的 ``$\\Omega$''区间, 则其不满足性质\\textcircled{2}, 必满足性质\\textcircled{1}, 即 $S \\subseteq I$.\\\\\n$f(x)=-x^2+2 x=-(x-1)^2+1$. 当 $0<m<1$时, $f(x)$ 在 $I$ 上严格增, 且 $f(m)-m=-m(m-1)>0$, 此时 $S=[0, f(m)]$ 不包含于 $I=[0, m]$, 不合题意;\\\\\n当 $1 \\le m \\le 2$ 时, $S=[f(0), f(1)]=[0,1] \\subseteq[0, m]$, 符合題意;\\\\\n当 $m>2$ 时, $f(m)<f(2)=f(0)=0$, 所以 $f(m) \\notin I$, 不合题意.\\\\\n综上, $m$ 的取值范围为 $[1,2]$.\\\\\n(3) 不妨设$x_1<x_2$, 由题意知$f(x_1)>f(x_2)$且$f(x_1)+x_1>f(x_2)+x_2$, 故$y=f(x)$是严格减函数, $y=f(x)+x$是严格减函数. 下证$y=f(x)-x$存在唯一零点. 首先, 假设$y=f(x)-x$不存在零点, 因为$y=f(x)-x$是连续函数, 故不妨设$f(x)-x>0$恒成立(恒小于$0$同理), 即$f(x)>x$恒成立, 于是$f(x)+x>2x$恒成立. 此时有$f(0)+0>0$, 令$M=f(0)>0$, 由$y=f(x)+x$是严格减函数可知$f(M)+M<f(0)+0$,结合$f(M)+M>2M$, $M>0$, $M=f(0)$可推出$2M<M$, 矛盾, 故$y=f(x)-x$存在零点.\\\\\n其次, 由于$y=f(x)-x$是严格减函数, 故$y=f(x)-x$存在唯一零点, 不妨设为$x_1$.\\\\\n\\textcircled{1} 考虑区间$[x_1+1,x_1+2]$,任取$x\\in[x_1+1,x_1+2]$, 有$f(x)\\le f(x_1+1)<f(x_1)=x_1<x_1+1$, 故$[x_1+1,x_1+2]$是$f(x)$的一个``$\\Omega$''区间.\\\\\n\\textcircled{2} $x_1$不属于$f(x)$的任何一个``$\\Omega$区间''. 假设$[a,b]$是$f(x)$的一个``$\\Omega$区间''且$x_1\\in [a,b]$, 因为$f(x_1)=x_1$, 故$[a,b]$满足性质1. 故$f(b),f(a)\\in [a,b]$,  因此$f(a)-f(b)\\leq b-a$, 从而$\\dfrac{f(b)-f(a)}{b-a}\\geq -1$ , 这与$\\dfrac{f(b)-f(a)}{b-a}<-1$矛盾. 因此假设不成立, 即$x_1$不属于$f(x)$的任何一个``$\\Omega$区间''.",
         "duration": -1,
         "usages": [
             "20240111\t2026届高一01班\t0.963\t0.451\t0.116",