LATE新高考數(shù)學期中試卷模板

%這是用latex(texlive2023+texstudio)書寫的一套高中數(shù)學期中試卷A4版，你下載后，在對應的地方進展修改，就可以為自己所用。文庫賬號HKMTLG\documentclass[title]{article}\usepackage{ctex}\usepackage{geometry}\geometry{a4paper,left=2.7cm,right=2.5cm,top=2.5cm,bottom=2.5cm} %landscape 頁面橫制,twocolumn雙欄%\geometry{a3paper,landscape,twocolumn,left=2.5cm,right=2.5cm,top=2.5cm,bottom=2.5cm}%landscape頁面橫制,twocolumn雙欄\usepackage{amsmath,amssymb,bm}\usepackage{tikz}\usepackage{xeCJKfntef}\usepackage[Symbol]{upgreek}\usepackage{enumerate}\setlength{\parindent}{4em}\usepackage{setspace}\usepackage{paralist}\usepackage{last}%頁碼\usepackage{color}%顏色\usepackage{tcolorbox}%盒子\tcbuselibrary{skins,breakable,listings,theorems}\usepackage{array}%表格間距\usepackage{multirow}%合并單元格\usepackage{booktabs}%調(diào)整表格線與上下內(nèi)容的間隔\usepackage{diagbox}\usepackage{pifont} %各種小符號\usepackage{graphicx}%插圖宏包\usepackage{fancyhdr}%設置頁眉頁腳\usepackage{ulem}\fancystyle{plain}{\style{fancy}}\style{fancy}\fancyhf{}\cfoot{高一上學期期中考試$\bullet$\textbf{數(shù)學}~~第~\the~頁(共\ref{Last}~頁〕}\renewcommand{\headrulewidth}{0pt}%把握頁眉與正文的分\線\renewcommand{\footrulewidth}{0pt} %把握頁腳的分隔線\usepackage{titlesec,titletoc}%定義章節(jié)格式\titleformat{\section}{\zihao{5}\bfseries}{\quad\chinese{section}、}{0em}{}\titleformat{\subsection}[block]{\zihao{5}\bfseries}{\quad(\chinese{subsection})}{0em}{}\newcommand{\biaoti}[2]{\centerline{\zihao{-2}齊市普高聯(lián)誼校2023\textasciitilde2023學年上學期期中考試\vspace{1cm}}\centerline{\heiti\zihao{2}高一數(shù)學\vspace{1cm}}}\newcommand{\onexzt}{選擇題:此題共$8$小題,每題$5$分,共$40$分.在每題給出的四個選項中，只有一項為哪一項符合題目要求的.\vspace{-0.2cm}}\newcommand{\twoxzt}{選擇題:此題共$4$小題,每題$5$分,共$20$分.在每題給出的四個選項中，有多項符合題目要求，全部選對的得$5$$0$分，局部選對的得$2$分.\vspace{-0.2cm}}\newcommand{\tkt}{填空題:此題共$4$小題,每題$5$分,共$20$分.\vspace{-0.2cm}}\newcommand{\jdt}{解答題:此題$6$小題，共$70$分.解同意寫出文字說明、證明過程或演算步驟.\vspace{-0.2cm}}\newcommand{\bkt}{必考題:共$60$分\vspace{-0.2cm}}%\newcommand{\xkt}{選考題:共$10$$22$$23$題中任選一題作答.假設多做,則按所做的第一題計分.\vspace{-0.3cm}}%4個選項，使用一個命令依據(jù)選項內(nèi)容長度自動排版\usepackage{ifthen}\newlength{\lab}\newlength{\lb}\newlength{\lc}\newlength{\ld}\newlength{\lhalf}\newlength{\lquarter}\newlength{\lmax}\newcommand{\xx}[4]{\mbox{}\\[.5pt]%\settowidth{\lab}{A.~#1~~~}\settowidth{\lb}{B.~#2~~~}\settowidth{\lc}{C.~#3~~~}\settowidth{\ld}{D.~#4~~~}\ifthenelse{\lengthtest{\lab>\lb}} {\setlength{\lmax}{\lab}} {\setlength{\lmax}{\lb}}\ifthenelse{\lengthtest{\lmax<\lc}} {\setlength{\lmax}{\lc}} {}\ifthenelse{\lengthtest{\lmax<\ld}} {\setlength{\lmax}{\ld}} {}\setlength{\lhalf}{0.5\linewidth}\setlength{\lquarter}{0.25\linewidth}\ifthenelse{\lengthtest{\lmax>\lhalf}} {\noindent{}A.~#1\\B.~#2\\C.~#3\\D.~#4} {\ifthenelse{\lengthtest{\lmax>\lquarter}} {\noindent\makebox[\lhalf][l]{A.~#1~~~}%\makebox[\lhalf][l]{B.~#2~~~}\\%\makebox[\lhalf][l]{C.~#3~~~}%\makebox[\lhalf][l]{D.~#4~~~}}%{\noindent\makebox[\lquarter][l]{A.~#1~~~}%\makebox[\lquarter][l]{B.~#2~~~}%\makebox[\lquarter][l]{C.~#3~~~}%\makebox[\lquarter][l]{D.~#4~~~}}}}\newcommand{\xz}{\hfill(\makebox[0.4cm][c]{})}\usepackage{ulem}\newcommand{\tk}{\uline{\makebox[2cm][c]{}}}\begin{document}\biaoti{2023}{理}%\thisstyle{doubleline}\noindent\uuline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\noindent\begin{flushleft}\setstretch{1.6}\textbf{{\Large考生留意：}}\\~~~~~~~~1.本試卷分選擇題和非選擇題兩局部，總分值150120分鐘。\\~~~~~~~~2.答題前,考生務必用直徑$0.5$毫米黑色墨水簽字筆將密封線內(nèi)工程填寫清楚。\\~~~~~~~~3.2B鉛筆把答題卡上\\~~~~~~~~~~~~對應題目的答案標號涂黑；非選擇題用直徑$0.5$毫米黑色墨水簽字筆在答題卡上各題的答\\~~~~~~~~~~~~題區(qū)內(nèi)作答，{\large\textbf{\CJKunderdot{超出答題區(qū)域書寫的答案無效，在試卷、草稿紙上作答無效。}}}\\~~~~~~~~4.本卷命題范圍：必修第一冊第一、二、三、章。\end{flushleft}\section{\onexzt}\begin{enumerate}\setstretch{2}\item $A=\left\lbrace x \:|\: 0 \leqslant x \leqslant 2 \right\rbrace $，$B=\left\lbrace x\:|\:x\leqslant2,x\in\textbf{Z} \right\rbrace $,則$A\capB=( ~~~~)$\xx{$\left[2,3\right]$}{$\left[0,3\right]$}{$\{0,1,2\}$} {$ \{ 0,1,2,3\} $}\item命題“$\forall x\in\textbf{R},x^2-2x+1\geqslant0$“的否認是〔~~~~〕\xx{$\exists~ x\in \textbf{R},x^2-2x+1\leqslant 0 $}{$\forall ~ x\in\textbf{R},x^2-2x+1\leqslant0 $}{$\exists~x\in\textbf{R},x^2-2x+1\leqslant0 $}{$\exists~x\in\textbf{R},x^2-2x+1<0 $}\item $f(x)=2x-3$,且$f(a)=3$，則a=〔~~~~〕\xx{$4$} {$3$} {$2$} {$1$}\item$a>b,a,b,c\inR$，則以下不等式成立的有〔~~~~〕\xx{$\verta\vert > \vertb\vert$} {$\dfrac{1}>\dfrac{1}{a}$} {$a^2>b^2$}{$a(c^2+1>b(c^2+1))$}\item集合$A=\{x^2+x-6=0\},B=\{ax-1=0\}$,假設$B\subseteqA$，則實數(shù)的數(shù)值構成的集合是\xx{$\{-\dfrac{1}{3},0,\dfrac{1}{2}\}$} {$ \{ -2,0\} $}{$ \{ -2,\dfrac{1}{2}\} $}{$ \{ 0,\dfrac{1}{2}\} $}\item 正實數(shù)$x,y$滿足$\dfrac{1}{x}+\dfrac{4}{y}=1$，則的$x+y$最小值為〔~~~~〕\xx{$6$} {$7$} {$8$} {$9$}\item函數(shù)$f(x)=\dfrac{\vertx\vert}{x^2+2}$的局部圖象大致是〔~~~~〕\begin{tikzpicture} [scale=0.5]\draw[-stealth](-3,0)--(3,0)node[below]{${\smallx}$};%x軸\draw[-stealth](0,-1)--(0,1)node[above]{${\smally}$}; %y軸\draw[-](0,1)--(0,-1)node[below]{~~~A};\fill(0,0)circle(1pt); %原點\draw[-](-2,0)--(-2,0.1);%畫短線刻度\draw[-] (2,0)--(2,0.1);%畫短線刻\foreach\xin{-2,2}\node[above]at(\x,0){\x};%標刻度值\draw[color=black,domain=-2.75:2.75] plot(\x,{-abs(\x)/((\x)^2+2)});象

%做函數(shù)圖\end{tikzpicture}\hspace{12em}\begin{tikzpicture} [scale=0.55]\draw[-stealth](-3,0)--(3,0)node[below]{$x$};\draw[-stealth](0,-1)--(0,1)node[above]{$y$};%y軸\draw[-](0,1)--(0,-1)node[below]{~~~B};\fill(0,0)circle(1pt); %原點\draw[-](-2,0)--(-2,0.1);%畫短線刻度\draw[-] (2,0)--(2,0.1);%畫短線刻\foreach\xin{-2,2}\node[below]at(\x,0){{{\small}\x}};%標刻度值

%x軸\draw[color=black,domain2.75:2.75] plot(\x,{-(\x)/((\x)^2+2)});%做函數(shù)圖象\end{tikzpicture}\begin{tikzpicture} [scale=0.55]\draw[-stealth](-3,0)--(3,0)node[below]{$x$}; %x軸\draw[-stealth](0,-1)--(0,1)node[above]{$y$};%y軸\draw[-](0,1)--(0,-1)node[below]{~~~C};\fill(0,0)circle(1pt); %原點\draw[-](-2,0)--(-2,0.1);%畫短線刻度\draw[-] (2,0)--(2,0.1);%畫短線刻\foreach\xin {-2,2}\node[below]at(\x,0){\x};%標刻度值\draw[color=black,domain=-2.75:2.75] plot(\x,{abs(\x)/((\x)^2+2)});象

%做函數(shù)圖\end{tikzpicture}\hspace{12em}\begin{tikzpicture} [scale=0.5]\draw[-stealth](-3,0)--(3,0)node[below]{$x$};\draw[-stealth](0,-1)--(0,1)node[above]{$y$};%y軸\draw[-](0,1)--(0,-1)node[below]{~~~D};\fill(0,0)circle(1pt); %原點\draw[-](-2,0)--(-2,0.1);%畫短線刻度\draw[-] (2,0)--(2,0.1);%畫短線刻\foreach\xin{-2,2}\node[below]at(\x,0){\x};%標刻度值

%x軸\draw[color=black,domain2.8:2.8] plot(\x,{\x/((\x)^2+2)});%做函數(shù)圖象\end{tikzpicture}\item函數(shù)$f(x)=ax^2+2\left( a\inR\right) $，假設對于任意$1<x_{1}<x_{2}<2$，都有$\dfrac{f(x_{1})-f(x_{2})}{x_{1}-x_{2}}>-2$，則實數(shù)$a$的取值范圍是〔~~~~〕\xx {(-$\infty$,$-\dfrac{1}{2} $] $\bigcup$ [0,+$\infty$),} {[$-\dfrac{1}{2}$,+$\infty$)} {[$-\dfrac{1}{2}$,0)} {(0,+$\infty$)}\end{enumerate}\section{\twoxzt}\begin{enumerate}\setcounter{enumi}{8}\setstretch{2}\item 命“$\forall 1\leqslantx\leqslant 2,x^2-2a\leqslant 0$“是真命題的一個充分不必要條件是\xx {$a\geqslant1$} {$a\geqslant3$}{$a\geqslant4$}{$a\leqslant4$}\item以下各組函數(shù)中表示同個函數(shù)的是\xx {$f(x)=\sqrt{x^2}$,g(x)=$\sqrt{x^3}$}{$f(x)=x-1$,g(x)=$\sqrt[3]{x^3}$} {$f(x)=\dfrac{x^2}{x}$,g(x)=x}{$f(x)=\lvert 1-x \rvert$ ,g(x)=$\sqrt{(x-1)^2}$}\item下結(jié)論中正確的選項是\xx{“$a>b$”是“$2a>a+b$”的充要條件}{函數(shù)$y=\sqrt{x^2+3}+\dfrac{2}{\sqrt{x^2+3}}$4}{命題“$\forallx>1,x^2-x>0$“的否認是“$\exists\x_{0}>1,x_{0}^2-x_{0}\leqslant0$“}{假設函數(shù)$y=-x^2+2ax-1 $有正值，則實數(shù)$a $的取值范圍是$a>1 $或$a<-1$}\item函數(shù)$f(x)=\begin{cases}\left( 4a-2 \right)x+3a,,&\mbox{$x>1$}\\-x^2+2ax+5,,&\mbox{$x\leqslant0$}\end{cases}$ 在\textbf{R}上單調(diào)遞增，則實數(shù)$a $的取值范圍可以是\xx{$0$} {$1$} {$2$} {$3$}\end{enumerate}\section{\tkt}\begin{enumerate}\setcounter{enumi}{12}\item集合$A=\left\lbrace 0,a,a^2\right\rbrace$,且$1\in A$則實數(shù)a=\tk.\item函數(shù)$f(x)=\left( m^2-2m+1\right)x^{m^2-m+1}$是冪函數(shù)，則實數(shù)$m$的取值為\tk.\item定義在\textbf{R}上的奇函數(shù)$f(x) $對任意$x\in$\textbf{R}滿足$f(x+3)=f(x)$，且$f(1)=3$則$f(8)=$\tk.\item 函數(shù)$ f(x)=x^2+2x$ ,g$ (x)=ax+1( a>1) $ , 對$\forall x_{1}\in[-2,1],\existsx_{0}\in[-2,1$],使g$(x_{1})=f(x_{0})$，則實數(shù)$a$的取值范圍是\tk.\end{enumerate}\section{\jdt}\begin{enumerate}\setcounter{enumi}{16}\item($10$分)\\集合$A=\{x\:|\:x\leqslant-1$或$ x\geqslant4\}$，$ B=\{ x\:|\: 0<x<5\}，C=\{ x\:|\:m-2\geqslantx\geqslant2m\} $.\begin{enumerate}[(1)]\item求$A\bigcapB$,$(\complement_{R}A)\bigcupB$;\item假設$B\bigcapC=C$，求實數(shù)$m$的取值范圍.\end{enumerate}\vspace{6em}\item($12$分)\\$p:(x+1)(2-x)\geqslant0 ,q:x^2+2mx-3m^2\leqslant0$,假設$p$是$q$的充分不必要條件，求實數(shù)$m$的取值范圍.\vspace{6em}\item($12$分)\\二次函數(shù)$f(x)$的最大值為$2$，且$f(0)=f(2)=0$.\begin{enumerate}[(1)]\item 求$f(x)$的解析式；\item 假設$f(x)$在區(qū)間[2m,m+3]上不單調(diào)，求實數(shù)$m$的取值范圍.\end{enumerate}\vspace{6em}\item($12$分)\\函數(shù)$f(x)=\dfrac{1}{ax+b}$，且$f(1)=\dfrac{1}{3},f(-1)=-1$\begin{enumerate}[(1)]\item求$a,b$的值\item試推斷函數(shù)$f(x)$在$(2,+\infty)$上的單調(diào)性，并證明；\item求函數(shù)$f(x)$在$x\in[2,6]$上的最大值和最小值.\end{enumerate}\vspace{12em}\item($12$分) \