```tex \documentclass{book}%文档格式 \RequirePackage[a4paper]{geometry} \geometry{ textwidth=138mm, textheight=215mm, left=15mm, right=10mm, top=10mm, bottom=5mm, headheight=2.17cm, headsep=4mm, footskip=12mm, heightrounded,} %设置页边距 \usepackage{amsmath}%数学公式 \usepackage{ctex}%输出汉字 \usepackage{graphicx}%插入图片 \usepackage{tikz}%画图用 \pagestyle{myheadings} \markboth{由文龙}{数列}%设置页眉及页码 \usepackage{multicol}%分栏 \usepackage{newtxmath}%接近于mathtype字体 \usepackage{ifthen} \newlength{\la} \newlength{\lb} \newlength{\lc} \newlength{\ld} \newlength{\lhalf} \newlength{\lquarter} \newlength{\lmax} \newcommand{\xx}[4]{\\[.5pt]% \settowidth{\la}{A.~#1~~~} \settowidth{\lb}{B.~#2~~~} \settowidth{\lc}{C.~#3~~~} \settowidth{\ld}{D.~#4~~~} \ifthenelse{\lengthtest{\la > \lb}} {\setlength{\lmax}{\la}} {\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~~~}}}}%选项自动识别长度 \newcommand{\kh}{\nolinebreak\dotfill\mbox{\raisebox{-1.8pt} {$\cdots$}(\hspace{0.3cm})}} \begin{document} \begin{multicols}{2}%分栏加中间线 \columnseprule=1pt%加中间线 \begin{enumerate} \item[一、] 选择题 \item 设$ f(x) = \ln (\sqrt {{{(x + 1)}^2}} + x + 1) - 2 $，若$ f(a) = 1 $，$ f(b) = - 5 $，则$ a+b= $\kh 　　 \xx{$ 2 $}{$ 0 $}{$ 1 $}{$ -2 $}%1 \item 设函数$ f(x) = \frac{{{x^5} + {{(x + 1)}^2}}}{{{x^2} + 1}} $在区间$ \left[ { - 12,12} \right] $上的最大值为$ M $，最小值为$ N $，则$ {(M + n - 1)^{2019}} $的值为\kh \xx{$ 1 $}{$ -1 $}{$ 2^{2019} $}{$ 0 $}%2 \item 已知函数$ f(x) = \frac{{{2^x}}}{{{2^x} - 1}} $，若$ f(-m)=2 $，则$ f(m)= $\kh \xx{$ -2 $}{$ -1 $}{$ 0 $}{$ \frac{1}{2} $}%3 \item 已知函数$ y=f(x)+x $是偶函数，且$ f(2)=1 $，则$ f(-2)= $\kh \xx{$ 2 $}{$ 3 $}{$ 4 $}{$ 5 $}%4 \item 已知$ f(x)=g(x)-4 $，函数$ g(x) $是定义在$ R $上的奇函数，若$ f(2017)=2017 $，则$ f(-2017)= $\kh \xx{$ -2017 $}{$ -2021 $}{$ -2025 $}{$ 2025 $}%5 \item 已知函数$ f(x) $是定义在$ R $上的偶函数，且在$ \left[ {0, + \infty } \right) $上是减函数，则满足$ f(a-1)>f(1) $的实数$ a $的取值范围是\kh \xx{$ ({\rm{2, + }}\infty ) $}{$ ( - \infty ,2) $}{$ ({\rm{0,2}}) $}{$({\rm{1,2}}) $}%6 \item 已知$ f(x) $为偶函数，其在$ ( - \infty ,0] $上为增函数，$ f(2)=0 $，满足不等式$ f(1-x)<0 xss=removed>f(2) $的解集为\kh \xx{$ (-1,1) $}{$ (-4,2) $}{$ ( - \infty , - 1) \cup ({\rm{1, + }}\infty ) $}{$ ( - \infty , - 4) \cup ({\rm{2, + }}\infty ) $}%8 \item 已知$ f(x) = \frac{{2x + 1}}{{x - 1}} + {(x - 1)^3} $，若$ f(2018)=a $，则$ f(-2016)= $\kh \xx{$ -a $}{$ 2-a $}{$ 4-a $}{$ 1-a$}%9 \item 已知函数$ f(x) = {e^{|x|}} + \cos x $，若$ f(2x-1)\le f(x) $，则实数$ x $的取值范围是\kh \xx{$ ( - \infty , \frac{1}{3}] \cup [{\rm{1, + }}\infty ) $}{$ [\frac{1}{3},1] $}{$ ( - \infty , \frac{1}{2}] $}{$ [{\rm{\frac{1}{2}, + }}\infty ) $} \item 已知函数$ f(x) = x\ln (x + \sqrt {{x^2} + 1} ) $，若$ f(2a-1)1 $}{$ a<0>1 $}{$ 0 0 $的解集为\kh \xx{$ ( - \infty , - 3] \cup (0,3) $}{$ ( - \infty , - 3) \cup (3{\rm{, + }}\infty ) $}{$ ( - 3,0) \cup (0,3) $}{$ ( - 3,0) \cup (3{\rm{, + }}\infty ) $} \item 已知函数$ f(x)=\sqrt{x-2} $，若$ f(2{a^2} - 5a + 4) < f xss=removed xss=removed xss=removed xss=removed>4 $的解集为\kh \xx{$ (-\frac{1}{4},+\infty) $}{$ (-\infty,-\frac{1}{4}) $}{$ (-\infty,0) $}{$ (0,+\infty) $} \item 设$ f(x) = {2^x} - {2^{ - x}} + \ln \frac{{1 + x}}{{1 - x}} + 1 $，若$ f(a)+f(1+a)>2 $，则$ a $的取值范围\kh \xx{$ (-\frac{1}{2},+\infty) $}{$ (-\frac{1}{2},1) $}{$ (-\frac{1}{2},0) $}{$ (0,\frac{1}{2}) $}! \item 已知定义在$ R $上的奇函数$ f(x) $，任意$ x_{1} ,x_{2}\in [0{\rm{, + }}\infty ) $，且$x_{1} \ne x_{2}$都有 $ \frac{{f({x_1}) - f({x_2})}}{{{x_1} - {x_2}}} < 0> 0} $，若$ f(2)=0 $，则$ (x-2)f(x)>0 $的解集为\kh \xx{$\left( {-\infty,-2}\right) \cup\left( {0,2}\right) \cup\left( {2,+\infty}\right) $}{$\left( {-2,0}\right) \cup\left( {0, + \infty}\right)$ }{$\left( {-\infty,-2}\right) \cup\left( {0,2}\right) $}{$\left( {-2,0}\right) \cup(0,2)$}# \item 已知函数$ f(x) $满足$ f(x)=f(-x+2) $，且$ f(x) $在$ ( - \infty , 1] $上单调递增，则\kh \xx{$ f(1) > f( - 1) > f(4) $}{$ f(-1) > f( 1) > f(4) $}{$ f(4) > f( 1) > f(-1) $}{$ f(1) > f( 4) > f(-1) $}$ \item 函数$ f(x) = \sqrt {{x^2} - x + 2} $的单调递增区间为\kh \xx{$ [2, + \infty ) $}{$ ( - \infty ,\frac{1}{2}] $}{$ [\frac{1}{2}, + \infty ) $}{$ ( - \infty ,-1] $}% \item 函数$ f(x) = {\log _{0.6}}({x^2} + 6x - 7) $的单调递减区间是\kh \xx{$ ( - \infty ,-7) $}{$ ( - \infty ,-3) $}{$ (-3, + \infty ) $}{$ (1, + \infty ) $}& \item 若定义在$ R $上的函数$ f(x) $满足：$ x \in ( - 3, - 1) $时，$ f(x+1)=e^{x} $，对于任意$ x\in R $，都有 $ f(x + 2) = \frac{1}{{f(x)}} $成立，$ f(2019)= $\kh \xx{$ e^{2} $}{$ e^{-2} $}{$ e $}{$ 1 $}' \item 若定义在$ R $上的函数$ f(x) $满足$ f(-x)=f(x) $，$ f(x+1)=f(1-x) $，且当$ x\in [0,1] $时，$ f(x) = {\log _2}(x + 1) $，则$ f（(2019)= $\kh \xx{$ 0 $}{$ 1 $}{$ -1 $}{$ 2 $}( \item 若定义在$ R $上的函数$ f(x) $满足$ f(-x)=-f(x) $，$ f(x)=f(x+2) $，且当$ x\in (-1,0) $时，$ f(x) = {2^x} + \frac{1}{3} $，则$ f({\log _2}6) = $\kh \xx{$ 1 $}{$ -1 $}{$ \frac{4}{5} $}{$ -\frac{4}{5} $}) \item 若定义在$ R $上的函数$ f(x) $满足$ f(x)=-f(x+1) $，且当$ 1\le x<2 $时，$ f(x) = {9^x} - 9 $，则$ f( - \frac{1}{2}) = $\kh \xx{$ 0 $}{$ -6 $}{$ 18 $}{$ -18 $}0 \item 已知函数$ f(x) $对于任意实数$ x $满足条件$ f(x + 2) = - \frac{1}{{f(x)}} $，若$ f(0)=\frac{1}{2} $，则$ f(2018)= $\kh \xx{$ -\frac{1}{2} $}{$ \frac{1}{2} $}{$ -2 $}{$ 2 $}1 \item 已知函数$ y=f(x) $满足$ f(x + 1) = \frac{1}{{f(x - 1)}} $和$ f(2-x)=f(x+1) $，且当$ x \in [\frac{1}{2},\frac{3}{2}] $时，$ f(x)=2x+2 $，则$ f(2018)= $\kh \xx{$ 0 $}{$ 2 $}{$ 4 $}{$ 5 $}2 \item 若定义在$ R $上的偶函数$ f(x) $，对任意的实数$ x $都有$ f(x+4)=-f(x)+2 $，且$ f(-3)=3 $，则$ f(2015)= $\kh \xx{$ -1 $}{$ 3 $}{$ 2015 $}{$ -4028 $}3 \item 若定义在$ R $上的函数$ f(x) $满足对任意$ x\in R $，有$ f(x + 2) = \frac{1}{{f(x)}} $，且$ f(x) $的图像关于直线$ x=1 $对称，当$ x\in [-1,1] $时，$ f(x)=x+1 $，则$ f(6)= $\kh \xx{$ 1 $}{$ 2 $}{$ 3 $}{$ 4 $}4 \item 设定义在$ R $上的函数$ f(x) $满足$ f(x) \cdot f(x + 2) = 13 $，若$ f(1)=2 $，则$ f(2015)= $\kh \xx{$ \frac{13}{3} $}{$ \frac{13}{2} $}{$ 13 $}{$ \frac{39}{2} $}5 \item 定义在$ R $上的偶函数$ f(x) $满足:对任意的实数$ x $都有$ f(-x)=f(x+2) $，$ f(-1)=2 $，$ f(2)=-1 $，则$ f(1) + f(2) + f(3) + \cdot \cdot \cdot + f(2017) $的值为\kh \xx{$ 2017 $}{$ 1010 $}{$ 1008 $}{$ 2 $}6 \item 已知函数$ f(x) $是定义在$ R $上的奇函数，$ f(\frac{3}{2}+x)=f(\frac{3}{2}-x) $，且$ x\in (-\frac{3}{2},0) $时，$ f(x) = {\log _2}( - 3x + 1) $，则$ f(2020)= $\kh \xx{$ 4 $}{$ log_{2}7 $}{$ 2 $}{$ -2 $}7 \item 定义在$ R $上的奇函数$ f(x) $满足$ f(1+x)=f(1-x) $，且当$ x\in [0,1] $时，$ f(x)=x(3-2x) $，则$ f(\frac{31}{2})= $\kh \xx{$ -1 $}{$ -\frac{1}{2}$}{$ \frac{1}{2} $}{$ 1 $}8 \item 已知函数$ f(x) = {\log _{\frac{1}{3}}}(3{x^2} - ax + 8) $在$ \left[ { - 1, + \infty } \right) $上是减函数，则实数$ a $的取值范围\kh \xx{$ ( - \infty , - 6] $}{$ [ - 11, - 6] $}{$ ( - 11, - 6] $}{$ ( - 11, + \infty )$}9 \item 已知函数$ f(x) = x + 2\sin (x - \frac{1}{2}) $，则$ f(\frac{1}{{2019}}) + f(\frac{2}{{2019}}) + \cdot \cdot \cdot + f(\frac{{2018}}{{2019}}) $的值等于\kh \xx{$ 2019 $}{$ 2018 $}{$ \frac{2019}{2} $}{$ 1009 $}@ \item 12 \xx{我我我我我我我我我我我}{我我我我我我我我我我我}{我我我我我我我我我我我}{我我我我我我我我我我我} \end{enumerate} \end{multicols}%分栏 加中间线 \end{document} ```