山西省高考真題及答案

一、單項(xiàng)選擇題1.已知集合\(A=\{x|x^2-3x+2=0\}\)，\(B=\{x|0<x<6,x\inN\}\)，則滿足\(A\subseteqC\subseteqB\)的集合\(C\)的個(gè)數(shù)為（）A.4B.8C.7D.16答案：B2.已知\(i\)為虛數(shù)單位，若復(fù)數(shù)\(z=\frac{1+2i}{2-i}\)，則\(z\)的共軛復(fù)數(shù)\(\overline{z}\)是（）A.\(i\)B.\(-i\)C.\(1+i\)D.\(1-i\)答案：B3.已知\(\sin(\frac{\pi}{6}+\alpha)=\frac{1}{3}\)，則\(\cos(\frac{2\pi}{3}-2\alpha)\)的值為（）A.\(\frac{7}{9}\)B.\(\frac{1}{3}\)C.\(-\frac{7}{9}\)D.\(-\frac{1}{3}\)答案：C4.已知\(a=\log_{3}2\)，\(b=\log_{5}3\)，\(c=0.5^{-0.2}\)，則\(a\)，\(b\)，\(c\)的大小關(guān)系為（）A.\(a<b<c\)B.\(a<c<b\)C.\(c<a<b\)D.\(c<b<a\)答案：A5.已知雙曲線\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0,b>0)\)的一條漸近線方程為\(y=\sqrt{3}x\)，且與橢圓\(\frac{x^{2}}{10}+\frac{y^{2}}{4}=1\)有公共焦點(diǎn)，則雙曲線的方程為（）A.\(\frac{x^{2}}{4}-\frac{y^{2}}{12}=1\)B.\(\frac{x^{2}}{12}-\frac{y^{2}}{4}=1\)C.\(x^{2}-\frac{y^{2}}{3}=1\)D.\(\frac{x^{2}}{3}-y^{2}=1\)答案：A6.執(zhí)行如圖所示的程序框圖，若輸入的\(x=2\)，則輸出的\(y\)值為（）A.2B.5C.11D.23答案：D7.已知函數(shù)\(f(x)=\sin(\omegax+\varphi)(\omega>0,|\varphi|<\frac{\pi}{2})\)的最小正周期為\(\pi\)，將該函數(shù)的圖象向左平移\(\frac{\pi}{6}\)個(gè)單位后，得到的圖象對(duì)應(yīng)的函數(shù)為偶函數(shù)，則\(f(x)\)的圖象（）A.關(guān)于點(diǎn)\((\frac{\pi}{12},0)\)對(duì)稱B.關(guān)于直線\(x=\frac{\pi}{12}\)對(duì)稱C.關(guān)于點(diǎn)\((\frac{5\pi}{12},0)\)對(duì)稱D.關(guān)于直線\(x=\frac{5\pi}{12}\)對(duì)稱答案：A8.已知某幾何體的三視圖如圖所示，則該幾何體的體積為（）A.\(\frac{16\pi}{3}\)B.\(\frac{8\pi}{3}\)C.\(\frac{4\pi}{3}\)D.\(\frac{2\pi}{3}\)答案：B9.已知\(a\)，\(b\)為單位向量，且\(a\cdotb=0\)，若\(c=2a-\sqrt{5}b\)，則\(\cos\langlea,c\rangle=\)（）A.\(\frac{\sqrt{5}}{3}\)B.\(\frac{\sqrt{5}}{5}\)C.\(\frac{2\sqrt{5}}{5}\)D.\(\frac{\sqrt{5}}{2}\)答案：C10.已知函數(shù)\(f(x)=\begin{cases}2^{x-1}-2,x\leq1\\-\log_{2}(x+1),x>1\end{cases}\)，且\(f(a)=-3\)，則\(f(6-a)=\)（）A.\(-\frac{7}{4}\)B.\(-\frac{5}{4}\)C.\(-\frac{3}{4}\)D.\(-\frac{1}{4}\)答案：A二、多項(xiàng)選擇題1.下列函數(shù)中，在區(qū)間\((0,+\infty)\)上單調(diào)遞增的是（）A.\(y=x^{\frac{1}{2}}\)B.\(y=2^{x}\)C.\(y=\log_{\frac{1}{2}}x\)D.\(y=\frac{1}{x}\)答案：AB2.已知向量\(\vec{a}=(1,m)\)，\(\vec=(3,-2)\)，且\((\vec{a}+\vec)\perp\vec\)，則\(m\)的值可以是（）A.-8B.-6C.6D.8答案：AD3.已知\(a\)，\(b\)，\(c\)為正實(shí)數(shù)，且\(a+b+c=1\)，則（）A.\(a^{2}+b^{2}+c^{2}\geq\frac{1}{3}\)B.\(ab+bc+ca\geq\frac{1}{3}\)C.\(\frac{1}{a}+\frac{1}+\frac{1}{c}\geq9\)D.\(\sqrt{a}+\sqrt+\sqrt{c}\leq\sqrt{3}\)答案：ACD4.已知函數(shù)\(f(x)=\sin(2x+\varphi)(|\varphi|<\frac{\pi}{2})\)的圖象向左平移\(\frac{\pi}{6}\)個(gè)單位后關(guān)于\(y\)軸對(duì)稱，則以下說(shuō)法正確的是（）A.\(\varphi=\frac{\pi}{6}\)B.\(f(x)\)在\([0,\frac{\pi}{6}]\)上單調(diào)遞增C.\(f(x)\)的一個(gè)對(duì)稱中心是\((\frac{5\pi}{12},0)\)D.\(f(x)\)在\([-\frac{\pi}{6},\frac{\pi}{3}]\)上的最小值為\(-\frac{1}{2}\)答案：ABC5.已知圓\(C\)：\((x-1)^{2}+(y-2)^{2}=25\)，直線\(l\)：\((2m+1)x+(m+1)y-7m-4=0\)，則以下說(shuō)法正確的是（）A.直線\(l\)恒過(guò)定點(diǎn)\((3,1)\)B.直線\(l\)與圓\(C\)相交C.當(dāng)直線\(l\)被圓\(C\)截得的弦長(zhǎng)最短時(shí)，直線\(l\)的方程為\(2x-y-5=0\)D.直線\(l\)被圓\(C\)截得的最長(zhǎng)弦長(zhǎng)為\(10\)答案：ABCD6.已知\(a\)，\(b\)是兩條不同的直線，\(\alpha\)，\(\beta\)是兩個(gè)不同的平面，則下列說(shuō)法正確的是（）A.若\(a\parallel\alpha\)，\(b\parallel\alpha\)，則\(a\parallelb\)B.若\(a\parallel\alpha\)，\(a\parallel\beta\)，則\(\alpha\parallel\beta\)C.若\(a\perp\alpha\)，\(b\perp\alpha\)，則\(a\parallelb\)D.若\(\alpha\perp\beta\)，\(a\subset\alpha\)，\(\alpha\cap\beta=b\)，\(a\perpb\)，則\(a\perp\beta\)答案：CD7.已知函數(shù)\(f(x)=x^{3}-3x^{2}+2\)，則（）A.\(f(x)\)有兩個(gè)極值點(diǎn)B.\(f(x)\)在\((-\infty,0)\)上單調(diào)遞增C.\(f(x)\)在\(x=2\)處取得極小值\(-2\)D.\(f(x)\)在\([0,2]\)上的最大值為\(2\)答案：ACD8.已知數(shù)列\(zhòng)(\{a_{n}\}\)為等差數(shù)列，\(S_{n}\)為其前\(n\)項(xiàng)和，若\(a_{1}=1\)，\(S_{3}=9\)，則（）A.\(a_{n}=2n-1\)B.\(a_{n}=n\)C.\(S_{n}=n^{2}\)D.\(S_{n}=\frac{n(n+1)}{2}\)答案：AC9.已知\(F_{1}\)，\(F_{2}\)是橢圓\(C\)：\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)\)的兩個(gè)焦點(diǎn)，\(P\)為橢圓\(C\)上一點(diǎn)，且\(\angleF_{1}PF_{2}=60^{\circ}\)，若\(\triangleF_{1}PF_{2}\)的面積為\(\sqrt{3}\)，則（）A.\(b=1\)B.\(a=2\)C.橢圓\(C\)的離心率為\(\frac{1}{2}\)D.\(|PF_{1}|+|PF_{2}|=4\)答案：ACD10.已知函數(shù)\(f(x)\)是定義在\(R\)上的奇函數(shù)，且當(dāng)\(x\geq0\)時(shí)，\(f(x)=x^{2}-2x\)，則（）A.\(f(-1)=1\)B.當(dāng)\(x<0\)時(shí)，\(f(x)=-x^{2}-2x\)C.\(f(x)\)的單調(diào)遞增區(qū)間為\((-\infty,-1)\)和\((1,+\infty)\)D.\(f(x)\)的值域?yàn)閈(R\)答案：BD三、判斷題1.若\(a>b\)，則\(ac^{2}>bc^{2}\)。（×）2.函數(shù)\(y=\sinx\)的圖象的一個(gè)對(duì)稱中心是\((\frac{\pi}{2},0)\)。（×）3.若直線\(l\)與平面\(\alpha\)內(nèi)的無(wú)數(shù)條直線垂直，則\(l\perp\alpha\)。（×）4.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比\(q=2\)，則\(\frac{a_{1}+a_{3}}{a_{2}+a_{4}}=\frac{1}{2}\)。（√）5.若\(\log_{a}2<\log_2\)，則\(0<a<b<1\)。（×）6.函數(shù)\(y=\sqrt{x^{2}-4}\)的定義域是\((-\infty,-2]\cup[2,+\infty)\)。（√）7.已知向量\(\vec{a}=(1,2)\)，\(\vec=(3,4)\)，則\(\vec{a}\cdot\vec=11\)。（√）8.橢圓\(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)的離心率為\(\frac{\sqrt{5}}{3}\)。（√）9.函數(shù)\(y=2^{x}\)與\(y=\log_{2}x\)的圖象關(guān)于直線\(y=x\)對(duì)稱。（√）10.若\(x\in[0,\frac{\pi}{2}]\)，則函數(shù)\(y=\sinx+\cosx\)的最大值為\(\sqrt{2}\)。（√）四、簡(jiǎn)答題1.已知\(\{a_{n}\}\)是等差數(shù)列，\(a_{1}=1\)，\(a_{3}+a_{4}=12\)。求數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式。答案：設(shè)等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)。因?yàn)閈(a_{3}+a_{4}=12\)，\(a_{1}=1\)，則\(a_{1}+2d+a_{1}+3d=12\)，即\(2+5d=12\)，解得\(d=2\)。所以\(a_{n}=a_{1}+(n-1)d=1+2(n-1)=2n-1\)。2.已知函數(shù)\(f(x)=\sin(2x+\frac{\pi}{3})\)，求\(f(x)\)的最小正周期和單調(diào)遞增區(qū)間。答案：對(duì)于函數(shù)\(f(x)=\sin(2x+\frac{\pi}{3})\)，其最小正周期\(T=\frac{2\pi}{\omega}\)（\(\omega=2\)），所以\(T=\pi\)。由\(2k\pi-\frac{\pi}{2}\leq2x+\frac{\pi}{3}\leq2k\pi+\frac{\pi}{2}(k\inZ)\)，得\(k\pi-\frac{5\pi}{12}\leqx\leqk\pi+\frac{\pi}{12}(k\inZ)\)，所以\(f(x)\)的單調(diào)遞增區(qū)間是\([k\pi-\frac{5\pi}{12},k\pi+\frac{\pi}{12}](k\inZ)\)。