貴州理科高考試題及答案

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.若復(fù)數(shù)\(z\)滿足\((1+i)z=1-3i\)，則\(z\)的實(shí)部為（）A.1B.-1C.2D.-2答案：B3.已知\(\vec{a}=(1,2)\)，\(\vec=(m,-1)\)，且\(\vec{a}\perp(\vec{a}+\vec)\)，則\(m\)的值為（）A.-1B.-3C.-7D.3答案：C4.執(zhí)行如圖所示的程序框圖，若輸入\(n=10\)，則輸出的\(S\)值為（）A.\(\frac{5}{11}\)B.\(\frac{10}{11}\)C.\(\frac{36}{55}\)D.\(\frac{72}{55}\)答案：A5.已知\(\sin\alpha=\frac{3}{5}\)，\(\alpha\in(\frac{\pi}{2},\pi)\)，則\(\tan(\alpha+\frac{\pi}{4})\)的值為（）A.\(\frac{1}{7}\)B.7C.-\(\frac{1}{7}\)D.-7答案：C6.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\(\frac{x^2}{100}+\frac{y^2}{64}=1\)有相同的焦點(diǎn)，則雙曲線的方程為（）A.\(\frac{x^2}{9}-\frac{y^2}{27}=1\)B.\(\frac{x^2}{18}-\frac{y^2}{54}=1\)C.\(\frac{x^2}{36}-\frac{y^2}{108}=1\)D.\(\frac{x^2}{15}-\frac{y^2}{45}=1\)答案：A7.一個(gè)正方體被一個(gè)平面截去一部分后，剩余部分的三視圖如圖所示，則截去部分體積與剩余部分體積的比值為（）A.\(\frac{1}{8}\)B.\(\frac{1}{7}\)C.\(\frac{1}{6}\)D.\(\frac{1}{5}\)答案：D8.函數(shù)\(f(x)=\frac{\lnx}{x}\)的圖象大致為（）A.B.C.D.答案：A9.已知\(f(x)\)是定義在\(R\)上的奇函數(shù)，當(dāng)\(x\geq0\)時(shí)，\(f(x)=x^2-3x\)，則函數(shù)\(g(x)=f(x)-x+3\)的零點(diǎn)的集合為（）A.\(\{1,3\}\)B.\(\{-3,-1,1,3\}\)C.\(\{2-\sqrt{7},1,3\}\)D.\(\{-2-\sqrt{7},1,3\}\)答案：D10.已知\(a=\log_{0.5}0.2\)，\(b=\log_{2}0.2\)，\(c=0.5^{0.2}\)，則\(a\)，\(b\)，\(c\)的大小關(guān)系為（）A.\(a>c>b\)B.\(a>b>c\)C.\(c>a>b\)D.\(c>b>a\)答案：A二、多項(xiàng)選擇題1.下列說(shuō)法正確的是（）A.命題“\(\forallx\inR,x^2+1>0\)”的否定是“\(\existsx\inR,x^2+1\leq0\)”B.若\(a,b\inR\)，則“\(a^2+b^2\neq0\)”是“\(a,b\)不全為\(0\)”的充要條件C.“\(x>1\)”是“\(x^2>1\)”的必要不充分條件D.若“\(p\landq\)”為假命題，則\(p,q\)均為假命題答案：AB2.已知函數(shù)\(f(x)=\sin(2x+\varphi)(0<\varphi<\frac{\pi}{2})\)的圖象經(jīng)過(guò)點(diǎn)\((0,\frac{3}{5})\)，則（）A.\(f(x)\)在\([0,\frac{\pi}{8}]\)上單調(diào)遞增B.\(f(x)\)在\([0,\frac{\pi}{4}]\)上的最大值為\(\frac{4}{5}\)C.直線\(x=\frac{\pi}{8}\)是\(f(x)\)圖象的一條對(duì)稱軸D.若把\(f(x)\)的圖象向左平移\(\frac{\pi}{8}\)個(gè)單位長(zhǎng)度，得到的圖象對(duì)應(yīng)的函數(shù)是偶函數(shù)答案：ABD3.已知\(a,b\)為正實(shí)數(shù)，且\(a+b=1\)，則（）A.\(a^2+b^2\geq\frac{1}{2}\)B.\(\frac{1}{a}+\frac{1}\geq4\)C.\(a^b\leqb^a\)D.\(\sqrt{a}+\sqrt\leq\sqrt{2}\)答案：ABD4.如圖，在正方體\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中，\(M,N\)分別為棱\(C_{1}D_{1},C_{1}C\)的中點(diǎn)，則下列說(shuō)法正確的是（）A.\(A_{1}M\parallel\)平面\(BDN\)B.平面\(BDN\perp\)平面\(CC_{1}D_{1}D\)C.異面直線\(A_{1}M\)與\(BD\)所成角的正切值為\(\sqrt{2}\)D.直線\(A_{1}M\)與平面\(ABCD\)所成角的余弦值為\(\frac{\sqrt{5}}{3}\)答案：ABC5.已知\(F\)是拋物線\(y^2=4x\)的焦點(diǎn)，\(P\)是拋物線上任意一點(diǎn)，點(diǎn)\(A(2,1)\)，則（）A.\(|PF|+|PA|\)的最小值為\(3\)B.\(|PF|-|PA|\)的最大值為\(\sqrt{2}\)C.以線段\(PF\)為直徑的圓與\(y\)軸相切D.以線段\(PF\)為直徑的圓與直線\(x=-1\)相切答案：AC6.已知函數(shù)\(f(x)=x^3-3x^2+2\)，則（）A.\(f(x)\)在\(x=1\)處取得極大值\(0\)B.\(f(x)\)在\(x=2\)處取得極小值\(-2\)C.\(f(x)\)的單調(diào)遞增區(qū)間為\((-\infty,0)\)和\((2,+\infty)\)D.\(f(x)\)的單調(diào)遞減區(qū)間為\((0,2)\)答案：ABCD7.已知\(\triangleABC\)的內(nèi)角\(A,B,C\)所對(duì)的邊分別為\(a,b,c\)，且\(a\sinA+b\sinB=c\sinC+\sqrt{2}a\sinB\)，則（）A.\(C=\frac{\pi}{4}\)B.\(\cosA\cosB=\frac{\sqrt{2}}{2}\sin(A+B)\)C.最大內(nèi)角的余弦值為\(\frac{\sqrt{2}}{2}\)D.若\(c=2\)，則\(\triangleABC\)面積的最大值為\(\sqrt{2}\)答案：ABD8.已知\(a_n\)是等差數(shù)列，\(S_n\)是其前\(n\)項(xiàng)和，若\(S_{11}=S_{20}\)，則（）A.\(a_{16}=0\)B.\(S_{31}=0\)C.當(dāng)\(a_1>0\)時(shí)，\(a_{10}+a_{11}>0\)D.當(dāng)\(a_1<0\)時(shí)，\(|a_{10}|<|a_{11}|\)答案：ABC9.已知\(f(x)\)是定義在\(R\)上的周期為\(4\)的奇函數(shù)，當(dāng)\(x\in(0,2)\)時(shí)，\(f(x)=x^2-2x\)，則（）A.\(f(2023)=-1\)B.\(f(x)\)在\((-2,0)\)上的解析式為\(f(x)=-x^2-2x\)C.\(f(x)\)的圖象關(guān)于直線\(x=2\)對(duì)稱D.\(f(x)\)在\((-4,4)\)上有\(zhòng)(5\)個(gè)零點(diǎn)答案：ABD10.已知函數(shù)\(f(x)=e^x-ax-1\)，則（）A.當(dāng)\(a=1\)時(shí)，\(f(x)\)在\((-\infty,0)\)上單調(diào)遞減B.當(dāng)\(a=0\)時(shí)，\(f(x)\)的圖象與\(x\)軸有兩個(gè)交點(diǎn)C.當(dāng)\(0<a<1\)時(shí)，\(f(x)\)在\((0,+\infty)\)上存在極小值D.當(dāng)\(a>1\)時(shí)，\(f(x)\)在\((0,+\infty)\)上存在零點(diǎn)答案：ACD三、判斷題1.若直線\(l\)與平面\(\alpha\)內(nèi)的無(wú)數(shù)條直線垂直，則\(l\perp\alpha\)。（×）2.若\(z_1,z_2\)為復(fù)數(shù)，則\(|z_1z_2|=|z_1|\cdot|z_2|\)。（√）3.函數(shù)\(y=\sin^2x\)的最小正周期為\(2\pi\)。（×）4.若\(\{a_n\}\)是等比數(shù)列，\(S_n\)是其前\(n\)項(xiàng)和，則\(S_n,S_{2n}-S_n,S_{3n}-S_{2n}\)仍成等比數(shù)列。（×）5.命題“若\(x^2<1\)，則\(-1<x<1\)”的逆否命題是“若\(x\geq1\)或\(x\leq-1\)，則\(x^2\geq1\)”。（√）6.若\(\vec{a}\cdot\vec=0\)，則\(\vec{a}=\vec{0}\)或\(\vec=\vec{0}\)。（×）7.函數(shù)\(y=\frac{1}{x}\)在其定義域上是減函數(shù)。（×）8.橢圓\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)\)的離心率\(e=\frac{c}{a}\)，其中\(zhòng)(c^2=a^2-b^2\)。（√）9.若\(f(x)\)是奇函數(shù)，則\(f(0)=0\)。（×）10.已知\(a,b\)為實(shí)數(shù)，則\(a^2+b^2\geq2ab\)。（√）四、簡(jiǎn)答題1.已知數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_1=1\)，\(a_{n+1}=2a_n+1\)。（1）證明數(shù)列\(zhòng)(\{a_n+1\}\)是等比數(shù)列；（2）求數(shù)列\(zhòng)(\{a_n\}\)的通項(xiàng)公式。答案：（1）由\(a_{n+1}=2a_n+1\)可得\(a_{n+1}+1=2(a_n+1)\)，又\(a_1+1=2\neq0\)，所以\(\frac{a_{n+1}+1}{a_n+1}=2\)，所以數(shù)列\(zhòng)(\{a_n+1\}\)是以\(2\)為首項(xiàng)，\(2\)為公比的等比數(shù)列。（2）由（1）知\(a_n+1=2\times2^{n-1}=2^n\)，所以\(a_n=2^n-1\)。2.已知函數(shù)\(f(x)=x^3-3x^2+2\)，求函數(shù)\(f(x)\)的單調(diào)區(qū)間和極值。答案：對(duì)\(f(x)=x^3-3x^2+2\)求導(dǎo)得\(f^\prime(x)=3x^2-6x=3x(x-2)\)。令\(f^\prime(x)=0\)，解得\(x=0\)或\(x=2\)。當(dāng)\(x<0\)或\(x>2\)時(shí)，\(f^\prime(x)>0\)，\(f(x)\)單調(diào)遞增；當(dāng)\(0<x<2\)時(shí)，\(f^\prime(x)<0\)，\(f(x)\)單調(diào)遞減。所以\(f(x)\)的極大值為\(f(0)=2\)，極小值為\(f(2)=-2\)，單調(diào)遞增區(qū)間為\((-\infty,0)\)和\((2,+\infty)\)，單調(diào)遞減區(qū)間為\((0,2)\)。3.在\(\triangleABC\)中，角\(A,B,C\)所對(duì)的邊分別為\(a,b,c\)，已知\(a=2\sqrt{3}\)，\(b=2\)，\(\cosA=-\frac{1}{2}\)。（1）求角\(B\)的大??；（2）求\(c\)的值。答案：（1）因?yàn)閈(\cosA=-\frac{1}{2}\)，\(0<A<\pi\)，所以\(A=\frac{2\pi}{3}\)。由正弦定理\(\frac{a}{\sinA}=\frac{\sinB}\)，可得\(\sinB=\frac{b\sinA}{a}=\frac{2\times\frac{\sqrt{3}}{2}}{2\sqrt{3}}=\frac{1}{2}\)。又\(a>b\)，所以\(A>B\)，所以\(B=\frac{\pi}{6}\)。（2）因?yàn)閈(A+B+C=\pi\)，所以\(C=\pi-A-B=\frac{\pi}{6}\)，所以\(c=b=2\)。4.已知橢圓\(C:\frac{x^2