數(shù)學(xué)山東高考真題及答案

一、單項(xiàng)選擇題1.設(shè)集合\(A=\{x|x^2-4x+3<0\}\)，\(B=\{x|2x-3>0\}\)，則\(A\capB=\)（）A.\((-3,-\frac{3}{2})\)B.\((-3,\frac{3}{2})\)C.\((1,\frac{3}{2})\)D.\((\frac{3}{2},3)\)答案：D2.已知\(a\inR\)，若\(a-1+(a-2)i\)（\(i\)為虛數(shù)單位）是實(shí)數(shù)，則\(a=\)（）A.\(1\)B.\(-1\)C.\(2\)D.\(-2\)答案：C3.已知向量\(\overrightarrow{a}=(1,2)\)，\(\overrightarrow=(2,-2)\)，\(\overrightarrow{c}=(1,\lambda)\)。若\(\overrightarrow{c}\parallel(2\overrightarrow{a}+\overrightarrow)\)，則\(\lambda=\)（）A.\(\frac{1}{2}\)B.\(\frac{2}{5}\)C.\(-2\)D.\(-\frac{2}{5}\)答案：A4.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：D5.某學(xué)校為了解\(1000\)名新生的身體素質(zhì)，將這些學(xué)生編號(hào)為\(1\)，\(2\)，\(\cdots\)，\(1000\)，從這些新生中用系統(tǒng)抽樣方法等距抽取\(100\)名學(xué)生進(jìn)行體質(zhì)測(cè)驗(yàn)。若\(46\)號(hào)學(xué)生被抽到，則下面\(4\)名學(xué)生中被抽到的是（）A.\(8\)號(hào)學(xué)生B.\(200\)號(hào)學(xué)生C.\(616\)號(hào)學(xué)生D.\(815\)號(hào)學(xué)生答案：C6.已知\(a=\log_27\)，\(b=\log_38\)，\(c=0.3^{0.2}\)，則\(a\)，\(b\)，\(c\)的大小關(guān)系為（）A.\(c<b<a\)B.\(a<b<c\)C.\(b<c<a\)D.\(c<a<b\)答案：A7.函數(shù)\(y=\frac{2x^3}{2^x+2^{-x}}\)在\([-6,6]\)的圖象大致為（）（選項(xiàng)略）答案：B8.在\(\triangleABC\)中，\(\cos\frac{C}{2}=\frac{\sqrt{5}}{5}\)，\(BC=1\)，\(AC=5\)，則\(AB=\)（）A.\(4\sqrt{2}\)B.\(\sqrt{30}\)C.\(\sqrt{29}\)D.\(2\sqrt{5}\)答案：A9.已知雙曲線\(C:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a>0,b>0)\)的離心率為\(\sqrt{2}\)，則點(diǎn)\((4,0)\)到\(C\)的漸近線的距離為（）A.\(\sqrt{2}\)B.\(2\)C.\(\frac{3\sqrt{2}}{2}\)D.\(2\sqrt{2}\)答案：D10.已知\(f(x)\)是\(R\)上最小正周期為\(2\)的周期函數(shù)，且當(dāng)\(0\leqx<2\)時(shí)，\(f(x)=x^3-x\)，則函數(shù)\(y=f(x)\)的圖象在區(qū)間\([0,6]\)上與\(x\)軸的交點(diǎn)的個(gè)數(shù)為（）A.\(6\)B.\(7\)C.\(8\)D.\(9\)答案：B二、多項(xiàng)選擇題1.下列函數(shù)中，既是偶函數(shù)又在\((0,+\infty)\)上單調(diào)遞增的是（）A.\(y=x^3\)B.\(y=|x|+1\)C.\(y=-x^2+1\)D.\(y=2^{|x|}\)答案：BD2.已知曲線\(C:mx^2+ny^2=1\)（）A.若\(m>n>0\)，則\(C\)是橢圓，其焦點(diǎn)在\(y\)軸上B.若\(m=n>0\)，則\(C\)是圓，其半徑為\(\sqrt{n}\)C.若\(mn<0\)，則\(C\)是雙曲線，其漸近線方程為\(y=\pm\sqrt{-\frac{m}{n}}x\)D.若\(m=0\)，\(n>0\)，則\(C\)是兩條直線答案：ACD3.已知\(a,b\inR\)，且\(a-3b+6=0\)，則\(2^a+\frac{1}{8^b}\)的最小值為（）A.\(\frac{1}{4}\)B.\(4\)C.\(\frac{5}{2}\)D.\(3\)答案：A4.下列說(shuō)法正確的是（）A.命題“\(\forallx\inR,x^2+x+1>0\)”的否定是“\(\existsx_0\inR,x_0^2+x_0+1\leq0\)”B.若\(a,b\inR\)，則“\(a^2>b^2\)”的充要條件是“\(a>b\)”C.函數(shù)\(y=\sinx+\frac{4}{\sinx}\)，\(x\in(0,\pi)\)的最小值為\(4\)D.“若\(\alpha=\frac{\pi}{6}\)，則\(\sin\alpha=\frac{1}{2}\)”的逆否命題為真命題答案：AD5.已知函數(shù)\(f(x)=\sin(\omegax+\varphi)(\omega>0,|\varphi|<\frac{\pi}{2})\)的圖象上相鄰兩個(gè)最高點(diǎn)的距離為\(\pi\)，則“\(\varphi=\frac{\pi}{6}\)”是“\(f(x)\)的圖象關(guān)于直線\(x=\frac{\pi}{3}\)對(duì)稱”的（）A.充分而不必要條件B.必要而不充分條件C.充分必要條件D.既不充分也不必要條件答案：A6.已知正方體\(ABCD-A_1B_1C_1D_1\)的棱長(zhǎng)為\(1\)，點(diǎn)\(E,F\)分別是棱\(BC,CC_1\)的中點(diǎn)，則（）A.\(A_1D\)與\(EF\)所成角為\(60^{\circ}\)B.直線\(A_1D\)與平面\(AEF\)平行C.三棱錐\(A-DEF\)的體積為\(\frac{1}{12}\)D.平面\(AEF\)截正方體所得的截面面積為\(\frac{9}{8}\)答案：ABD7.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\(x=1\)處取得極小值\(0\)C.\(f(x)\)在\(x=2\)處取得極大值\(-2\)D.\(f(x)\)在\(x=2\)處取得極小值\(-2\)答案：AD8.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\(\overrightarrow{PF_1}\perp\overrightarrow{PF_2}\)。若\(\trianglePF_1F_2\)的面積為\(9\)，則\(b=\)（）A.\(3\)B.\(4\)C.\(5\)D.\(6\)答案：A9.已知函數(shù)\(y=f(x)\)的定義域?yàn)閈(R\)，且\(f(-x)=-f(x)\)，當(dāng)\(x\in(0,1]\)時(shí)，\(f(x)=\frac{2^x}{4^x+1}\)。則（）A.\(f(x)\)在\((-1,0)\)上是減函數(shù)B.\(f(x)\)在\((0,1)\)上的最大值為\(\frac{\sqrt{2}}{2}\)C.\(f(x)\)的圖象關(guān)于點(diǎn)\((0,0)\)對(duì)稱D.\(f(2-x)=f(x)\)答案：AC10.已知數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_1=1\)，\(a_{n+1}=2a_n+1(n\inN^)\)，則（）A.數(shù)列\(zhòng)(\{a_n+1\}\)是等比數(shù)列B.數(shù)列\(zhòng)(\{a_n\}\)的通項(xiàng)公式為\(a_n=2^n-1\)C.數(shù)列\(zhòng)(\{a_n\}\)是遞增數(shù)列D.數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n=2^{n+1}-n-2\)答案：ABC三、判斷題1.若\(a,b\inR\)，且\(ab>0\)，則\(\frac{a}+\frac{a}\geq2\)。（）答案：√2.若直線\(l_1:ax+2y+6=0\)與直線\(l_2:x+(a-1)y+a^2-1=0\)平行，則\(a=-1\)或\(a=2\)。（）答案：×3.若\(\sin\alpha=\frac{1}{3}\)，\(\alpha\)為第二象限角，則\(\cos\alpha=-\frac{2\sqrt{2}}{3}\)。（）答案：√4.若向量\(\overrightarrow{a}=(1,2)\)，\(\overrightarrow=(x,1)\)，且\(\overrightarrow{a}+2\overrightarrow\)與\(2\overrightarrow{a}-\overrightarrow\)平行，則\(x=\frac{1}{2}\)。（）答案：√5.函數(shù)\(y=\log_2(x^2-4x+3)\)的單調(diào)遞增區(qū)間是\((2,+\infty)\)。（）答案：×6.若\(x\inR\)，則“\(x>1\)”是“\(x^2>1\)”的充分不必要條件。（）答案：√7.已知圓\(C:x^2+y^2-2x+4y-4=0\)，則圓心\(C\)的坐標(biāo)為\((1,-2)\)，半徑為\(3\)。（）答案：√8.若\(f(x)\)是定義在\(R\)上的奇函數(shù)，且\(f(x+4)=f(x)\)，則\(f(2020)=0\)。（）答案：√9.已知\(a=\log_32\)，\(b=\ln2\)，\(c=5^{-\frac{1}{2}}\)，則\(c<a<b\)。（）答案：√10.若數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n=2n^2-3n\)，則\(a_n=4n-5\)。（）答案：√四、簡(jiǎn)答題1.已知等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，\(a_3=5\)，\(S_4=16\)。求\(\{a_n\}\)的通項(xiàng)公式。答案：設(shè)等差數(shù)列\(zhòng)(\{a_n\}\)的首項(xiàng)為\(a_1\)，公差為\(d\)。由\(a_3=5\)可得\(a_1+2d=5\)；由\(S_4=16\)可得\(4a_1+\frac{4\times3}{2}d=16\)，即\(4a_1+6d=16\)。聯(lián)立方程解得\(a_1=1\)，\(d=2\)。所以\(a_n=a_1+(n-1)d=1+2(n-1)=2n-1\)。2.已知函數(shù)\(f(x)=x^3-3x^2+2\)，求\(f(x)\)的單調(diào)區(qū)間。答案：對(duì)\(f(x)=x^3-3x^2+2\)求導(dǎo)得\(f^\prime(x)=3x^2-6x\)。令\(f^\prime(x)=0\)，即\(3x^2-6x=0\)，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\(f(x)\)的單調(diào)遞增區(qū)間是\((-\infty,0)\)和\((2,+\infty)\)，單調(diào)遞減區(qū)間是\((0,2)\)。3.已知橢圓\(C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)\)的離心率為\(\frac{\sqrt{2}}{2}\)，且過(guò)點(diǎn)\((\sqrt{2},1)\)。求橢圓\(C\)的方程。答案：由離心率\(e=\frac{c}{a}=\frac{\sqrt{2}}{2}\)，可得\(c^2=\frac{1}{2}a^2\)，又\(a^2=b^2+c^2\)，所以\(a^2=2b^