1、模拟试卷(一),一、选择题(本大题共12小题，每小题5分，共60分) 1.设集合Ax|1x2，Bx|2x1，则AB等于 A. 0,2) B.0,1) C.(1,0 D.(1,0),解析 由题意得Bx|2x1x|x0，又Ax|1x2， ABx|0x20,2).故选A.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,故选B.,3.下列函数中，既是偶函数，又在(，0)上单调递增的是 A.f(x)2x2x B.f(x)x2
2、1 C.f(x)xcos x D.f(x)ln|x|,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 A中，f(x)2x2xf(x)，不是偶函数，A错； B中，f(x)(x)21x21f(x)，是偶函数，但在(，0)上单调递减，B错； C中，f(x)xcos(x)xcos xf(x)，不是偶函数，C错； D中，f(x)ln|x|ln|x|f(x)，是偶函数，且函数在(，0)上单调递增，故选D.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,4.设等比数列a
3、n的前n项和为Sn，且Snk2n3，则ak等于 A.4 B.8 C.12 D.16,解析 当n2时，anSnSn1k2n1； 当n1时，a1S12k3k211，解得k3, aka3323112. 故选C.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12
4、,13,14,15,16,17,18,19,20,21,22,结合各选项可得C符合题意. 故选C.,7.函数f(x) 有两个不同的零点，则实数a的取值范围是 A.a2 B.a2,故选C.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,8.(2019安徽省江淮名校试题)RtABC的斜边AB等于4，点P在以C为圆心，1为半径的圆上，则 的取值范围是,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
5、,16,17,18,19,20,21,22,9. (1x)5的展开式中x2的系数为 A.1 B.9 C.31 D.19,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,10.如图，B是AC上一点，分别以AB，BC，AC为直径作半圆.过B作BDAC，与半圆相交于D. AC6，BD ，在整个图形中随机取一点，则此点取自图中阴影部分的概率是,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17
6、,18,19,20,21,22,解析 连接AD，CD， 可知ACD是直角三角形，又BDAC，所以BD2ABBC，设ABx(0x6)，则有8x(6x)，得x2或x4，当x2时，AB2，BC4， 由此可得图中阴影部分的面积等于,故选C.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 由题意，矩形的对角线长相等，,4a2b2(b23a2)c2， 4a2(c2a2)(c24a2)c2， e48e240，,故选C.,1
7、,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,12.设正三棱锥PABC的每个顶点都在半径为2的球O的球面上，则三棱锥PABC体积的最大值为,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,故选C.,二、填空题(本大题共4小题，每小题5分，共
8、20分),1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,13.已知向量a，b的夹角为45，且|a|b|2，则a(a b)_.,0,14.若函数f(x)(a1)x3ax22x为奇函数，则曲线yf(x)在点(1，f(1)处的切线方程为_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,解析 f(x)(a1)x3ax22x为奇函数，则a0， f(x)x32x， f(x)3x22，f(1)31221，又f(1)1， 曲线yf(x)在点(1，f(1)处的切线方程为y1x1，即xy2
9、0.,21,22,xy20,15.(2019安徽省江淮名校联考)已知正数a，b满足ab1，则 的最大值为_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 令xa1，yb2，则xy4，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,16.设mR，若函数f(x)|x33xm|在x0， 上的最大值与最小值之差为2，则实数m的取值范围是_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,(，2
10、0，),1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,则g(x)3x233(x1)(x1)，,2m0或m0， 解得m2或m0. 实数m的取值范围为(，20，).,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,三、解答题(本大题共70分) 17.(10分)设Sn为等差数列an的前n项和，S981，a2a38. (1)求an的通项公式；,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,故an1(n1
11、)22n1(nN*).,(2)若S3，a14，Sm成等比数列，求S2m.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,即9m2272，解得m9，故S2m182324.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 在ABC中，根据正弦定理，,又ADCBBADB6060， 所以ADC120. 于是C1801203030，所以B60.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,(2)若B
12、D2DC，且AD ，求DC的长.,在ABD中，由余弦定理，得 AD2AB2BD22ABBDcos B，,故DC2.,19.(12分)如图，四边形ABCD为正方形，BEDF，且ABBEDF EC，AB平面BCE. (1)证明：平面AEC平面BDFE；,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,证明 四边形ABCD为正方形，ACBD.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,又AB平面BCE，ABBE. ABBCC，BE平面ABCD，BEAC. 又BEB
13、DB，AC平面BDFE， AC平面AEC，平面AEC平面BDEF.,(2)求二面角AFCD的余弦值.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 BE平面ABCD，BEDF，DF平面ABCD. 以D为坐标原点建立如图所示的空间直角坐标系Dxyz，令AB1， 则A(1,0,0)，C(0,1,0)，E(1,1,1)，F(0,0,1)，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,设平面AFC的法向量为n1(x1，y1，z1)，,令x11，则n1(1,1,1
14、). 易知平面FCD的一个法向量n2(1,0,0)，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,二面角AFCD为锐角，,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20.(12分)某中学为了解中学生的课外阅读时间，决定在该中学的1 200名男生和800名女生中按分层抽样的方法抽取20名学生，对他们的课外阅读时间进行问卷调查.现在按课外阅读时间的情况将学生分成三类：A类(不参加课外阅读)，B类(参加课外阅读，但平均每周参加课外阅读的时间不超过3小时)，C类(参加课外阅
15、读，且平均每周参加课外阅读的时间超过3小时).调查结果如下表： (1)求出表中x，y的值；,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,解 设抽取的20人中，男、女生人数分别为n1，n2，,21,22,所以x12534，y8332.,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,(2)根据表中的统计数据，完成下面的列联表，并判断是否有90%的把握认为“参加阅读与否”与性别有关；,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,1
16、6,17,18,19,解 列联表如下：,21,22,所以没有90%的把握认为“参加阅读与否”与性别有关.,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,(3)从抽出的女生中再随机抽取3人进一步了解情况，记X为抽取的这3名女生中A类人数和C类人数差的绝对值，求X的均值.,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,解 X的可能取值为0,1,2,3，,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21.(1
17、2分)在直角坐标系xOy中，直线yx4与抛物线C：x22py(p0)交于A，B两点，且OAOB. (1)求C的方程；,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,得x22px8p0，4p232p0， 设A(x1，y1)，B(x2，y2)，则x1x22p，x1x28p， 从而y1y2(x14)(x24)x1x24(x1x2)16.,2x1x24(x1x2)160， 即16p8p160，解得p2，故C的方程为x24y.,(2)试问：在x轴的正半轴上是否存在一点D，使得ABD的外心在C上？若存在，求出D的坐标；若不存在
18、，请说明理由.,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,解 设线段AB的中点为N(x0，y0)，,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,则线段AB的中垂线方程为y6(x2)， 即yx8.,从而ABD的外心P的坐标为(4,4)或(8,16). 假设存在点D(m,0)(m0)，设P的坐标为(4,4)，,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,若P的坐标为(8,16)，,20,1
19、,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,则P的坐标不可能为(8,16).,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,22.(12分)(2019安徽省江淮名校联考)已知函数f(x)exax2在x1处的切线方程为ybx1. (1)求a，b的值；,21,22,解 f(x)ex2ax，,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,(2)证明：当x0时ex2xx2ex1.,21,22,20,1,2,3,4,5,6,7,8,
20、9,10,11,12,13,14,15,16,17,18,19,证明 实际上是证明当x0时，f(x)exx2的图象在直线y(e2)x1的上方. 令g(x)exx2(e2)x1，x0，则g(x)ex2xe2， 令t(x)ex2xe2，则t(x)ex2， 所以g(x)在(0，ln 2)上单调递减，在(ln 2，)上单调递增；g(x)在xln 2处取唯一的极小值. 注意到g(0)3e0，g(1)0，而0ln 21， 所以g(ln 2)0，所以g(0)g(ln 2)0； 又因为g(x)在(0，ln 2)上单调递减， 所以存在唯一的x0(0，ln 2)，使g(x0)0；,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,因此当x(0，x0)或x(1，)时，g(x)0， 当x(x0，1)时，g(x)0时，不等式ex2xx2ex1成立.,21,22,
