鲁京津琼专用2020版高考数学大一轮复习第三章导数及其应用阶段自测卷二课件

上传人:hua****011 文档编号:107025 上传时间:2019-12-12 格式:PPTX 页数:52 大小:2.38MB
下载 相关 举报
鲁京津琼专用2020版高考数学大一轮复习第三章导数及其应用阶段自测卷二课件_第1页
第1页 / 共52页
鲁京津琼专用2020版高考数学大一轮复习第三章导数及其应用阶段自测卷二课件_第2页
第2页 / 共52页
鲁京津琼专用2020版高考数学大一轮复习第三章导数及其应用阶段自测卷二课件_第3页
第3页 / 共52页
鲁京津琼专用2020版高考数学大一轮复习第三章导数及其应用阶段自测卷二课件_第4页
第4页 / 共52页
鲁京津琼专用2020版高考数学大一轮复习第三章导数及其应用阶段自测卷二课件_第5页
第5页 / 共52页
点击查看更多>>
资源描述

1、,第三章 导数及其应用,阶段自测卷(二),一、选择题(本大题共12小题,每小题5分,共60分) 1.(2019沈阳东北育才学校联考)已知曲线yf(x)在x5处的切线方程是yx5,则f(5)与f(5)分别为 A.5,1 B.1,5 C.1,0 D.0,1,解析 由题意可得f(5)550,f(5)1, 故选D.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,2.已知函数f(x)xsin xax,且 1,则a等于 A.0 B.1 C.2 D.4,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18

2、,19,20,21,22,3.(2019淄博期中)若曲线ymxln x在点(1,m)处的切线垂直于y轴,则实数m等于 A.1 B.0 C.1 D.2,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 f(x)的导数为f(x)m 曲线yf(x)在点(1,m)处的切线斜率为km10,可得m1. 故选A.,4.已知f1(x)sin xcos x,fn1(x)是fn(x)的导函数,即f2(x)f1(x),f3(x)f2(x),fn1(x)fn(x),nN*,则f2 020(x)等于 A.sin xcos x B.sin xcos x

3、 C.sin xcos x D.sin xcos 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 f1(x)sin xcos x,f2(x)f1(x)cos xsin x, f3(x)f2(x)sin xcos x, f4(x)f3(x)cos xsin x,f5(x)f4(x)sin xcos xf1(x), fn(x)是以4为周期的函数, f2 020(x)f4(x)sin xcos x,故选B.,5.(2019四川诊断)已知函数f(x)的导函数为f(x),且满足f(x)2xf(e)ln x (其中e为自然对数的底

4、数),则f(e)等于 A.1 B.1 C.e D.e1,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 已知f(x)2xf(e)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 由题意知,函数的定义域为(0,),,7.(2019沈阳东北育才学校模拟)已知定义在(0,)上的函数f(x)x2m,g(x)6ln x4x,设两曲线yf(x)与yg(x)在公共点处的切线相同,则m值等于 A.5 B.3 C.3 D.5,解得x1,这就是切点的横坐标,代入

5、g(x)求得切点的纵坐标为4, 将(1,4)代入f(x)得1m4,m5. 故选D.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 依题意得,f(x)aexcos x0,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,9.(2019河北衡水中学调研)如图所示,某几何体由底面半径和高均为5的圆柱与半径为5的半球面对接而成,该封闭几何体内部放入一个

6、小圆柱体,且小圆柱体的上下底面均与外层圆柱的底面平行,则小圆柱体积的最大值为,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 小圆柱的高分为上下两部分,上部分同大圆柱一样为5,下部分深入底部半球内设为h(0h5), 小圆柱的底面半径设为r(0r5),由于r,h和球的半径5满足勾股定理, 即r2h252, 所以小圆柱体积Vr2(h5)(25h2)(h5)(0h5), 求导V(3h5)(h5),,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,10.(2019凉

7、山诊断)若对任意的0x1x2a都有x2ln x1x1ln x2x1x2成立,则a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,故函数在(0,1)上导数大于零,单调递增,在(1,)上导数小于零,单调递减. 由于x1x2且f(x1)f(x2),故x1,x2在区间(0,1)上,故a的最大值为1,故选B.,11.(2019洛阳、许昌质检)设函数yf(x),xR的导函数为f(x),且f(x)f(x), f(x)f(x),则下列不等式成立的是(注:e为自然对数的底数) A.f(0)e1f(1)e2f(2) B.e1f(1)f(

8、0)e2f(2) C.e2f(2)e1f(1)f(0) D.e2f(2)f(0)e1f(1),1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 设g(x)exf(x), g(x)exf(x)exf(x)ex(f(x)f(x), f(x)g(0)g(1), e1f(1) f(0)e2f(2),故选B.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,f(x)x2x2(x1)(x2), 所以函数f(x)在(2,1)上单调递增,在(,2),(1,)上单调递减, 又由

9、关于x的方程f(x2)m有四个不同的实数解,等价于函数f(x)的图象与直线ym在x(0,),上有两个交点,,可得f(x)x2xa, 则f(0)ba,f(0)a2,则b2 ,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,二、填空题(本大题共4小题,每小题5分,共20分) 13.(2019陕西四校联考)已知函数f(x)ln x2x24x,则函数f(x)的图象在x1处的切线方程为_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,xy30,解析 f(x)ln x2

10、x24x,,所求切线方程为y(2)x1,即xy30.,14.已知函数f(x)(xa)ln x(aR),若函数f(x)存在三个单调区间,则实数a的取值范围是_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,函数f(x)(xa)ln x(aR),若函数f(x)存在三个单调区间,则f(x)有两个变号零点, 即f(x)0有两个不等实根,即ax(ln x1)有两个不等实根,转化为ya与yx(ln x1)的图象有两个不同的交点.令g(x

11、)x(ln x1),,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,15.(2019山师大附中模拟)已知函数f(x)x32xex ,其中e是自然对数的 底数,f(a1)f(2a2) 0,则实数a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,当且仅当x0时等号成立,可得f(x)在R上递增,,可得f(x)为奇函数,则f(a1)f(2a2)0 ,即有f(2a2)0f(a1)f(1a), 即有2a21a ,,1,2,3,4,5,6,7,8

12、,9,10,11,12,13,14,15,16,17,18,19,20,21,22,16.(2019湖北黄冈中学、华师附中等八校联考)定义在R上的函数f(x)满足f(x)f(x),且对任意的不相等的实数x1,x20,)有 0成立,若关于x的不等式f(2mxln x3) 2f(3)f(2mxln x3)在x1,3上恒成立,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,解析 函数f(x)满足f(x)f(x), 函数f(x)为偶函数. 又f(2mxln x3)2f(3)f(2mxln x3) 2f(3)f

13、(2mxln x3), f(2mxln x3)f(3). 由题意可得函数f(x)在(,0)上单调递增,在0,)上单调递减. |2mxln x3|3对x1,3恒成立, 32mxln x33对x1,3恒成立,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,g(x)在1,e上单调递增,在(e,3上单调递减,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,h(x)在1,3上单调递减,,三、解答题(本大题共70分) 17.(10分)(2019辽宁重点高中联考)已知函数f

14、(x)x3mx2m2x1(m为常数,且m0)有极大值9. (1)求m的值;,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 f(x)3x22mxm2(xm)(3xm)0,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,当x变化时,f(x)与f(x)的变化情况如下表:,从而可知,当xm时,函数f(x)取得极大值9, 即f(m)m3m3m319,m2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,2

15、2,(2)若斜率为5的直线是曲线yf(x)的切线,求此直线方程.,解 由(1)知,f(x)x32x24x1, 依题意知f(x)3x24x45,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,即5xy10或135x27y230.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,18.(12分)(2019成都七中诊断)已知函数f(x)xsin x2cos xax2,其中a为常数. (1)若曲线yf(x)在x 处的切线斜率为2,求该切线的方程;,解 求导得f(x)xc

16、os xsin xa,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,(2)求函数f(x)在x0,上的最小值.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 对任意x0,f(x)xsin x0, 所以f(x)在0,内单调递减. 当a0时,f(x)f(0)a0, f(x)在区间0,上单调递减,故f(x)minf()a. 当a时,f(x)f()a0, f(x)在区间0,上单调递增,故f(x)minf(0)4. 当00,f()a0,且f(x)在区间0,上单调递减

17、,结合零点存在定理可知,存在唯一x0(0,),使得f(x0)0, 且f(x)在0,x0上单调递增,在x0,上单调递减.故f(x)的最小值等于f(0)4和f()a中较小的一个值.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,19.(12分)(2019武汉示范高中联考)已知函数f(x)4ln xmx21(mR). (1)若函数f(x)在点(1,f(1)处的切线与直线2xy10平行,求实数m的值;,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 f(x)4ln

18、xmx21, f(x) 2mx, f(1)42m, 函数f(x)在(1,f(1)处的切线与直线2xy10平行,f(1)42m2, m1.,(2)若对于任意x1,e,f(x)0恒成立,求实数m的取值范围.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 对于任意x1,e,f(x)0恒成立, 4ln xmx210,在x1,e上恒成立,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,令g(x)0,得1x ,,令g(x)0,得 xe,,当x变化时,g(x),g(x)

19、的变化如下表:,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20.(12分)已知函数f(x)ln xax(ax1),其中aR. (1)讨论函数f(x)的单调性;,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,解 依题意知,函数f(x)的定义域为(0,),,21,22,当a0时,f(x)ln x,函数f(x)在(0,)上单调递增;,20,1,2,3,4,5,6

20、,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,(2)若函数f(x)在(0,1内至少有1个零点,求实数a的取值范围.,21,22,由于当x0时,f(x)且f(1)a2a0知,函数f(x)在(0,1内无零点;,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,f(x)在(0,1内无零点;,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,

21、22,综上可得a的取值范围是1,0.,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21.(12分)(2019湖北黄冈中学、华师附中等八校联考)在工业生产中,对一正三角形薄钢板(厚度不计)进行裁剪可以得到一种梯形钢板零件,现有一边长为3(单位:米)的正三角形钢板(如图),沿平行于边BC的直线DE将ADE剪去,得到所需的梯形钢板BCED,记这个梯形钢板的周长为x (单位:米),面积为S(单位:平方米). (1)求梯形BCED的面积S关于它的周长x的函数关系式;,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,1

22、4,15,16,17,18,19,解 DEBC,ABC是正三角形, ADE是正三角形,ADDEAE,BDCE3AD, 则DE2(3AD)39ADx,,21,22,故梯形BCED的面积S关于它的周长x的函数关系式为,(2)若在生产中,梯形BCED的面积与周长之比 达到最大值时,零件才能符合使用要求,试确定这个梯形的周长x为多少时,该零件才可以在生产中使用?,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,f(x),f(x)随x的变化如下表:,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,

23、18,19,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,22.(12分)(2019衡水中学调研)已知函数f(x)kexx2(其中kR,e是自然对数的底数). (1)若k2,当x(0,)时,试比较f(x)与2的大小;,21,22,解 当k2时,f(x)2exx2,则f(x)2ex2x,令h(x)2ex2x,h(x)2ex2, 由于x(0,),故h(x)2ex20, 于是h(x)2ex2x在(0,)上为增函数, 所以

24、h(x)2ex2xh(0)20,即f(x)2ex2x0在(0,)上恒成立, 从而f(x)2exx2在(0,)上为增函数, 故f(x)2exx2f(0)2.,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,(2)若函数f(x)有两个极值点x1,x2(x1x2),求k的取值范围,并证明:0f(x1)1.,21,22,当x0,函数(x)单调递增且(x)0,函数(x)单调递增且(x)0; 当x1时,(x)0. 作出函数(x)的图象如图所示,,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,由于x1(0,1),所以0(x11)211, 所以0f(x1)1.,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,(x11)21,,

展开阅读全文
相关资源
相关搜索
资源标签

当前位置:首页 > 高中 > 高中数学 > 数学高考 > 一轮复习