（魯京津瓊專用）2020版高考數(shù)學復習第三章導數(shù)及其應用階段自測卷（二）課件.pptx

,第三章 導數(shù)及其應用,階段自測卷(二),一、選擇題(本大題共12小題，每小題5分，共60分) 1.(2019沈陽東北育才學校聯(lián)考)已知曲線yf(x)在x5處的切線方程是yx5，則f(5)與f(5)分別為 A.5，1 B.1,5 C.1,0 D.0，1,解析 由題意可得f(5)550，f(5)1， 故選D.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,2.已知函數(shù)f(x)xsin xax，且 1，則a等于 A.0 B.1 C.2 D.4,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,3.(2019淄博期中)若曲線ymxln x在點(1，m)處的切線垂直于y軸，則實數(shù)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)的導數(shù)為f(x)m 曲線yf(x)在點(1，m)處的切線斜率為km10，可得m1. 故選A.,4.已知f1(x)sin xcos x，fn1(x)是fn(x)的導函數(shù)，即f2(x)f1(x)，f3(x)f2(x)，fn1(x)fn(x)，nN*，則f2 020(x)等于 A.sin xcos x B.sin xcos x 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為周期的函數(shù)， f2 020(x)f4(x)sin xcos x，故選B.,5.(2019四川診斷)已知函數(shù)f(x)的導函數(shù)為f(x)，且滿足f(x)2xf(e)ln x (其中e為自然對數(shù)的底數(shù))，則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 由題意知，函數(shù)的定義域為(0，)，,7.(2019沈陽東北育才學校模擬)已知定義在(0，)上的函數(shù)f(x)x2m，g(x)6ln x4x，設兩曲線yf(x)與yg(x)在公共點處的切線相同，則m值等于 A.5 B.3 C.3 D.5,解得x1，這就是切點的橫坐標，代入g(x)求得切點的縱坐標為4， 將(1，4)代入f(x)得1m4，m5. 故選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 依題意得，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河北衡水中學調(diào)研)如圖所示，某幾何體由底面半徑和高均為5的圓柱與半徑為5的半球面對接而成，該封閉幾何體內(nèi)部放入一個小圓柱體，且小圓柱體的上下底面均與外層圓柱的底面平行，則小圓柱體積的最大值為,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 小圓柱的高分為上下兩部分，上部分同大圓柱一樣為5，下部分深入底部半球內(nèi)設為h(0h5)， 小圓柱的底面半徑設為r(0r5)，由于r，h和球的半徑5滿足勾股定理， 即r2h252， 所以小圓柱體積Vr2(h5)(25h2)(h5)(0h5)， 求導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涼山診斷)若對任意的0x1x2a都有x2ln x1x1ln x2x1x2成立，則a的最大值為,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,故函數(shù)在(0,1)上導數(shù)大于零，單調(diào)遞增，在(1，)上導數(shù)小于零，單調(diào)遞減. 由于x1x2且f(x1)f(x2)，故x1，x2在區(qū)間(0,1)上，故a的最大值為1，故選B.,11.(2019洛陽、許昌質(zhì)檢)設函數(shù)yf(x)，xR的導函數(shù)為f(x)，且f(x)f(x), f(x)f(x)，則下列不等式成立的是(注：e為自然對數(shù)的底數(shù)) A.f(0)e1f(1)e2f(2) B.e1f(1)f(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 設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)，故選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)， 所以函數(shù)f(x)在(2,1)上單調(diào)遞增，在(，2)，(1，)上單調(diào)遞減， 又由關于x的方程f(x2)m有四個不同的實數(shù)解，等價于函數(shù)f(x)的圖象與直線ym在x(0，)，上有兩個交點，,可得f(x)x2xa， 則f(0)ba，f(0)a2，則b2 ，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,二、填空題(本大題共4小題，每小題5分，共20分) 13.(2019陜西四校聯(lián)考)已知函數(shù)f(x)ln x2x24x，則函數(shù)f(x)的圖象在x1處的切線方程為_.,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 x2x24x，,所求切線方程為y(2)x1，即xy30.,14.已知函數(shù)f(x)(xa)ln x(aR)，若函數(shù)f(x)存在三個單調(diào)區(qū)間，則實數(shù)a的取值范圍是_.,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,函數(shù)f(x)(xa)ln x(aR)，若函數(shù)f(x)存在三個單調(diào)區(qū)間，則f(x)有兩個變號零點， 即f(x)0有兩個不等實根，即ax(ln x1)有兩個不等實根，轉(zhuǎn)化為ya與yx(ln x1)的圖象有兩個不同的交點.令g(x)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山師大附中模擬)已知函數(shù)f(x)x32xex ，其中e是自然對數(shù)的 底數(shù)，f(a1)f(2a2) 0，則實數(shù)a的取值范圍是_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,當且僅當x0時等號成立，可得f(x)在R上遞增，,可得f(x)為奇函數(shù)，則f(a1)f(2a2)0 ，即有f(2a2)0f(a1)f(1a)， 即有2a21a ，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,16.(2019湖北黃岡中學、華師附中等八校聯(lián)考)定義在R上的函數(shù)f(x)滿足f(x)f(x)，且對任意的不相等的實數(shù)x1，x20，)有 0成立，若關于x的不等式f(2mxln x3) 2f(3)f(2mxln x3)在x1,3上恒成立，則 實數(shù)m的取值范圍是_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 函數(shù)f(x)滿足f(x)f(x)， 函數(shù)f(x)為偶函數(shù). 又f(2mxln x3)2f(3)f(2mxln x3) 2f(3)f(2mxln x3)， f(2mxln x3)f(3). 由題意可得函數(shù)f(x)在(，0)上單調(diào)遞增，在0，)上單調(diào)遞減. |2mxln x3|3對x1,3恒成立， 32mxln x33對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上單調(diào)遞增，在(e,3上單調(diào)遞減，,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上單調(diào)遞減，,三、解答題(本大題共70分) 17.(10分)(2019遼寧重點高中聯(lián)考)已知函數(shù)f(x)x3mx2m2x1(m為常數(shù)，且m0)有極大值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,當x變化時，f(x)與f(x)的變化情況如下表：,從而可知，當xm時，函數(shù)f(x)取得極大值9， 即f(m)m3m3m319，m2.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,(2)若斜率為5的直線是曲線yf(x)的切線，求此直線方程.,解 由(1)知，f(x)x32x24x1， 依題意知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成都七中診斷)已知函數(shù)f(x)xsin x2cos xax2，其中a為常數(shù). (1)若曲線yf(x)在x 處的切線斜率為2，求該切線的方程；,解 求導得f(x)xcos 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)求函數(shù)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 對任意x0，f(x)xsin x0， 所以f(x)在0，內(nèi)單調(diào)遞減. 當a0時，f(x)f(0)a0， f(x)在區(qū)間0，上單調(diào)遞減，故f(x)minf()a. 當a時，f(x)f()a0， f(x)在區(qū)間0，上單調(diào)遞增，故f(x)minf(0)4. 當00，f()a0，且f(x)在區(qū)間0，上單調(diào)遞減，結合零點存在定理可知，存在唯一x0(0，)，使得f(x0)0， 且f(x)在0，x0上單調(diào)遞增，在x0，上單調(diào)遞減.故f(x)的最小值等于f(0)4和f()a中較小的一個值.,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武漢示范高中聯(lián)考)已知函數(shù)f(x)4ln xmx21(mR). (1)若函數(shù)f(x)在點(1，f(1)處的切線與直線2xy10平行，求實數(shù)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 xmx21， f(x) 2mx， f(1)42m， 函數(shù)f(x)在(1，f(1)處的切線與直線2xy10平行，f(1)42m2， m1.,(2)若對于任意x1，e，f(x)0恒成立，求實數(shù)m的取值范圍.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 對于任意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，,當x變化時，g(x)，g(x)的變化如下表：,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)已知函數(shù)f(x)ln xax(ax1)，其中aR. (1)討論函數(shù)f(x)的單調(diào)性；,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,解 依題意知，函數(shù)f(x)的定義域為(0，)，,21,22,當a0時，f(x)ln x，函數(shù)f(x)在(0，)上單調(diào)遞增；,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,(2)若函數(shù)f(x)在(0,1內(nèi)至少有1個零點，求實數(shù)a的取值范圍.,21,22,由于當x0時，f(x)且f(1)a2a0知，函數(shù)f(x)在(0,1內(nèi)無零點；,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內(nèi)無零點；,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,綜上可得a的取值范圍是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湖北黃岡中學、華師附中等八校聯(lián)考)在工業(yè)生產(chǎn)中，對一正三角形薄鋼板(厚度不計)進行裁剪可以得到一種梯形鋼板零件，現(xiàn)有一邊長為3(單位：米)的正三角形鋼板(如圖)，沿平行于邊BC的直線DE將ADE剪去，得到所需的梯形鋼板BCED，記這個梯形鋼板的周長為x (單位：米)，面積為S(單位：平方米). (1)求梯形BCED的面積S關于它的周長x的函數(shù)關系式；,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,解 DEBC，ABC是正三角形， ADE是正三角形，ADDEAE，BDCE3AD， 則DE2(3AD)39ADx，,21,22,故梯形BCED的面積S關于它的周長x的函數(shù)關系式為,(2)若在生產(chǎn)中，梯形BCED的面積與周長之比 達到最大值時，零件才能符合使用要求，試確定這個梯形的周長x為多少時，該零件才可以在生產(chǎn)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)隨x的變化如下表：,20,1,2,3,4,5,6,7,8