《(全國(guó)通用)2020版高考數(shù)學(xué)二輪復(fù)習(xí) 第四層熱身篇 專題檢測(cè)(二十)導(dǎo)數(shù)的幾何意義及簡(jiǎn)單應(yīng)用》由會(huì)員分享,可在線閱讀,更多相關(guān)《(全國(guó)通用)2020版高考數(shù)學(xué)二輪復(fù)習(xí) 第四層熱身篇 專題檢測(cè)(二十)導(dǎo)數(shù)的幾何意義及簡(jiǎn)單應(yīng)用(9頁(yè)珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、專題檢測(cè)(二十) 導(dǎo)數(shù)的幾何意義及簡(jiǎn)單應(yīng)用A組“633”考點(diǎn)落實(shí)練一、選擇題1.已知函數(shù)f(x)的導(dǎo)函數(shù)f(x)滿足下列條件:f(x)0時(shí),x2;f(x)0時(shí),1x2;f(x)0時(shí),x1或x2.則函數(shù)f(x)的大致圖象是()解析:選A根據(jù)條件知,函數(shù)f(x)在(1,2)上是減函數(shù).在(,1),(2,)上是增函數(shù),故選A.2.設(shè)函數(shù)f(x)xex1,則()A.x1為f(x)的極大值點(diǎn)B.x1為f(x)的極小值點(diǎn)C.x1為f(x)的極大值點(diǎn)D.x1為f(x)的極小值點(diǎn)解析:選D由題意得,f(x)(x1)ex,令f(x)0,得x1,當(dāng)x(,1)時(shí),f(x)0,當(dāng)x(1,)時(shí),f(x)0,則f(x)在
2、(,1)上單調(diào)遞減,在(1,)上單調(diào)遞增,所以x1為f(x)的極小值點(diǎn),故選D.3.已知直線ykx2與曲線yxln x相切,則實(shí)數(shù)k的值為()A.ln 2B.1C.1ln 2 D.1ln 2解析:選D由yxln x知yln x1,設(shè)切點(diǎn)為(x0,x0ln x0),則切線方程為yx0ln x0(ln x01)(xx0),因?yàn)榍芯€ykx2過定點(diǎn)(0,2),所以2x0ln x0(ln x01)(0x0),解得x02,故k1ln 2,選D.4.若x 是函數(shù)f(x)(x22ax)ex的極值點(diǎn),則函數(shù)yf(x)的最小值為()A.e B.0C.e D.e解析:選Cf(x)(x22ax)ex,f(x)(2x2
3、a)ex(x22ax)exx22(1a)x2aex,由已知得,f0,所以222a2a0,解得a1.所以f(x)(x22x)ex,所以f(x)(x22)ex,所以函數(shù)的極值點(diǎn)為,當(dāng)x時(shí),f(x)0;所以函數(shù)yf(x)是減函數(shù),當(dāng)x或x時(shí),f(x)0,函數(shù)yf(x)是增函數(shù).又當(dāng)x(,0)(2,)時(shí),x22x0,f(x)0,當(dāng)x(0,2)時(shí),x22x0,f(x)0,所以f(x)min在x(0,2)上,又當(dāng)x時(shí),函數(shù)yf(x)遞減,當(dāng)x時(shí),函數(shù)yf(x)遞增,所以f(x)minfe.5.已知函數(shù)f(x)(2xln xa)ex在(0,)上單調(diào)遞增,則實(shí)數(shù)a的最大值是()A.5ln 2 B.52ln 2
4、C.2ln 2 D.52ln 2解析:選Af(x)(2xln xa)ex,f(x)(2xln x2a)ex,x(0,).依題意,知x(0,)時(shí),f(x)0恒成立,即a2xln x2在(0,)上恒成立.設(shè)g(x)2xln x2,則g(x)2,x(0,).令g(x)0,得x或x1(舍去).令g(x)0,則0x,令g(x)0,則x,當(dāng)x時(shí),函數(shù)g(x)取得最小值,g(x)ming5ln 2,a5ln 2,即實(shí)數(shù)a的最大值是5ln 2.故選A.6.已知函數(shù)f(x)為偶函數(shù),當(dāng)x0時(shí),f(x)4x,設(shè)af(log30.2),bf(30.2),cf(31.1),則()A.cab B.abcC.cba D.
5、bac解析:選A因?yàn)楹瘮?shù)f(x)為偶函數(shù),所以af(log30.2)f(log30.2),cf(31.1)f(31.1).因?yàn)閘og3log30.2log3,所以2log30.21,所以1log30.22,所以31.13log30.2130.2.因?yàn)閥在(0,)上為增函數(shù),y4x在(0,)上為增函數(shù),所以f(x)在(0,)上為增函數(shù),所以f(31.1)f(log30.2)f(30.2),所以cab,故選A.二、填空題7.(2019江蘇高考)在平面直角坐標(biāo)系xOy中,點(diǎn)A在曲線yln x上,且該曲線在點(diǎn)A處的切線經(jīng)過點(diǎn)(e,1)(e為自然對(duì)數(shù)的底數(shù)),則點(diǎn)A的坐標(biāo)是_.解析:設(shè)A(m,n),則曲
6、線yln x在點(diǎn)A處的切線方程為yn(xm).又切線過點(diǎn)(e,1),所以有n1(me).再由nln m,解得me,n1.故點(diǎn)A的坐標(biāo)為(e,1).答案:(e,1)8.若函數(shù)f(x)xaln x不是單調(diào)函數(shù),則實(shí)數(shù)a的取值范圍是_.解析:由題意知f(x)的定義域?yàn)?0,),f(x)1,要使函數(shù)f(x)xaln x不是單調(diào)函數(shù),則需方程10在(0,)上有解,即xa,a0.答案:(,0)9.設(shè)定義在R上的函數(shù)yf(x)的導(dǎo)函數(shù)為f(x).如果存在x0a,b,使得f(b)f(a)f(x0)(ba)成立,則稱x0為函數(shù)f(x)在區(qū)間a,b上的“中值點(diǎn)”.那么函數(shù)f(x)x33x在區(qū)間2,2上的“中值點(diǎn)”
7、為_.解析:由f(x)x33x求導(dǎo)可得f(x)3x23,設(shè)x0為函數(shù)f(x)在區(qū)間2,2上的“中值點(diǎn)”,則f(x0)1,即3x31,解得x0.答案:三、解答題10.已知函數(shù)f(x)x2axaln x.(1)若曲線yf(x)在x2處的切線與直線x3y20垂直,求實(shí)數(shù)a的值;(2)若函數(shù)f(x)在2,3上單調(diào)遞增,求實(shí)數(shù)a的取值范圍.解:(1)f(x)2xa(x0),依題意有f(2)3,a2.(2)依題意有2x2axa0在x2,3上恒成立,即a在2,3上恒成立,0(x2,3), y在2,3上單調(diào)遞減,當(dāng)x2,3時(shí),8,實(shí)數(shù)a的取值范圍為8,).11.(2019重慶市七校聯(lián)合考試)設(shè)函數(shù)f(x),g
8、(x)a(x21)ln x(aR,e為自然對(duì)數(shù)的底數(shù)).(1)證明:當(dāng)x1時(shí),f(x)0;(2)討論g(x)的單調(diào)性.解:(1)證明:f(x),令s(x)ex1x,則s(x)ex11,當(dāng)x1時(shí),s(x)0,所以s(x)在(1,)上單調(diào)遞增,又s(1)0,所以s(x)0,從而當(dāng)x1時(shí),f(x)0.(2)g(x)2ax(x0),當(dāng)a0時(shí),g(x)0,g(x)在(0,)上單調(diào)遞減,當(dāng)a0時(shí),由g(x)0得x .當(dāng)x時(shí),g(x)0,g(x)單調(diào)遞減,當(dāng)x時(shí),g(x)0,g(x)單調(diào)遞增.12.已知函數(shù)f(x)asin xbcos x(a,bR),曲線yf(x)在點(diǎn)處的切線方程為yx.(1)求a,b的值
9、;(2)求函數(shù)g(x)在上的最小值.解:(1)由切線方程知,當(dāng)x時(shí),y0,所以fab0.因?yàn)閒(x)acos xbsin x.所以由切線方程知,fab1,所以a,b.(2)由(1)知,f(x)sin xcos xsin,所以函數(shù)g(x),g(x),設(shè)u(x)xcos xsin x,則u(x)xsin x0,故u(x)在上單調(diào)遞減,所以u(píng)(x)u(0)0,即g(x)0在上恒成立,所以g(x)在上單調(diào)遞減,所以函數(shù)g(x)在上的最小值為g.B組大題專攻強(qiáng)化練1.設(shè)f(x)xln xax2(2a1)x,aR.(1)令g(x)f(x),求g(x)的單調(diào)區(qū)間;(2)已知f(x)在x1處取得極大值,求實(shí)數(shù)
10、a的取值范圍.解:(1)由f(x)ln x2ax2a,可得g(x)ln x2ax2a,x(0,).則g(x)2a.當(dāng)a0時(shí),x(0,)時(shí),g(x)0,函數(shù)g(x)單調(diào)遞增;當(dāng)a0時(shí),x時(shí),g(x)0,函數(shù)g(x)單調(diào)遞增,x時(shí),g(x)0,函數(shù)g(x)單調(diào)遞減.所以當(dāng)a0時(shí),g(x)的單調(diào)增區(qū)間為(0,),當(dāng)a0時(shí),g(x)的單調(diào)增區(qū)間為,單調(diào)減區(qū)間為.(2)由(1)知,f(1)0.當(dāng)a0時(shí),f(x)單調(diào)遞增,所以當(dāng)x(0,1)時(shí),f(x)0,f(x)單調(diào)遞減;當(dāng)x(1,)時(shí),f(x)0,f(x)單調(diào)遞增.所以f(x)在x1處取得極小值,不符合題意.當(dāng)0a時(shí),1,由(1)知f(x)在內(nèi)單調(diào)遞增
11、,可得當(dāng)x(0,1)時(shí),f(x)0,x時(shí),f(x)0.所以f(x)在(0,1)內(nèi)單調(diào)遞減,在內(nèi)單調(diào)遞增,所以f(x)在x1處取得極小值,不符合題意.當(dāng)a時(shí),1,f(x)在(0,1)內(nèi)單調(diào)遞增,在(1,)內(nèi)單調(diào)遞減,所以當(dāng)x(0,)時(shí),f(x)0,f(x)單調(diào)遞減,不符合題意.當(dāng)a時(shí),01,當(dāng)x時(shí),f(x)0,f(x)單調(diào)遞增,當(dāng)x(1,)時(shí),f(x)0,f(x)單調(diào)遞減,所以f(x)在x1處取得極大值,符合題意.綜上可知,實(shí)數(shù)a的取值范圍為a.2.已知函數(shù)f(x)x2axln x(aR).(1)若函數(shù)f(x)是單調(diào)遞減函數(shù),求實(shí)數(shù)a的取值范圍;(2)若函數(shù)f(x)在區(qū)間(0,3)上既有極大值又
12、有極小值,求實(shí)數(shù)a的取值范圍.解:(1)f(x)2xa(x0),因?yàn)楹瘮?shù)f(x)是單調(diào)遞減函數(shù),所以f(x)0在(0,)恒成立,所以2x2ax10在(0,)恒成立,即a2x對(duì)(0,)恒成立,因?yàn)?x22,所以a2.(2)因?yàn)楹瘮?shù)f(x)在(0,3)上既有極大值又有極小值,所以f(x)0在(0,3)上有兩個(gè)相異實(shí)根,即2x2ax10在(0,3)上有兩個(gè)相異實(shí)根,令g(x)2x2ax1,則得即2a.所以實(shí)數(shù)a的取值范圍是.3.(2019全國(guó)卷)已知函數(shù)f(x)2x3ax22.(1)討論f(x)的單調(diào)性;(2)當(dāng)0a0,則當(dāng)x(,0)時(shí),f(x)0,當(dāng)x時(shí),f(x)0,故f(x)在(,0),單調(diào)遞增
13、,在單調(diào)遞減;若a0,f(x)在(,)單調(diào)遞增;若a0,當(dāng)x時(shí),f(x)0,故f(x)在,(0,)單調(diào)遞增,在單調(diào)遞減.(2)當(dāng)0a3時(shí),由(1)知,f(x)在單調(diào)遞減,在單調(diào)遞增,所以f(x)在0,1的最小值為f2,最大值為f(0)2或f(1)4a.于是m2,M所以Mm當(dāng)0a2時(shí),可知2a單調(diào)遞減,所以Mm的取值范圍是.當(dāng)2a3時(shí),單調(diào)遞增,所以Mm的取值范圍是.綜上,Mm的取值范圍是.4.已知常數(shù)a0,f(x)aln x2x.(1)當(dāng)a4時(shí),求f(x)的極值;(2)當(dāng)f(x)的最小值不小于a時(shí),求實(shí)數(shù)a的取值范圍.解:(1)由已知得f(x)的定義域?yàn)閤(0,),f(x)2.當(dāng)a4時(shí),f(x).當(dāng)0x2時(shí),f(x)2時(shí),f(x)0,即f(x)單調(diào)遞增.f(x)只有極小值,且在x2時(shí),f(x)取得極小值f(2)44ln 2.(2)f(x),當(dāng)a0,x(0,)時(shí),f(x)0,即f(x)在x(0,)上單調(diào)遞增,沒有最小值;當(dāng)a0得,x,f(x)在上單調(diào)遞增;由f(x)0得,x,f(x)在上單調(diào)遞減.當(dāng)a0時(shí),f(x)的最小值為faln2.根據(jù)題意得faln2a,即aln(a)ln 20.a0,ln(a)ln 20,解得a2,實(shí)數(shù)a的取值范圍是2,0).- 9 -