2019-2020年高考數(shù)學(xué)二輪復(fù)習(xí) 專題能力訓(xùn)練6 導(dǎo)數(shù)的簡(jiǎn)單應(yīng)用 文一、選擇題1.(xx河南鄭州第一次質(zhì)量預(yù)測(cè))已知曲線y=-3ln x的一條切線的斜率為2,則切點(diǎn)的橫坐標(biāo)為()A.3B.2C.1D.2.(xx四川雅安三診)設(shè)函數(shù)f(x)在R上可導(dǎo),其導(dǎo)函數(shù)為f(x),且函數(shù)y=(1-x)f(x)的圖象如圖所示,則下列結(jié)論中一定成立的是()A.函數(shù)f(x)有極大值f(2)和極小值f(1)B.函數(shù)f(x)有極大值f(-2)和極小值f(1)C.函數(shù)f(x)有極大值f(2)和極小值f(-2)D.函數(shù)f(x)有極大值f(-2)和極小值f(2)3.(xx廣東深圳第一次調(diào)研)若函數(shù)f(x)=x3+x2-ax在區(qū)間(1,+)上單調(diào)遞增,且在區(qū)間(1,2)上有零點(diǎn),則實(shí)數(shù)a的取值范圍是()A.B.C.D.(-,34.已知函數(shù)f(x)=-x3+ax2-4在x=2處取得極值,若m,n-1,1,則f(m)+f(n)的最小值是()A.-13B.-15C.10D.155.(xx四川成都三診改編)設(shè)定義在R上的函數(shù)f(x)滿足f(1)=1,f(x)>,其中f(x)是f(x)的導(dǎo)函數(shù),則不等式f(x3)<x3+的解集為()A.(-,1)B.(-1,1)C.(-1,+)D.(0,1)二、填空題6.若曲線y=ax2-ln x在點(diǎn)(1,a)處的切線平行于x軸,則a=.7.設(shè)函數(shù)f(x)在(0,+)內(nèi)可導(dǎo),且f(ex)=x+ex,則f(1)=.8.函數(shù)f(x)=x3-x2-3x-1的圖象與x軸的交點(diǎn)個(gè)數(shù)是.三、解答題9.設(shè)f(x)=ax2-6ln x,其中aR,曲線y=f(x)在點(diǎn)(1,f(1)處的切線與y軸相交于點(diǎn)(0,3).(1)確定a的值;(2)求函數(shù)f(x)的單調(diào)區(qū)間與極值.10.已知函數(shù)f(x)=x3+fx2-x+c.(1)求f的值;(2)求函數(shù)f(x)的單調(diào)區(qū)間;(3)設(shè)函數(shù)g(x)=f(x)-x3ex,若函數(shù)g(x)在區(qū)間-3,2上單調(diào)遞增,求實(shí)數(shù)c的取值范圍.11.設(shè)函數(shù)f(x)=aex+b(a>0).(1)求f(x)在0,+)內(nèi)的最小值;(2)設(shè)曲線y=f(x)在點(diǎn)(2,f(2)處的切線方程為y=x,求a,b的值.答案與解析專題能力訓(xùn)練6導(dǎo)數(shù)的簡(jiǎn)單應(yīng)用1.A解析:設(shè)切點(diǎn)坐標(biāo)為(x0,y0),且x0>0,由y=x-,得k=x0-=2,解得x0=3.2.D解析:當(dāng)x<-2時(shí),1-x>0,由圖知f(x)>0;當(dāng)-2<x<1時(shí),1-x>0,由圖知f(x)<0;當(dāng)1<x<2時(shí),1-x<0,由圖知f(x)<0;當(dāng)x>2時(shí),1-x<0,由圖知f(x)>0.綜上可知函數(shù)f(x)的極大值為f(-2),極小值為f(2).故選D.3.C解析:f(x)=x2+2x-a,且函數(shù)f(x)在(1,+)上單調(diào)遞增,x2+2x-a0在(1,+)上恒成立,即ax2+2x,x(1,+).設(shè)g(x)=x2+2x,x(1,+),則g(x)min=g(1)=3.故a3.又函數(shù)f(x)在區(qū)間(1,2)上有零點(diǎn),f(1)f(2)<0,即<0.<a<.由可知<a3.故選C.4.A解析:求導(dǎo)得f(x)=-3x2+2ax,由函數(shù)f(x)在x=2處取得極值知f(2)=0,即-34+2a2=0,可得a=3.故f(x)=-x3+3x2-4,f(x)=-3x2+6x.由此可得函數(shù)f(x)在區(qū)間(-1,0)上單調(diào)遞減,在區(qū)間(0,1)上單調(diào)遞增.故當(dāng)m-1,1時(shí),f(m)min=f(0)=-4.f(x)=-3x2+6x的圖象開口向下,且對(duì)稱軸為x=1,當(dāng)n-1,1時(shí),f(n)min=f(-1)=-9.因此f(m)+f(n)的最小值為-13.5.A6.解析:y=2ax-,y|x=1=2a-1=0.a=.7.2解析:令t=ex,則x=ln t,所以函數(shù)為f(t)=ln t+t,即f(x)=ln x+x.所以f(x)=+1.故f(1)=+1=2.8.3解析:f(x)=x2-2x-3=(x+1)(x-3),函數(shù)f(x)在(-,-1)和(3,+)上是增函數(shù),在區(qū)間(-1,3)上是減函數(shù),由f(x)極小值=f(3)=-10<0,f(x)極大值=f(-1)=>0知函數(shù)f(x)的圖象與x軸的交點(diǎn)個(gè)數(shù)為3.9.解:(1)f(x)=2ax-(x>0),則f(1)=2a-6.又f(1)=a,所以f(x)在(1,f(1)處的切線方程為y-a=(2a-6)(x-1).令x=0,得y=6-a.由題可知6-a=3,即a=3.(2)由(1)知f(x)=(x>0).令f(x)>0,則x>1,令f(x)<0,則0<x<1.所以f(x)遞增區(qū)間為(1,+),遞減區(qū)間為(0,1).故f(x)有極小值f(1)=3,無(wú)極大值.10.解:(1)設(shè)f=a,則f(x)=x3+ax2-x+c,f(x)=3x2+2ax-1.于是f=3+2a-1=-1=.故有ff,解得f=-1.(2)由(1)知f(x)=x3-x2-x+c,則f(x)=3x2-2x-1=(3x+1)(x-1).當(dāng)x變化時(shí),f(x),f(x)的變化情況如下表:x-1(1,+)f(x)+0-0+f(x)有極大值有極小值所以函數(shù)f(x)的單調(diào)遞增區(qū)間是,(1,+);單調(diào)遞減區(qū)間是.(3)由于g(x)=f(x)-x3ex=(-x2-x+c)ex,則g(x)=(-x2-3x+c-1)ex.若函數(shù)g(x)在區(qū)間-3,2上單調(diào)遞增,則g(x)0在區(qū)間-3,2上恒成立,即g(x)=(-x2-3x+c-1)ex0在區(qū)間-3,2上恒成立,ex>0,-x2-3x+c-10在區(qū)間-3,2上恒成立.令h(x)=-x2-3x+c-1,則解得c11.故函數(shù)g(x)在區(qū)間-3,2上單調(diào)遞增時(shí)c的取值范圍為c11.11.解:(1)f(x)=aex-,當(dāng)f(x)>0,即x>-ln a時(shí),f(x)在(-ln a,+)上遞增;當(dāng)f(x)<0,即x<-ln a時(shí),f(x)在(-,-ln a)上遞減.當(dāng)0<a<1時(shí),-ln a>0,f(x)在(0,-ln a)上遞減,在(-ln a,+)上遞增,從而f(x)在0,+)內(nèi)的最小值為f(-ln a)=2+b;當(dāng)a1時(shí),-ln a0,f(x)在0,+)上遞增,從而f(x)在0,+)內(nèi)的最小值為f(0)=a+b.(2)依題意知f(2)=ae2-,解得ae2=2或ae2=-(舍去).所以a=,代入原函數(shù)可得2+b=3,即b=.故a=,b=.