2022年高考數(shù)學(xué)一輪復(fù)習(xí)單元質(zhì)檢3導(dǎo)數(shù)及其應(yīng)用含解析新人教A版

1、PAGE PAGE 11單元質(zhì)檢三導(dǎo)數(shù)及其應(yīng)用(時間:100分鐘滿分:150分)一、選擇題(本大題共12小題,每小題5分,共60分)1.如果一個物體的運動方程為s=1-t+t2,其中s的單位是米,t的單位是秒,那么物體在3秒末的瞬時速度是()A.7米/秒B.6米/秒C.5米/秒D.8米/秒答案:C解析:根據(jù)瞬時速度的意義,可得3秒末的瞬時速度是v=s|t=3=(-1+2t)|t=3=5.2.設(shè)曲線y=x+1x-1在點(3,2)處的切線與直線ax+y+3=0垂直,則a等于()A.2B.-2C.12D.-12答案:B解析:因為y=x+1x-1的導(dǎo)數(shù)為y=-2(x-1)2,所以曲線在點(3,2)處的

2、切線斜率k=-12.又因為直線ax+y+3=0的斜率為-a,所以-a-12=-1,解得a=-2.3.若函數(shù)y=ex+mx有極值,則實數(shù)m的取值范圍是()A.m0B.m1D.m0,若y=ex+mx有極值,則必須使y的值有正有負(fù),故m0.4.已知函數(shù)f(x)=-x3+ax2-x-1在R上是減函數(shù),則實數(shù)a的取值范圍是()A.(-,-33,+)B.-3,3C.(-,-3)(3,+)D.(-3,3)答案:B解析:由題意,知f(x)=-3x2+2ax-10在R上恒成立,故=(2a)2-4(-3)(-1)0,解得-3a3.5.函數(shù)f(x)=x2+x-ln x的零點的個數(shù)是()A.0B.1C.2D.3答案:

3、A解析:由f(x)=2x+1-1x=2x2+x-1x=0,得x=12或x=-1(舍去).當(dāng)0 x12時,f(x)12時,f(x)0,f(x)單調(diào)遞增.則f(x)的最小值為f12=34+ln20,所以無零點.6.設(shè)函數(shù)f(x)=x3+(a-1)x2+ax,若f(x)為奇函數(shù),則曲線y=f(x)在點(0,0)處的切線方程為()A.y=-2xB.y=-xC.y=2xD.y=x答案:D解析:因為f(x)為奇函數(shù),所以f(-x)=-f(x),即-x3+(a-1)x2-ax=-x3-(a-1)x2-ax,解得a=1,則f(x)=x3+x.由f(x)=3x2+1,得在(0,0)處的切線斜率k=f(0)=1.

4、故切線方程為y=x.7.已知當(dāng)x12,2時,a1-xx+ln x恒成立,則a的最大值為()A.0B.1C.2D.3答案:A解析:令f(x)=1-xx+lnx,則f(x)=x-1x2.當(dāng)x12,1時,f(x)0.f(x)在區(qū)間12,1內(nèi)單調(diào)遞減,在區(qū)間(1,2上單調(diào)遞增,在x12,2上,f(x)min=f(1)=0,a0,即a的最大值為0.8.已知函數(shù)f(x)=ln x+tan 02的導(dǎo)函數(shù)為f(x),若方程f(x)=f(x)的根x0小于1,則的取值范圍為()A.4,2B.0,3C.6,4D.0,4答案:A解析:f(x)=lnx+tan,f(x)=1x.令f(x)=f(x),得lnx+tan=1

5、x,即tan=1x-lnx.設(shè)g(x)=1x-lnx,顯然g(x)在區(qū)間(0,+)內(nèi)單調(diào)遞減,而當(dāng)x0時,g(x)+,故要使?jié)M足f(x)=f(x)的根x0g(1)=1.又02,4,2.9.已知f(x),g(x)分別是定義在R上的奇函數(shù)和偶函數(shù),當(dāng)x0,且g(3)=0,則不等式f(x)g(x)0的解集是()A.(-3,0)(3,+)B.(-3,0)(0,3)C.(-,-3)(3,+)D.(-,-3)(0,3)答案:D解析:當(dāng)x0,即f(x)g(x)0,當(dāng)x0時,f(x)g(x)為增函數(shù),又g(x)是偶函數(shù),且g(3)=0,g(-3)=0,f(-3)g(-3)=0.故當(dāng)x-3時,f(x)g(x)0

6、時,f(x)g(x)為增函數(shù),且f(3)g(3)=0,故當(dāng)0 x3時,f(x)g(x)0,解得x344,令f(x)344,故f(x)在區(qū)間0,344內(nèi)遞增,在區(qū)間344,+內(nèi)遞減,故f(x)的最大值是f344,a=344.11.若函數(shù)f(x)=x33-a2x2+x+1在區(qū)間12,3內(nèi)有極值點,則實數(shù)a的取值范圍是()A.2,52B.2,52C.2,103D.2,103答案:C解析:若f(x)=x33-a2x2+x+1在區(qū)間12,3內(nèi)有極值點,則f(x)=x2-ax+1在區(qū)間12,3內(nèi)有零點,且零點不是f(x)的圖象頂點的橫坐標(biāo).由x2-ax+1=0,得a=x+1x.因為x12,3,y=x+1x

7、的值域是2,103,當(dāng)a=2時,f(x)=x2-2x+1=(x-1)2,不合題意.所以實數(shù)a的取值范圍是2,103,故選C.12.(2021全國,文12)設(shè)a0,若x=a為函數(shù)f(x)=a(x-a)2(x-b)的極大值點,則()A.abC.aba2答案:D解析:因為f(x)=a(x-a)2(x-b),所以f(x)=2a(x-a)(x-b)+a(x-a)2=a(x-a)(2x-2b)+(x-a)=a(x-a)3x-(a+2b)=3a(x-a)x-a+2b3.由f(x)=0,解得x=a或x=a+2b3.若a0,則由x=a為函數(shù)f(x)的極大值點,可得a+2b3a,化簡得ba.此時在區(qū)間-,a+2b

8、3和(a,+)內(nèi),f(x)0,函數(shù)f(x)單調(diào)遞增.此時a(a-b)0,即a20,則由x=a為函數(shù)的極大值點可得aa+2b3,化簡得a0,函數(shù)f(x)單調(diào)遞增;在區(qū)間a,a+2b3內(nèi),f(x)0,函數(shù)f(x)單調(diào)遞減.此時a(a-b)0,即a2ab.綜上可得a2ab.故選D.二、填空題(本大題共4小題,每小題5分,共20分)13.已知f(x),g(x)分別是定義在R上的偶函數(shù)和奇函數(shù),且f(x)-g(x)=ex+x2+1,則函數(shù)h(x)=2f(x)-g(x)在點(0,h(0)處的切線方程是.答案:x-y+4=0解析:f(x)-g(x)=ex+x2+1,f(x)+g(x)=e-x+x2+1.f(

9、x)=ex+e-x+2x2+22,g(x)=e-x-ex2.h(x)=2f(x)-g(x)=ex+e-x+2x2+2-e-x-ex2=32ex+12e-x+2x2+2.h(x)=32ex-12e-x+4x,即h(0)=32-12=1,又h(0)=4,所求的切線方程是x-y+4=0.14.已知函數(shù)f(x)=ax3+3x2-x+1在區(qū)間(-,+)內(nèi)是減函數(shù),則實數(shù)a的取值范圍是.答案:(-,-3解析:由題意可知f(x)=3ax2+6x-10在R上恒成立,則a0,=62+43a0,解得a-3.15.函數(shù)f(x)=e|x-1|,函數(shù)g(x)=ln x-x+a,若x1,x2使得f(x1)g(x2)成立,

10、則a的取值范圍是.答案:(2,+)解析:由題意,若x1,x2使得f(x1)g(x2)成立,可轉(zhuǎn)化為f(x)min0),當(dāng)x(0,1)時,g(x)0,則函數(shù)g(x)單調(diào)遞增;當(dāng)x(1,+)時,g(x)1,解得a2,即實數(shù)a的取值范圍是(2,+).16.已知函數(shù)f(x)=xln x+12x2,x0是函數(shù)f(x)的極值點,給出以下幾個命題:0 x01e;f(x0)+x00.其中真命題是.(填出所有真命題的序號)答案:解析:由已知得f(x)=lnx+x+1(x0),不妨令g(x)=lnx+x+1(x0),由g(x)=1x+1,當(dāng)x(0,+)時,有g(shù)(x)0總成立,所以g(x)在區(qū)間(0,+)內(nèi)單調(diào)遞增

11、,且g1e=1e0,又x0是函數(shù)f(x)的極值點,所以f(x0)=g(x0)=0,即g1eg(x0),所以0 x01e,即命題成立,則命題錯;因為lnx0+x0+1=0,所以f(x0)+x0=x0lnx0+12x02+x0=x0(lnx0+x0+1)-12x02=-12x020,所以f(x)=0有兩個不相等的實數(shù)根.所以不存在實數(shù)a,使得f(x)是在區(qū)間(-,+)內(nèi)的單調(diào)函數(shù).18.(12分)已知f(x)=x3-12x2-2x+5.(1)求f(x)的單調(diào)區(qū)間;(2)過點(0,a)可作y=f(x)的三條切線,求a的取值范圍.解:(1)f(x)=3x2-x-2=(3x+2)(x-1),故f(x)在

12、區(qū)間-,-23,(1,+)內(nèi)單調(diào)遞增,在區(qū)間-23,1內(nèi)單調(diào)遞減.(2)設(shè)切點為(x0,f(x0),則切線的方程為y-f(x0)=f(x0)(x-x0),即y-(x03-12x02-2x0+5)=(3x02-x0-2)(x-x0).又點(0,a)在切線上,故a-x03-12x02-2x0+5=(3x02-x0-2)(0-x0),即a=-2x03+12x02+5.令g(x)=-2x3+12x2+5,由已知得y=a的圖象與g(x)=-2x3+12x2+5的圖象有三個交點,g(x)=-6x2+x,令g(x)=0,得x1=0,x2=16,g(x1)=5,g(x2)=51216,故a的取值范圍為5,51

13、216.19.(12分)已知函數(shù)f(x)=ex-ax(a為常數(shù))的圖象與y軸交于點A,曲線y=f(x)在點A處的切線斜率為-1.(1)求a的值及函數(shù)f(x)的極值;(2)證明:當(dāng)x0時,x2ex.答案:(1)解由f(x)=ex-ax,得f(x)=ex-a.因為f(0)=1-a=-1,所以a=2.所以f(x)=ex-2x,f(x)=ex-2.令f(x)=0,得x=ln2.當(dāng)xln2時,f(x)ln2時,f(x)0,f(x)單調(diào)遞增,所以當(dāng)x=ln2時,f(x)取得極小值,極小值為f(ln2)=2-2ln2=2-ln4,f(x)無極大值.(2)證明令g(x)=ex-x2,則g(x)=ex-2x.由

14、(1),得g(x)=f(x)f(ln2)=2-ln40,故g(x)在R上單調(diào)遞增.因為g(0)=10,所以當(dāng)x0,g(x)g(0)0,即x2ex.20.(12分)已知函數(shù)f(x)=ln x-x.(1)判斷函數(shù)f(x)的單調(diào)性;(2)函數(shù)g(x)=f(x)+x+12x-m有兩個零點x1,x2,且x11.答案:(1)解函數(shù)f(x)的定義域為(0,+),f(x)=1x-1=1-xx.令f(x)0,解得0 x1;令f(x)1.故函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,1),單調(diào)遞減區(qū)間為(1,+).(2)證明根據(jù)題意得g(x)=lnx+12x-m(x0).因為x1,x2是函數(shù)g(x)=lnx+12x-m的兩

15、個零點,所以lnx1+12x1-m=0,lnx2+12x2-m=0.兩式相減,可得lnx1x2=12x2-12x1,即lnx1x2=x1-x22x2x1,故x1x2=x1-x22lnx1x2,因此x1=x1x2-12lnx1x2,x2=1-x2x12lnx1x2.令t=x1x2,其中0t1,則x1+x2=t-12lnt+1-1t2lnt=t-1t2lnt.構(gòu)造函數(shù)h(t)=t-1t-2lnt(0t1),則h(t)=(t-1)2t2.因為0t0恒成立,故h(t)h(1),即t-1t-2lnt1,故x1+x21.21.(12分)已知函數(shù)f(x)=(ex-e)ex+ax2,aR.(1)討論f(x)的

16、單調(diào)性;(2)若f(x)有兩個零點,求a的取值范圍.解:(1)由題意得f(x)=x(ex+1+2a),當(dāng)a0時,ex+1+2a0,若x(-,0),則f(x)=x(ex+1+2a)0,函數(shù)f(x)單調(diào)遞增;當(dāng)-e2a0,函數(shù)f(x)單調(diào)遞增,若x(ln(-2a)-1,0),則f(x)=x(ex+1+2a)0,函數(shù)f(x)單調(diào)遞增;當(dāng)a=-e2時,f(x)=x(ex+1+2a)0恒成立,函數(shù)f(x)單調(diào)遞增;當(dāng)a0,函數(shù)f(x)單調(diào)遞增,若x(0,ln(-2a)-1),則f(x)=x(ex+1+2a)0,函數(shù)f(x)單調(diào)遞增.(2)當(dāng)a=0時,f(x)=(ex-e)ex有唯一零點x=1,不符合題意

17、;由(1)知,當(dāng)a0時,若x(-,0),則函數(shù)f(x)單調(diào)遞減,若x(0,+),則函數(shù)f(x)單調(diào)遞增,當(dāng)x-時,f(x)+;當(dāng)x+時,f(x)+,f(0)=-e0必有兩個零點;當(dāng)-e2a0時,若x(-,ln(-2a)-1),則函數(shù)f(x)單調(diào)遞增,若x(ln(-2a)-1,0),則函數(shù)f(x)單調(diào)遞減,若x(0,+),則函數(shù)f(x)單調(diào)遞增,f(ln(-2a)-1)=-2a(ln(-2a)-1)+a(ln(-2a)-1)20,f(0)=-e0,函數(shù)f(x)至多有一個零點;當(dāng)a=-e2時,函數(shù)f(x)單調(diào)遞增,函數(shù)f(x)至多有一個零點;當(dāng)a-e2時,若x(-,0),則函數(shù)f(x)單調(diào)遞增,若

18、x(0,ln(-2a)-1),則函數(shù)f(x)單調(diào)遞減,若x(ln(-2a)-1,+),則函數(shù)f(x)單調(diào)遞增,f(0)=-e0時,函數(shù)f(x)有兩個零點.22.(12分)已知1a2,函數(shù)f(x)=ex-x-a,其中e=2.718 28是自然對數(shù)的底數(shù).(1)證明:函數(shù)y=f(x)在(0,+)上有唯一零點;(2)記x0為函數(shù)y=f(x)在(0,+)上的零點,證明:a-1x02(a-1);x0f(ex0)(e-1)(a-1)a.答案:證明(1)因為f(0)=1-a0,所以y=f(x)在(0,+)上存在零點.因為f(x)=ex-1,所以當(dāng)x0時,f(x)0,故函數(shù)f(x)在0,+)上單調(diào)遞增,所以函數(shù)y=f(x)在(0,+)上有唯一零點.(2)令g(x)=ex-12x2-x-1(x0),g(x)=ex-x-1=f(x)+a-1,由知函數(shù)g(x)在0,+)上單調(diào)遞增,故當(dāng)x0時,g(x)g(0)=0,所以函數(shù)g(x)在0,+)上單調(diào)遞增,故g(x)g(0)=0.由g(2(a-1)0,得f(2(a-1)=e2(a-1)-2(a-1)-a0