高考數(shù)學(xué)（人教A版）一輪復(fù)習(xí)單元質(zhì)檢三導(dǎo)數(shù)及其應(yīng)用

單元質(zhì)檢三導(dǎo)數(shù)及其應(yīng)用(時(shí)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(gè)物體的運(yùn)動(dòng)方程為s=1t+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(shí)速度是()A.7米/秒 B.6米/秒 C.5米/秒 D2.設(shè)曲線y=在點(diǎn)(3,2)處的切線與直線ax+y+3=0垂直,則a等于()A.2 B.-2 C. D3.若函數(shù)y=ex+mx有極值,則實(shí)數(shù)m的取值范圍是()A.m>0 B.m<0 C.m>1 D.m<14.已知函數(shù)f(x)=x3+ax2x1在R上是減函數(shù),則實(shí)數(shù)a的取值范圍是()A.(∞,]∪[,+∞)B.[]C.(∞,)∪(,+∞)D.()5.函數(shù)f(x)=x2+xlnx的零點(diǎn)的個(gè)數(shù)是()A.0 B.1 C.2 D.6.若f(x)=aexex為奇函數(shù),則f(x1)<e的解集為()A.(∞,0) B.(∞,2) C.(2,+∞) D.(0,+∞)7.已知當(dāng)x∈時(shí),a≤+lnx恒成立,則a的最大值為()A.0 B.1 C.2 D.8.已知函數(shù)f(x)=lnx+tanα的導(dǎo)函數(shù)為f'(x),若方程f'(x)=f(x)的根x0小于1,則α的取值范圍為()A. B.C. D.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)x<0時(shí),f'(x)g(x)+f(x)g'(x)>0,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)10.(2017遼寧撫順重點(diǎn)校一模)已知函數(shù)f(x)=x2的最大值為f(a),則a等于()A. B. C. D.11.若函數(shù)f(x)=x2+x+1在區(qū)間內(nèi)有極值點(diǎn),則實(shí)數(shù)a的取值范圍是()A. B. C. D.12.(2017河北唐山三模)已知函數(shù)f(x)=x3+ax2+bx有兩個(gè)極值點(diǎn)x1,x2,且x1<x2,若x1+2x0=3x2,函數(shù)g(x)=f(x)f(x0),則g(x)()A.恰有一個(gè)零點(diǎn) B.恰有兩個(gè)零點(diǎn)C.恰有三個(gè)零點(diǎn) D.至多兩個(gè)零點(diǎn)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(diǎn)(0,h(0))處的切線方程是.14.已知函數(shù)f(x)=ax3+3x2x+1在區(qū)間(∞,+∞)內(nèi)是減函數(shù),則實(shí)數(shù)a的取值范圍是.

15.已知f(x)=x36x2+9xabc,a<b<c,且f(a)=f(b)=f(c)=0,現(xiàn)給出如下結(jié)論:①f(0)f(1)<0;②f(0)f(1)>0;③f(0)f(3)>0;④f(0)f(3)<0;⑤f(1)f(3)>0;⑥f(1)f(3)<0.其中正確的結(jié)論是.(填序號(hào))

16.(2017福建龍巖一模)若實(shí)數(shù)a,b,c,d滿足=1,則(ac)2+(bd)2的最小值為.

三、解答題(本大題共6小題,共70分)17.(10分)設(shè)函數(shù)f(x)=6x3+3(a+2)x2+2ax.(1)若f(x)的兩個(gè)極值點(diǎn)為x1,x2,且x1x2=1,求實(shí)數(shù)a的值;(2)是否存在實(shí)數(shù)a,使得f(x)是(∞,+∞)內(nèi)的單調(diào)函數(shù)?若存在,求出a的值;若不存在,說(shuō)明理由.18.(12分)已知f(x)=x3x22x+5.(1)求f(x)的單調(diào)區(qū)間;(2)過(guò)點(diǎn)(0,a)可作y=f(x)的三條切線,求a的取值范圍.19.(12分)已知函數(shù)f(x)=exax(a為常數(shù))的圖象與y軸交于點(diǎn)A,曲線y=f(x)在點(diǎn)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)x>0時(shí),x2<ex.20.(12分)已知函數(shù)f(x)=lnxx.(1)判斷函數(shù)f(x)的單調(diào)性;(2)函數(shù)g(x)=f(x)+x+m有兩個(gè)零點(diǎn)x1,x2,且x1<x2.求證:x1+x2>1.21.(12分)已知函數(shù)f(x)=exx2+a,x∈R的圖象在x=0處的切線方程為y=bx.(e≈2.71828)(1)求函數(shù)f(x)的解析式;(2)當(dāng)x∈R時(shí),求證:f(x)≥x2+x;(3)若f(x)>kx對(duì)任意的x∈(0,+∞)恒成立,求實(shí)數(shù)k的取值范圍.22.(12分)(2017江蘇,20)已知函數(shù)f(x)=x3+ax2+bx+1(a>0,b∈R)有極值,且導(dǎo)函數(shù)f'(x)的極值點(diǎn)是f(x)的零點(diǎn).(極值點(diǎn)是指函數(shù)取極值時(shí)對(duì)應(yīng)的自變量的值)(1)求b關(guān)于a的函數(shù)關(guān)系式,并寫出定義域;(2)證明:b2>3a(3)若f(x),f'(x)這兩個(gè)函數(shù)的所有極值之和不小于,求a的取值范圍.答案：1.C解析:根據(jù)瞬時(shí)速度的意義,可得3秒末的瞬時(shí)速度是v=s'|t=3=(1+2t)|t=3=5.2.B解析:因?yàn)閥=的導(dǎo)數(shù)為y'=,所以曲線在點(diǎn)(3,2)處的切線斜率k=.又因?yàn)橹本€ax+y+3=0的斜率為a,所以a·=1,解得a=2.3.B解析:求導(dǎo)得y'=ex+m,由于ex>0,若y=ex+mx有極值,則必須使y'的值有正有負(fù),故m<0.4.B解析:由題意,知f'(x)=3x2+2ax1≤0在R上恒成立,故Δ=(2a)24×(3)×(1)≤0,解得≤a≤5.A解析:由f'(x)=2x+1=0,得x=或x=1(舍去).當(dāng)0<x<時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x>時(shí),f'(x)>0,f(x)單調(diào)遞增.則f(x)的最小值為f+ln2>0,所以無(wú)零點(diǎn).6.D解析:∵f(x)在R上為奇函數(shù),∴f(0)=0,即a1=0.∴a=1.∴f(x)=exex,∴f'(x)=exex<0.∴f(x)在R上單調(diào)遞減.∴由f(x1)<e=f(1),得x1>1,即x>0.∴f(x1)<e的解集為(0,+∞).7.A解析:令f(x)=+lnx,則f'(x)=.當(dāng)x∈時(shí),f'(x)<0;當(dāng)x∈(1,2]時(shí),f'(x)>0.∴f(x)在區(qū)間內(nèi)單調(diào)遞減,在區(qū)間(1,2]上單調(diào)遞增,∴在x∈上,f(x)min=f(1)=0,∴a≤0,即a的最大值為0.8.A解析:∵f(x)=lnx+tanα,∴f'(x)=.令f(x)=f'(x),得lnx+tanα=,即tanα=lnx.設(shè)g(x)=lnx,顯然g(x)在(0,+∞)內(nèi)單調(diào)遞減,而當(dāng)x→0時(shí),g(x)→+∞,故要使?jié)M足f'(x)=f(x)的根x0<1,只需tanα>g(1)=1.又0<α<,∴α∈.9.D解析:∵當(dāng)x<0時(shí),f'(x)g(x)+f(x)g'(x)>0,即[f(x)g(x)]'>0,∴當(dāng)x<0時(shí),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(shí),f(x)g(x)<0;∵f(x)g(x)是奇函數(shù),∴當(dāng)x>0時(shí),f(x)g(x)為增函數(shù),且f(3)g(3)=0,故當(dāng)0<x<3時(shí),f(x)g(x)<0.故選D.10.B解析:∵f'(x)=2x,∴f'(1)=f'(1)2,解得f'(1)=,∴f(x)=x2,f'(x)=,令f'(x)>0,解得x<,令f'(x)<0,解得x>,故f(x)在內(nèi)遞增,在內(nèi)遞減,故f(x)的最大值是f,a=.11.C解析:若f(x)=x2+x+1在區(qū)間內(nèi)有極值點(diǎn),則f'(x)=x2ax+1在區(qū)間內(nèi)有零點(diǎn),且零點(diǎn)不是f'(x)的圖象頂點(diǎn)的橫坐標(biāo).由x2ax+1=0,得a=x+.因?yàn)閤∈,y=x+的值域是,當(dāng)a=2時(shí),f'(x)=x22x+1=(x1)2,不合題意.所以實(shí)數(shù)a的取值范圍是,故選C.12.B解析:由已知g(x)=f(x)f(x0)=x3+ax2+bx(+a+bx0)=(xx0)[x2+(x0+a)x++ax0+b],∵f'(x)=3x2+2ax+b,∴代入上式可得g(x)=(xx0)(xx1)2,所以g(x)恰有兩個(gè)零點(diǎn).13.xy+4=0解析:∵f(x)g(x)=ex+x2+1,∴f(x)+g(x)=ex+x2+1.∴f(x)=,g(x)=.∴h(x)=2f(x)g(x)=ex+ex+2x2+=ex+ex+2x2+2.∴h'(x)=exex+4x,即h'(0)==1,又h(0)=4,∴所求的切線方程是xy+4=0.14.(∞,3]解析:由題意可知f'(x)=3ax2+6x1≤0在R上恒成立,則解得a≤3.15.①③⑥解析:∵f(x)=x36x2+9xabc,∴f'(x)=3x212x+9=3(x1)(x3).∴當(dāng)1<x<3時(shí),f'(x)<0;當(dāng)x<1或x>3時(shí),f'(x)>0.∴f(x)的單調(diào)遞增區(qū)間為(∞,1)和(3,+∞),單調(diào)遞減區(qū)間為(1,3).∴f(x)極大值=f(1)=16+9abc=4abc,f(x)極小值=f(3)=2754+27abc=abc.∵f(x)=0有三個(gè)解a,b,c,∴a<1<b<3<c,∴f(1)=4abc>0,且f(3)=abc<0.∴0<abc<4.∵f(0)=abc,∴f(0)<0,∴f(0)f(1)<0,f(0)f(3)>0,f(1)f(3)<0.16.解析:由已知條件,得b=lna+2a2,d=3c2,令f(x)=lnx+2x2,g(x)=3x2,所求最小值可轉(zhuǎn)化為兩個(gè)函數(shù)f(x)與g(x)圖象上的點(diǎn)之間的距離的最小值的平方,f'(x)=+4x,設(shè)與直線y=3x2平行且與曲線f(x)相切的切點(diǎn)為P(x0,y則+4x0=3,x0>0,解得x0=1,可得切點(diǎn)P(1,2),切點(diǎn)P(1,2)到直線y=3x2的距離d=.所以(ac)2+(bd)2的最小值為d2=.故答案為.17.解:(1)由題意知f'(x)=18x2+6(a+2)x+2a因?yàn)閤1,x2是f(x)的兩個(gè)極值點(diǎn),所以f'(x1)=f'(x2)=0.所以x1x2==1,所以a=9.(2)因?yàn)棣?36(a+2)24×18×2a=36(a2+4)>所以f'(x)=0有兩個(gè)不相等的實(shí)數(shù)根.所以不存在實(shí)數(shù)a,使得f(x)是(∞,+∞)內(nèi)的單調(diào)函數(shù).18.解:(1)f'(x)=3x2x2=(3x+2)(x1),故f(x)在,(1,+∞)內(nèi)單調(diào)遞增,在內(nèi)單調(diào)遞減.(2)設(shè)切點(diǎn)為(x0,f(x0)),則切線的方程為yf(x0)=f'(x0)(xx0),即y(2x0+5)=(3x02)(xx0).又點(diǎn)(0,a)在切線上,故a=(3x02)(0x0),即a=2+5.令g(x)=2x3+x2+5,由已知得y=a的圖象與g(x)=2x3+x2+5的圖象有三個(gè)交點(diǎn),g'(x)=6x2+x,令g'(x)=0,得x1=0,x2=,g(x1)=5,g(x2)=5,故a的取值范圍為.19.(1)解:由f(x)=exax,得f'(x)=exa.因?yàn)閒'(0)=1a=1,所以a=2.所以f(x)=ex2x,f'(x)=ex2.令f'(x)=0,得x=ln2.當(dāng)x<ln2時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x>ln2時(shí),f'(x)>0,f(x)單調(diào)遞增,所以當(dāng)x=ln2時(shí),f(x)取得極小值,極小值為f(ln2)=22ln2=2ln4,f(x)無(wú)極大值.(2)證明:令g(x)=exx2,則g'(x)=ex2x.由(1),得g'(x)=f(x)≥f(ln2)=2ln4>0,故g(x)在R上單調(diào)遞增.因?yàn)間(0)=1>0,所以當(dāng)x>0,g(x)>g(0)>0,即x2<ex.20.(1)解:函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=1=.令f'(x)>0,解得0<x<1;令f'(x)<0,解得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+m(x>0).因?yàn)閤1,x2是函數(shù)g(x)=lnx+m的兩個(gè)零點(diǎn),所以lnx1+m=0,lnx2+m=0.兩式相減,可得ln,即ln,故x1x2=,因此x1=,x2=.令t=,其中0<t<1,則x1+x2=.構(gòu)造函數(shù)h(t)=t2lnt(0<t<1),則h'(t)=.因?yàn)?<t<1,所以h'(t)>0恒成立,故h(t)<h(1),即t2lnt<0,可知>1,故x1+x2>1.21.(1)解:∵f(x)=exx2+a,∴f'(x)=ex2x.由已知,得解得∴函數(shù)f(x)的解析式為f(x)=exx21.(2)證明:令φ(x)=f(x)+x2x=exx1,則φ'(x)=ex1.由φ'(x)=0,得x=0.當(dāng)x∈(∞,0)時(shí),φ'(x)<0,φ(x)單調(diào)遞減;當(dāng)x∈(0,+∞)時(shí),φ'(x)>0,φ(x)單調(diào)遞增.故φ(x)min=φ(0)=0,從而f(x)≥x2+x.(3)解:f(x)>kx對(duì)任意的x∈(0,+∞)恒成立?>k對(duì)任意的x∈(0,+∞)恒成立.令g(x)=,x>0,則g'(x)==.由(2)可知當(dāng)x∈(0,+∞)時(shí),exx1>0恒成立,由g'(x)>0,得x>1;由g'(x)<0,得0<x<1.故g(x)的遞增區(qū)間為(1,+∞),遞減區(qū)間為(0,1),即g(x)min=g(1)=e2.故k<g(x)min=g(1)=e2,即實(shí)數(shù)k的取值范圍為(∞,e2).22.(1)解:由f(x