(福建專(zhuān)版)2019高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)規(guī)范練15導(dǎo)數(shù)與函數(shù)的小綜合文

(福建專(zhuān)版)2019高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)規(guī)范練15導(dǎo)數(shù)與函數(shù)的小綜合文LtDPage4。。內(nèi)部文件，版權(quán)追溯內(nèi)部文件，版權(quán)追溯課時(shí)規(guī)范練15導(dǎo)數(shù)與函數(shù)的小綜合基礎(chǔ)鞏固組1.函數(shù)f(x)=(x-3)ex的單調(diào)遞增區(qū)間是()A.(-∞,2) B.(0,3)C.(1,4) D.(2,+∞)2.(2017山東煙臺(tái)一模,文9)已知函數(shù)f(x)=ax3+bx2+cx+d的圖象如圖所示,則下列結(jié)論成立的是()A.a>0,b>0,c>0,d<0017,設(shè)a=f(20.5),b=f(log43),c=f(logπ3),則a,b,c的大小關(guān)系是.

10.設(shè)函數(shù)f'(x)是奇函數(shù)f(x)(x∈R)的導(dǎo)函數(shù),f(-1)=0,當(dāng)x>0時(shí),xf'(x)-f(x)<0,則使得f(x)>0成立的x的取值范圍是.

11.(2017山東泰安一模,文14)已知函數(shù)f(x)是定義在R上的奇函數(shù),若g(x)=f(x+1)+5,g'(x)為g(x)的導(dǎo)函數(shù),對(duì)?x∈R,總有g(shù)'(x)>2x,則g(x)<x2+4的解集為.

綜合提升組12.(2017廣西南寧一模)已知函數(shù)f(x)=-x2-6x-3,g(x)=2x3+3x2-12x+9,m<-2,若?x1∈[m,-2),?x2∈(0,+∞),使得f(x1)=g(x2)成立,則m的最小值為()A.-5 B.-4 C.-25 D.-313.定義在(0,+∞)內(nèi)的函數(shù)f(x)滿(mǎn)足f(x)>0,且對(duì)?x∈(0,+∞),2f(x)<xf'(x)<3f(x)恒成立,其中f'(x)為f(x)的導(dǎo)函數(shù),則()A.116B.1C.14D.13<f(1)14.(2017河北邯鄲二模,文16)若函數(shù)f(x)=(x2-ax+a+1)ex(a∈N)在區(qū)間(1,3)內(nèi)只有1個(gè)極值點(diǎn),則曲線f(x)在點(diǎn)(0,f(0))處切線的方程為.

創(chuàng)新應(yīng)用組15.(2017安徽淮南一模,文12)如果定義在R上的函數(shù)f(x)滿(mǎn)足:對(duì)于任意x1≠x2,都有x1f(x1)+x2f(x2)≥x1f(x2)+x2f(x1),則稱(chēng)f(x)為“H函數(shù)”.給出下列函數(shù):①y=-x3+x+1;②y=3x-2(sinx-cosx);③y=1-ex;④f(x)=ln其中“H函數(shù)”為()A.3 B.2 C.1 D.0?導(dǎo)學(xué)號(hào)24190734?16.(2017安徽合肥一模,文16)已知函數(shù)f(x)=-x3+3x2-ax-2a,若存在唯一的正整數(shù)x0,使得f(x0)>0,則a的取值范圍是.

答案：1.D函數(shù)f(x)=(x-3)ex的導(dǎo)數(shù)為f'(x)=[(x-3)ex]'=ex+(x-3)ex=(x-2)ex.由導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系,得當(dāng)f'(x)>0時(shí),函數(shù)f(x)單調(diào)遞增,此時(shí)由不等式f'(x)=(x-2)ex>0,解得x>2.2.C由題圖可知f(0)=d>0,排除選項(xiàng)A,B;∵f'(x)=3ax2+2bx+c,且由題圖知(-∞,x1),(x2,+∞)是函數(shù)的遞減區(qū)間,可知a<0,排除D.故選C.3.B因?yàn)楹瘮?shù)f(x)=x3-3x2+x的極大值點(diǎn)為m,極小值點(diǎn)為n,所以m,n為f'(x)=3x2-6x+1=0的兩根.由根與系數(shù)的關(guān)系可知m+n=-(-6)3=24.C設(shè)g(x)=f(x)ex,則g'(∵f(x)<f'(x),∴g'(x)>0,即函數(shù)g(x)在定義域內(nèi)單調(diào)遞增.∵f(0)=2,∴g(0)=f(0)=2,∴不等式f(x)>2ex等價(jià)于g(x)>g(0).∵函數(shù)g(x)在定義域內(nèi)單調(diào)遞增,∴x>0,∴不等式的解集為(0,+∞),故選C.5.B函數(shù)f(x)=exx的定義域?yàn)閤≠0,x∈R,當(dāng)x>0時(shí),函數(shù)f'(x)=xex-exx2,可得函數(shù)的極值點(diǎn)為x=1,當(dāng)x∈(0,1)時(shí),函數(shù)是減函數(shù),當(dāng)x>1時(shí),函數(shù)是增函數(shù),并且f(x)>0,選項(xiàng)B,D滿(mǎn)足題意.當(dāng)x<0時(shí),函數(shù)f(x)=ex6.B∵xf'(x)+2f(x)=1x∴x2f'(x)+2xf(x)=1x令g(x)=x2f(x),則g'(x)=2xf(x)+x2f'(x)=1x>0,∴函數(shù)g(x)在(0,+∞)內(nèi)單調(diào)遞增∴g(2)=4f(2)<g(e)=e2f(e)<g(3)=9f(3),∴f(2)9<f7.B∵f(x)=x(lnx-ax),∴f'(x)=lnx-2ax+1,由題意可知f'(x)在(0,+∞)內(nèi)有兩個(gè)不同的零點(diǎn),令f'(x)=0,得2a=lnx+1x,設(shè)g(x)=lnx+1x,則g'(x)=-lnxx2,∴g(x)∵當(dāng)x→0時(shí),g(x)→-∞,當(dāng)x→+∞時(shí),g(x)→0,而g(x)max=g(1)=1,∴只需0<2a<1,即0<a<128.(0,1)∪(2,3)由題意知f'(x)=-x+4-3x=-由f'(x)=0得x1=1,x2=3,可知1,3是函數(shù)f(x)的兩個(gè)極值點(diǎn).則只要這兩個(gè)極值點(diǎn)有一個(gè)在區(qū)間(t,t+1)內(nèi),函數(shù)f(x)在區(qū)間[t,t+1]上就不單調(diào),由t<1<t+1或t<3<t+1,得0<t<1或2<t<3.9.a>c>b∵方程f'(x)=0無(wú)解,∴f'(x)>0或f'(x)<0恒成立,∴f(x)是單調(diào)函數(shù);由題意得?x∈(0,+∞),f(f(x)-log2015x)=2017,且f(x)是定義在(0,+∞)的單調(diào)函數(shù),則f(x)-log2015x是定值.設(shè)t=f(x)-log2015x,則f(x)=t+log2015x,∴f(x)是增函數(shù).又0<log43<logπ3<1<20.5,∴a>c>b.故答案為a>c>b.10.(-∞,-1)∪(0,1)當(dāng)x>0時(shí),令F(x)=f(則F'(x)=xf'(x∴當(dāng)x>0時(shí),F(x)=f(x∵f(x)為奇函數(shù),且由f(-1)=0,得f(1)=0,故F(1)=0.在區(qū)間(0,1)內(nèi),F(x)>0;在(1,+∞)內(nèi),F(x)<0,即當(dāng)0<x<1時(shí),f(x)>0;當(dāng)x>1時(shí),f(x)<0.又f(x)為奇函數(shù),∴當(dāng)x∈(-∞,-1)時(shí),f(x)>0;當(dāng)x∈(-1,0)時(shí),f(x)<0.綜上可知,f(x)>0的解集為(-∞,-1)∪(0,1).11.(-∞,-1)∵f(x)是定義在R上的奇函數(shù),∴f(x)的圖象過(guò)原點(diǎn),∵g(x)=f(x+1)+5,∴g(x)的圖象過(guò)點(diǎn)(-1,5).令h(x)=g(x)-x2-4,∴h'(x)=g'(x)-2x.∵對(duì)?x∈R,總有g(shù)'(x)>2x,∴h(x)在R上是增函數(shù),又h(-1)=g(-1)-1-4=0,∴g(x)<x2+4的解集為(-∞,-1).12.A∵g(x)=2x3+3x2-12x+9,∴g'(x)=6x2+6x-12=6(x+2)(x-1),則當(dāng)0<x<1時(shí),g'(x)<0,函數(shù)g(x)遞減,當(dāng)x>1時(shí),g'(x)>0,函數(shù)g(x)遞增,∴當(dāng)x>0時(shí),g(x)min=g(1)=2.∵f(x)=-x2-6x-3=-(x+3)2+6≤6,作函數(shù)y=(x)的圖象,如圖所示,當(dāng)f(x)=2時(shí),方程兩根分別為-5和-1,則m的最小值為-5,故選A.13.B令g(x)=f(x)x2,則g'(x)=xf'(∵?x∈(0,+∞),2f(x)<xf'(x)<3f(x)恒成立,∴0<xf'(∴g'(x)>0,∴函數(shù)g(x)在(0,+∞)內(nèi)單調(diào)遞增,∴f(1)1<f(2)4,又∴f(1)令h(x)=f(x)x3,x∈(0,+∞),則h'(∵?x∈(0,+∞),2f(x)<xf'(x)<3f(x)恒成立,∴h'(x)=xf'(x∴函數(shù)h(x)在(0,+∞)內(nèi)單調(diào)遞減,∴f(1)又f(x)>0,∴18綜上可得18<f(1)14.x-y+6=0∵f'(x)=ex[x2+(2-a)x+1],若f(x)在(1,3)內(nèi)只有1個(gè)極值點(diǎn),∴f'(1)·f'(3)<0,即(a-4)(3a-16)<0,解得4<a<163∵a∈N,∴a=5.故f(x)=ex(x2-5x+6),f'(x)=ex(x2-3x+1),故f(0)=6,f'(0)=1,故切線方程是y-6=x,故答案為x-y+6=0.15.B根據(jù)題意,對(duì)于x1f(x1)+x2f(x2)≥x1f(x2)+x2f(x1),則有f(x1)(x1-x2)-f(x2)(x1-x2)≥0,即[f(x1)-f(x2)](x1-x2)≥0,分析可得,若函數(shù)f(x)為“H函數(shù)”,則函數(shù)f(x)為增函數(shù)或常數(shù)函數(shù).對(duì)于①,y=-x3+x+1,有y'=-3x2+1,不是增函數(shù)也不是常數(shù)函數(shù),則其不是“H函數(shù)”;對(duì)于②,y=3x-2(sinx-cosx),有y'=3-2(sinx+cosx)=3-22sinx+π4,y=3x-2(sinx-cosx)為增函數(shù),則其是“H函數(shù)”;對(duì)于③,y=1-ex=-ex+1,是減函數(shù),則其不是“H函數(shù)”;對(duì)于④,f(x)=lnx(x≥1),0(x<1),當(dāng)x<1時(shí),f(x)是常數(shù)函數(shù),當(dāng)x≥1時(shí),f(x)故“H函數(shù)”有2個(gè),故選B.16.23,1由題意設(shè)g(x)=-x3+3x