2023-2024學(xué)年高中數(shù)學(xué)人教A版2019選擇性課后習(xí)題第五章　一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用培優(yōu)課構(gòu)造函數(shù)法解決導(dǎo)數(shù)問題

培優(yōu)課——構(gòu)造函數(shù)法解決導(dǎo)數(shù)問題必備知識基礎(chǔ)練1.已知函數(shù)f(x)(x∈R)滿足f(1)=1,且f'(x)<12,則f(x)<x2+1A.{x|1<x<1} B.{x|x<1}C.{x|x<1或x>1} D.{x|x>1}2.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,xf'(x)f(x)<0,則使得f(x)>0成立的x的取值范圍是()A.(∞,1)∪(0,1) B.(1,0)∪(1,+∞)C.(∞,1)∪(1,0) D.(0,1)∪(1,+∞)3.已知函數(shù)f(x)的定義域為R,f'(x)為f(x)的導(dǎo)函數(shù),且f(x)+(x1)f'(x)>0,則下列式子正確的是()A.f(1)=0 B.f(x)<0C.f(x)>0 D.(x1)f(x)<04.已知定義在R上的函數(shù)f(x)滿足f(x)<f'(x),則下列式子一定成立的是()A.f(2020)>ef(2021) B.f(2020)<ef(2021)C.ef(2020)>f(2021) D.ef(2020)<f(2021)5.(多選題)已知f(x)為(0,+∞)上的可導(dǎo)函數(shù),且(x+1)·f'(x)>f(x),則下列不等式一定成立的是()A.3f(4)<4f(3) B.4f(4)>5f(3)C.3f(3)<4f(2) D.3f(3)>4f(2)6.已知f(x)是定義在0,π2上的函數(shù),其導(dǎo)函數(shù)為f'(x),fπ3=23,且當(dāng)x∈0,π2時,f'(x)sinx+f(x)cosx>0,則不等式f(x)sinx<3的解集為.

7.設(shè)函數(shù)f'(x)是定義在(0,π)上的函數(shù)f(x)的導(dǎo)函數(shù),有f'(x)cosxf(x)sinx>0,若a=12fπ3,b=0,c=32f5π6,則a,b,c的大小關(guān)系是.8.若對任意的x∈[e,+∞),都有xlnx≥axa,求實數(shù)a的取值范圍.9.已知函數(shù)f(x)=12x22alnx+(a2)x(1)當(dāng)a=1時,求函數(shù)f(x)在[1,e]上的最小值和最大值.(2)是否存在實數(shù)a,對任意的x1,x2∈(0,+∞),且x1≠x2,都有f(x2)-f(x1)x2-x關(guān)鍵能力提升練10.設(shè)f(x),g(x)是定義域為R的恒大于0的可導(dǎo)函數(shù),且f'(x)g(x)f(x)g'(x)<0,則當(dāng)a<x<b時,下列式子一定正確的是()A.f(x)g(x)>f(b)g(b)B.f(x)g(b)>f(b)g(x)C.f(x)g(a)>f(a)g(x)D.f(x)g(x)>f(a)g(a)11.已知函數(shù)f(x)的定義域為R,其導(dǎo)函數(shù)為f'(x),且3f(x)f'(x)>0在R上恒成立,則下列不等式一定成立的是()A.f(1)<e3f(0) B.f(1)<e2f(0)C.f(1)>e3f(0) D.f(1)>e2f(0)12.設(shè)函數(shù)f(x)的定義域為R,f'(x)是其導(dǎo)函數(shù),若3f(x)+f'(x)>0,f(0)=1,則不等式f(x)>e3x的解集是()A.(0,+∞) B.(1,+∞)C.(∞,0) D.(0,1)13.定義域為π2,π2的函數(shù)f(x)滿足f(x)+f(x)=0,其導(dǎo)函數(shù)為f'(x),當(dāng)0≤x<π2時,有f'(x)cosx+f(x)sinx<0成立,則關(guān)于x的不等式f(x)<2fπ4·cosx的解集為(A.π2,π4∪π4,B.π4,C.π4,0∪0,π4D.π4,0∪π4,π14.已知函數(shù)f(x)是定義在R上的奇函數(shù),f(1)=0,當(dāng)x>0時,有xf'(x)-f(x)x2>0,則不等式x2f15.已知函數(shù)f(x)=axlnx(a∈R)(1)討論f(x)的單調(diào)性;(2)若x1,x2是方程f(x)=2的兩個不同實根,證明:x1+x2>2e學(xué)科素養(yǎng)創(chuàng)新練16.設(shè)函數(shù)f(x)=aex,x∈R.(1)當(dāng)a=1時,過原點作y=f(x)的切線,求切線方程;(2)若不等式xf(x)x+2>lnx對于x∈(0,+∞)恒成立,求a的取值范圍;(3)在(1)的條件下,證明:xx2x2lnx<1+e2e3f(x參考答案培優(yōu)課——構(gòu)造函數(shù)法解決導(dǎo)數(shù)問題1.D構(gòu)造函數(shù)h(x)=f(x)x2-12,所以h'(x)=f'(x)12<0,故h(x)在R上是減函數(shù),且h(1)=f(1)12-12=0,故h(2.A構(gòu)造函數(shù)h(x)=f(x)x,因為f(x)為奇函數(shù),所以h(x)為偶函數(shù),又因為h'(x)=xf'(x)-f(x)x2,且當(dāng)x>0時,xf'(x)f(x)<0,所以h(x)在(0,+∞)上是減函數(shù),根據(jù)對稱性知h(x)在(∞,0)上是增函數(shù),又f(1)=0,所以f(1)=0,數(shù)形結(jié)合可知,使得f(x3.C令g(x)=(x1)f(x),則g'(x)=f(x)+(x1)f'(x)>0,所以g(x)在R上是增函數(shù),又因為g(1)=0,所以當(dāng)x>1時,g(x)=(x1)f(x)>0;當(dāng)x<1時,g(x)=(x1)f(x)<0.所以當(dāng)x≠1時,f(x)>0.又f(1)+(11)f'(1)=f(1)>0,所以ABD錯誤,C正確.4.A依題意得f(x)+f'(x)<0,令g(x)=exf(x),則g'(x)=ex[f(x)+f'(x)]<0在R上恒成立,所以函數(shù)g(x)=exf(x)在R上是減函數(shù),所以g(2020)>g(2021),即e2020f(2020)>e2021f(2021)?f(2020)>ef(2021).5.BD由(x+1)f'(x)>f(x),得(x+1)f'(x)f(x)>0,令g(x)=f(則g'(x)=(x+1∴g(x)在(0,+∞)上是增函數(shù),∴g(2)<g(3)<g(4),則f(即4f(2)<3f(3),5f(3)<4f(4),故選BD.6.x0<x<π3因為當(dāng)x∈0,π2時,f'(x)sinx+f(x)cosx>所以[f(x)sinx]'>0,x∈0,π2,令g(x)=f(x)sinx,則當(dāng)x∈0,π2時,g'(x)>0,g(x)在0,π2上是增函數(shù),因為fπ3=23,所以gπ3=fπ3sinπ3=3,不等式f(x)sinx<3,即g(x)<gπ3.因為g(x)在0,π2上是增函數(shù),所以原不等式的解集為x0<x<π7.a<b<c設(shè)函數(shù)g(x)=f(x)cosx,則g'(x)=f'(x)cosxf(x)sinx,因為f'(x)cosxf(x)sinx>0,所以g'(x)>0,所以g(x)在(0,π)上是增函數(shù),a=12fπ3=fπ3cosπ3=gπ3,b=0=fπ2cosπ2=gπ2,c=32f5π6=f5π6cos5π6=g5所以a<b<c.8.解若對任意的x∈[e,+∞),都有xlnx≥axa,等價于a≤xlnxx-1在[e,令h(x)=xlnxx-1,則h'(x)=x-ln令m(x)=xlnx1,則當(dāng)x≥e時,m'(x)=11x>即m(x)在[e,+∞)上是增函數(shù),故m(x)≥m(e)=e2>0,所以h'(x)>0,所以h(x)=xlnxx-1在[e,h(x)min=h(e)=ee-1,所以a即實數(shù)a的取值范圍是∞,ee-19.解(1)當(dāng)a=1時,f(x)=12x22lnxx則f'(x)=x2x1=x2-x-∴當(dāng)x∈[1,2)時,f'(x)<0;當(dāng)x∈(2,e]時,f'(x)>0.∴f(x)在[1,2)上是減函數(shù),在(2,e]上是增函數(shù).∴當(dāng)x=2時,f(x)取得最小值,其最小值為f(2)=2ln2.又f(1)=12,f(e)=e22e2,f(e)f(1)=e22e2+12=∴f(x)max=f(1)=12(2)假設(shè)存在實數(shù)a,對任意的x1,x2∈(0,+∞),且x1≠x2,都有f(x2不妨設(shè)0<x1<x2,∵f(x∴f(x2)ax2>f(x1)ax1.令g(x)=f(x)ax,則由此可知g(x)在(0,+∞)上是增函數(shù),又g(x)=12x22alnx+(a2)xax=12x22alnx2則g'(x)=x2ax2=由此可得g'(x)≥0在(0,+∞)上恒成立,只需12a≥0,解得a≤12即a的取值范圍是∞,12.10.B設(shè)F(x)=f(則F'(x)=f'(由f'(x)g(x)f(x)g'(x)<0,得F'(x)<0,所以F(x)在R上是減函數(shù),因為a<x<b,所以f(b)g(b)<f(x)g(x)<f(a)g(a),又f(x),g(x11.A令g(x)=f(則g'(x)=f'(因為3f(x)f'(x)>0在R上恒成立,所以g'(x)<0在R上恒成立,故g(x)在R上是減函數(shù),所以g(1)<g(0),即f(1)e3<f(0)12.A令g(x)=e3xf(x),則g'(x)=3e3xf(x)+e3xf'(x),因為3f(x)+f'(x)>0,所以3e3xf(x)+e3xf'(x)>0,所以g'(x)>0,所以函數(shù)g(x)=e3xf(x)在R上是增函數(shù),又f(x)>e3x可化為e3xf(x)>1,且g(0)=e3×0f(0)=1,所以g(x)>g(0),解得x>0,所以不等式f(x)>e3x的解集是(0,+∞).13.B∵f(x)+f(x)=0且x∈π2,π∴f(x)是奇函數(shù).設(shè)g(x)=f(x)cosx,則當(dāng)0g'(x)=f'(x∴g(x)在0,π2上是減函數(shù).又f(x)是奇函數(shù),∴g(x)=f(x∴g(x)在π2,0上是減函數(shù),從而g(x)在π2,π2∵不等式f(x)<2fπ4·cosx,∴f(x即g(x)<gπ4,∴π4<x<π214.(1,0)∪(1,+∞)令g(x)=f(x)x則g'(x)=xf'(∵當(dāng)x>0時,xf'(x)-f(x)x2∴g(x)在(0,+∞)上是增函數(shù).又f(1)=0,∴g(1)=f(1)=0,∴在(0,+∞)上,g(x)>0的解集為(1,+∞),g(x)<0的解集為(0,1).∵f(x)為奇函數(shù),∴g(x)為偶函數(shù),∴在(∞,0)上,g(x)>0的解集為(∞,1),g(x)<0的解集為(1,0).由x2f(x)>0,得f(x)>0(x≠0).又f(x)>0的解集為(1,0)∪(1,+∞),∴不等式x2f(x)>0的解集為(1,0)∪(1,+∞).15.(1)解因為f(x)=axlnx所以f'(x)=ax2-①當(dāng)a≥0時,f'(x)<0在(0,+∞)上恒成立,故f(x)在(0,+∞)上是減函數(shù).②當(dāng)a<0時,由f'(x)>0,得0<x<a;由f'(x)<0,得x>a.即f(x)在(0,a)上是增函數(shù),在(a,+∞)上是減函數(shù).綜上,當(dāng)a≥0時,f(x)在(0,+∞)上是減函數(shù);當(dāng)a<0時,f(x)在(0,a)上是增函數(shù),在(a,+∞)上是減函數(shù).(2)證明因為f(x1)=f(x2)=2,所以ax1lnx12=0,ax2lnx即x1lnx1+2x1a=0,x2lnx2+2x2a=0.設(shè)g(x)=xlnx+2xa,則g'(x)=lnx+3,故g(x)在0,1e3上是減函數(shù),在1e3,+∞上是增函數(shù).由題意,設(shè)0<x1<1e3<x欲證x1+x2>2e3,只需證x2>2e又x2∈1e3,+∞,2e3x1∈1e3,+∞,g(x)在1e3,+∞上是增函數(shù),故只需證g(x2)>g2e3x因為g(x1)=g(x2),所以只需證g(x1)>g2e3x1對任意的x1∈0,1e3恒成立即可,即x1lnx1+2x1a>2e3x1ln2e3x1+22e3x1a,整理得x1lnx1+2x1>2e3x1ln2e3x1+4e32x即x1lnx12e3x1ln2e3x1+4x14e3>設(shè)h(x)=xlnx2e3xln2e3x+4x4e3,x∈0,1e則h'(x)=lnx+ln2e3x+6=ln2xe3x2+6因為0<x<1e3,所以0<2xe3x2<1e6,所以h'(x)=ln2xe3x2+6<0,所以h(x則h(x)>h1e3=0.所以x1+x2>2e316.(1)解當(dāng)a=1時,f(x)=ex,則f'(x)=ex,設(shè)切點坐標(biāo)為(x0,y0),則切線斜率為ex0,切線方程為yex0=ex0(xx0),將(0,0)代入切線方程,解得x(2)解不等式xf(x)x+2>lnx對于x∈(0,+∞)恒成立,則a>lnx+x-2xex,令g(x)=lnx+x-2xex,令m(x)=3lnxx,則m'(x)=1x1<所以m(x)單調(diào)遞減,m(2)>0,m(3)<0,所以?x1∈(2,3),m(x1)=0,所以當(dāng)x∈(0,x1)時,g(x)單調(diào)遞增;當(dāng)x∈(x1,+∞)時,g(x)單調(diào)遞減.所以g