高中數(shù)學(xué)第五章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用培優(yōu)課函數(shù)的單調(diào)性與導(dǎo)數(shù)關(guān)系的應(yīng)用課后習(xí)題新人教A版選擇性必修第二冊(cè)

培優(yōu)課——函數(shù)的單調(diào)性與導(dǎo)數(shù)關(guān)系的應(yīng)用必備知識(shí)基礎(chǔ)練1.已知f(x)=2cos2x+1,x∈(0,π),則f(x)的單調(diào)遞增區(qū)間是()A.(π,2π) B.(0,π)C.π2,π 2.若f(x)=13x3-ax2的單調(diào)遞減區(qū)間是[0,2],則正數(shù)a的值是(A.1 B.2 C.3 D.43.已知函數(shù)f(x)=exx,當(dāng)1<x<3時(shí),下列關(guān)系正確的是(A.f(x)<f(x)<f2(x) B.f(x)<f(x)<f2(x)C.f2(x)<f(x)<f(x)D.f2(x)<f(x)<f(x)4.若函數(shù)f(x)=x3-ax2-x+6在區(qū)間(0,1)內(nèi)單調(diào)遞減,則實(shí)數(shù)a的取值范圍是()A.[1,+∞) B.a=1C.(-∞,1] D.(0,1)5.已知在R上可導(dǎo)的函數(shù)f(x)的圖象如圖所示,則不等式(x2-2x-3)·f'(x)>0的解集為()A.(-∞,-2)∪(1,+∞)B.(-∞,-2)∪(1,2)C.(-∞,-1)∪(-1,0)∪(2,+∞)D.(-∞,-1)∪(-1,1)∪(3,+∞)6.(多選題)已知函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且f'(x)<f(x)對(duì)任意的x∈R恒成立,則()A.f(ln2)<2f(0) B.f(2)<e2f(0)C.f(ln2)>2f(0) D.f(2)>e2f(0)7.若函數(shù)y=-43x3+bx有三個(gè)單調(diào)區(qū)間,則b的取值范圍是.8.若函數(shù)f(x)=(x2+mx)ex的單調(diào)遞減區(qū)間是-32,1,則實(shí)數(shù)m的值為.

9.已知函數(shù)f(x)=x3+ax2-a2x+2.(1)若a=1,求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)若a>0,求函數(shù)f(x)的單調(diào)區(qū)間.關(guān)鍵能力提升練10.設(shè)f(x),g(x)分別是定義在R上的奇函數(shù)和偶函數(shù),g(x)不為0,當(dāng)x<0時(shí),f'(x)g(x)-f(x)g'(x)>0,且f(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)11.若函數(shù)f(x)=2x2-lnx在定義域內(nèi)的一個(gè)子區(qū)間(k-1,k+1)上不是單調(diào)函數(shù),則實(shí)數(shù)k的取值范圍是()A.-12,32 B.-1C.1,32 D.1,3212.已知函數(shù)f(x)=xlnx+x(x-a)2(a∈R).若存在x∈12,2,使得f(x)>xf'(x)成立,則實(shí)數(shù)a的取值范圍是()A.94,+∞ B.32,+∞ C.(2,+∞) D.(3,+∞)13.若函數(shù)f(x)=x3+x2+mx+1是R內(nèi)的單調(diào)函數(shù),則實(shí)數(shù)m的取值范圍為.

14.已知f(x)是定義在R上的奇函數(shù),且f(2)=0,若當(dāng)x>0時(shí),xf'(x)+f(x)>0,則不等式xf(x)>0的解集是.

15.(2021山東煙臺(tái)期中)已知函數(shù)f(x)=exsinx-ax在(-π,0)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是.

16.試討論函數(shù)f(x)=kx-lnx的單調(diào)區(qū)間.學(xué)科素養(yǎng)創(chuàng)新練17.已知函數(shù)f(x)=lnx,g(x)=12ax2+2x(a≠0)(1)若函數(shù)h(x)=f(x)-g(x)存在單調(diào)遞減區(qū)間,則實(shí)數(shù)a的取值范圍是;

(2)若函數(shù)h(x)=f(x)-g(x)在[1,4]上單調(diào)遞減,則實(shí)數(shù)a的取值范圍是.

18.已知函數(shù)y=ax與y=-bx在(0,+∞)內(nèi)都是減函數(shù),試確定函數(shù)y=ax3+bx2+5的單調(diào)區(qū)間

參考答案培優(yōu)課——函數(shù)的單調(diào)性與導(dǎo)數(shù)關(guān)系的應(yīng)用1.C∵f(x)=2cos2x+1=2+cos2x,x∈(0,π),∴f'(x)=-2sin2x.令f'(x)>0,則sin2x<0.又x∈(0,π),∴0<2x<2π.∴π<2x<2π,即π2<x<π2.Af'(x)=x2-2ax,令f'(x)≤0,由于a>0,解得0≤x≤2a,故2a=2,即a=1.3.A由題意得f'(x)=(x-1)exx2,當(dāng)1<x<3時(shí),f'(x)>0,所以又1<x<x<3,所以f(x)<f(x).由f(x)在(1,3)上單調(diào)遞增,可知當(dāng)x∈(1,3)時(shí),f(x)>f(1)=e,所以f2(x)>f(x).綜上,f(x)<f(x)<f2(x).4.Af'(x)=3x2-2ax-1.因?yàn)閒(x)在區(qū)間(0,1)內(nèi)單調(diào)遞減,所以不等式3x2-2ax-1≤0在區(qū)間(0,1)內(nèi)恒成立.所以f'(0)≤0,f'(1)≤0.所以a≥1.故選A.5.D原不等式?f即x解得x<-1或x>3或-1<x<1.6.AB令g(x)=f(x)ex(x∈R),因?yàn)閒'(x)<f(x),所以g'(x故g(x)在R上單調(diào)遞減,而ln2>0,2>0,故g(ln2)<g(0),g(2)<g(0),即f(所以f(ln2)<2f(0),f(2)<e2f(0).7.(0,+∞)若函數(shù)y=-43x3+bx有三個(gè)單調(diào)區(qū)間,則其導(dǎo)數(shù)y'=-4x2+b=0有兩個(gè)不相等的實(shí)數(shù)根,所以b>08.-32f'(x)=[x2+(m+2)x+m]ex因?yàn)閒(x)的單調(diào)減區(qū)間是-32,1,所以方程f'(x)=0的兩個(gè)根分別為x1=-32,x2=1,即f'(-39.解(1)∵a=1,∴f(x)=x3+x2-x+2,∴f'(x)=3x2+2x-1,∴f'(1)=4.又f(1)=3,∴切點(diǎn)坐標(biāo)為(1,3),∴所求切線方程為y-3=4(x-1),即4x-y-1=0.(2)f(x)的定義域?yàn)镽,f'(x)=3x2+2ax-a2=(x+a)·(3x-a),由f'(x)=0得x=-a或x=a3又a>0,由f'(x)<0,得-a<x<a3由f'(x)>0,得x<-a或x>a3故f(x)的單調(diào)遞減區(qū)間為-a,a3,單調(diào)遞增區(qū)間為-∞,-a和a3,+∞.10.D令F(x)=f(x)g(x)(g(x)恒不為0),則F(x)為奇函數(shù),∵當(dāng)x<0時(shí),F'(x)>0,∴F(x)在(-∞,0)內(nèi)為增函數(shù).又F(3)=f(3)g(3)=0,∴∴當(dāng)x<-3時(shí),F(x)<0;當(dāng)-3<x<0時(shí),F(x)>0.又F(x)為奇函數(shù),∴當(dāng)0<x<3時(shí),F(x)<0;當(dāng)x>3時(shí),F(x)>0.而不等式f(x)g(x)<0和f(x)g∴不等式f(x)g(x)<0的解集為(-∞,-3)∪(0,3).11.C由題意,得函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=4x-1x令f'(x)=0,解得x=12或x=-12(舍去當(dāng)0<x<12時(shí),f'(x)<0,函數(shù)f(x)在區(qū)間0,12上單調(diào)遞減;當(dāng)x>12時(shí),f'(x)>0,函數(shù)f(x)在區(qū)間12,+∞上單調(diào)遞增.因?yàn)楹瘮?shù)f(x)在區(qū)間(k-1,k+1)上不是單調(diào)函數(shù),所以k-1<12<k+1,解得-12<k<又k-1≥0,所以1≤k<32.故選C12.C設(shè)g(x)=f(x)x,則g'(x)=xf'(x)-f(x)x2,存在x∈12,2,使得f(x)>xf'(x)成立等價(jià)于存在x∈12∵g(x)=f(x)x=lnx+(∴g'(x)=1x+2(x-a)由g'(x)<0得a>x+12∴a>x+12xmin,x∈12,2,又x+12x≥2x·12x=2,當(dāng)且僅當(dāng)x=22∴a>2.故選C.13.13,+∞f'(x)=3x2因?yàn)閒(x)是R內(nèi)的單調(diào)函數(shù),所以f'(x)≥0恒成立或f'(x)≤0恒成立.因?yàn)閷?dǎo)函數(shù)的二次項(xiàng)系數(shù)3>0,所以只能有f'(x)≥0恒成立.所以Δ=4-12m≤0,故m≥13經(jīng)檢驗(yàn),當(dāng)m=13時(shí),只有一個(gè)點(diǎn)使f'(x)=0,符合題意,故實(shí)數(shù)m的取值范圍是114.(-∞,-2)∪(2,+∞)由題意設(shè)g(x)=xf(x),則g'(x)=xf'(x)+f(x).∵當(dāng)x>0時(shí),xf'(x)+f(x)>0,∴g(x)在(0,+∞)上單調(diào)遞增.∵f(x)是定義在R上的奇函數(shù),∴g(x)是定義在R上的偶函數(shù).又f(2)=0,則g(2)=2f(2)=0,∴不等式xf(x)>0等價(jià)于g(x)>0=g(2),∴|x|>2,解得x<-2或x>2,∴不等式xf(x)>0的解集是(-∞,-2)∪(2,+∞).15.(-∞,-e-π2]由題意f'(x)=ex(sinx+cosx)-a≥0在(-π即a≤ex(sinx+cosx)在(-π,0)上恒成立,令g(x)=ex(sinx+cosx),x∈(-π,0),則g'(x)=2excosx,易知x∈-π,-π2時(shí),g'(x)<0;x∈-π2,0時(shí),g'(x)>0,故g(x)在-π,-π2上單調(diào)遞減,在-π2,0上單調(diào)遞增,故g(x)min=g-π2=-e-π2,故a≤-e16.解函數(shù)f(x)=kx-lnx的定義域?yàn)?0,+∞),f'(x)=k-1x當(dāng)k≤0時(shí),kx-1<0,∴f'(x)<0,則f(x)在(0,+∞)上單調(diào)遞減.當(dāng)k>0時(shí),由f'(x)<0,即kx-1解得0<x<1k由f'(x)>0,即kx-1x>0,解得∴當(dāng)k>0時(shí),f(x)的單調(diào)遞減區(qū)間為0,1k,單調(diào)遞增區(qū)間為1k,+∞.綜上所述,當(dāng)k≤0時(shí),f(x)的單調(diào)遞減區(qū)間為(0,+∞),無(wú)單調(diào)遞增區(qū)間;當(dāng)k>0時(shí),f(x)的單調(diào)遞減區(qū)間為0,1k,單調(diào)遞增區(qū)間為1k,+∞.17.(1)(-1,0)∪(0,+∞)(2)-716,0∪(0,+∞)(1)由題知h(x)=lnx-12ax2-2x,x∈(0,+∞),所以h'(x)=1x-ax-2由h(x)在(0,+∞)上存在單調(diào)遞減區(qū)間,可得當(dāng)x∈(0,+∞)時(shí),1x-ax-2<0有解,即a>1x設(shè)G(x)=1x2?2x(x>0),所以只要a>G(而G(x)=1x-12-1,所以G(x)min=-1.因?yàn)閍≠0,所以-1<a<0或a>0.(2)由h(x)在[1,4]上單調(diào)遞減,得當(dāng)x∈[1,4]時(shí),h'(x)=1x-ax-2≤0恒成立,即a≥1x設(shè)H(x)=1x2?2x,x∈[1,4],所以a≥H(x)max,而H(x)=1x-12-1,