浙江專用2025版高考數(shù)學大一輪復習課時133.2導數(shù)與函數(shù)單調(diào)性夯基提能作業(yè)

PAGEPAGE13.2導數(shù)與函數(shù)單調(diào)性A組基礎(chǔ)題組1.函數(shù)y=4x2+1xA.(0,+∞) B.1C.(-∞,-1) D.-∞,-答案B由y=4x2+1x得y'=8x-1x2,令y'>0,即8x-1x2>0,解得x>12,∴函數(shù)y=4x22.已知m是實數(shù),函數(shù)f(x)=x2(x-m),若f'(-1)=-1,則函數(shù)f(x)的單調(diào)增區(qū)間是()A.-43C.-∞,-43,(0,+∞) D.答案C由題意得f'(x)=3x2-2mx,∴f'(-1)=3+2m=-1,解得m=-2,∴f'(x)=3x2+4x,令f'(x)>0,解得x<-43故f(x)的單調(diào)增區(qū)間為-∞,-43.已知函數(shù)f(x)=x2+2cosx,若f'(x)是f(x)的導函數(shù),則函數(shù)f'(x)的圖象大致是()答案A令g(x)=f'(x)=2x-2sinx,則g'(x)=2-2cosx,易知g'(x)≥0,所以函數(shù)f'(x)在R上單調(diào)遞增.4.若冪函數(shù)f(x)的圖象過點22,1A.(-∞,0) B.(-∞,-2)C.(-2,-1) D.(-2,0)答案D設(shè)冪函數(shù)f(x)=xα,因為圖象過點22,12,所以12=22α,α=2,所以f(x)=x2,故g(x)=exx2,則g'(x)=exx2+2e5.若函數(shù)f(x)=x+alnx不是單調(diào)函數(shù),則實數(shù)a的取值范圍是()A.[0,+∞) B.(-∞,0]C.(-∞,0) D.(0,+∞)答案Cf'(x)=1+ax=x+6.已知函數(shù)f(x)=ax+lnx,則當a<0時,f(x)的單調(diào)遞增區(qū)間是,單調(diào)遞減區(qū)間是.

答案0,-1a解析由已知得f(x)的定義域為(0,+∞).當a<0時,因為f'(x)=a+1x=ax+f'(x)<0,當0<x<-1a時,f'(x)>0,所以f(x)的單調(diào)遞增區(qū)間為0,-17.若f(x)=xsinx+cosx,則f(-3),fπ2,f(2)的大小關(guān)系是答案f(-3)<f(2)<fπ2解析函數(shù)f(x)為偶函數(shù),因此f(-3)=f(3).又f'(x)=sinx+xcosx-sinx=xcosx,當x∈π2所以f(x)在區(qū)間π2所以fπ28.已知函數(shù)f(x)=12x2+2ax-lnx,若f(x)在區(qū)間13,答案43解析由題意得f'(x)=x+2a-1x≥0在13,2上恒成立,即2a≥-x+1x在13,2上恒成立,∵-9.已知函數(shù)f(x)=lnx+a2x2解析由已知得f'(x)=1x+ax-(a+1),則f'(1)=0.而f(1)=ln1+a2-(a+1)=-a∴曲線y=f(x)在x=1處的切線方程為y=-a2∴-a2∴f(x)=lnx+x2-3x,f'(x)=1x由f'(x)=1x+2x-3=2x2由f'(x)=1x+2x-3<0,得1∴f(x)的單調(diào)遞增區(qū)間為0,1210.已知函數(shù)f(x)=ax3+x2(a∈R)在x=-43(1)確定a的值;(2)若g(x)=f(x)ex,探討g(x)的單調(diào)性.解析(1)對f(x)求導得f'(x)=3ax2+2x.因為f(x)在x=-43所以f'-43=3a×169+2×-43解得a=12,經(jīng)檢驗,滿意(2)由(1)知,g(x)=12x3所以g'(x)=32x2+2xex=12x3+52x令g'(x)=0,解得x=0或x=-1或x=-4.當x<-4時,g'(x)<0,故g(x)在(-∞,-4)上為減函數(shù);當-4<x<-1時,g'(x)>0,故g(x)在(-4,-1)上為增函數(shù);當-1<x<0時,g'(x)<0,故g(x)在(-1,0)上為減函數(shù);當x>0時,g'(x)>0,故g(x)在(0,+∞)上為增函數(shù).綜上,g(x)在(-∞,-4)和(-1,0)上為減函數(shù),在(-4,-1)和(0,+∞)上為增函數(shù).11.已知函數(shù)f(x)=x2+alnx.(1)當a=-2時,求函數(shù)f(x)的單調(diào)遞減區(qū)間;(2)若函數(shù)g(x)=f(x)+2x解析(1)由題意知,函數(shù)的定義域為(0,+∞),當a=-2時,f'(x)=2x-2x=2由f'(x)<0得0<x<1,故f(x)的單調(diào)遞減區(qū)間是(0,1).(2)由題意得g'(x)=2x+ax-2因函數(shù)g(x)在[1,+∞)上單調(diào),故:①若g(x)為[1,+∞)上的單調(diào)增函數(shù),則g'(x)≥0在[1,+∞)上恒成立,即a≥2x-2x2在[1,+∞)上恒成立,設(shè)φ(x)=2x-2x∵φ(x)在[1,+∞)上單調(diào)遞減,∴在[1,+∞)上,φ(x)max=φ(1)=0,∴a≥0.②若g(x)為[1,+∞)上的單調(diào)減函數(shù),則g'(x)≤0在[1,+∞)上恒成立,易知其不行能成立.∴實數(shù)a的取值范圍是[0,+∞).B組提升題組1.已知f(x)的定義域為(0,+∞),f'(x)為f(x)的導函數(shù),且滿意f(x)<-xf'(x),則不等式f(x+1)>(x-1)f(x2-1)的解集是()A.(0,1) B.(1,+∞)C.(1,2) D.(2,+∞)答案D因為f(x)+xf'(x)<0,所以(xf(x))'<0,所以xf(x)在(0,+∞)上為減函數(shù),又因為f(x+1)>(x-1)f(x2-1),所以(x+1)f(x+1)>(x2-1)f(x2-1),所以0<x+1<x2-1,所以x>2.2.若定義在R上的函數(shù)f(x)滿意f(x)+f'(x)>1,f(0)=4,則不等式f(x)>3eA.(0,+∞) B.(-∞,0)∪(3,+∞)C.(-∞,0)∪(0,+∞) D.(3,+∞)答案A由f(x)>3ex+1,得exf(x)>3+ex構(gòu)造函數(shù)F(x)=exf(x)-ex-3,得F'(x)=exf(x)+exf'(x)-ex=ex[f(x)+f'(x)-1],由f(x)+f'(x)>1,ex>0,可知F'(x)>0,所以F(x)在R上單調(diào)遞增,又因為F(0)=e0f(0)-e0-3=f(0)-4=0,所以F(x)>0的解集為(0,+∞),即f(x)>3e3.已知函數(shù)f(x)=a(x-lnx)+2x-1解析易知f(x)的定義域為(0,+∞),f'(x)=a-ax-2x2+2當a≤0時,若x∈(0,1),則f'(x)>0,f(x)單調(diào)遞增,若x∈(1,+∞),則f'(x)<0,f(x)單調(diào)遞減.當a>0時,f'(x)=a(①當0<a<2時,2a當x∈(0,1)或x∈2a當x∈1,②當a=2時,2a在區(qū)間(0,+∞