浙江省2020版高考数学大一轮复习导数与函数单调性夯基提能作业.docx

3.2导数与函数单调性A组基础题组1.函数y=4x2+1x的单调递增区间为()A.(0,+)B.12,+C.(-,-1)D.-,-12答案B由y=4x2+1x得y=8x-1x2,令y0,即8x-1x20,解得x12,函数y=4x2+1x在12,+上单调递增.故选B.2.已知m是实数,函数f(x)=x2(x-m),若f (-1)=-1,则函数f(x)的单调增区间是() A.-43,0B.0,43C.-,-43,(0,+)D.-,-43(0,+)答案C由题意得f (x)=3x2-2mx,f (-1)=3+2m=-1,解得m=-2,f (x)=3x2+4x,令f (x)0,解得x0,故f(x)的单调增区间为-,-43,(0,+).3.已知函数f(x)=x2+2cos x,若f (x)是f(x)的导函数,则函数f (x)的图象大致是()答案A令g(x)=f (x)=2x-2sin x,则g(x)=2-2cos x,易知g(x)0,所以函数f (x)在R上单调递增.4.若幂函数f(x)的图象过点22,12,则函数g(x)=exf(x)的单调递减区间为()A.(-,0)B.(-,-2)C.(-2,-1)D.(-2,0)答案D设幂函数f(x)=x,因为图象过点22,12,所以12=22,=2,所以f(x)=x2,故g(x)=exx2,则g(x)=exx2+2exx=ex(x2+2x),令g(x)0,得-2x0,故函数g(x)的单调递减区间为(-2,0).5.若函数f(x)=x+aln x不是单调函数,则实数a的取值范围是()A.0,+)B.(-,0C.(-,0)D.(0,+)答案Cf (x)=1+ax=x+ax,若f(x)=x+aln x不是单调函数,则f (x)=0在(0,+)内有解,所以a0,故选C.6.已知函数f(x)=ax+ln x,则当a0时, f(x)的单调递增区间是,单调递减区间是.答案0,-1a;-1a,+解析由已知得f(x)的定义域为(0,+).当a-1a时,f (x)0,当0x0,所以f(x)的单调递增区间为0,-1a,单调递减区间为-1a,+.7.若f(x)=xsin x+cos x,则f(-3), f2, f(2)的大小关系是.答案f(-3)f(2)f2解析函数f(x)为偶函数,因此f(-3)=f(3).又f (x)=sin x+xcos x-sin x=xcos x,当x2,时, f (x)f(2)f(3)=f(-3).8.已知函数f(x)=12x2+2ax-ln x,若f(x)在区间13,2上是增函数,则实数a的取值范围是.答案43,+解析由题意得f (x)=x+2a-1x0在13,2上恒成立,即2a-x+1x在13,2上恒成立,-x+1xmax=83,2a83,即a43.9.已知函数f(x)=ln x+a2x2-(a+1)x.若曲线y=f(x)在x=1处的切线方程为y=-2,求f(x)的单调区间.解析由已知得f (x)=1x+ax-(a+1),则f (1)=0.而f(1)=ln 1+a2-(a+1)=-a2-1,曲线y=f(x)在x=1处的切线方程为y=-a2-1.-a2-1=-2,解得a=2.f(x)=ln x+x2-3x, f (x)=1x+2x-3.由f (x)=1x+2x-3=2x2-3x+1x0,得0x1,由f (x)=1x+2x-30,得12x1,f(x)的单调递增区间为0,12和(1,+), f(x)的单调递减区间为12,1.10.已知函数f(x)=ax3+x2(aR)在x=-43处取得极值.(1)确定a的值;(2)若g(x)=f(x)ex,讨论g(x)的单调性.解析(1)对f(x)求导得f (x)=3ax2+2x.因为f(x)在x=-43处取得极值,所以f -43=3a169+2-43=16a3-83=0,解得a=12,经检验,满足题意.(2)由(1)知,g(x)=12x3+x2ex,所以g(x)=32x2+2xex+12x3+x2ex=12x3+52x2+2xex=12x(x+1)(x+4)ex.令g(x)=0,解得x=0或x=-1或x=-4.当x-4时,g(x)0,故g(x)在(-,-4)上为减函数;当-4x0,故g(x)在(-4,-1)上为增函数;当-1x0时,g(x)0时,g(x)0,故g(x)在(0,+)上为增函数.综上,g(x)在(-,-4)和(-1,0)上为减函数,在(-4,-1)和(0,+)上为增函数.11.已知函数f(x)=x2+aln x.(1)当a=-2时,求函数f(x)的单调递减区间;(2)若函数g(x)=f(x)+2x在1,+)上单调,求实数a的取值范围.解析(1)由题意知,函数的定义域为(0,+),当a=-2时, f (x)=2x-2x=2(x+1)(x-1)x,由f (x)0得0x1,故f(x)的单调递减区间是(0,1).(2)由题意得g(x)=2x+ax-2x2,因函数g(x)在1,+)上单调,故:若g(x)为1,+)上的单调增函数,则g(x)0在1,+)上恒成立,即a2x-2x2在1,+)上恒成立,设(x)=2x-2x2.(x)在1,+)上单调递减,在1,+)上,(x)max=(1)=0,a0.若g(x)为1,+)上的单调减函数,则g(x)0在1,+)上恒成立,易知其不可能成立.实数a的取值范围是0,+).B组提升题组1.已知f(x)的定义域为(0,+), f (x)为f(x)的导函数,且满足f(x)(x-1)f(x2-1)的解集是() A.(0,1)B.(1,+)C.(1,2)D.(2,+)答案D因为f(x)+xf (x)0,所以(xf(x)(x-1)f(x2-1),所以(x+1)f(x+1)(x2-1)f(x2-1),所以0x+12.2.若定义在R上的函数f(x)满足f(x)+f (x)1, f(0)=4,则不等式f(x)3ex+1(e为自然对数的底数)的解集为()A.(0,+)B.(-,0)(3,+)C.(-,0)(0,+)D.(3,+)答案A由f(x)3ex+1,得exf(x)3+ex,构造函数F(x)=exf(x)-ex-3,得F (x)=exf(x)+exf (x)-ex=exf(x)+f (x)-1,由f(x)+f (x)1,ex0,可知F (x)0,所以F(x)在R上单调递增,又因为F(0)=e0f(0)-e0-3=f(0)-4=0,所以F(x)0的解集为(0,+),即f(x)3ex+1的解集为(0,+).3.已知函数f(x)=a(x-ln x)+2x-1x2(aR),讨论f(x)的单调性.解析易知f(x)的定义域为(0,+),f (x)=a-ax-2x2+2x3=(ax2-2)(x-1)x3.当a0时,若x(0,1),则f (x)0, f(x)单调递增,若x(1,+),则f (x)0时, f (x)=a(x-1)x3x-2ax+2a,当0a1,当x(0,1)或x2a,+时, f (x)0