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y'>0,即8x-1x2>0,解得x>12,∴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.(-43,0)B.(0,43)C.(-∞,-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x<-43或x>0,故f(x)的单调增区间为(-∞,-43),(0,+∞).3.已知函数f(x)=x2+2cosx,若f'(x)是f(x)的导函数,则函数f'(x)的图象大致是()1答案A令g(x)=f'(x)=2x-2sinx,则g'(x)=2-2cosx,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)=xα,因为图象过点(√22,12),所以12=(√22)α,α=2,所以f(x)=x2,故g(x)=exx2,则g'(x)=exx2+2exx=ex(x2+2x),令g'(x)<0,得-2-1a时,f'(x)<0,当00,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)=xsinx+cosx,则f(-3),f(π2),f(2)的大小关系是.答案f(-3)f(2)>f(3)=f(-3).8.已知函数f(x)=12x2+2ax-lnx,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-1x≥0在[13,2]上恒成立,即2a≥-x+1x在[13,2]上恒成立, (-x+1x)max=83,∴2a≥83,即a≥43.9.已知函数f(x)=lnx+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)=ln1+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)=lnx+x2-3x,f'(x)=1x+2x-3.由f'(x)=1x+2x-3=2x2-3x+1x>0,得01,由f'(x)=1x+2x-3<0,得120,故g(x)在(-4,-1)上为增函数;当-10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+alnx.(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,+∞),4当a=-2时,f'(x)=2x-2x=2(x+1)(x-1)x,由f'(x)<0得0
