专题对点练7 导数与不等式及参数范围1.已知函数f(x)=12x2+(1-a)x-alnx.(1)讨论f(x)的单调性;(2)设a<0,若对∀x1,x2∈(0,+∞),|f(x1)-f(x2)|≥4|x1-x2|,求a的取值范围.2.设函数f(x)=(1-x2)ex.(1)求f(x)的单调区间;(2)当x≥0时,f(x)≤ax+1,求a的取值范围.3.(2022北京,文19)设函数f(x)=[ax2-(3a+1)x+3a+2]ex.(1)若曲线y=f(x)在点(2,f(2))处的切线斜率为0,求a;(2)若f(x)在x=1处取得极小值,求a的取值范围.4.已知函数f(x)=lnx+ax2+(2a+1)x.(1)讨论f(x)的单调性;(2)当a<0时,证明f(x)≤-34a-2.4\n专题对点练7答案1.解(1)f(x)的定义域为(0,+∞),f'(x)=x+1-a-ax=x2+(1-a)x-ax=(x+1)(x-a)x,若a≤0,则f'(x)>0,此时f(x)在(0,+∞)内单调递增;若a>0,则由f'(x)=0得x=a,当0<x<a时,f'(x)<0,当x>a时,f'(x)>0,此时f(x)在(0,a)内单调递减,在(a,+∞)内单调递增.(2)不妨设x1≤x2,而a<0,由(1)知,f(x)在(0,+∞)内单调递增,∴f(x1)≤f(x2),|f(x1)-f(x2)|≥4|x1-x2|⇔4x1-f(x1)≥4x2-f(x2),令g(x)=4x-f(x),则g(x)在(0,+∞)内单调递减,∵g'(x)=4-f'(x)=4-x+1-a-ax=ax-x+3+a,∴g'(x)=ax-x+3+a≤0对∀x∈(0,+∞)恒成立,∴a≤x2-3xx+1对∀x∈(0,+∞)恒成立,∴a≤x2-3xx+1min.又x2-3xx+1=x+1+4x+1-5≥2(x+1)·4x+1-5=-1,当且仅当x+1=4x+1,即x=1时,等号成立.∴a≤-1,故a的取值范围为(-∞,-1].2.解(1)f'(x)=(1-2x-x2)ex.令f'(x)=0得x=-1-2,x=-1+2.当x∈(-∞,-1-2)时,f'(x)<0;当x∈(-1-2,-1+2)时,f'(x)>0;当x∈(-1+2,+∞)时,f'(x)<0.所以f(x)在(-∞,-1-2),(-1+2,+∞)内单调递减,在(-1-2,-1+2)内单调递增.(2)f(x)=(1+x)(1-x)ex.当a≥1时,设函数h(x)=(1-x)ex,h'(x)=-xex<0(x>0),因此h(x)在[0,+∞)内单调递减,而h(0)=1,故h(x)≤1,所以f(x)=(x+1)h(x)≤x+1≤ax+1.当0<a<1时,设函数g(x)=ex-x-1,g'(x)=ex-1>0(x>0),所以g(x)在[0,+∞)内单调递增,而g(0)=0,故ex≥x+1.当0<x<1时,f(x)>(1-x)(1+x)2,(1-x)(1+x)2-ax-1=x(1-a-x-x2),取x0=5-4a-12,则x0∈(0,1),(1-x0)(1+x0)2-ax0-1=0,故f(x0)>ax0+1.当a≤0时,取x0=5-12,则x0∈(0,1),f(x0)>(1-x0)(1+x0)2=1≥ax0+1.综上,a的取值范围是[1,+∞).3.解(1)因为f(x)=[ax2-(3a+1)x+3a+2]ex,所以f'(x)=[ax2-(a+1)x+1]ex.所以f'(2)=(2a-1)e2.由题设知f'(2)=0,即(2a-1)e2=0,解得a=12.(2)(方法一)由(1)得f'(x)=[ax2-(a+1)x+1]ex=(ax-1)(x-1)ex.若a>1,则当x∈1a,1时,f'(x)<0;当x∈(1,+∞)时,f'(x)>0.所以f(x)在x=1处取得极小值.4\n若a≤1,则当x∈(0,1)时,ax-1≤x-1<0,所以f'(x)>0.所以1不是f(x)的极小值点.综上可知,a的取值范围是(1,+∞).(方法二)由(1)得f'(x)=(ax-1)(x-1)ex.当a=0时,令f'(x)=0,得x=1.f'(x),f(x)随x的变化情况如下表:x(-∞,1)1(1,+∞)f'(x)+0-f(x)↗极大值↘∴f(x)在x=1处取得极大值,不合题意.当a>0时,令f'(x)=0,得x1=1a,x2=1.①当x1=x2,即a=1时,f'(x)=(x-1)2ex≥0,∴f(x)在R上单调递增,∴f(x)无极值,不合题意.②当x1>x2,即0<a<1时,f'(x),f(x)随x的变化情况如下表:x(-∞,1)11,1a1a1a,+∞f'(x)+0-0+f(x)↗极大值↘极小值↗∴f(x)在x=1处取得极大值,不合题意.③当x1<x2,即a>1时,f'(x),f(x)随x的变化情况如下表:x-∞,1a1a1a,11(1,+∞)f'(x)+0-0+f(x)↗极大值↘极小值↗∴f(x)在x=1处取得极小值,即a>1满足题意.当a<0时,令f'(x)=0,得x1=1a,x2=1.f'(x),f(x)随x的变化情况如下表:x-∞,1a1a1a,11(1,+∞)f'(x)-0+0-f(x)↘极小值↗极大值↘∴f(x)在x=1处取得极大值,不合题意.综上所述,a的取值范围为(1,+∞).4.(1)解f(x)的定义域为(0,+∞),f'(x)=1x+2ax+2a+1=(x+1)(2ax+1)x.若a≥0,则当x∈(0,+∞)时,f'(x)>0,故f(x)在(0,+∞)单调递增.若a<0,则当x∈0,-12a时,f'(x)>0;当x∈-12a,+∞时,f'(x)<0.故f(x)在0,-12a单调递增,在-12a,+∞单调递减.(2)证明由(1)知,当a<0时,f(x)在x=-12a取得最大值,最大值为f-12a=ln-12a-1-14a.4\n所以f(x)≤-34a-2等价于ln-12a-1-14a≤-34a-2,即ln-12a+12a+1≤0.设g(x)=lnx-x+1,则g'(x)=1x-1.当x∈(0,1)时,g'(x)>0;当x∈(1,+∞)时,g'(x)<0.所以g(x)在(0,1)单调递增,在(1,+∞)单调递减.故当x=1时,g(x)取得最大值,最大值为g(1)=0.所以当x>0时,g(x)≤0.从而当a<0时,ln-12a+12a+1≤0,即f(x)≤-34a-2.4
