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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-aln x.(1)讨论f(x)的单调性;(2)设a0,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x0时,f(x)ax+1,求a的取值范围.3.(2018北京,文19)设函数f(x)=ax2-(3a+1)x+3a+2ex.(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)=ln x+ax2+(2a+1)x.(1)讨论f(x)的单调性;(2)当a0,此时f(x)在(0,+)内单调递增;若a0,则由f(x)=0得x=a,当0xa时,f(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x1x2,而a0,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+a0对x(0,+)恒成立,ax2-3xx+1对x(0,+)恒成立,ax2-3xx+1min.又x2-3xx+1=x+1+4x+1-52(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,+)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a1时,设函数h(x)=(1-x)ex,h(x)=-xex0),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+1ax+1.当0a0(x0),所以g(x)在0,+)内单调递增,而g(0)=0,故exx+1.当0x(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a0时,取x0=5-12,则x0(0,1),f(x0)(1-x0)(1+x0)2=1ax0+1.综上,a的取值范围是1,+).3.解 (1)因为f(x)=ax2-(3a+1)x+3a+2ex,所以f(x)=ax2-(a+1)x+1ex.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+1ex=(ax-1)(x-1)ex.若a1,则当x1a,1时,f(x)0.所以f(x)在x=1处取得极小值.若a1,则当x(0,1)时,ax-1x-10.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a0时,令f(x)=0,得x1=1a,x2=1.当x1=x2,即a=1时,f(x)=(x-1)2ex0,f(x)在R上单调递增,f(x)无极值,不合题意.当x1x2,即0a1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.当x11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a1满足题意.当a0,故f(x)在(0,+)单调递增.若a0;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a0;当x(1,+)时,g(x)0时,g(x)0.从而当a0时,ln-12a+12a+10,即f(x)-34a-2.
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