课时作业(十四)1.D [解析] 依题意有f'(x )=1x ·√2x -2×12×(2x )-12·lnx 2x,故f'12=2+ln21=2+ln 2,故选D .2.A [解析] 当x=1时,f (1)=-2+0=-2,所以切点为(1,-2),由题得f'(x )=-2+1x,所以f'(1)=-2+11=-1,所以切线方程为y+2=-1×(x-1),即x+y+1=0,故选A .3.A [解析] 由题意,f'(x )=2x+2f'(1),则f'(1)=2+2f'(1),解得f'(1)=-2,故f (x )=x 2-4x.故选A .4.B [解析] f'(x )=-sin x-f'π2,令x=π2,得f'π2=-12,即f (x )=cos x+12x.f (0)=1,f'(0)=12,所以l 的方程为y=12x+1,结合选项可知直线2x+y+1=0与直线l 垂直.故选B . 5.32[解析] ∵f'(x )=2x -x ,f'(1)=-12,又∵f (1)=1,∴切点是(1,1),∴切线方程是y-1=-12(x-1),将点(0,a )代入,解得a=12+1=32.6.D [解析] 令f (x )=x 3-4x+4,则f'(x )=3x 2-4,f'(1)=-1,设切线的倾斜角为α,则tan α=-1,可得α=135°.故选D .7.A [解析] 由题意,得f'(x )=ln x+1,∴f'(1)=1,又f (1)=a ,∴切线方程为y=x-1+a.∵切线过原点,∴0=0-1+a ,解得a=1.故选A .8.A [解析] 由题意知,函数f (x )是定义在R 上的奇函数,可得f (0)=0,即f (0)=-m=0,解得m=0,即当x ≤0时,函数f (x )=x 3-2x ,则f'(x )=3x 2-2,所以f'(-2)=3×(-2)2-2=10,由奇函数的导函数为偶函数,可知f'(-2)=f'(2)=10,即曲线y=f (x )在点P (2,f (2))处的切线斜率为10.故选A .9.B [解析] 由y=2x ln x ,得y'=2×ln x+2x×1x=2ln x+2,所以y'|x=e =2+2=4,且y|x=e =2e,所以切线方程为y-2e =4(x-e),即y=4x-2e,此切线与x 轴、y 轴的交点坐标分别为e 2,0,(0,-2e),所以切线与坐标轴围成的三角形面积S=12×e 2×2e =e 22.故选B .10.C [解析] 设直线与曲线切于点(x 0,y 0)(x 0≠0),则切线的斜率k=y 0-1x 0-1=x 03-1x 0-1=x 02+x 0+1,又∵y'=3x 2,∴y'|x=x 0=3x 02,∴2x 02-x 0-1=0,解得x 0=1或x 0=-12,∴过点P (1,1)与曲线y=x 3相切的直线方程为3x-y-2=0或3x-4y+1=0.故选C .11.C [解析] y'=1+1x,当x=1时,切线的斜率k=2,切线方程为y=2(x-1)+1=2x-1,因为它与抛物线相切.所以ax 2+(a+2)x+1=2x-1有唯一解,即ax 2+ax+2=0,故{a ≠0,a 2-8a =0,解得a=8.故选C .12.3 [解析] ∵f (x )=(x 2-a )ln x ,∴f'(x )=2x ln x+x 2-ax,∴f'(1)=1-a=-2,得a=3.13.5 [解析] 将点P (1,4)代入y=ax+b x 2,得a+b=4.函数y=ax+b x 2的导函数为y'=a-2b x3,由曲线在点P 处的切线与直线x+y+3=0垂直,得曲线在点P 处的切线的斜率k=y'|x=1=a-2b=1,联立{a +b =4,a -2b =1,得{a =3,b =1,所以a+2b=5.14.解:(1)由题意得f'(x )=x 2-4x+3, 则f'(x )=(x-2)2-1≥-1,即过曲线C 上任意一点的切线斜率的取值范围是[-1,+∞). (2)设曲线C 的其中一条切线的斜率为k ,则由(2)中条件并结合(1)中结论可知{k ≥-1,-1k≥-1,解得-1≤k<0或k ≥1,所以-1≤x 2-4x+3<0或x 2-4x+3≥1, 解得x ∈(-∞,2-√2]∪(1,3)∪[2+√2,+∞). 15.解:(1)由题意,知f'(x )=1x +1x2,所以f'(1)=2,所以切线方程为y+1=2(x-1), 即2x-y-3=0.(2)由已知,得g (x )=x ln x ,切点坐标为(e,e), 由g'(x )=ln x+1,得g'(e)=2,所以l 2的方程为y-e =2(x-e),即y=2x-e ①.所以直线l 1的斜率为-12,故l 1的方程为y=-12x+2e ②,联立①②,得直线l 1与l 2交点的坐标为65e,75e ,又l 2与x 轴的交点为e 2,0,l 1与x 轴的交点为(4e,0),此封闭图形为三角形,底边m=4e -e 2=7e 2,高h=75e, 所以三角形面积S=12mh=12×7e 2×75e =4920e 2.16.B [解析] 设P 1(x 1,f (x 1)),P 2(x 2,f (x 2)),当0<x<1时,f'(x )=-1x ,当x>1时,f'(x )=1x,故不妨设x 1∈(0,1),x 2∈(1,+∞),故l 1:y=-1x 1(x-x 1)-ln x 1,整理得到l 1:y=-1x 1x-ln x 1+1,l 2:y=1x 2(x-x 2)+ln x 2,整理得到l 2:y=1x 2x+ln x 2-1,所以A (0,1-ln x 1),B (0,ln x 2-1),所以|AB|=|2-ln(x 1x 2)|.因为l 1⊥l 2,所以-1x 1·1x 2=-1,即x 1x 2=1,所以|AB|=2.故选B . 17.2√e [解析] 设P (x 0,y 0),f'(x )=2x+2a ,g'(x )=4a 2x.由题意知,f (x 0)=g (x 0),f'(x 0)=g'(x 0),即x 02+2ax 0=4a 2ln x 0+b ,①2x 0+2a=4a 2x 0,②由②得x 0=a 或x 0=-2a (舍),将x 0=a 代入①,得b=3a 2-4a 2ln a ,a ∈(0,+∞).令h (a )=3a 2-4a 2ln a ,a ∈(0,+∞),则h'(a )=6a-8a ln a-4a=2a (1-4ln a ),当a ∈(0,e 14)时,h'(a )>0,当a ∈(e 14,+∞)时,h'(a )<0.∴h (a )的最大值是h (e 14)=3√e -4√e ln e 14=2√e ,即实数b 的最大值为2√e .课时作业(十五)1.D [解析] 由函数y=12x 2-4ln x ,得y'=x-4x =x 2-4x,x>0.令y'>0,即x 2-4>0且x>0,解得x>2,所以函数y=12x 2-4ln x 的单调递增区间为[2,+∞),故D 正确.2.A [解析] 由题意,函数f (x )=x e|x |,可得其定义域为R,又由f (-x )=-x e |-x |=-xe|x |=-f (x ),即f (-x )=-f (x ),所以函数f (x )是奇函数.当x ∈(-∞,-1)时,f (x )=x e -x =x ·e x ,则f'(x )=e x +x e x =(1+x )e x ,则f'(x )<0,所以函数f (x )在(-∞,-1)上单调递减,故选A .3.C [解析] 由题意可知当x ∈(-∞,0)∪(2,+∞)时,f'(x )>0,函数f (x )是增函数;当x ∈(0,2)时,f'(x )<0,函数f (x )是减函数.结合选项知函数f (x )的图像只能是C .4.B [解析] ∵f (-x )=(-x )2-cos(-x )=x 2-cos x=f (x ),∴f (x )为偶函数,∴f (0.5)=f (-0.5).f'(x )=2x+sin x ,当0<x<π2时,f'(x )=2x+sin x>0,∴函数f (x )在0,π2上单调递增,又0<0.5<0.6<π2,∴f (0)<f (0.5)<f (0.6),即f (0)<f (-0.5)<f (0.6).故选B .5.[0,+∞) [解析] 由题得f'(x )=3x 2+a ≥0在R 上恒成立,即a ≥-3x 2恒成立,故a ≥0,所以a 的取值范围是[0,+∞).6.C [解析] 由题意,得f'(x )=6x 2-6mx+6,由已知条件知当x ∈[1,+∞)时,f'(x )≥0恒成立,设g (x )=6x 2-6mx+6,则g (x )≥0在[1,+∞)上恒成立.方法一:当Δ=36(m 2-4)≤0,即-2≤m ≤2时,g (x )≥0在[1,+∞)上恒成立;当Δ=36(m 2-4)>0,即m<-2或m>2时,则需{m2<1,g (1)=12-6m ≥0,解得m<2,∴m<-2.∴综上,得m ≤2,∴实数m的取值范围是(-∞,2].方法二:问题转化为m ≤x+1x在[1,+∞)上恒成立,而函数y=x+1x≥2,当且仅当x=1时,等号成立,故m ≤2,故选C . 7.A [解析] 构造函数g (x )=f (x )x,则g'(x )=xf '(x )-f (x )x 2.由题意知当x>0时,g'(x )>0,∴函数y=g (x )在(0,+∞)上单调递增.∵π>e,∴g (π)>g (e),即f (π)π>f (e )e,即e f (π)>πf (e).故选A .8.B [解析] 整理f (x )=x [f'(x )-ln x ],得f'(x )=f (x )x+ln x ,因为函数f (x )在(0,+∞)上单调递增,所以f'(x )≥0在(0,+∞)上恒成立,即f (x )x +ln x ≥0,所以f(1e )1e+ln 1e≥0,整理得f1e≥1e.故选B .9.C [解析] 由题意,∀x ∈-π2,π2,f'(x )=e x (-sin x+cos x-a )≤0恒成立,即a ≥cos x-sin x=√2cos x+π4恒成立.当x ∈-π2,π2时,x+π4∈-π4,3π4,∴cos x+π4∈-√22,1,∴√2cos x+π4∈-1,√2,∴实数a 的取值范围是[√2,+∞).故选C .10.B [解析] 函数f (x )=e x -e -x +sin 2x ,定义域为R,f (-x )=e -x -e x +sin(- 2x )=-(e x -e -x +sin 2x )=-f (x ),∴f (x )为R 上的奇函数.又f'(x )=e x +e -x +2cos 2x ≥2+2cos 2x ,当且仅当e x =1,即x=0时等号成立,又2+2cos 2x ≥0恒成立,∴f'(x )≥0恒成立,∴f (x )为R 上的增函数,又f (2x 2-1)+f (x )>0,得f (2x 2-1)>-f (x )=f (-x ),∴2x 2-1>-x ,即2x 2+x-1>0,解得x<-1或x>12,∴x 的取值范围是(-∞,-1)∪12,+∞.故选B .11.C [解析] 不等式e x f (x )>e x -1⇒e x f (x )-e x +1>0,设g (x )=e x f (x )-e x +1,则求不等式e x f (x )>e x -1的解集等价于求不等式g (x )>0的解集.∵g'(x )=e x f (x )+e x f'(x )-e x =e x [f (x )+f'(x )-1],由f (x )>1-f'(x ),得f (x )+f'(x )-1>0,∴g'(x )>0,∴函数g (x )是R 上的增函数.∵f (0)=0,∴g (0)=0,则g (x )>0=g (0),∴x>0.故选C . 12.(-∞,e] [解析] 由题意,得f'(x )=e x -a x≥0在[1,2]上恒成立,则a ≤(x e x )min .令g (x )=x e x ,x ∈[1,2],则g'(x )=(x+1)e x ,当x ∈[1,2]时,g'(x )>0,∴g (x )在[1,2]上单调递增,∴g (x )min =g (1)=e,故a ≤e .13.0,215∪13,+∞ [解析] ∵f (x )=xa -2x 2+ln x ,∴f'(x )=1a -4x+1x .①当函数f (x )在区间[1,2]上单调递增时,不等式f'(x )≥0在[1,2]上恒成立,即1a-4x+1x≥0,1a≥4x-1x在[1,2]上恒成立,由于函数y=4x-1x在区间[1,2]上单调递增,∴y max =4×2-12=152,∴1a ≥152,∵a>0,∴0<a ≤215;②当函数f (x )在区间[1,2]上单调递减时,不等式f'(x )≤0在区间[1,2]上恒成立,即1a-4x+1x≤0,1a≤4x-1x在[1,2]上恒成立,由于函数y=4x-1x在区间[1,2]上单调递增,∴y min =4×1-11=3,∴1a≤3,∵a>0,∴a ≥13.综上,实数a 的取值范围是0,215∪13,+∞.14.解:(1)由题意可得f'(x )=x-a+a -1x ,∵f'(2)=2-a+a -12=0,∴a=3. (2)∵函数f (x )=12x 2-ax+(a-1)ln x ,其中a>1,∴f (x )的定义域为(0,+∞),f'(x )=x-a+a -1x =(x -1)(x+1-a )x =(x -1)[x -(a -1)]x. 令f'(x )=0,得x 1=1,x 2=a-1.①当a-1=1,即a=2时,f'(x )=(x -1)2x≥0恒成立,故f (x )在(0,+∞)上单调递增.②当0<a-1<1,即1<a<2时, 由f'(x )<0,得a-1<x<1; 由f'(x )>0,得0<x<a-1或x>1.故f (x )在(a-1,1)上单调递减,在(0,a-1),(1,+∞)上单调递增.③当a-1>1,即a>2时,由f'(x )<0,得1<x<a-1;由f'(x )>0,得0<x<1,或x>a-1. 故f (x )在(1,a-1)上单调递减,在(0,1),(a-1,+∞)上单调递增. 综上可得,当a=2时,f (x )在(0,+∞)上单调递增;当1<a<2时,f (x )在(a-1,1)上单调递减,在(0,a-1),(1,+∞)上单调递增; 当a>2时,f (x )在(1,a-1)上单调递减,在(0,1),(a-1,+∞)上单调递增. 15.解:(1)由题意得f (x )的定义域为(0,+∞). 当a=-1时,f (x )=-ln x+12x 2+1,则f'(x )=-1x+x=x 2-1x(x>0), 令f'(x )>0,解得x>1,∴f (x )的单调递增区间为(1,+∞). (2)f'(x )=a x+x+(a+1)=x 2+(a+1)x+a x =(x+a )(x+1)x(x>0). ①当a ≥0时,f'(x )>0在(0,+∞)上恒成立,∴f (x )在(0,+∞)上单调递增,可知a ≥0满足题意; ②当a<0时,-a>0,∴当x ∈(0,-a )时,f'(x )<0,当x ∈(-a ,+∞)时,f'(x )>0,∴f (x )在(0,-a )上单调递减,在(-a ,+∞)上单调递增,不满足题意. 综上所述,实数a 的取值范围为[0,+∞). 16.解:(1)由题意得f'(x )=e ax +1-2x+1,x ∈(-1,+∞), 由f'(1)=e,知e a =e,∴a=1.(2)由题知,当x ∈(0,+∞)时,f'(x )=e ax +1-2x+1<0有解, 当x ∈[1,+∞)时,f'(x )=e ax +1-2x+1>0恒成立, 不存在单调递减区间; 当x ∈(0,1)时,f'(x )=e ax +1-2x+1<0有解 等价于ln1-x1+x-ax>0有解. 设φ(x )=ln 1-x1+x-ax ,x ∈(0,1),则φ'(x )=2x 2-1-a ,x ∈(0,1), 当x ∈(0,1)时,2x 2-1<-2. ①当a ≥-2时,φ'(x )=2x 2-1-a<0恒成立, 则φ(x )=ln1-x1+x-ax 在(0,1)上单调递减, 所以φ(x )<0恒成立,不符合题意;②当a<-2时,0<a+2a<1,当x ∈0,√a+2a时,φ'(x )=2x 2-1-a>0, 则φ(x )=ln1-x1+x-ax 在0,√a+2a上单调递增,所以φ(x )>0,即ln 1-x1+x-ax>0. 综上所述,a<-2.课时作业(十六)1.A [解析] 当x<-2时,f'(x )<0,当-2<x<12时,f'(x )>0,当12<x<2时,f'(x )<0,当x>2时,f'(x )>0,所以-2为f (x )的极小值点,12为f (x )的极大值点,2为f (x )的极小值点,所以只有A 正确. 2.D [解析] 由题意得f'(x )=1-lnxx 2,令f'(x )>0得0<x<e,令f'(x )<0得x>e,故f (x )在(0,e)上单调递增,在(e,+∞)上单调递减,所以f (x )有极大值,也是最大值,最大值为f (e)=1e,无极小值和最小值,故选D .3.C [解析] f'(x )=1-2sin x ,令f'(x )=0,得x=π6或x=5π6,所以f (x )=x+2cos x 在区间0,π6上是增函数,在区间π6,5π6上是减函数,在5π6,π上是增函数.故x=5π6是函数f (x )的极小值点,故选C .4.B [解析] f'(x )=3x 2-6,令f'(x )>0,解得x>√2或x<-√2,令f'(x )<0,解得-√2<x<√2,所以f (x )在(-∞,-√2)上单调递增,在(-√2,√2)上单调递减,在(√2,+∞)上单调递增,故x=√2是函数f (x )的极小值点,故a=√2,故选B .5.2-2ln 2 [解析] f'(x )=e x -2,令f'(x )=0,得x=ln 2,所以f (x )在(-∞,ln 2)上单调递减,在(ln 2,+∞)上单调递增,所以f (x )min =f (ln 2)=2-2ln 2.6.C [解析] 因为f (x )=kx-ln x ,所以f'(x )=k-1x,又f (x )=kx-ln x 的极值点为x=2,所以f'(2)=0,即k=12.经验证可知,k=12符合题意,故选C .7.A [解析] 由题意知,当x ∈(0,2)时,f (x )的最大值为-1.当x ∈(0,2)时,f'(x )=1x-a ,令f'(x )=0,得x=1a,当0<x<1a时,f'(x )>0;当x>1a时,f'(x )<0.所以f (x )max =f 1a=-ln a-1=-1,解得a=1.8.AB [解析]由f'(x )的图像知,当-1≤x<0或2<x<4时,f'(x )>0,函数f (x )为增函数,当0<x<2或4<x ≤5时,f'(x )<0,函数f (x )为减函数,故当x=0时,函数f (x )取得极大值,当x=4时,函数f (x )取得极大值,即函数f (x )有2个极大值点,故A 正确;函数f (x )在[0,2]上是减函数,故B 正确;若x ∈[-1,t ]时,f (x )的最大值是2,则t 满足0≤t ≤5,即t 的最大值是5,故C 错误;由y=f (x )-a=0得f (x )=a ,若f (2)≤1,则当1<a<2时,f (x )=a 有4个根,若1<f (2)<2,则当1<a<2时,f (x )=a 不一定有4个根,还有可能有2个或3个根,故函数y=f (x )-a 有4个零点不一定正确,故D 错误.故选AB .9.C [解析] 存在x 1,x 2∈R,使得f (x 2)≤g (x 1)成立,则f (x )min ≤g (x )max ,由题得f'(x )=e x +x e x =(x+1)e x ,所以函数f (x )在(-∞,-1)上单调递减,在(-1,+∞)上单调递增,所以f (x )min =f (-1)=-1e.又g (x )max =g (-1)=a ,所以a ≥-1e.故选C .10.(e -1)x+y+1=0 [解析] 由题得f'(x )=1x-k ,则f'1e=e -k=0,所以k=e .f'(1)=1-k=1-e,f (1)=-k=-e,故切点为(1,-e),所以所求切线方程为y+e =(1-e)(x-1),即(e -1)x+y+1=0.11.e 2 [解析] f'(x )=a-1x ,x ∈(0,e],当a ≤1e时,f'(x )≤0,∴f (x )在(0,e]上是减函数,f (x )最小值=f (e)=a e -1=3,解得a=4e(舍去).当a>1e时,当0<x<1a时,f'(x )<0,f (x )单调递减,当1a<x ≤e 时,f'(x )>0,f (x )单调递增,∴f (x )最小值=f (x )极小值=f 1a=1-ln 1a=3,得a=e 2,符合题意.故答案为e 2.12.e -1 [解析] 由题得k ≤e x -lnx -1x,令h (x )=e x -lnx -1x(x>0),则h'(x )=e x (x -1)+lnxx 2,可知当x ∈(0,1)时,h (x )单调递减,当x ∈(1,+∞)时,h (x )单调递增,故当x=1时,h (x )取到最小值e -1,故k 的最大值为e -1. 13.(-∞,e +1e -14] [2e -2+ln22,3e -92+ln33) [解析] 若对于任意的x 1∈[12,e],存在x 2∈[12,e],使f'(x 1)≤g (x 2),则只需函数f'(x )在[12,e]上的最大值小于等于函数g (x )在[12,e]上的最大值.f'(x )=x 2-2e x+a 的图像的对称轴为直线x=e,易知函数f'(x )在[12,e]上的最大值为f'(12)=14-e +a.g'(x )=1-lnxx 2,令g'(x )=0,解得x=e,当x ∈(0,e)时,g'(x )>0,g (x )为增函数,故函数g (x )在[12,e]上的最大值为g (e)=1e,∴14-e +a ≤1e,∴a ≤e+1e -14,即实数a 的取值范围为(-∞,e +1e -14].不等式f (x )+16x 3<xg (x ),即为12x 3-e x 2+ax<ln x ,∴a<lnx x +e x-12x 2有且仅有一个整数解.令h (x )=lnx x +e x-12x 2,则h'(x )=1-lnxx 2+e -x ,易知函数h'(x )在(0,+∞)上为减函数,且h'(e)=0,∴当x ∈(0,e)时,h'(x )>0,函数h (x )单调递增,当x ∈(e,+∞)时,h'(x )<0,函数h (x )单调递减,∴函数h (x )在x=e 处取得最大值,易知h (4)<h (2)<h (3),结合函数h (x )的图像可知,要使a<lnx x +e x-12x 2有且仅有一个整数解,只需h (2)≤a<h (3),可得2e -2+ln22≤a<3e -92+ln33. 14.解:(1)由题可知f'(x )=2a+b x 2+1x ,∵函数f (x )在x=1,x=12处取得极值,∴f'(1)=0,f'12=0,即{2a +b +1=0,2a +4b +2=0,解得{a =-13,b =-13.(2)由(1)可得f (x )=-23x+13x+ln x , 令f'(x )=-23-13x 2+1x>0,得-(2x-1)(x-1)>0, 即(2x-1)(x-1)<0,∴12<x<1,∴f(x)在12,1上单调递增,在0,12,(1,+∞)上单调递减.又∵x∈[14,1],∴f(x)在[14,12]上单调递减,在12,1上单调递增,∴f(x)在[14,1]上的极小值为f12=13-ln 2,又f14=76-ln 4,f(1)=-13,且f14-f(1)=96-ln 4>0,∴f(x)max=f14=76-ln 4,∴要使对任意x∈[14,1],f(x)<c恒成立,则c>76-ln 4.15.解:(1)当a=1时,g(x)=f(x)x=x ln x,可得g'(x)=1+ln x,令g'(x)=0,可得x=1e.当x∈0,1e 时,g'(x)<0,g(x)单调递减,∴单调递减区间为0,1e;当x∈1e ,+∞时,g'(x)>0,g(x)单调递增,∴单调递增区间为1e,+∞.∴当x=1e 时,g(x)取得极小值-1e.(2)f'(x)=2x(ln x+ln a)+x,f'(x)x2=2x(lnx+lna)+xx2≤1,即2ln x+2ln a+1≤x,2ln a≤x-2ln x-1对任意的x>0恒成立,设m(x)=x-2ln x-1(x>0),可得m'(x)=x-2x,令m'(x)=0,可得x=2,当0<x<2时,m'(x)<0,函数单调递减,当x>2时,m'(x)>0,函数单调递增,∴当x=2时,m(x)有最小值,且最小值为m(2)=1-2ln 2,∴2ln a≤1-2ln 2,得0<a≤√e2.16.解:(1)f(x)的定义域为(-1,+∞),f'(x)=a ln(x+1)-2x.由f(x)是减函数,得对任意的x∈(-1,+∞),f'(x)=a ln(x+1)-2x≤0恒成立.设g(x)=a ln(x+1)-2x,则g'(x)=-2[x-(a2-1)]x+1,由a>0知a2-1>-1.当x∈(-1,a2-1)时,g'(x)>0;当x∈(a2-1,+∞)时,g'(x)<0.∴g(x)在(-1,a2-1)上单调递增,在(a2-1,+∞)上单调递减,∴当x=a2-1时g(x)取得最大值.又∵g(0)=0,∴对任意的x∈(-1,+∞),g(x)≤g(0)恒成立,即g(x)的最大值为g(0),∴a2-1=0,解得a=2.(2)证明:由f (x )是减函数,且f (0)=0,可得当x>0时,f (x )<0,∴f (n )<0(n ∈N *),即2(n+1)ln(n+1)<n 2+2n , 两边同除以2(n+1)2,得ln (n+1)n+1<12·n n+1·n+2n+1,即a n <12·n n+1·n+2n+1. 故T n =a 1a 2a 3…a n <12n ·(12×23×34×…×n n+1)(32×43×54×…×n+2n+1)=12n+1·n+2n+1, ∴ln[(n+2)T n ]<ln [(n+2)22n+1(n+1)]=2ln(n+2)-ln(n+1)-(n+1)ln 2①.记h (x )=2ln(x+2)-ln(x+1)-(x+1)ln 2+x 2-1,x ∈[1,+∞), 则h'(x )=2x+2-1x+1-ln 2+12=x x 2+3x+2-ln 2+12=1x+2x+3-ln 2+12. ∵y=x+2x 在[2,+∞)上单调递增, ∴h'(x )在[2,+∞)上单调递减.又h'(2)=16-ln 2+12=13×(2-3ln 2)=13×(2-ln 8)<0,∴当x ∈[2,+∞)时,h'(x )<0恒成立, ∴h (x )在[2,+∞)上单调递减,即当x ∈[2,+∞)时,h (x )≤h (2)=2ln 4-ln 3-3ln 2=ln 2-ln 3<0,∴当n ≥2时,h (n )<0.又∵h (1)=2ln 3-ln 2-2ln 2-12=ln 98-ln √e <0,∴当n ∈N *时,h (n )<0,即2ln(n+2)-ln(n+1)-(n+1)ln 2<1-n2②. 由①②可得,ln[(n+2)T n ]<1-n 2.专题突破训练(一)1.解:(1)当a=-12时,f'(x )=-x+1-lnxx 2,所以切线的斜率k=f'(e)=-1e,又因为f (e)=-1+1e ,所以切线方程为y=-1ex+1e. (2)由题意知当x ∈(0,+∞)时,f'(x )=2ax+1-lnxx 2≥0,即2a ≥lnx -1x, 令g (x )=lnx -1x(x>0),则g'(x )=2-lnxx 2,令g'(x )>0,得x ∈(0,e 2),令g'(x )<0,得x ∈(e 2,+∞),所以g (x )max =g (e 2)=1e2, 所以a ≥12e -2.2.解:(1)依题意,f (x )=e x -mx ,f'(x )=e x -m.①若m ≤0,则f'(x )>0,函数f (x )在R 上单调递增. ②若m>0,令f'(x )=0,得x=ln m.当x<ln m 时,f'(x )<0,函数f (x )在(-∞,ln m )上单调递减, 当x>ln m 时,f'(x )>0,函数f (x )在(ln m ,+∞)上单调递增.综上所述,当m ≤0时,函数f (x )在R 上单调递增;当m>0时,函数f (x )在(-∞,ln m )上单调递减,在(ln m ,+∞)上单调递增. (2)依题意,当x>0时,e x -mx ≥x 2+1恒成立,即 m ≤e x x-x-1x对任意x>0恒成立. 令g (x )=e x x-x-1x(x>0),则 g'(x )=e x (x -1)x 2-1+1x 2=(e x -x -1)(x -1)x 2. 由(1)可知,当m=1时,f (x )=e x -x 在(0,+∞)上单调递增, 故f (x )>f (0)=1,即e x -x>1,得e x -x-1>0, 所以方程g'(x )=0有唯一解x=1.当0<x<1时,g'(x )<0,g (x )在(0,1)上单调递减, 当x>1时,g'(x )>0,g (x )在(1,+∞)上单调递增, 所以g (x )min =g (1)=e -2,所以m ≤e -2. 3.解:(1)函数f (x )可化为f (x )={x -lnx -a ,x ≥a ,a -x -lnx ,0<x <a ,当0<x<a 时,f'(x )=-1-1x<0,从而f (x )在(0,a )上单调递减. 当x ≥a 时,f'(x )=1-1x =x -1x,此时要考虑a 与1的大小.若a ≥1,则f'(x )≥0,故f (x )在[a ,+∞)上单调递增,若0<a<1,则当a ≤x<1时,f'(x )<0,当x>1时,f'(x )>0,故f (x )在[a ,1)上单调递减,在(1,+∞)上单调递增,而f (x )在x=a 处连续,所以当a>0时,f (x )在(0,a )上单调递减,当a ≥1时,f (x )在[a ,+∞)上单调递增,当0<a<1时,f (x )在[a ,1)上单调递减,在(1,+∞)上单调递增.(2)由(1)可知当a=1,x>1时,x-1-ln x>0,即ln x<x-1,所以lnx x <1-1x,所以 ln 2222+ln 3232+…+ln n 2n 2<1-122+1-132+…+1-1n 2=n-1-122+132+…+1n 2<n-1-12×3+13×4+…+1n (n+1)=n-1-12-1n+1=(n-1)-n -12(n+1)=2n 2-2-n+12(n+1)=(n -1)(2n+1)2(n+1),即ln2222+ln 3232+…+ln n 2n 2<(n -1)(2n+1)2(n+1). 4.解:(1)由f (x )=e x -ax 2,得f'(x )=e x -2ax.因为曲线y=f (x )在点(1,f (1))处的切线与直线x+(e -2)y=0垂直,所以f'(1)=e -2a=e -2,所以a=1,即f (x )=e x -x 2,f'(x )=e x -2x.令g (x )=e x -2x ,则g'(x )=e x -2.当x ∈(-∞,ln 2)时,g'(x )<0,g (x )单调递减; 当x ∈(ln 2,+∞)时,g'(x )>0,g (x )单调递增. 所以g (x )min =g (ln 2)=2-2ln 2>0, 所以f'(x )>0,f (x )单调递增.所以f (x )的单调递增区间为(-∞,+∞),无减区间.(2)证明:由(1)知f (x )=e x -x 2,f (1)=e -1,所以曲线y=f (x )在点(1,f (1))处的切线的方程为y-(e -1)=(e -2)(x-1),即y=(e -2)x+1.令h (x )=e x -x 2-(e -2)x-1,则h'(x )=e x -2x-(e -2)=e x -e -2(x-1), 易知h'(1)=0,令m (x )=h'(x ),则m'(x )=e x -2, 当x ∈(-∞,ln 2)时,m'(x )<0,h'(x )单调递减; 当x ∈(ln 2,+∞)时,m'(x )>0,h'(x )单调递增.因为h'(1)=0,所以h'(x )min =h'(ln 2)=4-e -2ln 2<0,因为h'(0)=3-e >0,所以存在x 0∈(0,1),当x ∈(0,x 0)时,h'(x )>0,h (x )单调递增;当x ∈(x 0,1)时,h'(x )<0,h (x )单调递减;当x ∈(1,+∞)时,h'(x )>0,h (x )单调递增. 又h (0)=h (1)=0,所以当x>0时,h (x )≥0,即e x -x 2-(e -2)x-1≥0, 所以e x -(e -2)x-1≥x 2. 令φ(x )=ln x-x ,则φ'(x )=1x-1=1-xx,所以当x ∈(0,1)时,φ'(x )>0,φ(x )单调递增,当x ∈(1,+∞)时,φ'(x )<0,φ(x )单调递减,所以φ(x )≤φ(1)=-1,即ln x+1≤x.因为x>0,所以x (ln x+1)≤x 2,所以当x>0时,e x -(e -2)x-1≥x (ln x+1), 即当x>0时,e x -e x-1≥x (ln x-1). 5.解:(1)∵F (x )=g (x )f (x )=ax 2+x+1e x, ∴F'(x )=-ax 2+(2a -1)x e x =-ax(x -2a -1a )e x.①若a=12,则F'(x )=-ax 2e x≤0,∴F (x )在R 上单调递减.②若a>12,则2a -1a>0, 当x<0或x>2a -1a时,F'(x )<0,当0<x<2a -1a时,F'(x )>0, ∴F (x )在(-∞,0),2a -1a,+∞上单调递减,在0,2a -1a上单调递增.③若0<a<12,则2a -1a<0, 当x<2a -1a或x>0时,F'(x )<0,当2a -1a<x<0时,F'(x )>0, ∴F (x )在-∞,2a -1a ,(0,+∞)上单调递减,在2a -1a,0上单调递增.(2)证明:∵0<a ≤12,∴ax 2+x+1≤12x 2+x+1. 设h (x )=e x -12x 2-x-1,则h'(x )=e x -x-1.设p (x )=h'(x )=e x -x-1,则p'(x )=e x -1,当x ∈(0,+∞)时,p'(x )>0恒成立.∴h'(x )在(0,+∞)上单调递增.又∵h'(0)=0,∴当x ∈(0,+∞)时,h'(x )>0,∴h (x )在(0,+∞)上单调递增, ∴h (x )>h (0)=0,∴e x -12x 2-x-1>0,即e x >12x 2+x+1. ∴e x >12x 2+x+1≥ax 2+x+1,∴所以f (x )>g (x )在(0,+∞)上恒成立. 6.解:(1)由题得f'(x )=x-2+a x =x 2-2x+ax,其中x>0.令g (x )=x 2-2x+a ,x>0,其图像的对称轴为x=1,Δ=4-4a. 若a ≥1,则Δ≤0,此时g (x )≥0,则f'(x )≥0,所以f (x )在(0,+∞)上单调递增.若0<a<1,则Δ>0,此时x2-2x+a=0在R上有两个根x1=1-√1-a,x2=1+√1-a,且0<x1<1<x2,所以当x∈(0,x1)时,g(x)>0,则f'(x)>0,f(x)单调递增;当x∈(x1,x2)时,g(x)<0,则f'(x)<0,f(x)单调递减;当x∈(x2,+∞)时,g(x)>0,则f'(x)>0,f(x)单调递增.综上,当a≥1时,f(x)在(0,+∞)上单调递增;当 0<a<1时,f(x)在(0,1-√1-a)上单调递增,在(1-√1-a,1+√1-a)上单调递减,在(1+√1-a,+∞)上单调递增.(2)证明:由(1)知,当0<a<1时,f(x)有两个极值点x1,x2,且x1+x2=2,x1x2=a,所以f(x1)+f(x2)=12x12-2x1+a ln x1+12x22-2x2+a ln x2=12(x12+x22)-2(x1+x2)+a(ln x1+ln x2)=12[(x1+x2)2-2x1x2]-2(x1+x2)+a ln(x1x2)=12(22-2a)-4+a ln a=a ln a-a-2.令h(x)=x ln x-x-2,0<x<1,则只需证明-3<h(x)<-2,由于h'(x)=ln x<0,故h(x)在(0,1)上单调递减,所以h(x)>h(1)=-3.又当0<x<1时,ln x-1<-1,x(ln x-1)<0,故h(x)=x ln x-x-2=x(ln x-1)-2<-2,所以对任意的0<x<1,-3<h(x)<-2.综上,-3<f(x1)+f(x2)<-2.专题突破训练(二)1.解:(1)函数f(x)的定义域为(0,+∞),f'(x)=(x+√m)(x-√m)x.当0<x<√m时,f'(x)<0,函数f(x)单调递减;当x>√m时,f'(x)>0,函数f(x)单调递增.综上,函数f(x)的单调递增区间是(√m,+∞),单调递减区间是(0,√m).(2)令F(x)=f(x)-g(x)=-12x2+(m+1)x-m ln x,x>0,题中问题等价于求函数F(x)的零点个数.易知F'(x)=-(x-1)(x-m)x,当m=1时,F'(x)≤0,函数F(x)为减函数,因为F(1)=32>0,F(4)=-ln 4<0,所以F(x)有唯一零点;当m>1时,0<x<1或x>m时,F'(x)<0,1<x<m时,F'(x)>0,所以函数F(x)在(0,1),(m,+∞)上单调递减,在(1,m)上单调递增,因为F(1)=m+12>0,F(2m+2)=-m ln(2m+2)<0,所以F(x)有唯一零点.综上,函数F(x)有唯一零点,即函数f(x)与g(x)的图像总有一个交点.2.解:(1)由f (x )=e x +ax-a ,得f'(x )=e x +a.∵函数f (x )在x=0处取得极值,∴f'(0)=e 0+a=0,∴a=-1,∴f (x )=e x -x+1,f'(x )=e x -1.∴当x ∈(-∞,0)时,f'(x )<0,f (x )单调递减;当x ∈(0,+∞)时,f'(x )>0,f (x )单调递增.易知f (x )在[-2,0)上单调递减,在(0,1]上单调递增,且f (-2)=1e2+3,f (1)=e,f (-2)>f (1),∴f (x )在[-2,1]上的最大值是1e 2+3.(2)f'(x )=e x +a.①当a>0时,f'(x )>0,f (x )在R 上单调递增,且当x>1时,f (x )=e x +a (x-1)>0, 当x<0时,取x=-1a,则f -1a<1+a -1a-1=-a<0,∴函数f (x )存在零点,不满足题意.②当a<0时,令f'(x )=e x +a=0,则x=ln(-a ). 当x ∈(-∞,ln(-a ))时,f'(x )<0,f (x )单调递减; 当x ∈(ln(-a ),+∞)时,f'(x )>0,f (x )单调递增. ∴当x=ln(-a )时,f (x )取得极小值,也是最小值.函数f (x )不存在零点,等价于f [ln(-a )]=e ln(-a )+a ln(-a )-a=-2a+a ln(-a )>0,解得-e 2<a<0. 综上所述,所求实数a 的取值范围是(-e 2,0). 3.解:(1)证明:当a=0时,f (x )=e x -x. 令g (x )=f (x )-x=e x -x-x=e x -2x , 则g'(x )=e x -2,当g'(x )=0时,x=ln 2,当x<ln 2时,g'(x )<0,当x>ln 2时,g'(x )>0, 所以g (x )在(-∞,ln 2)上单调递减,在(ln 2,+∞)上单调递增, 所以x=ln 2是g (x )的极小值点,也是最小值点, 即g (x )min =g (ln 2)=e ln 2-2ln 2=2ln e 2>0, 故当a=0时,f (x )>x 恒成立. (2)f'(x )=e x -1,由f'(x )=0得x=0.当x<0时,f'(x )<0,当x>0时,f'(x )>0,所以f (x )在(-∞,0)上单调递减,在(0,+∞)上单调递增, 所以x=0是函数f (x )的极小值点,也是最小值点, 即f (x )min =f (0)=1-a.当1-a>0,即a<1时,f (x )没有零点. 当1-a=0,即a=1时,f (x )只有一个零点.当1-a<0,即a>1时,因为f (-a )=e -a -(-a )-a=e -a >0,所以f (x )在(-a ,0)上有一个零点; 由(1)得e x >2x ,令x=a ,得e a >2a ,所以f (a )=e a -a-a=e a -2a>0,于是f (x )在(0,a )上有一个零点, 因此,当a>1时,f (x )有两个零点. 综上,当a<1时,f (x )没有零点; 当a=1时,f (x )只有一个零点; 当a>1时,f (x )有两个零点.4.解:(1)当a=2时,f (x )=(2x 2+x+2)e x ,则f'(x )=(2x 2+5x+3)e x =(x+1)(2x+3)e x . 令f'(x )=0,解得x=-1或x=-32.当x 变化时,f (x ),f'(x )的变化情况如下表所示:x-∞,-32-32-32,-1-1 (-1,+∞)f'(x ) + 0 - 0 + f (x ) 单调递增 极大单调递减 极小单调递增值 值∴f (x )极大值=f -32=5e -32,f (x )极小值=f (-1)=3e -1.(2)∵a=1,∴f (x )=(x 2+x+2)e x ,设h (x )=f (x )-x-4=(x 2+x+2)e x -x-4,则h'(x )=(x 2+3x+3)e x -1, 令φ(x )=(x 2+3x+3)e x -1,则φ'(x )=(x 2+5x+6)e x , 令φ'(x )=(x 2+5x+6)e x =0,解得x=-2或x=-3. 当x 变化时,φ'(x ),φ(x )的变化情况如下表所示:x (-∞,-3) -3 (-3,-2)-2 (-2,+∞) φ'(x ) + 0- 0 +φ(x ) 单调递增极大值 单调递减 极小值单调递增 ∴φ(x )极大值=φ(-3)=3e 3-1<0,φ(x )极小值=φ(-2)=1e 2-1<0. ∵φ(-1)=1e -1<0,φ(0)=2>0,∴存在x 0∈(-1,0),使得φ(x 0)=0,∴当x ∈(-∞,x 0)时,φ(x )<0,当x ∈(x 0,+∞)时,φ(x )>0,∴h (x )在(-∞,x 0)上单调递减,在(x 0,+∞)上单调递增. ∵h (-4)=14e 4>0,h (-3)=8e 3-1<0,h (0)=-2<0,h (1)=4e -5>0,由零点的存在性定理可知,h (x )=0的根x 1∈(-4,-3),x 2∈(0,1),∴t=-4,0. 5.证明:(1)当a=1时,f (x )=(x-2)ln x+x-1,∴f'(x )=ln x+x -2x +1, 若f (x )的图像与x 轴相切,设切点为(x 0,0),∴f (x 0)=(x 0-2)ln x 0+x 0-1=0, f'(x 0)=ln x 0+x 0-2x 0+1=0, 解得x 0=1或x 0=4(舍去),∴x 0=1,∴切点为(1,0),故f (x )的图像与x 轴相切. (2)令f (x )=(x-2)ln x+ax-1=0,∴a=1x -(x -2)lnx x =1x -ln x+2lnxx, 设g (x )=1x-ln x+2lnxx,x>0, ∴g'(x )=-1x 2-1x +2(1-lnx )x 2=1-x -2lnxx2, 令h (x )=1-x-2ln x ,易知h (x )在(0,+∞)上为减函数.∵h (1)=1-1-2ln 1=0,∴当x ∈(0,1)时,g'(x )>0,函数g (x )单调递增, 当x ∈(1,+∞)时,g'(x )<0,函数g (x )单调递减,∴g (x )max =g (1)=1,当x →0时,g (x )→-∞,当x →+∞时,g (x )→-∞,∴当a<1时,y=g (x )的图像与直线y=a 有两个交点, 即当a<1时,f (x )存在两个零点.6.解:(1)由题可得g (x )=1x-2a ln x+3x ,定义域为(0,+∞),∴g'(x )=-1x 2-2ax +3=3x 2-2ax -1x 2. ∵a=1, ∴g'(x )=3x 2-2x -1x 2, 令g'(x )=3x 2-2x -1x 2=0,得x=1或x=-13(舍去).令g'(x )=3x 2-2x -1x 2>0,得x>1或x<-13,结合定义域得x>1,令g'(x )=3x 2-2x -1x 2<0得-13<x<1,结合定义域得0<x<1.∴g (x )的单调递增区间为(1,+∞),单调递减区间为(0,1). (2)函数f (x )有唯一的零点等价于1x=2a ln x 有唯一的实数根,显然a ≠0,则1x=2a ln x 有唯一的实数根等价于关于x 的方程1a=2x ln x 有唯一的实数根. 构造函数φ(x )=2x ln x (x>0),则φ'(x )=2(1+ln x ), 令φ'(x )=2(1+ln x )=0,解得x=1e;令φ'(x )>0,解得x>1e,则函数φ(x )在1e,+∞上单调递增; 令φ'(x )<0,解得0<x<1e,则函数φ(x )在0,1e上单调递减;∴φ(x )的极小值为φ1e=-2e.如图,作出函数φ(x )的大致图像,则要使方程1a=2x ln x 有唯一实数根,只需要直线y=1a与φ(x )的图像有唯一交点,∴1a =-2e 或1a >0,解得a=-e2或a>0, 故实数a 的取值范围为{-e 2}∪(0,+∞)。