2021新高考数学二轮复习：专题突破练8 函数的单调性、极值点、极值、最值

专题突破练8函数的单调性、极值点、极值、最
值
,其中a为常数.
1.设函数f(x)=a ln x+x-1
x+1
(1)若a=0,求曲线y=f(x)在点(1,f(1))处的切线方程;
(2)讨论函数f(x)的单调性.
2.已知函数f(x)=e x-ax2-bx-1,其中a,b∈R,e=2.718 28…为自然对数的底数.设g(x)是函数f(x)的导函数,求函数g(x)在区间[0,1]上的最小值.
3.(2020山东济南三模,21)已知函数f(x)=a ln(x+b)-√x.
(1)若a=1,b=0,求f(x)的最大值;
(2)当b>0时,讨论f(x)极值点的个数.
4.(2020山西太原三模,21)已知函数f (x )=ln x+kx. (1)当k=-1时,求函数f (x )的极值点;
(2)当k=0时,若f (x )+b x
-a ≥0(a ,b ∈R )恒成立,求e a-1-b+1的最大值.
5.(2020山东烟台模拟,22)已知函数f (x )=a 2
x 2-x (ln x-b-1),a ,b ∈R . (1)略;
(2)若f (x )在(0,+∞)上单调递增,且c ≤e 2a+b ,求c 的最大值.
6.已知函数f (x )=x 3+32
x 2-4ax+1(a ∈R ).
(1)若函数f (x )有两个极值点,且都小于0,求a 的取值范围; (2)若函数h (x )=a (a-1)ln x-x 3+3x+f (x ),求函数h (x )的单调区间.
7.(2020山东济宁6月模拟,22)已知函数f (x )=x-a ln x.
(1)若曲线y=f (x )+b (a ,b ∈R )在x=1处的切线方程为x+y-3=0,求a ,b 的值; (2)求函数g (x )=f (x )+a+1x (a ∈R )的极值点;
(3)设h (x )=1
a f (x )+a e x -x a +ln a (a>0),若当x>a 时,不等式h (x )≥0恒成立,求a 的最小值.
8.已知函数f(x)=ax2+x ln x(a为常数,a∈R,e为自然对数的底数,e=2.718 28…).
(1)若函数f(x)≤0恒成立,求实数a的取值范围;
(2)若曲线y=f(x)在点(e,f(e))处的切线方程为y=(2e+2)x-e2-e,k∈Z且k<f(x)
x-1
对任意x>1都成立,求k的最大值.
专题突破练8函数的单调性、
极值点、极值、最值
1.解(1)当a=0时,f(x)=x-1
x+1
,x∈(0,+∞).
此时f'(x)=
2
(x+1)2
,于是f'(1)=1
2
,f(1)=0,所以曲线y=f(x)在点(1,f(1))处的切线方程为y-
0=1
2
(x-1),
即x-2y-1=0.
(2)函数f(x)的定义域为(0,+∞),f'(x)=a
x +2
(x+1)2
=ax2+2(a+1)x+a
x(x+1)2
.
①当a≥0时,f'(x)>0,所以函数f(x)在(0,+∞)上单调递增.
②当a<0时,令g(x)=ax2+2(a+1)x+a,则Δ=4(a+1)2-4a2=4(2a+1).
(ⅰ)当a≤-1
2
时,Δ≤0,所以g(x)≤0,于是f'(x)≤0,所以函数f(x)在(0,+∞)上单调递减.
(ⅱ)当-1
2
<a<0时,Δ>0,此时g(x)=0有两个不相等的实数根,分别是
x1=-(a+1)+√2a+1
a ,x2=-(a+1)-√2a+1
a
,x1<x2.下面判断x1,x2是否在定义域(0,+∞)内.由韦达定
理{x1+x2=-2(a+1)
a
>0,
x1x2=1>0,
可得0<x1<x2.
当0<x<x 1或x>x 2时,有g (x )<0,f'(x )<0,所以函数f (x )单调递减;
当x 1<x<x 2时,有g (x )>0,f'(x )>0,所以函数f (x )单调递增.综上所述,当a ≥0时,函数f (x )在(0,+∞)上单调递增;当a ≤-1
2时,函数f (x )在(0,+∞)上单调递减;当-1
2<a<0时,函数f (x )在0,
-(a+1)+√2a+1
a
,
-(a+1)-√2a+1
a
,+∞上单调递减,在
-(a+1)+√2a+1a ,
-(a+1)-√2a+1
a
上单调递增.
2.解 g (x )=f'(x )=e x -2ax-b ,g'(x )=e x -2a.因为x ∈[0,1],所以1-2a ≤g'(x )≤e -2a.
①若1-2a ≥0,即a ≤1
时,有g'(x )=e x -2a ≥0,所以函数g (x )在区间[0,1]上单调递增,于是[g (x )]min =g (0)=1-b.
②若1<2a<e,即1
2<a<e
2时,当0≤x<ln(2a )时,g'(x )=e x -2a<0,当ln(2a )≤x ≤1时,g'(x )=e x -2a ≥0,所以函数g (x )在区间[0,ln(2a ))上单调递减,在区间[ln(2a ),1]上单调递增,于是[g (x )]min =g [ln(2a )]=2a-2a ln(2a )-b.
③若e -2a ≤0,即a ≥e
2时,有g'(x )=e x -2a ≤0,所以函数g (x )在区间[0,1]上单调递减,于是[g (x )]min =g (1)=e -2a-b.
综上所述,g (x )在区间[0,1]上的最小值为[g (x )]min =
{
1-b ,a ≤1
2,
2a -2aln (2a )-b ,12<a <e
2
,e -2a -b ,a ≥e 2.
3.解 (1)当a=1,b=0时,f (x )=ln x-√x ,
此时,函数f (x )定义域为(0,+∞),
f'(x )=1
x −2√x =2-√x
2x ,
由f'(x )>0,得0<x<4, 由f'(x )<0,得x>4,
所以f (x )在(0,4)上单调递增,在(4,+∞)上单调递减, 所以f (x )max =f (4)=2ln 2-2.
(2)当b>0时,函数f (x )定义域为[0,+∞),f'(x )=a
x+b −2√x =√x -2√
x (x+b ),
①当a ≤0时,f'(x )<0对任意的x ∈(0,+∞)恒成立, 所以此时f (x )极值点的个数为0个;
②当a>0时,设h (x )=-x+2a √x -b ,
(ⅰ)当4a2-4b≤0,即0<a≤√b时,f'(x)≤0对任意的x∈(0,+∞)恒成立,即f'(x)在(0,+∞)上无变号零点,所以,此时f(x)极值点的个数为0个;
(ⅱ)当4a2-4b>0,即a>√b时,记方程h(x)=0的两根分别为x1,x2,
则√x1+√x2=2a>0,√x1x2=b>0,所以x1,x2都大于0,即f'(x)在(0,+∞)上有2个变号零点,所以,此时f(x)极值点的个数为2个.
综上所述,a≤√b时,f(x)极值点的个数为0个;a>√b时,f(x)极值点的个数为2个.
4.解(1)f(x)定义域为(0,+∞),当k=-1时,f(x)=ln x-x,f'(x)=1
x
-1,令f'(x)=0,得x=1,当f'(x)>0时,解得0<x<1,当f'(x)<0时,解得x>1,所以f(x)在(0,1)上单调递增,在(1,+∞)上单调递减,所以f(x)有极大值点为x=1,无极小值点.
(2)当k=0时,f(x)+b
x -a=ln x+b
x
-a.
若f(x)+b
x
-a≥0(a,b∈R)恒成立,则ln x+b
x
-a≥0(a,b∈R)恒成立,
所以a≤ln x+b
x
恒成立,
令g(x)=ln x+b
x
,则g'(x)=x-b
x2
,
由题意b>0,函数g(x)在(0,b)上单调递减,在(b,+∞)上单调递增,
所以g(x)min=g(b)=ln b+1,
所以a≤ln b+1,所以a-1≤ln b,
所以e a-1≤b,e a-1-b+1≤1,故当且仅当e a-1=b时,e a-1-b+1的最大值为1.
5.解(1)略
(2)因为f(x)在(0,+∞)上单调递增,即f'(x)=ax+b-ln x≥0在(0,+∞)上恒成立,设
h(x)=ax+b-ln x,则h'(x)=a-1
x
,
①若a=0,则h'(x)<0,则h(x)在(0,+∞)上单调递减,显然f'(x)=b-ln x≥0在(0,+∞)上不恒成立;
②若a<0,则h'(x)<0,h(x)在(0,+∞)上单调递减,当x>max{-b
a
,1}时,ax+b<0,-ln x<0,故h(x)<0,f(x)单调递减,不符合题意;
③若a>0,当0<x<1
a 时,h'(x)<0,h(x)单调递减,当x>
1
a
时,h'(x)>0,h(x)单调递增,
所以h(x)min=h(1
a
)=1+b+ln a,
由h(x)min≥0得2a+b≥2a-1-ln a, 设m(x)=2x-1-ln x,x>0,
则m'(x)=2-1 x ,
当0<x<1
时,m'(x )<0,m (x )单调递减;
当x>1
2时,m'(x )>0,m (x )单调递增,所以m (x )≥m (1
2)=ln 2,
所以2a+b ≥ln 2.又c ≤e 2a+b ,所以c ≤2,即c 的最大值为2.
6.解 (1)由f (x )有两个极值点且都小于0,得f'(x )=3x 2+3x-4a=0有两个不相等的负实根,∴{Δ=9+48a >0,x 1+x 2=-1<0,x 1x 2=-43a >0, 解得-316<a<0.
故a 的取值范围为-3
16,0.
(2)由h (x )=a (a-1)ln x+32x 2-(4a-3)x+1,x>0, 则h'(x )=a (a -1)
x +3x-(4a-3)=1x (3x-a )[x-(a-1)].
令(3x-a )[x-(a-1)]=0,得x=a
3或x=a-1,令a
3=a-1,得a=3
2.
①当{a
≤0,
a -1≤0,
即a ≤0时,在(0,+∞)上h'(x )>0恒成立;
②当{0<a ,a -1≤0,
即0<a ≤1时,当x>a 3时,h'(x )>0,当0<x<a
3时,h'(x )<0;
③当a 3>a-1>0,即1<a<32,当0<x<a-1或x>a 3时,h'(x )>0,当a-1<x<a
3时,h'(x )<0; ④当a
3=a-1>0,即a=3
2时,h'(x )>0恒成立;
⑤当a-1>a
3>0,即a>3
2,当0<x<a
3或x>a-1时,h'(x )>0,当a
3<x<a-1时,h'(x )<0. 综上所述:当a ≤0或a=3
2时,h (x )在(0,+∞)上单调递增; 当0<a ≤1时,h (x )在a
3,+∞上单调递增,在0,a
3上单调递减; 当1<a<3
2时,h (x )在(0,a-1),a
3,+∞上单调递增,在a-1,a
3上单调递减; 当a>3
2时,h (x )在0,a
3,(a-1,+∞)上单调递增,在a
3,a-1上单调递减.
7.解 (1)由f (x )=x-a ln x ,得y=f (x )+b=x-a ln x+b ,∴y'=f'(x )=1-a
x .由已知可得{f '(1)=-1,f (1)+b =2,即
{
1-a =-1,
1+b =2,
∴a=2,b=1.
(2)∵g (x )=f (x )+a+1
=x-a ln x+a+1
,∴g'(x )=1-a
−a+1
2=
(x+1)[x -(a+1)]
2
(x>0),所以当
a+1≤0,即a ≤-1时,g'(x )>0,g (x )在(0,+∞)上为增函数,无极值点;
当a+1>0,即a>-1时,则有当0<x<a+1时,g'(x )<0,当x>a+1时,g'(x )>0,∴g (x )在(0,a+1)上为减函数,在(a+1,+∞)上为增函数,所以x=a+1是g (x )极小值点,无极大值点;
综上可知,当a ≤-1时,函数g (x )无极值点,当a>-1时,函数g (x )的极小值点是a+1,无极大值点.
(3)h (x )=1
a f (x )+a e x -x
a +ln a=a e x -ln x+ln a (a>0),由题意知,当x>a 时,a e x -ln x+ln a ≥0恒成立,
又不等式a e x -ln x+ln a ≥0等价于a e x ≥ln x
a , 即e x ≥1a ln x
a ,即x e x ≥x a ln x
a ,
①
由x>a>0知:x
a >1,ln x
a >0,所以①式等价于ln(x e x )≥ln (x
a ln x
a ), 即x+ln x ≥ln x
a +ln (ln x
a ), 设φ(x )=x+ln x (x>0),
则原不等式即为φ(x )≥φ(ln x
a ), 又φ(x )=x+ln x (x>0)在(0,+∞)上为增函数, 所以原不等式等价于x ≥ln x a ,
②
又②式等价于e x ≥x
a ,亦即a ≥x
e x (x>a>0),设F (x )=x
e x (x>0), 则F'(x )=1-x
e x ,∴F (x )在(0,1)上为增函数,在(1,+∞)上为减函数,
又x>a>0,∴当0<a<1时,F (x )在(a ,1)上为增函数,在(1,+∞)上为减函数,
∴F (x )≤F (1)=1
e ,要使原不等式恒成立,须使1
e ≤a<1, 当a ≥1时,则F (x )在(a ,+∞)上为减函数,F (x )<F (1)=1
e ,
要使原不等式恒成立,须使a ≥1
e ,
∴a ≥1时,原不等式恒成立.
综上可知:a 的取值范围是[1
e ,+∞),所以a 的最小值为1
e .
8.解 (1)函数f (x )≤0恒成立,即ax 2+x ln x ≤0恒成立,可得a ≤-lnx
恒成立.
设g (x )=-lnx
,g'(x )=lnx -1
2.
当0<x<e 时,g'(x )<0,g (x )递减;当x>e 时,g'(x )>0,g (x )递增, 可得当x=e 时g (x )取得最小值,且g (e)=-ln e
e =-1
e ,所以a ≤-1
e .
(2)f (x )的导数为f'(x )=2ax+1+ln x ,曲线y=f (x )在点(e,f (e))处的切线斜率为2a e +2=2e +2,
可得a=1,即f (x )=x 2+x ln x.
又由k<f (x )
x -1对任意
x>1都成立,可得
k<x 2+xlnx x -1对
x>1恒成立.
设
h (x )=x 2+xlnx x -1,x>1,h'(x )=x 2-x -lnx -1(x -1)
2. 设k (x )=x 2-x-ln x-1,x>1, k'(x )=2x-1-1
x =
(x -1)(2x+1)
x
>0, 可得k (x )在(1,+∞)内递增,由k (1.8)=0.44-ln 1.8. 由1.8>√e 可得ln 1.8>1
2,
即有k (1.8)<0,k (2)=1-ln 2>0,则存在m ∈(1.8,2),使得k (m )=0, 则1<x<m ,k (x )<0,h'(x )<0,即h (x )在(1,m )内递减; 当x>m 时,k (x )>0,h'(x )>0,h (x )在x>m 递增, 可得h (x )min =h (m )=
m 2+mlnm
m -1
. 又k (m )=m 2-m-ln m-1=0, 即有m 2-1=m+ln m ,
可得h (x )min =m 2+m 在(1.8.2)递增,可得h (x )min ∈(5.04,6),由k<h (x )min ,k ∈Z ,故k 的最大值为5.。