突破训练一(新考案)

突破训练一
1.(2020届福州高三期末考试)已知函数f (x )=eln x-ax (a ∈R).
(1)讨论f (x )的单调性;
(2)当a=e 时,证明:xf (x )-e x +2e x ≤0.
【解析】(1)f'(x )=e x
-a (x>0),
①若a ≤0,则f'(x )>0,f (x )在(0,+∞)上单调递增; ②若a>0,则当0<x<e
a 时,f'(x )>0,当x>e
a 时,f'(x )<0,
故f (x )在(0,e a
)上单调递增,在(e a
,+∞)上单调递减. (2)因为x>0,所以只需证明f (x )≤e x x
-2e,
当a=e 时,由(1)知,f (x )在(0,1)上单调递增,在(1,+∞)上单调递减, 所以f (x )max =f (1)=-e . 记g (x )=e x x
-2e(x>0), 则g'(x )=
(x -1)e x
x 2, 所以当0<x<1时,g'(x )<0,g (x )单调递减,当x>1时,g'(x )>0,g (x )单调递增, 所以g (x )min =g (1)=-e .
综上,当x>0时,f (x )≤g (x ),即f (x )≤e x x
-2e,即xf (x )-e x +2e x ≤0.
2.(2020届武汉调研)已知函数f (x )=ln x-a(x -1)
x
(a ∈R).
(1)求函数f (x )的单调区间.
(2)证明:不等式(x+1)ln x>2(x-1)对∀x ∈(1,2)恒成立.
【解析】(1)f (x )的定义域为(0,+∞),f'(x )=
x -a x 2
. ①若a ≤0,则f'(x )>0,故f (x )的单调递增区间为(0,+∞);
②若a>0,则当x>a 时,f'(x )>0,当0<x<a 时,f'(x )<0,故f (x )的单调递增区间为(a ,+∞),单调递减区间为(0,a ). (2)因为x ∈(1,2),所以x+1>0,
所以要证原不等式成立, 即证ln x>
2(x -1)
x+1对∀x ∈(1,2)恒成立, 令g (x )=ln x-2(x -1)
x+1
,则g'(x )=
(x -1)2
x(x+1)2
≥0,
所以g (x )在(0,+∞)上单调递增, 所以当x ∈(1,2)时,g (x )>g (1)=ln 1-2(1-1)
1+1
=0, 所以ln x>
2(x -1)
x+1
对∀x ∈(1,2)恒成立, 所以(x+1)ln x>2(x-1)对∀x ∈(1,2)恒成立.
3.已知函数f (x )=ln(x+a )-x 2-x 在x=0处取得极值.
(1)求实数a 的值.
(2)证明:对于任意的正整数n ,不等式2+34+49
+…+
n+1
n 2
>ln(n+1)都成立.
【解析】(1)因为f'(x )=
1
x+a
-2x-1, 又因为x=0为f (x )的极值点, 所以f'(0)=1a
-1=0,解得a=1. 经检验,a=1满足题意,故a=1. (2)由(1)知f (x )=ln(x+1)-x 2-x ,x>-1, 则f'(x )=
1x+1-2x-1=-x(2x+3)x+1.
令f'(x )>0,得-1<x<0;令f'(x )<0,得x>0. 当x 变化时,f'(x ),f (x )的变化情况如下表:
x (-1,0)
0 (0,+∞)
f'(x ) + 0 - f (x )
↗
极大值
↘
所以f (x )≤f (0)=0,即ln(x+1)≤x 2+x (当且仅当x=0时取等号).
令x=1n
,则ln (1n
+1)<1n
2+1n
, 即ln
n+1n <n+1
n 2, 所以ln 21
+ln 32
+…+ln n+1n
<2+34+…+n+1
n 2. 即2+34+49
+…+
n+1
n 2
>ln(n+1). 4.(2020届山东烟台模拟)设f (x )=e x (x -a -1
x ),g (x )=a ln x.
(1)当a>1时,讨论函数F (x )=
f(x)
e x
-g (x )的单调性. (2)求证:当a=0时,不等式f (x )>2√e 对任意x ∈(0,+∞)都成立.
【解析】(1)F (x )=x-a -1
x
-a ln x (x>0), F'(x )=1+a -1x 2-a x =x 2-ax+(a -1)
x 2
=
(x -1)[x -(a -1)]
x 2
(x>0). ①当0<a-1<1,即1<a<2时,
F'(x )与F (x )随x 的变化情况如下表:
x (0,a-1)
a-1 (a-1,1) 1 (1,+∞)
F'(x ) + 0 - 0 + F (x )
↗
极大值
↘
极小值
↗
由表知F (x )在(0,a-1),(1,+∞)上单调递增,在(a-1,1)上单调递减.
②当a-1=1,即a=2时,F'(x )=
(x -1)2
x 2
≥0恒成立,
故F (x )在(0,+∞)上单调递增.
③当a-1>1,即a>2时,F'(x )与F (x )随x 的变化情况如下表:
x
(0,1)
1
(1,a-1)
a-1 (a-1,+∞)
F'(x ) + 0 - 0 + F (x )
↗
极大值
↘
极小值
↗
由表知F (x )在(0,1),(a-1,+∞)上单调递增,在(1,a-1)上单调递减.
综上所述,当1<a<2时,F (x )在(0,a-1),(1,+∞)上单调递增,在(a-1,1)上单调递减; 当a=2时,F (x )在(0,+∞)上单调递增;
当a>2时,F (x )在(0,1),(a-1,+∞)上单调递增,在(1,a-1)上单调递减.
(2)当a=0时,f (x )=e x (x +1x
),
f'(x )=e x (x +1x )+e x (1-1
x 2)=
e x (x 3+x 2+x -1)
x 2
(x>0),
令h (x )=x 3+x 2+x-1(x>0),则h'(x )=3x 2+2x+1=3(x +13)2+23
>0,
∴h (x )=x 3+x 2+x-1在(0,+∞)上单调递增.
又h (1
2
)=-18
<0,h (1)=2>0,
∴h (x )在(0,+∞)上存在唯一零点x 0,且x 0∈(1
2,1), ∴当0<x<x 0时,h (x )<0,f'(x )=
e x h(x)
x 2
<0; 当x>x 0时,h (x )>0,f'(x )=
e x h(x)
x 2
>0. ∴f (x )在(0,x 0)上单调递减,在(x 0,+∞)上单调递增. ∴当x>0时,f (x )≥e x 0(x 0+1
x 0
)>e 1
2(x 0+1
x 0
)>2√e .。