已知函数f(x)=ax+bx+c(a>0)的图象在点(1,f(1))处的切线方程为y=x-1.(1)试用a表示出b,c;(2)若f(x)≥lnx在[1,+∞)上恒成立,求a的取值范围;(3)证明:1+

题目简介

已知函数f(x)=ax+bx+c(a>0)的图象在点(1,f(1))处的切线方程为y=x-1.(1)试用a表示出b,c;(2)若f(x)≥lnx在[1,+∞)上恒成立,求a的取值范围;(3)证明:1+

题目详情

已知函数f(x)=ax+
b
x
+c(a>0)的图象在点(1,f(1))处的切线方程为y=x-1.
(1)试用a表示出b,c;
(2)若f(x)≥lnx在[1,+∞)上恒成立,求a的取值范围;
(3)证明:1+
1
2
+
1
3
+…+
1
n
>ln(n+1)+
n
2(n+1)
(n≥1).
题型:解答题难度:中档来源:广州三模

答案

(1)∵f(x)=ax+class="stub"b
x
+c(a>0)

f′(x)=a-class="stub"b
x2
⇒f′(1)=a-b=1⇒b=a-1

∴f(1)=a+a-1+c=2a-1+c.
又∵点(1,f(1))在切线y=x-1上,
∴2a-1+c=0⇒c=1-2a,
b=a-1
c=1-2a

(2)∵f(x)=ax+class="stub"a-1
x
+1-2a(a>0)

f(x)≥lnx在[1,+∞]上恒成立,
设g(x)=f(x)-lnx,则g(x)=f(x)-lnx≥0在[1,+∞]上恒成立,
∴g(x)min≥0,
又∵g′(x)=a-class="stub"a-1
x2
-class="stub"1
x
=
a(x2-1)-(x-1)
x2
=
a(x-1)(x-class="stub"1-a
a
)
x2

而当class="stub"1-a
a
=1
时,a=class="stub"1
2

1°当class="stub"1-a
a
≤1
a≥class="stub"1
2
时,
g'(x)≥0在[1,+∞]上恒成立,
g(x)min=g(1)=2a-1≥0⇒a≥class="stub"1
2

2°当class="stub"1-a
a
>1
0<a<class="stub"1
2
时,
g'(x)=0时x=class="stub"1-a
a

1≤x<class="stub"1-a
a
时,g'(x)<0,
x>class="stub"1-a
a
时,g'(x)>0;
g(x)min=g(class="stub"1-a
a
)≥0
①,
又∵g(class="stub"1-a
a
)≤g(1)=2a-1<0
与①矛盾,不符题意,故舍.
∴综上所述,a的取值范围为:[class="stub"1
2
,+∞).

(3)证明:由(1)可知a≥class="stub"1
2
时,f(x)≥lnx在[1,+∞]上恒成立,
则当a=class="stub"1
2
时,class="stub"1
2
(x-class="stub"1
x
)≥lnx
在[1,+∞]上恒成立,
令x依次取class="stub"2
1
,class="stub"3
2
,class="stub"4
3
,class="stub"5
4
,class="stub"6
5
class="stub"n+1
n
时,
则有class="stub"1
2
×(class="stub"2
1
-class="stub"1
2
)≥lnclass="stub"2
1
class="stub"1
2
×(class="stub"3
2
-class="stub"2
3
)≥lnclass="stub"3
2


class="stub"1
2
×(class="stub"n+1
n
-class="stub"n
n+1
)≥lnclass="stub"n+1
n

由同向不等式可加性可得
class="stub"1
2
[(class="stub"2
1
+class="stub"3
2
+class="stub"4
3
+…+class="stub"n+1
n
)-(class="stub"1
2
+class="stub"2
3
+class="stub"3
4
+…+class="stub"n
n+1
)]≥ln(n+1)

class="stub"1
2
[(1+class="stub"1
2
+class="stub"1
3
+…+class="stub"1
n
+n)-(n-class="stub"1
2
-class="stub"1
3
-class="stub"1
4
-…-class="stub"1
n+1
)]≥ln(n+1)

也即class="stub"1
2
[2(1+class="stub"1
2
+class="stub"1
3
+…+class="stub"1
n
)+class="stub"1
n+1
-1]≥ln(n+1)

也即1+class="stub"1
2
+class="stub"1
3
+…+class="stub"1
n
>ln(n+1)+class="stub"n
2(n+1)
(n≥1).
解法二:①当n=1时左边=1,右边=ln2+class="stub"1
4
<1,不等式成立;
②假设n=k时,不等式成立,就是1+class="stub"1
2
+class="stub"1
3
+…+class="stub"1
k
>ln(k+1)+class="stub"k
2(k+1)
(k≥1).
那么1+class="stub"1
2
+class="stub"1
3
+…+class="stub"1
k
+class="stub"1
k+1
>ln(k+1)+class="stub"k
2(k+1)
+class="stub"1
k+1

=ln(k+1)+class="stub"k+2
2(k+1)

由(2)知:当a≥class="stub"1
2
时,有f(x)≥lnx  (x≥1)
a=class="stub"1
2
有f(x)=class="stub"1
2
(x-class="stub"1
x
)≥lnx
  (x≥1)
令x=class="stub"k+2
k+1
class="stub"1
2
(class="stub"k+2
k+1
-class="stub"k+1
k+2
)≥lnclass="stub"k+2
k+1
=ln(k+2)-ln(k+1)

ln(k+1)+class="stub"k+2
2(k+1)
≥ ln(k+2)+class="stub"k+1
2(k+2)

∴1+class="stub"1
2
+class="stub"1
3
+…+class="stub"1
k
+class="stub"1
k+1
ln(k+2)+class="stub"k+1
2(k+2)

这就是说,当n=k+1时,不等式也成立.
根据(1)和(2),可知不等式对任何n∈N*都成立.

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