已知函数f(x)=ax+lnx,a∈R.(I)当a=-1时,求f(x)的最大值;(II)对f(x)图象上的任意不同两点P1(x1,x2),P(x2,y2)(0<x1<x2),证明f(x)图象上存在点P

题目简介

已知函数f(x)=ax+lnx,a∈R.(I)当a=-1时,求f(x)的最大值;(II)对f(x)图象上的任意不同两点P1(x1,x2),P(x2,y2)(0<x1<x2),证明f(x)图象上存在点P

题目详情

已知函数f(x)=ax+lnx,a∈R.
(I)当a=-1时,求f(x)的最大值;
(II)对f(x)图象上的任意不同两点P1(x1,x2),P(x2,y2)(0<x1<x2),证明f(x)图象上存在点P0(x0,y0),满足x1<x0<x2,且f(x)图象上以P0为切点的切线与直线P1P2平等;
(III)当a=
3
2
时,设正项数列{an}满足:an+1=f'(an)(n∈N*),若数列{a2n}是递减数列,求a1的取值范围.
题型:解答题难度:中档来源:黄州区模拟

答案

(Ⅰ)当a=-1时,f(x)=-x+lnx,f′(x)=-1+class="stub"1
x
=
-x +1
x

对于x∈(0,1),有f'(x)>0,∴f(x)在区间(0,1]上为增函数,
对于x∈(1,+∞),有f'(x)<0,∴f(x)在区间(1,+∞)上为减函数,.
∴fmax(x)=f(1)=-1;
(II)直线P1P2的斜率为 k=
ax2+lnx2-ax1-lnx1
x2-x1
=a+
lnx2-lnx1
x2-x1

由(1)知-x+lnx≤-1,当且仅当x=1时取等号,
-
x2
x1
+ln
x2
x1
<-1⇒ln
x2
x1
x2
x1
-1⇒lnx2-lnx1
x2-x1
x1
lnx2-lnx1
x2-x1
<class="stub"1
x1

同理,由 -
x1
x2
+ln
x1
x2
<-1
,可得
lnx2-lnx1
x2-x1
>class="stub"1
x2

故P1P2的斜率 k∈(a+class="stub"1
x2
,a+class="stub"1
x1
)

又在x∈(x1,x2)上,f(x)=a+class="stub"1
x
∈(a+class="stub"1
x2
,a+class="stub"1
x1
)

所以f(x)图象上存在点P0(x0,y0),满足x1<x0<x2,且f(x)图象上以P0为切点的切线与直线P1P2平行;
(III)f(x)=class="stub"3
2
x+lnx
,f′(x)=class="stub"3
2
+class="stub"1
x
,∴an+1=class="stub"3
2
+class="stub"1
an

a3=class="stub"3
2
+class="stub"1
a2
,a4=class="stub"3
2
+class="stub"1
a3
=class="stub"3
2
+class="stub"1
class="stub"3
2
+class="stub"1
a2
=
13a2+6
2(3a2+2)
<a2⇒2a22-3a2-2>0,
⇒(2a2+1)(a2-1)>0⇒a2>2⇒class="stub"3
2
+class="stub"1
a1
>2
⇒0<a1<2,
下面我们证明:当0<a1<2时,a2n+2<a2n,且a2n>2(n∈N+)
事实上,当n=1时,0<a1<2⇒a2=class="stub"3
2
+class="stub"1
a1
>2

a4-a2=
13a2+6
2(3a2+2)
-a2=-
3(2a2+1)(a2-2)
2(3a2+2)
<0
⇒a4<a2,结论成立.
若当n=k时结论成立,即a2k+2<a2k,且a2k>2,则
a2k+2=class="stub"3
2
+class="stub"1
a2k
>2
⇒a2k+4=class="stub"3
2
+class="stub"1
a2k+2
>2

a2k+4-a2k+2=
13a2k+2+6
2(3a2k+2+2)
-a2k+2=-
3(2a2k+2+1)(a2k+2-2)
2(3a2k+2+2)
<0

⇒a2k+4<a2k+2,
由上述证明可知,a1的取值范围是(0,2).

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