设g(x)=px-qx-2f(x),其中f(x)=lnx,且g(e)=qe-pe-2.(e为自然对数的底数)(I)求p与q的关系;(Ⅱ)若g(x)在其定义域内为单调函数,求p的取值范围;(Ⅲ)证明:①

题目简介

设g(x)=px-qx-2f(x),其中f(x)=lnx,且g(e)=qe-pe-2.(e为自然对数的底数)(I)求p与q的关系;(Ⅱ)若g(x)在其定义域内为单调函数,求p的取值范围;(Ⅲ)证明:①

题目详情

g(x)=px-
q
x
-2f(x)
,其中f(x)=lnx,且g(e)=qe-
p
e
-2
.(e为自然对数的底数)
(I)求p与q的关系;
(Ⅱ)若g(x)在其定义域内为单调函数,求p的取值范围;
(Ⅲ)证明:
①f(1+x)≤x(x>-1);
ln2
22
+
ln3
32
+…+
lnn
n2
2n2-n-1
4(n+1)
(n∈N,n≥2).
题型:解答题难度:中档来源:乐山二模

答案

(I)由题意g(x)=px-class="stub"q
x
-2lnx

又g(e)=pe-class="stub"q
e
-2
,∴pe-class="stub"q
e
-2=qe-class="stub"q
e
-2

(p-q)e+(p-q)class="stub"1
e
=0
,∴(p-q)(e+class="stub"1
e
)=0

e+class="stub"1
e
≠0
,∴p=q
(II)由(I)知:g(x)=px-class="stub"p
x
-2lnx
g′(x)=p+class="stub"p
x2
-class="stub"2
x
=
px2-2x+p
x2

令h(x)=px2-2x+p.要使g(x)在(0,+∞)为单调函数,只需h(x)在(0,+∞)满足:
h(x)≥0或h(x)≤0恒成立.
①p=0时,h(x)=-2x,∵x>0,∴h(x)<0,∴g'(x)=-class="stub"2x
x2
<0

∴g(x)在(0,+∞)单调递减,∴p=0适合题意.
②当p>0时,h(x)=px2-2x+p图象为开口向上抛物线,
称轴为x=class="stub"1
p
∈(0,+∞).∴h(x)min=p-class="stub"1
p
.只需p-class="stub"1
p
≥0,即p≥1时h(x)≥0,g′(x)≥0,
∴g(x)在(0,+∞)单调递增,∴p≥1适合题意.
③当p<0时,h(x)=px2-2x+p图象为开口向下的抛物线,其对称轴为x=class="stub"1
p
∉(0,+∞),
只需h(0)≤0,即p≤0时h(0)≤(0,+∞)恒成立.
∴g′(x)<0,∴g(x)在(0,+∞)单调递减,∴p<0适合题意.
综上①②③可得,p≥1或p≤0.
(III)证明:①即证:lnx-x+1≤0(x>0),
设k(x)=lnx-x+1,则k'(x)=class="stub"1
x
-1=class="stub"1-x
x

当x∈(0,1)时,k′(x)>0,∴k(x)为单调递增函数;
当x∈(1,∞)时,k′(x)<0,∴k(x)为单调递减函数;
∴x=1为k(x)的极大值点,∴k(x)≤k(1)=0.即lnx-x+1≤0,
所以lnx≤x-1得证.
②由①知lnx≤x-1,又x>0,
class="stub"lnx
x
≤class="stub"x-1
x
=1-class="stub"1
x
∵n∈N*,n≥2时,令x=n2,
lnn2
n2
≤1-class="stub"1
n2

class="stub"lnn
n2
≤class="stub"1
2
(1-class="stub"1
n2
)

class="stub"ln2
22
+class="stub"ln3
32
++class="stub"lnn
22
≤class="stub"1
2
(1-class="stub"1
22
+1-class="stub"1
32
++1-class="stub"1
n2
)

=class="stub"1
2
[(n-1)]-(class="stub"1
22
+class="stub"1
32
++class="stub"1
n2
)]<class="stub"1
2
[(n-1)-(class="stub"1
2×3
+class="stub"1
3×4
++class="stub"1
n(n+1)
)]

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

=class="stub"1
2
[n-1-(class="stub"1
2
-class="stub"1
n+1
)]
=
2n2-n-1
4(n+1)

所以得证.

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