已知函数f(x)=x2+1x+c的图象关于原点对称.(1)求f(x)的表达式;(2)n≥2,n∈N时,求证:[f(1)-1]|[f(22)-22]+…+[f(n2)-n2]<2;(3)对n≥2,n∈N

题目简介

已知函数f(x)=x2+1x+c的图象关于原点对称.(1)求f(x)的表达式;(2)n≥2,n∈N时,求证:[f(1)-1]|[f(22)-22]+…+[f(n2)-n2]<2;(3)对n≥2,n∈N

题目详情

已知函数f(x)=
x2+1
x+c
的图象关于原点对称.
(1)求f(x)的表达式;
(2)n≥2,n∈N时,求证:[f(1)-1]|[f(22)-22]+…+[f(n2)-n2]<2;
(3)对n≥2,n∈N,x>0,求证[f(x)]n-f(xn)≥2n-2.
题型:解答题难度:中档来源:不详

答案

∵f(x)图象关于原点对称
∴f(x)是奇函数,代入特值,f(1)=-f(-1),求得c=0
f(x)=
x2+1
x

(2)∵n≥2,n∈N
f(n2)-n2=class="stub"1
n2
<class="stub"1
(n-1)n
=class="stub"1
n-1
-class="stub"1
n
(n≥2)

[f(1)-1]+…+[f(n2)-n2]<1+(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n-1
-class="stub"1
n
)<2

(3)[f(x)]n-f(xn)=(x+class="stub"1
x
)n-(xn+class="stub"1
xn
)

=
C1n
xn-1class="stub"1
x
+
C2n
xn-2(class="stub"1
x
)2+…+
Cn-1n
x(class="stub"1
x
)n-1

=class="stub"1
2
[(
C1n
xn-1class="stub"1
x
+
Cn-1n
x(class="stub"1
x
)n-1)+(
C2n
xn-2(class="stub"1
x
)2+
Cn-2n
x2(class="stub"1
x
)n-1)+…
+(
Cn-1n
x(class="stub"1
x
)n-1+
C1n
xn-1(class="stub"1
x
))]


class="stub"1
2
[
Cn1
2
xn-1class="stub"1
x
x(class="stub"1
x
)
n-1
  +
Cn2
•2
xn-2class="stub"1
x 2
x2(class="stub"1
x
) n-2
+…+
Cnn-1
•2
(class="stub"1
x
) n-1x n-1
]
=2n-2
∴[f(x)]n-f(xn)≥2n-2.

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