函数13

(1)

f(x)=ax^2+bx+c
a∈N+,a>0
b>2a,-b/2a<-1,f(x)对称轴在x=-1左侧

令x=sint,t∈R,则x∈[-1,1]
在区间[-1,1]上,
f(x)最小=f(-1)=a-b+c=-4....(1)
f(x)最大=f(1)=a+b+c=2......(2)
(1)+(2): a+c=-1
b=3
b>2a,a<b/2=3/2
a∈N+,a=1
c=-2

f(x)=x^2+3x-2
f(x)最小=[4(-2)-9]/4=-17/4

(2)
g(x)=4x与y轴交点(0,0)
h(x)=2(x^2+1)与y轴交点(0,2)
f(x)=ax^2+bx+c与y轴交点(0,c)
x∈R,g(x)<=f(x)<=h(x)恒成立
0<=c<=2,
c∈Z,c=0或1或2

令g(x)=h(x)
4x=2(x^2+1),x^2-2x+1=0,方程有两等根x=1
g(x),h(x)相切于点(1,4)
x∈R,g(x)<=f(x)<=h(x)恒成立,
f(x)与g(x)切于点(1,4)
a+b+c=4
令ax^2+bx+c=4x,ax^2+(b-4)x+c=0
△=(b-4)^2-4ac=0

i)c=0,
a+b=4
(b-4)^2=0
b=4
a=0 (不合题意)

ii)c=1
a+b=3
(b-4)^2-4a=0
b=2
a=1

f(x)=x^2+2x+1
令f(x)<2(x^2+1),x^2-2x+1>0,
x≠1,即存在x0,使得f(x0)<2(x^2+1)成立

iii)c=2
a+b=2
(b-4)^2-8a=0
b=0,a=2
f(x)=2x^2+2=2(x^2+1)
f(x)<2(x^2+1),无解,不合题意

综上c=1