1
g1(x)=ax^2+(2b-1)x+4c
△=(2b-1)^2-4a*4c>0----------------[1]
g2(x)=ax^2+(2b+1)x+4c
△=(2b+1)^2-4a*4c>0----------------[2]
[1]+[2]化簡(jiǎn)
2
ƖF(0)Ɩ是不是說(shuō)絕對(duì)值?
|4c|=1
|a+2b+4c|=1
|a-2b+4c|=1
因?yàn)閎>0所以a+2b+4c>a-2b+4c
|4c|=1
a+2b+4c=1
a-2b+4c=-1
...b=1/2
因?yàn)閍=-4c>0所以c<0,c=-1/4
...a=1
3
g1(x)=ax^2+(2b-1)x+(3-b)
=a(x+(b-1/2))^2+(3-b-a*(b-1/2)^2)
a>0時(shí)g(x)>=(3-b-a*(b-1/2)^2)>=0
a<=(3-b)/(b-1/2)^2
g2(x)=(3-b)/(b-1/2)^2 ---------b∈[0,2]
t=3-b∈[1,3]
g3(t)=t/(5/2-t)^2
因?yàn)閠>0所以g3(t)=1/(t-5+25/4t)
因?yàn)閠+25/4t∈[5,29/4]所以g3(t)∈[4/9,正無(wú)窮]
a∈[4/9,正無(wú)窮]