由f(x+1)=f(1-x)得：x=1為對稱軸,即-b/(2a)=1,得：b=-2a
f(x)=0只有一個(gè)零點(diǎn),則b^2-4ac=0,得：c=b^2/4a=4a^2/(4a)=a
所以f(x)=a(x^2-2x+1)=a(x-1)^2
1) a=1,f(x)=(x-1)^2,
y=log2 (x-1)^2
當(dāng)x>1時(shí),y單調(diào)增
2)y=√(f(x)+1)=√[a(x-1)^2+1]=√[ax^2-2ax+a+1]
定義域?yàn)镽,則ax^2-2ax+a+1>=0恒成立
故首先須a>0,其次須delta=4a^2-4a(a+1)0