Question

求解题，谢谢。

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Answer 1

case 1: x1,x2 <1
[f(x1)-f(x2)]/(x1-x2) 独立于a

case 2: x1, x2 ≥ 1
[f(x1)-f(x2)] /(x1-x2) >0
(3-a)(x1-x2)/(x1-x2) >0
a<3

case 3: x1<1 , x2≥ 1
[f(x1)-f(x2)] /(x1-x2) >0
f(x1) -f(x2) <0
-(x1-1)^2 - (3-a)x2 - 4a <0
(x1-1)^2 + (3-a)x2 + 4a >0
(3-a)x2 + 4a >0
(x2-4)a < 3x2
-∞ <a<∞

case 4: x2<1 , x1≥ 1
[f(x1)-f(x2)] /(x1-x2) >0
f(x1) -f(x2) >0
-(x1-1)^2 - (3-a)x2 - 4a >0
(x1-1)^2 + (3-a)x2 + 4a <0
(3-a)x2 + 4a <0
(x2-4)a < 3x2
a > 3x2/(x2-4)
= 3 + 12/(x2-4)
a<3

[f(x1)-f(x2)] /(x1-x2) >0
case 1 and case 2 and case 3 and case 4 and case 5
a<3 and -∞ <a<∞ and a<3
=> a<3

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f(x) = (1/2)(x-1)^2 +a
min f(x) = f(1) = a
=> a=1

f(x) = (1/2)(x-1)^2 +1
max f(x) = f(b)= b
(1/2)(b-1)^2 +1 =b
(b-1)^2 +2 =2b
b^2-4b +3 =0
(b-1)(b-3)=0
b=1 or 3
a+b= 2 or 4