已知函数f(x)＝x2-2ax+2a,其中a为常数，且a∈R.(1)若函数f(x)没有零点，已知函数f(x)＝x2-2ax+2a,其中a为常数，且a∈R.(1)若函数f(x)没有零点，求a的取值范围;.(2)若x∈[-1，2]时，f(x)≧-2恒

Question

已知函数f(x)＝x2-2ax+2a,其中a为常数，且a∈R. (1)若函数f(x)没有零点，
已知函数f(x)＝x2-2ax+2a,其中a为常数，且a∈R.
(1)若函数f(x)没有零点，求a的取值范围;
.(2)若x∈[-1，2]时，f(x)≧-2恒成立，求a的取值范围.

Answer 1

(1)
△<0
4a^2-8a<0
a(a-2)<0
0(2)
f(x)=x^2-2ax+2a
f'(x) = 2x -2a =0
x=a
f''(x) =2 >0 (min)
case 1: -1≤a≤ 2
min f(x) = f(a)
= a^2 -2a^2 +2a
=-a^2+2a
f(a) ≥-2
-a^2 +2a ≥-2
a^2-2a-2 ≤0
1-√3≤a≤1+√3
solution for case 1: 1-√3≤a≤2
case 2: a> 2
min f(x) = f(2)
= 4-4a+2a
= 4-2a
f(x) ≥-2
4-2a≥-2
a≤3
solution for case 2: 2case 3: a< -1
min f(x) = f(-1)
=1+3a
f(x) ≥-2
1+3a ≥-2
a≥-1
no solution for case 3
f(x) ≥-2
case 1 or case 2 or case 3
1-√3≤a≤2 or 21-√3≤a≤3