(2) According to the conditions, for any real number X, f 1(x)≤f2(x) is a constant, and for any real number X, it is transformed into | x-2a+1| | x-a |≤1,and then the value range of parameter A can be obtained by solving with absolute inequality. That is, f 1(x)≤f2(x) is constant for any real number x, that is, e | x-2a+1|≤ e | x-a |+1is constant for any real number x,
∴| x-2a+1|≤| x-a |+1,that is | x-2a+ 1 | | x-a |≤ 1 holds for any real number X. 。
And | x-2a+1|| x-a |≤| (x-2a+1)-(x-a) | = |-a+1| is a constant for any real number, so it is only necessary.
|-a+ 1|≤ 1, the solution is 0 ≤ A ≤ 2, and the range of ∴a is 0≤a≤2. Combined with the original problem, 1≤a≤2.
(3):? G(x)=f 1(x), f 1(x) ≤ F2(x) ```` F2 (x), f 1(x) > F2 (x), while f1(x).
Let F 1(x)=|x-2a+ 1|, F2 (x) = | x-a |+ 1,1≤ a ≤ 6 ∴ 2a-1≥ a.
∴g(x)min=f2(a)= 1,g(x)min=e 1=e
Honey? ? ~ ~ ~, give points! ! ! (Senior 1 student of Xuancheng Middle School, a century-old prestigious school)