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Mathematics in senior one. . . . . . . .
Solution: 1. From the meaning of the question, when x > 2, f (x) = (x-a) * (x-2); X & ltF(x)= -(x-a)*(x-2) When =2;

Therefore, a < 2 points.

At this time, the monotonous interval: negative infinity to 2 is increasing? ; 2 to 2/(a+2) is a negative number? ; A increases to positive infinity.

Similarly, one

When a=2, negative infinity and positive infinity increase. Done.

Two: if n exists, f(m)≤g(n) only needs f (m) min.

While g(n) increases from 0 to 2, and g (n) max = g (2) = 4;

If a & gt=2, f(m) increases from 0 to 1, and f (m) min = f (0) =-2a.

Similarly, if 2>a & gt=0, then f (m) min.

In a word, a & gt=-3 satisfies the condition.

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