1, the derivative formula of f (x) is =x*x+2x+m, and the straight line y = 3-x; The slope of the straight line is-1; Then x*x+2x+m= 1 is that the tangent is perpendicular to the straight line and has a unique solution, so b * b-4ac = 0；; There are 4-4 * (m-1) = 0; Get m=2.

(2)mOA-2 *(OA+AB)+OA+AC = 0；

Simplify to get (m-1) * OA = 2ab-AC; Can you know A through painting? b？ C*** line, so 2AB-AC is also on the original line, and the vectors should be equal, but the length and direction are different. Pay attention to the direction of OA Is it used? Answer? b？ C direction is different, only m- 1=0.

So m =1;

(3)f(x) is the increasing function in the interval [- 1, 1], and f(x)

So t*t-2at+ 1= > 1, we can know that a belongs to [- 1, 1], so let a be the unknown and t be the parameter; You can know-2t * a+t * t >; =0, this is a straight line with a slope of -2t and a domain of-1 and 1.

In two cases?

Answer.-2t > 0，-2t * 1+t * t & gt； =0; -2t *( 1)+t * t & gt； =0? Get t

B.-2t <; 0，-2t * 1+t * t & gt； =0; -2t *( 1)+t * t & gt； =0 to obtain t & gt=2;

C, -2t=0 gives t = 0；; Clearly established

When combining (negative infinity, -2) and {t=0} and [2, positive infinity)