Question

（1）若函数f(x)=mx²+x+5在[2，＋∞）上是增函数，则m的取值范围是

(2）在平面直角坐标系xoy种，点A(-1,-2),B(2,3),C(-2,-1)
①求以线段AB,AC为邻边的平行四边形两条对角线的长
②设实数t满足（向量AB-t向量OC)×向量OC=0，求t的值
麻烦我要详细的过程。谢谢。

Answer 1

(1)
f(x)=mx²+x+5
f'(x) = 2mx+1 > 0 ( for x∈[2，＋∞）)
4m+1>0
m> -1/4
(2)
①
A(-1,-2),B(2,3),C(-2,-1)
AB = OB - OA =( 3,5)
AC = OC - OA = (-1,1)

BC = BA+AC = -AB +AC = ( -4, -4)
|BC| = √(16+16) = 4√2
let the 4th point of //gram be D
AD = AB + AC = (2, 6)
|AD| = √(4+36) = 2√10
平行四边形两条对角线的长:|AD|, |BC| = 2√10, 4√2
②
(AB-tOC).OC=0
(3+2t,5+t).(-2,-1)=0
-2(3+2t)-(5+t) = 0
-11-5t=0
t = -11/5

Answer 2

(1)∵函数f(x)=mx²+x+5在[2，＋∞）上是增函数
∴函数对称轴0<x=-1/m<2
∴-1/2<m<0