Question

高一数学！急！！！！！！

1.k为何值时，二次不等式x^2-(k+2)x+4≥0的解集是R

2.求函数y=√-x^2+3x =4-1／√x-2 的定义域。

Answer 1

1,x^2-(k+2)x+4≥0
判别式<=0
(k+2)^2-16<=0
(k+2)^2<=16
-4<=k+2<=4
-6<=k<=2

2,-x^2+3x>=0
x^2-3x<=0
0<=x<=3
x-2>0
x>2

2<x<=3