A:f(2)<f(﹣2)<f(0)
B:f(0)<f(2)<f(﹣2)
C:f(﹣2)<f(0)<f(2)
D:f(2)<f(0)<f(﹣2)
(I)证明:BC⊥AB1;
(II)若OC=OA,求直线CD与平面ABC所成角.
函数f(x)=log5(x-1)的零点是( )