根据已知条件知函数f(x)在(-2,0)上单调递减;
(Ⅰ)y=x2+1
x∈(-2,0)时,y′=2x<0,∴函数y=x2+1在(-2,0)上单调递减.
(Ⅱ)y=|x|+1
x∈(-2,0)时,y=-x+1,y′=-1<0,∴该函数在(-2,0)上单调递减;
(Ⅲ)y=
2x+1,x≥0
x3+1,x<0
x∈(-2,0)时,y=x3+1,y′=3x2>0,∴该函数在(-2,0)上单调递增;
(Ⅳ)y=sinx
x∈(-2,?
)时,y′=cosx<0;x∈(?π 2
,0)时,y′=cosx>0.π 2
∴在区间(-2,0)上和f(x)的单调性相同的是(Ⅰ)(Ⅱ),个数为2.
故选:C.