(1)∵f1()=0∉(0,1), ∴f(x)在D1上不封闭; ∵f2(x)=-(x+)2+在(0,1)上是减函数, ∴0<f2(1)<f2(x)<f2(0)=1, ∴f2(x)∈(0,1)⇒f2(x)在D1上封闭; ∵f3(x)=2x-1在(0,1)上是增函数,∴0=f3(0)<f3(x)<f3(1)=1, ∴f3(x)∈(0,1)⇒f3(x)在D1上封闭; ∵f4(x)=cosx在(0,1)上是减函数,∴cos1=f4(1)<f4(x)<f4(0)=1, ∴f4(x)∈(cos1,1)⊂(0,1)⇒f4(x)在D1上封闭; (2)f(x)=5-,假设f(x)在D2上封闭,对a+10讨论如下: 若a+10>0,则f(x)在(1,2)上为增函数,故应有⇒⇒a=2 若a+10=0,则f(x)=5,此与f(x)∈(1,2)不合, 若a+10<0,则f(x)在(1,2)上为减函数,故应有⇒,无解, 综上可得,a=2时f(x)在D2上封闭. |