解答:解:(Ⅰ)當(dāng)a=2時(shí),
f(x)=x|x-2|-2=,①當(dāng)x≥2時(shí),f(x)=x
2-2x-2=(x-1)
2-3,
∴f(x)在(2,+∞)上單調(diào)遞增;
②當(dāng)x<2時(shí),f(x)=-x
2+2x-2=-(x-1)
2-1,
∴f(x)在(1,2)上單調(diào)遞減,在(-∞,1)上單調(diào)遞增;
綜上所述,f(x)的單調(diào)遞增區(qū)間是(-∞,1)和(2,+∞),單調(diào)遞減區(qū)間是(1,2).
(Ⅱ)(1)當(dāng)a=0時(shí),f(x)=x|x|,函數(shù)y=f(x)的零點(diǎn)為x
0=0;
(2)當(dāng)a>0時(shí),
f(x)=x|x-a|-a=,
故當(dāng)x≥a時(shí),
f(x)=(x-)2--a,二次函數(shù)對(duì)稱軸
x=<a,
∴f(x)在(a,+∞)上單調(diào)遞增,f(a)<0;
當(dāng)x<a時(shí),
f(x)=-(x-)2+-a,二次函數(shù)對(duì)稱軸
x=<a,
∴f(x)在
(,a)上單調(diào)遞減,在
(-∞,)上單調(diào)遞增;
∴f(x)的極大值為
f()=-()2+a×-a=-a,
1°當(dāng)
f()<0,即0<a<4時(shí),函數(shù)f(x)與x軸只有唯一交點(diǎn),即唯一零點(diǎn),
由x
2-ax-a=0解之得函數(shù)y=f(x)的零點(diǎn)為
x0=或
x0=(舍去);
2°當(dāng)
f()=0,即a=4時(shí),函數(shù)f(x)與x軸有兩個(gè)交點(diǎn),即兩個(gè)零點(diǎn),分別為x
1=2和
x2==2+2;
3°當(dāng)
f()>0,即a>4時(shí),函數(shù)f(x)與x軸有三個(gè)交點(diǎn),即有三個(gè)零點(diǎn),
由-x
2+ax-a=0解得,
x=,
∴函數(shù)y=f(x)的零點(diǎn)為
x=和
x0=.
綜上可得,當(dāng)a=0時(shí),函數(shù)的零點(diǎn)為0;
當(dāng)0<a<4時(shí),函數(shù)有一個(gè)零點(diǎn),且零點(diǎn)為
;
當(dāng)a=4時(shí),有兩個(gè)零點(diǎn)2和
2+2;
當(dāng)a>4時(shí),函數(shù)有三個(gè)零點(diǎn)
和
.