解答:解:(1)根據(jù)題意,可得
∵當(dāng)x>0時,拋物線經(jīng)過點(1,0)、(3,0),
∴設(shè)函數(shù)解析式為f(x)=a(x-1)(x-3),(a≠0),
由點(0,3)在拋物線上,得
f(0)=a×(-1)×(-3)=3a=3,解之得a=1..
∴當(dāng)x>0時,f(x)=x
2-4x+3;
當(dāng)x<0時,則-x>0,可得f(-x)=(-x)
2-4(-x)+3=x
2+4x+3,
∵f(x)是奇函數(shù),∴當(dāng)x>0時,f(x)=-f(-x)=-x
2-4x-3,
結(jié)合當(dāng)x=0時,f(0)=0,可得
f(x)= | x2-4x+3,x>0 | 0,x=0 | -x2-4x-3,x<0 |
| |
.
(2)由函數(shù)的圖象,可得函數(shù)的單調(diào)增區(qū)間為(-∞,-2]和[2,+∞);函數(shù)的單調(diào)減區(qū)間為[-2,0)和(0,2].
(3)由A={(x,y)|y=f(x)},B={(x,y)|y=t,x∈R,t∈R},
可得A∩B的元素的個數(shù),即為直線y=t與函數(shù)y=f(x)圖象公共點的個數(shù).
∵當(dāng)x>0時,f(x)=x
2-4x+3在x=2時有最小值-1;當(dāng)x>0時,f(x)=-x
2-4x-3在x=-2時有最大值1
∴根據(jù)函數(shù)y=f(x)的圖象,平移直線y=t可得
①當(dāng)t>1或t<-1時,直線y=t與函數(shù)圖象有2個公共點;
②當(dāng)t=±1時,直線y=t與函數(shù)圖象有3個公共點;
③當(dāng)t=0時,直線y=t與函數(shù)圖象有5個公共點;
④當(dāng)-1<t<1且t≠0時,直線y=t與函數(shù)圖象有4個公共點
由此可得滿足A∩B有4個元素的實數(shù)t的取值范圍為(-1,0)∪(0,1).