解答:
解:令t=x
2-1,則t≥-1,
此時(shí)y=f(x)=|t|
2-2|t|-1,
其圖象如下圖所示:
由圖可知:
當(dāng)a=-2時(shí),若y=|t|
2-2|t|-1=a,則t=-1,或t=1,此時(shí)f(x)=|x
2-1|
2-2|x
2-1|-1=a有三個(gè)根,即函數(shù)f(x)=|x
2-1|
2-2|x
2-1|-1的圖象與直線y=a有3個(gè)交點(diǎn),
當(dāng)-2<a<-1時(shí),若y=|t|
2-2|t|-1=a,有三個(gè)大于-1的t值,此時(shí)f(x)=|x
2-1|
2-2|x
2-1|-1=a有6個(gè)根,即函數(shù)f(x)=|x
2-1|
2-2|x
2-1|-1的圖象與直線y=a有6個(gè)交點(diǎn),
當(dāng)a=-1時(shí),若y=|t|
2-2|t|-1=a,則有兩個(gè)大于-1的t值,此時(shí)f(x)=|x
2-1|
2-2|x
2-1|-1=a有4個(gè)根,即函數(shù)f(x)=|x
2-1|
2-2|x
2-1|-1的圖象與直線y=a有4個(gè)交點(diǎn),
當(dāng)a>-1時(shí),若y=|t|
2-2|t|-1=a,則有一個(gè)大于-1的t值,此時(shí)f(x)=|x
2-1|
2-2|x
2-1|-1=a有2個(gè)根,即函數(shù)f(x)=|x
2-1|
2-2|x
2-1|-1的圖象與直線y=a有2個(gè)交點(diǎn),
綜上所述,函數(shù)f(x)=|x
2-1|
2-2|x
2-1|-1的圖象與直線y=a有六個(gè)交點(diǎn),a的取值范圍為(-2,-1)