21. (本小題滿分14分)
(1) 當(dāng)| t | £ 2時,由x⊥y得:x.y = – k + (t2
– 3 ) t = 0,
得k = f (t ) = t3 –
3t ( | t | £ 2 )
當(dāng)| t | > 2時, 由x∥y得: k =
所以k = f (t ) =
5分
(2) 當(dāng)| t
| £
2時, f `(t ) =3 t2
– 3 , 由f `(t
) < 0 , 得3 t2
– 3 < 0
解得 –1 < t < 1 ,
當(dāng)| t | > 2時, f `(t ) = = >
0
∴函數(shù)f (t )的單調(diào)遞減區(qū)間是(–1, 1).
4分
(3) 當(dāng)| t
| £
2時, 由f `(t ) =3 t2 – 3 =0得 t = 1或t = – 1
∵ 1 <| t | £ 2時, f `(t ) > 0
∴ f (t)極大值= f
(–1) = 2, f (t)極小值= f (1) = –2
又 f ( 2 ) = 8 – 6 =
2, f (–2) = –8 + 6 = –2
當(dāng) t > 2 時, f (t ) =< 0 ,
又由f `(t ) >
0知f (t )單調(diào)遞增, ∴ f (t ) > f (2) = –2,
即當(dāng) t > 2 時, –2 < f (t ) < 0,
同理可求, 當(dāng)t < –2時, 有0 < f
(t ) < 2,
綜合上述得, 當(dāng)t = –1或t = 2時, f ( t )取最大值2
當(dāng)t = 1或t = –2時, f ( t )取最小值–2
5分