21. (本小题满分14分)
(1) 当| t | £ 2时,由x⊥y得:x.y = – k + (t2
– 3 ) t = 0,
得k = f (t ) = t3 –
3t ( | t | £ 2 )
当| t | > 2时, 由x∥y得: k =
所以k = f (t ) =
5分
(2) 当| t
| £
2时, f `(t ) =3 t2
– 3 , 由f `(t
) < 0 , 得3 t2
– 3 < 0
解得 –1 < t < 1 ,
当| t | > 2时, f `(t ) = = >
0
∴函数f (t )的单调递减区间是(–1, 1).
4分
(3) 当| t
| £
2时, 由f `(t ) =3 t2 – 3 =0得 t = 1或t = – 1
∵ 1 <| t | £ 2时, f `(t ) > 0
∴ f (t)极大值= f
(–1) = 2, f (t)极小值= f (1) = –2
又 f ( 2 ) = 8 – 6 =
2, f (–2) = –8 + 6 = –2
当 t > 2 时, f (t ) =< 0 ,
又由f `(t ) >
0知f (t )单调递增, ∴ f (t ) > f (2) = –2,
即当 t > 2 时, –2 < f (t ) < 0,
同理可求, 当t < –2时, 有0 < f
(t ) < 2,
综合上述得, 当t = –1或t = 2时, f ( t )取最大值2
当t = 1或t = –2时, f ( t )取最小值–2
5分