摘要： (1) 当| t | £ 2时,由x⊥y得:x·y = – k + (t2 – 3 ) t = 0, 得k = f (t ) = t3 – 3t 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. 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 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 `单调递增, ∴ f = –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

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