Question

（08年杭州市質(zhì)檢一）(14分) 已知向量x = (1，t² 3 ) ,  y = (k ，t) (其中實(shí)數(shù)k和t不同時(shí)為零)，當(dāng)| t | £ 2時(shí), 有 x⊥y ,當(dāng)| t | > 2時(shí),有x∥y.

(1) 求函數(shù)關(guān)系式k = f (t ) ;

(2) 求函數(shù)f (t )的單調(diào)遞減區(qū)間;

(3) 求函數(shù)f (t )的最大值和最小值.

Answer 1

解析： (1) 當(dāng)| t | £ 2時(shí),由x⊥y得：x?y = k + (t² 3 ) t = 0,

得k = f (t ) = t³ 3t   (  | t | £ 2  )

當(dāng)| t | > 2時(shí), 由x∥y得： k =  

所以k = f (t ) =                         5分

(2) 當(dāng)| t | £ 2時(shí), f `(t ) =3 t<sup>2</sup> 3 ,  由f `(t ) < 0 , 得3 t² 3 < 0

解得 1 < t < 1 ,

當(dāng)| t | > 2時(shí), 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(shí), 由f `(t ) =3 t² 3 =0得 t = 1或t = 1

∵  1 <| t |  £ 2時(shí),  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(shí), 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(shí), 2 < f (t ) < 0,

同理可求, 當(dāng)t < 2時(shí),  有0 < f (t ) < 2,

綜合上述得, 當(dāng)t = 1或t = 2時(shí), f ( t )取最大值2

當(dāng)t = 1或t = 2時(shí), f ( t )取最小值2                    5分