②当时,函数是二次函数,设其解析式为,
∵点在函数图象上,
∴解得
·················································(4分)
综上.·········································(5分)
(2)由(1)得当时, ,∴。
结合图象可得若方程有三个不同解,则。
∴实数的取值范围.··············································(8分)
(3)当时,由得 解得 ;··················(9分)
当时,由得,
整理得 解得或(舍去)·······················(11分)
综上得满足的的取值集合是···························(12分).
20.(12分)解:(1)函数为奇函数.证明如下:
定义域为,又··············(3分)