2?2?0?1?0?cos?m,n??(2)2?02?02332??3?(2)2?12???3??15. 515, 5即二面角A?CC1?B的余弦值为20.(14分)解:⑴由已知
?(x,y)?m(1,0)?(m?1)(0,?1),
?x?m ???y?1?m?x?y?1,即点P的轨迹方程为x?y?1?0.
?x?y?1? ⑵由?x2得:(b2?a2)x2?2a2x?a2?a2b2?0. y2?2?2?1b?a点P轨迹与双曲线C交于相异两点M,N,?b?a?0,且??4a?4(b?a)(?a?ab)?0,(*)设M(x1,y1),N(x2,y2)2a2a2?a2b2,x1x2??2. 则:x1?x2??2b?a2b?a2?以MN为直径的圆经过原点,?OM?ON?0,即x1x2?y1y2?0,
22422222
2a22(a2?a2b2)2222?x1x2?(1?x1)(1?x2)?0.得1?2??0,即:b?a?2ab?0.222b?ab?a①
?e?3,
a2?b2?e??3,
a2?b2?2a2. ②
12. ∴由①、②解得a?,b?2212经检验a?,b?符合(*)式,
22?双曲线C的方程为4x2?2y2?1.
221.(14分)
22.(16分)解:(1) 由f(x?1)??11?f(x), 得f(x?2)??f(x?1)f(x)所以f(x)是周期为2的函数.
∴f(x)?f(2?x)?0即为f(x)?f(?x)?0, 故f(x)是奇函数.
(2)当x∈(,1)时, f(x)?f[1?(x?1)]??12111??1?x.
f(x?1)f(1?x)3所以, 当x∈(2k?11,2k?1)(k?Z)时,f(x)?f(x?2k)?2k?1?x. 23
222(3) log3f(x)?x?kx?2k即为x?2k?1?x?kx?2k,亦即x?(k?1)x?1?0.
令g(x)?x?(k?1)x?1(k是正整数),则g(x)在(2k?21,2k?1)上单调递增, 2k3??0, 2412∴x?(k?1)x?1?0在(2k?,2k?1)上无解,
21从而不存在正整数k,使得当x∈(2k?,2k?1)时,
2而g(2k?)?2k?22不等式log3f(x)?x?kx?2k有解.
12