1
综上,当a<时,不等式的解集为{x|x1-a},
211当a=时,不等式的解集为{x|x∈R,x≠},
221
22.解:(Ⅰ)由f(?x)?f(x)得:x2?2ax?2?x2?2ax?2,所以a?0 (Ⅱ)(1)当?a??5,即a?5时, f(x)在[?5,5]上递增,
f(x)min?f(?5)?27?10a??3解得a?3,与条件不符舍去;
(2)?5??a?5,即?5?a?5时,f(x)min?f(?a)??a2?2??3 解得:a??5,符合条件;
(3)当?a?5,即a??5时,f(x)在[?5,5]上递减,
f(x)min?f(5)?27?10a??3 解得a??3,与条件不符舍去;
综合(1)(2)(3),a??5
相关推荐: