设函数f(x)?x?a?5x.
(1)当a??1时,求不等式f(x)?5x?3的解集; (2)若x??1时有f(x)?0,求a的取值范围. 【解析】(1)当a??1时,不等式f(x)?5x?3, ∴x?1?5x?5x?3, ∴x?1?3,∴?4?x?2.
∴不等式f(x)?5x?3的解集为[?4,2]. (2)若x??1时,有f(x)?0, ∴x?a?5x?0,即x?a??5x,
∴x?a??5x,或x?a?5x,∴a?6x,或a??4x,
∵x??1,∴6x??6,?4x?4,∴a??6,或a?4. ∴a的取值范围是(??,?6]U[4,??).
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