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,??).
2020-2021学年高三数学(理科)全国卷高考适应性考试及单解析
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?
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