23. [选修4-5:不等式选讲]
2已知函数f?x???x?ax?4,g?x??x?1?x?1.
(1)当a?1时,求不等式f?x?≥g?x?的解集;
1?,求a的取值范围. (2)若不等式f?x?≥g?x?的解集包含??1,2【解析】(1)当a?1时,f?x???x?x?4,是开口向下,对称轴x?1的二次函数. 2?2x,x?1?g?x??x?1?x?1??2,?1≤x≤1,
??2x,x??1?当x?(1,??)时,令?x2?x?4?2x,解得x?17?1 2g?x?在?1,???上单调递增,f?x?在?1,???上单调递减 ?17?1?1,fx≥gx??解集为?∴此时???. ?2??1?时,g?x??2,f?x?≥f??1??2. 当x???1,当x????,?1?时,g?x?单调递减,f?x?单调递增,且g??1??f??1??2. ?17?1?综上所述,f?x?≥g?x?解集??1,?.
2??1?恒成立. (2)依题意得:?x2?ax?4≥2在??1,1?恒成立. 即x2?ax?2≤0在??1,2??1?a?1?2≤0则只须?,解出:?1≤a≤1. 2????1??a??1??2≤01?. 故a取值范围是??1,
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23.[选修4-5:不等式选讲]2已知函数f?x???x?ax?4,g?x??x?1?x?1.(1)当a?1时,求不等式f?x?≥g?x?的解集;1?,求a的取值范围.(2)若不等式f?x?≥g?x?的解集包含??1,2【解析】(1)当a?1时,f?x???x?x?4,是开口向下,对称轴x?1的二次函数.2?2x,x?1?
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