第一章 集合与函数概念 单元检测(人教A版必修1)

参考答案及评分标准

一、选择题 1-5:DCADC 6-10:ABCDA 11-12:CA 二、填空题 13．??1,1? 14．解：??x?1?0?x?1???xx?1?,定义域为?1,???或xx?1?

?x?0?x?0??15．解：f?x?在R上的减函数且f?2m?1??f?1?m?

2m?1?1?m?m?16．解：M?N

22?2?，实数m的取值范围是???,?或m?

3?3?3①N??时2a?1?a?1?a?2 ②N??由右图得

-3 4 ?2a?1≤a?1?a≤2???2a?1??3??a??1??1?a?2 ?a?1≤4?a≤3??综上所述a??1或??1,??? 三、解答题

2a-1 a+1 17．解：（Ⅰ）函数f?x?的定义域为R，关于原点对称．············································ 2分 ∵ f??x????x??2??x???x3?2x??x3?x??f?x?， ········································· 5分 ∴ 函数f?x?为奇函数． ·································································································· 6分 （Ⅱ）函数g?x?的定义域为???,0?∵ g?x??x?4?∴ g??x??13??·· 2分 ?0,???，关于原点对称． ································

1， ········································································································· 4分 x44??x??1?g?x?． ······················································································ 5分 x4∴ 函数g?x?为偶函数． ·································································································· 6分

18．解：（Ⅰ）∵ A??x3≤x?6?，

∴ AB??x5?x?6?， ································································································· 3分 ∴ CR(AB)??xx?5或x?6?． ··················································································· 6分 （Ⅱ）∵ B??xx??1或x?5?，

∴ CRB??x?1?x?5?， ································································································ 9分

∴

?CRB?··················································································· 12分 A??x?1?x?6?． ·

19．解：（Ⅰ）函数f?x?在?0,???上为增函数，下证之． ··········································· 1分 设x1,x2是?0,???上的任意两个实数，有x1?x2，则 ······················································ 2分 ?3??3?333?x1?x2?． ············································ 5分 f?x1??f?x2???2????2?????xxxxxx?1??2?2112∵ 0?x1?x2，

∴ x1?x2?0,x1x2?0， ···································································································· 6分 ∴ f?x1??f?x2??0，即f?x1??f?x2?， ····································································· 7分 ∴ 函数f?x?在?0,???上为增函数． ············································································· 8分 （Ⅱ）由（Ⅰ）可知函数f?x?在?2,5?上为增函数． ····················································· 9分

71∴ f?x?max?f?5??,f?x?min?f?2??． ······························································· 12分

5220．解：（Ⅰ）∵ f?a??1?a1·········································································· 2分 ??， ·

1?a3∴ a?2． ·························································································································· 4分 （Ⅱ）∵ f(x)?1?x， 1?x1?1?x?x?1??1?x， ·∴ f???················································································· 7分

1x?11?x?x?1?x1??1?∴ f????f?x?． ·········································································································· 8分

x???1?（Ⅲ）由（Ⅱ）可知f???f?x??0． ········································································ 10分

x???1?∴ f???f?2012??0,f2012???1????f?2011??0,2011???1?,f???f?2??0． ·············· 11分

2??又∵ f?1??0，∴ 原式?0． ···················································································· 12分

（3）设x＜0，则-x＞0， ∵当x?0 时，f(x)?x?6x

∴f(?x)?(?x)?6(?x)?x?6x ················································································ 8分

22221．解：（1）画出函数图象 ································································································ 3分 108642-8-6-4-20-2-4-6-82468yx-10（2）f（x）单调增区间为(??,?3),(3,??) ·································································· 5分

∵函数f（x）为R上的奇函数，

∴f(?x)??f(x) ··············································································································· 9分

f(?x)??f(x)?x2?6x

···································································································· 10分 f(x)??x2?6x,x?0·

2??x?6x,x?0,∴ f(x)??2 ······················································································· 12分

???x?6x,x?0.a?a2a??2，其对称轴为x?． ·22．解：（Ⅰ）f?x????x???·································· 2分

2?42?2∵ 函数f?x?不是单调函数， ∴ ?5?a················································································································· 5分 ?5， ·

2（说明：本步若取等号，扣1分） ∴ ?10?a?10，

∴ 实数a的取值范围为??10,10?． ················································································· 6分 （Ⅱ）①当

a

······················································································ 7分 ≤0，即a≤0时， ·

2

∴ f?x?min?f?5???25?5a?2?5a?23，即g?a??5a?23． ································· 9分 ②当

a································································································· 10分 ?0，即a?0时， ·

2································ 12分 f?x?min?f??5???25?5a?2??5a?23，即g?a???5a?23．·

?5a?23,a≤0,综上所述，g?a??? ·················································································· 14分

?5a?23,a?0.?