广西三新学术联盟高一1月期末联考
二、多项选择题:本题共4小题,每小题5分,共20分在每小愿给出的四个选项中,有多项符合题目要
求全部选对的得5分,有选错得0分,部分选对得3分
数学试题
9.下列结论正确的是
是第二象限角
本试卷满分150分,考试用时
注意事项
1.答卷前,考生务必将自己的姓名、准考证号填写在答题卡上
2.回答选择题时,选出每小题答象后,用铅笔把答题卡上对应题目的答案标号涂黑。如需改动,用橡度
C.圆心角为的扇形的弧长为x,则该形面积为号
干净后,再选涂其它答案标号。回答非选择题时,将答案写在答题卡上,写在本试舂上无效。
D.若角a为锐角,则角2a为纯角
3.考试结后,将本议卷和答题卡一并交曰
10如图所示的某池塘中的浮萍蔓延的面积y(m)与时间月)的关系为;y=d有以下几个判断,正确
选择题:本题共8小题,每小题5分,共40分在每小题给出的四个选项中,只有一项是符合题目要
求的
设集合A={x|x2-5x-6≥0},B={2,-16},则A∩B=
C.在第4个月,浮萍面积超过sor
23,6}
3,6
-
2,-1.3,6
D.若浮萍蔓延到2m,4m2,8m3所经过的时间分别为l1,l2,则22=+
2.函数f(x)=si
的最小正周期为
()
1l.下列各式最值正确的有
的最小值为
3.命题:“vx∈R,x2-x+2≥0”的否定是
B.当b>0时.b,的最小值为2
第10题图)
A.xgR,x2-x+2≥0
B.Vx∈R,x2-x+2<0
C.3x∈R,x2-x+2≥0
D.3r∈R,x2-x+2<0
C.当a>0.b>0时,(a+-)(b+)的最小值为4
4,函数f(x)=e+x-3的零点所在的区间为
D.当x<0时,x+一的最大值为-2
B.(0,亏)
12已知()是周期为4的奇函数,且当05≤2时,()-120,.设g()=(+(x+)
5.已知a-lg,2,b-ke,3,c=e"),则下列判断正确的是
D. aA.g(2022)=
B函数y=g(x)为周期函数
6.函数y=
的图象大致为
C.函数y=g(x)的最大值为2
D.函数y=g(x)的图象既有对称轴又有对称中心
三、填空题:本题共4小题,每小题5分,共20分
13.化简:132-1g25+(x-23+y8
D
4.已知幂函数y=∫(x)的图象经过点(2,4),则f(5)=
7.已知si
u()
则cos=-2a
)
5.若tan6=-2,则当
3
6已知函数f(x2s{3x+3,其中x6
若∫(x)的值域题/-4、√3
2
则实数m的取
8.已知函数f(x)满足对任意的实数x,恒有f(x)+f(-x)=2,函数g(x)=元若f(x)与g(x)的图象
值范围是
有3个不同的交点(x,y)(x,))(x,)其中16
25
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器广西三新学术联盟高一 1月期末联考
数学试题 参考答案
一、选择题
题号 1 2 3 4 5 6 7 8 9 10 11 12
答案 B A D C C D B B AC ACD BCD ABD
二、填空题
2 2 , 5 13． 3 14．25 15． 16．
5 9 18
三、解答题
17．解：（1）∵ A {x | x 3或 x 1}，
∴ RA x 3 x 1 ，·························································································· 1分
又 a 1即B x 0 x 5 ，················································································ 2分
∴ RA B x 0 x 1 .·················································································· 4分
（2）∵ A B A B A，··············································································· 5分
∴又 A {x | x 3或 x 1}
当 B 时， a 1 2a 3，解得 a 4；····································································7分
a 1 2a 3 a 1 2a 3
当 B 时， 或 ，解得 4 a 3或 a 2 .····························· 9分
2a 3 3 a 1 1
综上，a的取值范围为 a 3或 a 2 .········································································10分
sin 2x 2sin2 x 2sin x cos x 2sin2x
18 ．解： 1 tan x 1 sin x ··························································2分
cos x
2sin x cos x(cos x sin x)
······················································································· 3分
cos x sin x
2sin x cos x sin(x )
4
···························································································4分cos(x )
4
cos( x) 3 17 x 7 ， ，··········································································· 6分
4 5 12 4
4
sin(x ) ，······························································································· 8分
4 5
2sin x cos x 7 ．······························································································ 10分
25
2sin xcos xsin(x )
4 28 ，
cos(x ) 75
4
2
sin 2x 2sin x 28 ．······················································································ 12分
1 tan x 75
19．解：（1）设 f n 为前 n年的总盈利额，单位：万元；············································ 1分
由题意可得 f n 95n 10n2 5n 90 10n2 100n 90 10 n 1 n 9 ，·················3分
广西三新学术联盟高一 1月期末联考 数学答案 第 1页 共 4页
由 f n 0得1 n 9，又 n N*，···········································································4分
所以该设备从第 2年开始实现总盈利；······································································ 5分
（2）方案二更合理，理由如下：············································································· 6分
方案一：由（1）知，总盈利额 f n 10n2 100n 90 10 n 5 2 160，···················· 7分
当 n 5时， f n 取得最大值160；此时处理掉设备，则总利润为160 20 180万元；········8分
f n 10n2 100n 90 9 9
方案二：由（1）可得，平均盈利额为 10 n 100 100 2 0 n 40，n n n n
························································································································· 10分
9
当且仅当 n ，即 n 3时，等号成立；即 n 3时，平均盈利额最大，此时 f n 120，
n
此时处理掉设备，总利润为120 60 180万元；························································· 11分
综上，两种方案获利都是180万元，但方案二仅需要三年即可，故方案二更合适. ··············12分
20．解：（1）由函数 f (x) 2sin x
3
sin x
1
cos x 3 3 sin2 x sin x cos x 3
2 2
2 2
························································································································· 1分
3 (1 cos 2x) 1 sin 2x 3 ··········································································· 3分
2 2 2
sin 2x
···································································································· 4分
3
2k 2x 3 令 2k , k Z，解得 k
5 x 11 k ,k Z ，····················· 5分
2 3 2 12 12
所以 f x 5 11 单调减区间 k , k , k Z．····················································· 6分 12 12
（2）将函数 f (x)的图象向左平移 个单位，得到 y sin(2x )，···························· 7分
3 3
将函数 f x 1 的图象上每个点的横坐标缩小为原来的 ，得到 y g(x) sin(4x )，····· 8分
2 3
因为 0 x ，可得0 4x ，则 4x+ 4 ，·············································· 9分
4 3 3 3
3
可得 sin 4x+
1，··············································································· 11分2 3
3
所以 g x 在 0, 上值域为 ,1 ．··································································· 12分 4 2
21．解：（1）令 x y 0，得 f (0) f (0) f (0)，所以 f (0) 0，···································1分
令 y x，得 f (0) f (x) f ( x)，即0 f (x) f ( x)，所以 f ( x) f (x)，···················· 2分
所以函数 f x 是R上的奇函数. ·············································································· 3分
（2）任取 x1, x2 R ,且 x1 x2 ，则 f (x1) f (x2) f (x1) f ( x2) f (x1 x2) ，···················· 4分
因为当 x 0时， f (x) 0 ，而 x1 x2 ，即 x1 x2 0，所以 f (x1 x2 ) 0，······················5分
所以 f (x1) f (x2 )，所以 f x 在R上的单调递减. ······················································ 6分
1 1
（3）由（1）知 f x 是R上的奇函数，所以 f ( 1) f (1) ，所以 f (1) ，
2 2
所以 f (2) f (1 1) f (1)
1 1
f (1) 1，··························································· 7分
2 2
f (mx2 2所以不等式 x) f x x 1 1可化为 f (mx2 x) f (x2 x 1) f (2)，
即 f (mx2 x) f (2) f (x2 x 1)，所以 f (mx2 x) f (x2 x 3)，······························· 8分
广西三新学术联盟高一 1月期末联考 数学答案 第 2页 共 4页
由（2）知， f x 在 R上的单调递减，所以mx2 x x2 x 3，···································· 9分
故问题转化为mx2 x2 2x 3对于任意的 x [1, 4]恒成立，
即m 1
2 3
2 对于任意的 x [1, 4]恒成立，······························································ 10分x x
1 1 1
令 t 则 t ,14 m 3t
2 2t 1对任意的 t ,14 恒成立············································ 11分x
令 g(t) 3t 2
1
2t 1，其对称轴为 t ，
3
g(t) 1 2 2所以 min g( ) ，所以m 3 .···········································································12分3 3
x 2
22．解：（1）由 ＞0，得 x 2或 x 2． ······················································· 1分
x 2
∴ f (x)的定义域为（-∞，-2）∪（2，+∞）；····························································· 2分
（2）假设存在实数a，使得当 f (x)的定义域为[m,n]时，值域为[1 loga n,1 loga m]，
由m n且1 loga n,1 log m
m 2
a 及 loga 有意义，················································3分m 2
可知 2 m n，·································································································· 4分
又1 loga n 1 loga m，可得 0＜a＜1．································································ 5分
t(x) =1
4
在（2，+∞）上为增函数，
x 2
又∵0 a 1，
∴ f (x)在（2，+∞）上为减函数，···········································································6分
f m log
m 2
a log m 1 m 2 a
∴ ，······································································7分
f n n 2 loga loga n 1 n 2
m,n log x 2 log x 1 x 2∴即 是方程 a a 的两个实数根，即 ax在（2，+∞）上有两个互异实数根m,n，x 2 x 2
于是问题转化为关于 x的方程 ax2 (2a 1)x 2 0在（2，+∞）上有两个不同的实数根m,n，
························································································································· 9分
令 g(x) ax2 (2a 1)x 2，
(2a 1)2 8a＞0
2a 1
则 ＞2 ，····················································································· 11分
2a
g 2 8a＞0
0 a 3 2 2解得 ．
2
又∵0 a 1 3 2 2，故存在这样的实数a (0, )符合题意．····································· 12分
2
广西三新学术联盟高一 1月期末联考 数学答案 第 3页 共 4页
