2024 年春期高中一年级期中质量评估数学试题
参考答案
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该大题扣 2分．
一、选择题
题号 1 2 3 4 5 6 7 8
答案 B D B A D C B C
二、选择题
题号 9 10 11 12
答案 BC AD AC ACD
三、填空题（本大题共 4小题，每小题 5分，共 20分）
3 7 1 2
13．[0, ) ( ,2 ] 14． y sin(6x )
4 4 2 3
15 3． 16 2 2．
3 3
四、解答题（本大题共 6小题，共 70分．解答应写出文字说明、证明过程或演算步骤）
17．【解析】
（1）因为点Q在单位圆上且0 x2 ( 5，所以2 )
2 1 x 2 5 ，得 .········2分
5 5
由三角函数定义知， sin 5 ,cos 2 5 ，·············································4分
5 5
2sin 5cos
故 12 .··········································································5分
3sin 2cos
(2)由题意: sin sin( 2 5 ) cos ，···············································7分
2 5
cos cos( ) sin 5 ，························································9分
2 5
故 P( 5 , 2 5 ) .····················································································10分
5 5
18．【解析】
1
(1)NC BC BN a BD ··································································3分
3
高一数学参考答案 第 1 页（共 4 页）
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1 a (AD AB) a 1 (a 2 1
b) a b ··························6分
3 3 3 3
(2)因为MC MB BC 1 b a ,······························································8分
2
2 1 2
且由（1）知 NC (a b)，所以 NC MC，······································10分
3 2 3
所以 NC / /MC，又C为公共点, 所以M ,N ,C三点共线. ·······························12分
19．【解析】
(1)设D点坐标为 (x, y)，则 AB (1, 2), AD (x 1, y)，·······························2分
x 1 2y 5 x 2 x 2
2 2 ，解得 或 ，···········································5分
(x 1) y 10 y 1 y 3
即点D的坐标为 (2,1)或 ( 2,3) ··································································6分
(2)由向量 c与a+b共线，令 c t(a b),t R，则b c ta (1 t)b，··················7分
r r 1
而向量 a，b 为单位向量，且 a b ，2
2 2 2
于是得 b c ta (1 t)b ta 2t(1 t)a b (1 t)2b t2 t 1 ·············8分
(t 1 3 3 )2 ，·············································································10分
2 4 2
1
(当且仅当 t 时取“=”) ········································································11分
2
所以 | b c | 3的最小值为 .······································································12分
2
20．【解析】
2 2 2
(1)（方法一：）由余弦定理得： cos B a c b ,又由题知: 2c cos B 2a b，···1分
2ac
a2 2 2
所以 2c ( c b ) 2a b，化简得 a2 b2 c2 ab，·································3分
2ac
2
cosC a b
2 c2 1
所以： ，·································································5分
2ab 2
π
因为C (0,
π )，故C .··········································································6分
2 3
（方法二：）由正弦定理得: 2sinC cos B 2sin A sin B，·································1分
sin B 2sin(B C) 2sinC cos B
sin B 2sin B cosC ··································· ··············································3分
高一数学参考答案 第 2 页（共 4 页）
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因为 sin B 0
1
所以： cosC ，··················································································5分
2
C π
π
因为 (0, )，故C .··········································································6分
2 3
(2)由余弦定理： c2 a2 b2 2ab cosC，
整理得 a2 5a 6 0，解得 a 2或 a 3 .····················································8分
当 a 2时， a2 c2 b2，最大角 B是钝角， ABC为钝角三角形，舍去；··········9分
当 a 3时， a2 c2 b2，最大角 B是锐角， ABC为锐角三角形，符合题意. ·····10分
1 1
所以 S ABC absin C 5 3
3 15 3
.···············································12分△ 2 2 2 4
21．【解析】
5π 5π
(1) 由题对任意 x R ，都有 f (x) f ，故当 x 时， f (x)取得最大值. ·······1分
12 12
π
因为 f (x)
在 ,
5π π
是单调函数，且 f (x)的图象关于点 ,0 对称,
6 12 6
T 5π π π
所以 得T π，所以
2
2 ················································3分
4 12 6 4 T
f x x 5π 5π π又因为函数 在 时取得最大值，所以 2kπ,k Z，
12 6 2
π 2kπ,k Z π π即 ．因为 2 ，所以
3 ，··········································5分3
所以： f (x) sin 2x
π
3 .·········································································6
分
π 2π
(2)因为 x 0,
π
，令 t 2x
π
，则 t ,
2 3 3 3
y sin t π 2π 在 , 内的图象如图所示，···················7分 3 3
由题函数 g(x)
π
f (x) m x 0, 在 有两个零点 x , x , 2 1 2
y sin t y m π 2π即 与 在 ,
内有两个交点 t1, t2 ，3 3 ······································9
分
3
数形结合可得： m 1， t1 t2 2x
π π 5π
1 2x2 π，即 x1 x2 3 3 2
，······11分
6
f x x 5π π π π 3所以 1 2 sin
sin π
sin .····························12分
3 3 3 3 2
22．【解析】
(1) S = 1 S = 1 (1 1 3 3由题可知： CGD ACD S ABC )= =
3 ,·························2分
3 3 2 6 2 4
1
即 CG GD sin 60 = 3，所以GD=1，·················································3分
2 4
从而 GCD为等边三角形，则 BDG 120 , BD CD 1 ,························4分
高一数学参考答案 第 3 页（共 4 页）
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因为G为 ABC的重心，所以G为线段 AD的三等分点，所以 AD 3
在 ABD中，由余弦定理得： AB2 AD2 BD2 2AD BD cos BDG
32 12 2 3 1 cos120 13 ,··························································5分
所以 AB 13 .··················································································6分
(2)（方法一:） 由GB GC ,且D为 BC中点,则BD CD GD ,
不妨设 BD CD GD x ,则 AD 3x,BC 2x ,····································7分
在 ABC中, 由余弦定理: (2x)2 b2 c2 2bc cos BAC① , ··························8分
2 2 2 2 2 2
又因为 ADB ADC x (3x) c x (3x) b ,易得: 0② ·········9分
2 x (3x) 2 x (3x)
由①②解得:
2 2
cos BAC 2 (b c ) 2 b c 4 ( ) , (当且仅当b c时取“=”) ····················11分
5 bc 5 c b 5
4
故 cos BAC 的最小值为 .································································12分
5
2 2 1 2 1
（方法二:） GB AB AG AB AD AB (AB AC) AB AC ,
3 3 2 3 3
同理：GC 2 AC 1 AB ,··································································8分
3 3
由GB GC，得GB GC 0 ,
2 1
即GB GC ( AB AC) (2 AC 1 AB)
3 3 3 3
5 2AB AC 2 AB 2
2
AC 0 5,即 bc cos BAC 2 c 2 2 b 2，····················10分
9 9 9 9 9 9
2 2
cos BAC 2b 2c 2 b c 4 b c 4 ( ) ，当且仅当b c时，取“=”·····11分
5bc 5 c b 5 c b 5
4
故 cos BAC 的最小值为 .··········································································12分
5
高一数学参考答案 第 4 页（共 4 页）
{#{QQABQYAAggAAAJJAABgCEQEiCgEQkBEAACoOBEAIIAAASQFABCA=}#}2024年春期高中一年级期中质量评估
7.若a,B,0∈(0，罗)，且cosa=tana,cosB=B,cos0-sin0,则a,A,0的大小是
数学
试题
A.aB
B.a<<0
C.B
D.B<0<
注意事项：
8.已知f(x)=sin(x十p),其中w>0,01.本试卷分第I卷（选择题）和第Ⅱ卷（非选择题）两部分，考生做题时将答案答在答题
象如下图，则f(答)=
卡的指定位置上，在本试卷上答题无效.
2.答题前，考生务必先将自己的姓名、准考证号填写在答题卡上。
A.-1
3.选择题答案使用2B铅笔填涂，非选择题答案使用0.5毫米的黑色中性（签字）笔或碳
素笔书写，字体工整，笔迹清楚
吗
4,请按照题号在各题的答题区域（黑色线框）内作答，超出答题区城书写的答案无效
5.保持卷面清洁，不折叠、不破损
c.2
第I卷选择题（共60分）
n-号
毁
一、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
二、多选题（本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目
符合题目要求的.)
要求，全部选对的得5分，部分选对的得2分，有选错的得0分.)
戴
1.与角一2024°4'终边相同的角是
9.下列等式恒成立的是
A.-4044
B.-2244
C.31556
D.67556'
A.sin(ta)=sina
B.cos(a-)=sina
2.已知A(1,2),B(4,3),C(x,6),若AB∥AC,则x=
A.10
B.11
C.12
D.13
Csin(-受+e）=coa
D.tan(π+a)=-tang
3.在扇形AOB中，∠AOB=2,弦AB=2,则扇形AOB的面积是
10.已知向量a=(1,2),b1=2,a+b1=√7，则
A动
1
B.(sin1)
C.sin1
D.(sin1)2
4.在梯形ABCD中，∠BAD-∠CDA=90°，AD=5,则AD·BC
Aa在b上的投影数量是一己
BB在a上的授影向量是(-得，-5)
5
A.25
B.15
C.10
D.5
C.a与b夹角的正弦值是25
5
D.(4a+b)⊥b
5.在△ABC与△AB,C中，已知AB=AB=x,BC=B,C=3,∠C=∠C=5,若
△ABC≌△AB,C1,则
11.设函数f(x)=Asin(az十p)(其中A>0,w>0,-一元<9<0).若f(x)在[晋，受]上具有单
A.xE0,2]
B.x∈(0W3)
调性，且f(受)=管)=-f(答)=瓦，则
C.x∈[V3,十∞)
D.x[,+∞)U(2)
A.A=2
Ba-号
6,小娟，小明两个人共提一桶水匀速前进，已知水和水桶总重力为亡，两
人手臂上的拉力分别为，，且1=|1，与的夹角为0，下
Cp=-受
D当xe[-晋，]时)∈[-2w2
列结论中正确的是
12.在△ABC中，AB=2,AC=3,∠BAC=60°，则
.0越小越费力，0越大越省力
B始终有1=1-阿
A,△ABC的周长是5十√7
BBC边上的中线长
C.当9=时，1=
D.当0=罗时，1=G创
C.BC边上的角平分线长6
5
D.BC边上的高长3y
7
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