河南省南阳市2023-2024学年高一下学期期末质量评估数学试题(含答案)

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河南省南阳市2023-2024学年高一下学期期末质量评估数学试题(含答案)

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2024年春期高中一年级期终质量评估
数学试题参考答案
评分说明:
本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题
的主要考查内容比照评分标准制定相应的评分细则.
一、选择题
题号 1 2 3 4 5 6 7 8
答案 C A C C D B D D
二、选择题
9 10 11
CD ABC AC
三、填空题
4 5
12 5. ,1 或 ,0 13.3 1 3
14.
3 5
四、解答题
15.【解析】
(1) 由题意,可设 z a 3i( a R),则 a2 9 1 a,
解之得 a 4,即 z 4 3i.··························································4分
(1 3i)(3 4i) (1 3i)(3 4i) i(1 3i)(3 4i)
故 3 i.····················6分
z 4 3i 3 4i
0 x 1 1

(2) 由题意得 1 ,······························································8分
cos x 1 2
1 x 0

解之得 ·················································11分
2k x 2k ,k Z 3 3
故 x的取值范围为( 1,0)····································································13分
16.【解析】
2 ( 1 )
(1) sin 2 2sin cos
2sin cos 2 tan 7
2
7 .·············5分
cos sin2 1 tan2 1 ( 1 )2 25
7
高一数学参考答案 第 1 页(共 5 页)
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(2) 因为 为锐角, sin 10 cos 3 10 ,可得 ,····································6分
10 10
cos2 1 2sin2 4 sin 2 1 cos 2 3由 ,可得 ,
5 5 ··································7分
tan 2 sin 2 3所以 ,···································································8分
cos 2 4
1 3
tan 2 tan tan 2

7 4于是 11 3 ,·····························11分1 tan tan 2 1


7 4
3 π π
又因为 tan 2 0,所以0 2 ,而 π,
4 2 2
可得0 2 π,········································································13分

所以: 2 .··········································································15分
4
17.【解析】
(1)易知 S ABC S ABD S BCD ,····························································2分
1
即 AB BC sin( 1 1 ) AB BD sin BD BC sin ···············4分
2 2 2
1 AB BC BD sin( ) sin sin 上式两端同除以 得: ·················6分
2 BD BA BC
(2) 方法一:由 AB AC,C 72 ,结合(1)不妨设 36 ,则 BD BC ,于是:
sin72 sin36 sin36
·······························································8分
BD BA BC
2cos36 1 1
即: ·······························································10分
BC BA BC
BC sin36 1
即:2cos36 1 1 ······························12分
BA sin72 2cos36
1
5 1
解得:cos36 (负值舍去)·················································15分
4
高一数学参考答案 第 2 页(共 5 页)
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方法二:由题不妨设 AB AC 1, BC x, 36 ,则 BD AD x,
1 x
于是CD 1 x,显然 ABC∽ BCD,故 ,·······················10分
x 1 x
所以 x2 5 1 x 1 0,解得 x ,···············································12分
2
即BD AD 5 1 .在 ABD 中,由余弦定理得:
2
cos36 1
2 x2 x2 1 1 5 1
,··································15分
2 1 x 2x 5 1 4
方法三:因为 cos36 sin 54 3sin18 4sin318 (*)
即1 2sin 218 3sin18 4sin318 ,
即 (sin18 1)(4sin218 2sin18 1) 0,由于 sin18 1 0,
从而只有4sin218 2sin18 5 1 1 0,解得 sin18 ,
4
代入(*)可得cos36 5 1 .·························································15分
4
(注:考生用方法三的也给分)
18.【解析】
(1)因为 PA⊥平面 ABCD,所以 PA BC ,·············································1分
又底面 ABCD为矩形,所以 AB BC , PA AB A ,
所以 BC 平面 PAB ,·······································································3分
所以 BC AE ,又在等腰三角形PAB 中,点 E是PB中点,
所以 AE PB ,················································································5分
且PB BC B,所以 AE 平面 PBC ,··············································6分
AE 平面 PBC ,所以 AE PC .························································7分
(2)三棱锥 P ABQ的体积是定值, ·······················································8分
高一数学参考答案 第 3 页(共 5 页)
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证明如下:
过M 作MH / /PA交 AD于H ,连接 NH .·············································9分
PM AH PM AN AH AN
由于MH / /PA,所以 ,又 ,故 ,········10分
DM HD DM CN HD CN
所以HN / /CD ,
而CD / /AB,所以HN / /AB,HN 平面 PAB, AB 平面 PAB ,
所以HN / /平面 PAB,····································································11分
同理:MH / /平面 PAB ,··································································12分
又HN NH H ,所以平面MNH / /平面 PAB .···································13分
由于Q为MN上任意一点,故Q到平面 PAB 的距离可转化为平面MNH 与平面PAB 间距
离,即可转化为 AH 的长, ··································································14分
AH 2 AD 4易知 ,·······································································15分
3 3
1 1 1 4 2
于是V三棱锥P ABQ V三棱锥Q PAB S PAB AH 1 1 ,··········16分3 3 2 3 9
2
即:三棱锥P ABQ的体积是定值,且定值为 .···································17分
9
19.【解析】
(1)连接MA,MB,MC S S S S 1 1 1,则 MAB MCB MAC cr ar br2 2 2
故 r 2S .··············································································4分
a b c
(2) S 1 3①因为 ac sin B ac,······················································5分
2 4
B , cosB a
2 c2 b2
又因为 由 得, a2 c23 64 ac
, ·················6分
2ac
1 2 (a c 8)(a c 8)
故ac [(a c) 64] , (*)·························8分
3 3
由(1)知:
高一数学参考答案 第 4 页(共 5 页)
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r 2S 3 ac ,由(*)式可得,
a c b 2 a c 8
r 3 (a c 8).·········································································9分
6
2
由(*)得 (a c) 64 3ac 3 (a c)2,故 a c 16,·····················10分
4
当且仅当 a c 8时取“=”.····· ······················································11分
r 3因此, (a c 8) 4 3.即 r 的最大值为 4 3.·······················12分
6
(2)②不妨设另两个切点分别为 P,Q,
AT m,TC n,
则 AP m,CQ n,
AM AC | AC | | AT | 8 | AT | ······················································13分
显然,BP BQ,·········································································14分
即 c m a n,又m n b 8,
a c
故可求得: AT 4,····························································15分
2
3
由①知 r (a c 8),可得a c 10,
6
又a2 c2 64 ac,得 a c 2 13,···········································16分
故 AM AC | AC | | AT | 8 | AT | 32 8 13.·······························17分
高一数学参考答案 第 5 页(共 5 页)
{#{QQABDYYUggioQJIAAAgCEQWoCgGQkACACagOABAEIAAAgQFABAA=}#}2024年春期高中一年级期终质量评估
数学试题
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,只有一项是符合题目要求的.)
1.( )
A. B. C. D.
2.已知:,其中为虚数单位,则( )
A.1 B. C. D.2
3.如图是底面半径为1的圆锥,将其放倒在水平面上,使圆锥在此平面内绕圆锥顶点滚动,当这个圆锥在水平面内首次转回原位置时,圆锥本身恰好滚动了3周,则滚动过程中该圆锥上的点到水平面的距离最大值为( )
A. B.2 C. D.
4.已知:,,,若,则与的夹角为( )
A.30° B.60° C.120° D.150°
5.在平面直角坐标系中,平面向量,将绕原点逆时针旋转得到向量,则向量在向量上的投影向量是( )
A. B.
C. D.
6.如图,一个三棱锥容器的三条侧棱上各有一个小洞,,,经测量知,这个容器最多可盛原来水的( )
A. B. C. D.
7.在数学史上,为了三角计算的简便并且更加追求计算的精确性,曾经出现过下列两种三角函数:定义为角的正矢,记作;定义为角的余矢,记作,则下列命题正确的是( )
A.函数的对称中心为
B.若,则的最大值为
C.若,且,则圆心角为,半径为3的扇形的面积为
D.若,则
8.如图,在直角梯形中,已知,,,,现将沿折起到的位置,使二面角的大小为45°,则此时三棱锥的外接球表面积是( )
A. B. C. D.
二、选择题(本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.全部选对的得6分,部分选对的得部分分,有选错的得0分.)
9.下列有关复数内容表述正确的是( )
A.若复数满足,则一定为纯虚数
B.对任意的复数均满足:
C.设在复数范围内方程的两根为,,则
D.对任意两个复数,,若,则,至少有一个为0
10.已知函数,且,则( )
A. B.函数是偶函数
C.函数的图像关于直线对称 D.函数在区间上单调递减
11.如图,在正三棱锥中,底面边长为,侧棱长为,点,分别为侧棱,上的异于端点的动点.则下列说法正确的是( )
A.若,则不可能存在这样的点,使得
B.若,,则
C.若平面,则
D.周长的最小值是
三、填空题(本大题共3小题,每小题5分,共15分.)
12.已知向量,,点是线段的三等分点,则点的坐标是___________.
13.如图,在中,,,,的角平分线交于,交过点且与平行的直线于点,则___________.
14.设为函数图象上任意一点,则的最大值是___________.
四、解答题(本大题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤.)
15.(本小题满分13分)
(1)已知复数满足,求;
(2)设,复数在复平面内对应的点在第三象限,求的取值范围.
16.(本小题满分15分)
已知为锐角,为钝角,且,.
(1)求的值;
(2)求的值.
17.(本小题满分15分)
在中,,.
(1)求证:;
(2)若,,求的值.
18.(本小题满分17分)
如图,平面,底面为矩形,,点是棱的中点.
(1)求证:;
(2)若,分别是,上的点,且,为上任意一点,试判断:三棱锥的体积是否为定值?若是,请证明并求出该定值;若不是,请说明理由.
19.(本小题满分17分)
已知在中,角,,所对应的边分别为,,.圆与的边及,的延长线相切(即圆为的一个旁切圆),圆与边相切于点.记的面积为,圆的半径为.
(1)求证:;
(2)若,,
①求的最大值;
②当时,求的值.

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