惠州市2024届高三第一次调研考试数学试题含解析

图1图2
惠州市2024届高三第一次调研考试试题
数学
全卷满分150分，时间120分钟．注意事项：
1．答题前，考生务必将自己的姓名、准考证号、座位号、学校、班级等考生信息填写在答题卡上。

2．作答单项及多项选择题时，选出每个小题答案后，用2B 铅笔把答题卡上对应题目的答案信息点涂黑。

如需改动，用橡皮擦干净后，再选涂其它答案，写在本试卷上无效。

3．非选择题必须用黑色字迹签字笔作答，答案必须写在答题卡各题指定的位置上，写在本试卷上无效。

一、单项选择题：本题共8小题，每小题满分5分，共40分。

在每小题给出的四个选项中，只有一项符合题目要求，选对得5分，选错得0分。

1.已知集合{}*
|,6U x x x =∈≤N ，{}1,2,3A =，{}3,5B =，求()U A B = ð（
）
团用数学软件制作“蚊香”模型，画法如下：在水平直线上取长度为1的线段AB ，作一个等边三角形ABC ，然后以点B 为圆心，AB 为半径逆时针画圆弧交线段CB 的延长线于点D ，由此得到第1段圆弧 AD ，再以点C 为圆心，CD 为半径逆时针画圆弧交线段AC 的延长线于点E ，再以点A 为圆心，AE 为半径逆时针画圆弧……以此类推，当得到如图2所示的“蚊香”恰好有11段圆弧时，则该“蚊香”的长度为（）A ．14π
B ．18π
C ．30π
D ．44π
多项符合题目要求。

全部选对得5分，部分选对得2分，有选错的得0分。

三、填空题：本题共4小题，每题5分，共20分。

四、解答题：本题共6小题，共70分。

解答应写出文字说明、证明过程或演算步骤。

18.（本小题满分12分）
设等差数列{}n a 的公差为d ，且12d a =，59a =．（1）求数列{}n a 的通项公式；
（2）设数列{}n b 满足112223
32
n n n
n a b a b a b ++++=-
，求{}n b 的前n 项和n S ．19．（本小题满分12分）
如图，在五面体ABCDE 中，AD ⊥平面ABC ，//AD BE ，2AD BE =，AB BC =.（1）问：在线段CD 上是否存在点P ，使得PE ⊥平面ACD ？若存在，请指出点P 的位置，并证明；若不存在，请说明理由．
（2）若AB =
，2AC =，2AD =，
求平面ECD 与平面ABC 夹角的余弦值．
21．（本小题满分12分）
已知椭圆22
22:1(0)x y C a b a b
+=>>的左顶点为A ，上顶点为B ，右焦点为()0,1F ，O 为坐
标原点，线段OA 的中点为D ，且BD DF =．
（1）求C 的方程；
（2）已知点M N 、均在直线2=x 上，以MN 为直径的圆经过O 点，圆心为点T ，直线AM AN 、分别交椭圆C 于另一点P Q 、，证明直线PQ 与直线OT 垂直．
22．（本小题满分12分）
惠州市2024届高三第一次调研考试数学试题参考答案与评分细则
一、单项选择题：本题共8小题，每小题满分5分，共40分题号12345678答案
B
C
A
C
D
B
D
A
1.【解析】由已知可得{}1,2,3,5A B ⋃=，{}1,2,3,4,5,6U =，所以(){}6,4=B A C U ，故选：B ．
5.【解析】由弧长公式r l ⋅=α得：r l ⋅=
31，r l 232⋅=，r l 333⋅=，...，r l 11311⋅=，其中1==AB r ，()ππ
44113213
211321=+⋅⋅⋅+++=
+⋅⋅⋅+++=∴l l l l L 蚊香的长度故选：D.
二、多项选择题：本题共4小题，每小题满分5分，共20分。

在每小题给出的四个选项中，有多项符合题目要求。

全部选对得5分，部分选对得2分，有选错的得0分。

2
25624π⎛⎫
=- ⎪ ⎪⎝⎭，h 为球缺的高，
三、填空题：本题共4小题，每小题5分，共20分。

15.【解析1】
BP AD λ=，[]0,1λ∈，则()
2
AD AP AD AB BP AD AB AD
λ⋅=⋅+=⋅+22cos120442AD AP λλ⋅=⨯⨯︒+=-
，因为[]0,1λ∈，所以22AD AP -≤⋅≤ ．
【解析2】如图示，以C 为原点，BC
为x 轴正方向，过C 垂直向上方向为y 轴建立平面直角坐标系.
因为菱形ABCD 的边长为2，60ABC ∠=︒，则()2,0B
-,()0,0C ,(D ,(A -.
因为点P 在BC 边上（包括端点），所以(),0P t ,其中[]2,0t ∈-.
所以()2,0AD = ,(1,AP t =+ ,所以22AD AP t ⋅=+
.
因为[]2,0t ∈-，所以[]222,2AD AP t ⋅=+∈-
.故答案为：[]22-,
．【注1】答案用不等式表示时使用了含未定义参数的，得0分。

如：22a -≤≤，则认为a 为未定义
的参数。

【注2】本题可用临界值的投影向量点积可到范围的边界值。

【注】答案用不等式表示时使用了含未定义参数的，得0分。

如：2
05
λ<<，则认为λ为未定义的参数。

四、解答题：本题共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．（本小题满分10分，其中第一小问5分，第二小问5分）【解析】（1）由正弦定理
sin sin a b
A B
=得sin sin cos sin B B A A B +=················1分
sin 0B >，
∴cos 1A A -=·························································2分∴得1sin 62
A π⎛⎫-
= ⎪⎝
⎭·····································································3分
()0,A π∈，∴5,666A πππ⎛
⎫⎛⎫-∈- ⎪ ⎪⎝⎭⎝⎭········4分【注：无范围的说明，本得分点不得分】
∴6
6
π
π
=
-
A ，即3
π
=
A .·····································································5分
（2）【解法1】由余弦定理A bc c b a cos 2222-+=，·························································1分
代入已知得c
c 416212-+=·······································································2分
解得舍）或(15-==c c ······················································································3分所以1
sin 2
ABC S bc A =
······················································································4分14522
=⨯⨯⨯=所以ABC ∆的面积为··················································································5分
【解法2】由正弦定理
sin sin a b A B =得4
sin sin 3
B π=
，得27sin 7B =，·························1分由a b >，则0,
2B π⎛⎫∈ ⎪⎝⎭所以21cos 7
B ==
··········2分【注：无范围的说明，本得分
点不得分】
所以()sin sin sin cos sin cos C A B A B B A
=+=
+1=
277214
=
⨯+⨯·························································3分所以1
sin 2
ABC S ab C =
(4)
分
1574214
=⨯=所以ABC ∆
的面积为···········································································5分18．（本小题满分12分，其中第一小问5分，第二小问7分）【解析】（1）依题意有⎩⎨
⎧=+==9
42151
d a a a d ·······················································1分
解得2
,11==d a ··········································3分【注：每个结果正确各1分】
于是()1121
n a a n d n =+-=-······························4分【注：公式正确即可给分】
∴数列{}n a 的通项公式是21
n a n =-··········································5分
（2） N n *∈，11222332n n n
n a b a b a b ++++=-
当2n ≥时，1122111
21
32n n n n a b a b a b ---++++=- ···········································1分两式相减得：212n n n
n a b -=···································································2分
而当1n =时，1112a b =
满足上式，因此N n *∈，21
2
n n n n a b -=··························3分由（1）知21n a n =-，于是得1
2n n
b =
，·····················································4分∴21
212111=
=--n n n n b b ∴{}n b 是以
21为首项，2
1
为公比的等比数列······5分【注：无说明首项及公比，本得分点不得分】∴1211122112
n
n n
S b b b ⎡⎤⎛⎫-⎢⎥ ⎪⎝⎭⎢⎥⎣
⎦=+++=- ·······················6分【注：公式正确即可给分】
得112n
n S ⎛⎫
=- ⎪⎝⎭
··················································································7分
19.（本小题满分12分，其中第一小问6分，第二小问6分）
【解析】（1）当P 为线段CD 的中点时，PE ⊥平面ACD.·············································1分
证明如下：
【解法1】分别取AC 、CD 中点O 、P ，连接OB 、PE 、OP .
在ACD ∆中，O 、P 分别是AC 、CD 的中点，∴AD OP 2
1//······························2分
又 //AD BE ，2AD BE =，即AD BE 2
1
//
，∴OP //∴四边形OBEP 是平行四边形，PE
BO //·························································3分
又 ABC AD 平面⊥，ABC OB 面⊂，∴OB AD ⊥，则有AD PE ⊥······················4分由AB BC =知OB AC ⊥，则有AC
PE ⊥且A AD AC = ，ACD AC 面⊂，ACD AD 面⊂·····5分【注：未列举全三个条件，本得分点不得分】
∴PE ⊥平面ACD
····························································································6分
【解法2】取AC 中点O ，连接OB ，作//Oz AD ，由已知可知,,Oz OB AC 两两垂直，
故可建立如下图所示的空间直角坐标系O xyz -··························2分令22AD BE a ==，OB c =，OA OC b ==，取CD 中点P 则(0,,2)D b a -，(0,,0)C b ，(,0,)E c a ，()
a P ,0,0所以(0,2,2)CD
b a =- ，(,,)CE
c b a =-
，()0,0,c PE =············3分
若(,,)m x y z = 是面CDE 的一个法向量，即220
m CD by az m CE cx by az ⎧⋅=-+=⎪⎨⋅=-+=⎪⎩
令z b =，则(0,,)m a b =
···································································4分
0=⋅m PE ，·························································································5分
∴PE ⊥平面ACD .···················································································6分
（2）【解法1】在平面ABED 中分别延长DE 、AB 交于点F ，并连接CF
·····················1分
由AD BE 2
1
//
，知点B 是AF 中点，又O 是AC 的中点∴FC BO //，而由（1）知ACD OB 平面⊥，∴ACD FC 平面⊥····································2分∴ACD ∠就是平面ECD 与平面ABC 的夹角·····························································3分
在ACD Rt ∆中,2AC =，2AD =22=∴DC ···························································4分
∴2
2
2
22cos =
=
∠ACD ·······················································································5分所以平面ECD 与平面ABC 的夹角的余弦值为
2
2.················································6分
【解法2】取AC 中点O ，连接OB ，作//Oz AD ，由已知可知,,Oz OB AC 两两垂直，故可建立如下图所示的空间直角坐标系O xyz -·······························1分【注：若第一问已经建系，则不重复给分】
()0,1,0-A ，(0,1,2)D -，(0,1,0)C
，E 所以()2,2,0-=CD ，(
)
1,1,2-=
DE ········································2分
若()l k j n ,,=是面ECD 的一个法向量，即⎪⎩⎪⎨
⎧=-+=⋅=+-=⋅0
20
220l k j DE n l k CD n 令1=l ，则0=j ，1=k ，即()1,1,0-=n ······························································3分而平面ABC 的一个法向量为()
2,0,0=AD ··························································4分
设平面ECD 与平面ABC 的夹角为θ
，则22
2
2200cos =⨯++=
=
θ····················5分所以平面ECD 与平面ABC 的夹角的余弦值为
2
2
.··················································6分
【解法3】由题可知，AD ⊥平面ABC ，//AD BE ，BE ⊥平面ABC
·····························1分
所以∆CDE 在底面的投影为∆ABC
·····························2分【注：无本步骤，本得分点不得分】
在等腰∆CDE 中，CE =DE =2，CD =2，2222
1
E C =⨯⨯=
∆D S ，··························3分同理在等腰∆ABC 中，AB =BC =3，CA =2，22222
1
B C =⨯⨯=
∆A S ························4分设平面ECD 与平面ABC 夹角为θ，则22
S S cos DEC ABC =
=
∆∆θ········································5分所以平面ECD 与平面ABC 的夹角的余弦值为
2
2
.····················································6分20.（本小题满分12分，其中第一小问6分，第二小问6分）【解析】（1）根据题意知，X 的可能取值为15,0,15,30
-···········································1分
可得()150.40.50.750.15
P X =-=⨯⨯=()00.60.50.750.40.50.750.40.50.250.425
P X ==⨯⨯+⨯⨯+⨯⨯=······························2分
()150.40.50.250.60.50.250.60.50.750.35P X ==⨯⨯+⨯⨯+⨯⨯=()300.60.50.250.075P X ==⨯⨯=·······························································3分
∴随机变量X 的分布列为
21.（本小题满分12分，其中第一小问5分，第二小问7分）
【解析】（1）由题意知：1c =，(,0)2
a
D ··········································································1分
∴C 的方程1
43
+=·························································································5分
（2）【解法1】令(2,)M m ，(2,)N n ，则(2,
2
m n
T +，而(2,0)A -∴可设直线:(2)4
m
AM y x =
+··········································································1分
联立()⎪⎪⎩
⎪⎪⎨
⎧+==+2413
42
2x m y y x 整理得2222(12)44480m x m x m +++-=·····································2分
若11(,)P x y ，则由韦达定理得：2
12
4212
m x m -=-+∴2122(12)
12
m x m -=+，则1
21212m y m =+······································································3分
同理，直线:(2)4n BN y x =+，联立()
⎪⎪⎩
⎪⎪⎨
⎧+==+2413
42
2x n y y x 若22(,)Q x y ，可得2222(12)
12
n x n -=+，则221212n y n =+············4分
∴1222126(12)()12
24()4()
PQ y y mn n m mn k x x n m n m ----=
==--+··············5分
又 OM ON ⊥，∴
12
2-=⋅n
m ∴4mn =-，∴4PQ k n m =-
+，而4
OT n m
k +=·············································································6分∴1-=⋅OT PQ k k ，则直线PQ 与直线OT 垂直，得证.·····················································7分
【解法2】令(2,)M m ，(2,)N n ，则(2,
2
m n
T +由MN 为直径0,=⋅⊥∴ON OM ON OM 即所以4+mn =04
)2(20m
m k AM =
---=
则可设直线)2(4
AM +=
x m
y l ：·············································································1分联立椭圆方程()⎪⎪⎩
⎪⎪⎨
⎧+==+2413
42
2x m y y x 整理得2222(12)44480m x m x m +++-=·····················2分
若11(,)P x y ，则由韦达定理得：2
12
4212
m x m -=-+∴2122(12)
12
m x m -=+，则1
21212m y m =+···································································3分
1212,12)12(2(222++-∴m m m m P 同理1212,12)12(2(222++-∴n n
n n Q ···································4分
)12121212,12)12(212)12(2(222222+-++--+-=∴n n
m m n n m m PQ ，·········································5分
12612612612612)12(412)12(42222222222+-
+-+++++--+-=⋅∴n mn
n m n n n mn m m n n QT PQ 0
2212
2241222422
22=-=++-++=⋅∴m m n n QT PQ ··················································6分
OT PQ OT PQ ⊥⊥∴即,···················································································7分。