2022―2023 学年度东莞市高三第一学期教学质量检查数学参考答案

2022―2023学年度东莞市第一学期教学质量检查高三数学参考答案
1．【答案】A 【解析】由220x x --<，得(2)(1)0x x -+<，得12x -<<，即{|12}B x x =-<<， 则{|1}x x A B >-= ，故选：A ．
2．【答案】C 【解析】21i i(1i)
1i i i
z ++===-
，故z ==．故选：C ． 3．【答案】D 【解析】因为(4,3)AB = ，(3,)AC t = ，所以(1,3)BC AC AB t =-=--
，因为1BC = ，
所以22
(1)(3)1t -+-=，解得3t =，则(3,3)AC = ，所以433321AB AC ⋅=⨯+⨯= ．故选：D ．
4．【答案】A
【解析】由题意设三次函数的解析式为()(2)()f x ax x x b =-+，即32()(2)2f x ax a b x abx =+--，
2()32(2)2f x ax a b x ab '=+--，(0)21(2)124(2)22f ab f a a b ab =-=-⎧∴⎨=+--=''⎩，解得142
a b ⎧=⎪
⎨⎪=⎩，
∴3
11()(2)(2)44
f x x x x x x =-+=-，故选：A ．
5．【答案】C 【解析】(1,0)F ，准线方程为1x =，设(,)P P P x y ，由3PF =可得13P x +=，
2P x ∴=，代入24y x =得2
8P y =
，即(2,P ±
，故OP =
==，故选：C
6．【答案】C 【解析】由分步乘法计数原理，所有可能的传球情况有33327⨯⨯=种，其中仅第1次传球给乙的情况有326⨯=种，仅第1次传球给乙的情况有236⨯=种，仅第3次传球给乙的情况有224
⨯=种，所以所求概率为66416
2727
P ++==．故选：C 7．【答案】B 【解析】体积最大时，沿上下底面直径所在平面作出剖面图如图所示，显然此时圆F 与等腰梯形ABCD 的上底以及两腰相切，则建立如图所示直角坐标系，
由题意得(10)B ,
，C
，则BC k == 则直线BC 所在直线方程为3(1)y x =-
0y -=， 设(0,)F t ，体积最大时球的半径为R ，
则R EF t ==，则点F 到直线BC 的距离等于半径R ，
t
，解得t =
0t <<
t ∴=
，此时EF ==
，则3344
ππ33
V R =
=⨯=，故选：B ． 8．【答案】B 【解析】令()ln 1f x x x =++，则()f x 在定义域(0,)+∞内单调递增，e ln 1a a b b +=++ ，即()(e )1(e )a a f b f f =-<，e a b ∴<，A 错误，B 正确；
令3a =-，则3()e e 30a f b a -=+=-<，且(1)20f =>，01b ∴<<，此时1b a >+，C 错误； 令0a =，则()e 1a f b a =+=，且(1)21f =>，01b ∴<<，此时1b a <+，D 错误；故选：B ．
9．【答案】AC
【解析】二项式2023
(1
，展开式中，通项公式为12023
C
r
r
r T +=，
该二项展开式中二项式系数和为20232，令1x =得各项系数和为20232023(11)2+=，二项展开式中二项式系数和与各项系数和相等，A 选项正确；由二项式展开式的通项公式可知，r 为偶数时，对应的项为有理项，B 选项错误；该二项展开式中的常数项是012023C 1T ==，C 选项正确；
该二项展开式中含x
的项为2
2
32023
20232022C 2T x ⨯==
，系数是20232022
2
⨯，D 选项错误．
10．【答案】BD 【解析】对于A
，π()sin cos 4f x x x x ⎛
⎫=+=+ ⎪⎝
⎭，周期2π2π1T =
=．
当π0,
2x ⎛⎫∈ ⎪
⎝
⎭时，3,444
x πππ
⎛⎫+∈ ⎪⎝⎭，所以()f x 在π0,2⎛⎫
⎪⎝⎭
上先增后减，A 错误；
对于B
，π()sin cos 4f x x x x ⎛
⎫=-=- ⎪⎝⎭
，周期2π2π1T =
=． 当π0,2x ⎛⎫∈ ⎪⎝⎭时，,444x πππ⎛⎫-∈- ⎪⎝⎭，所以()f x 在π0,2⎛⎫
⎪⎝⎭
上单调递增，B 正确；
对于C ，1()sin cos sin 22f x x x x ==，最小正周期2π
π2
T =
=，满足()(2π)f x f x =+， 当π0,2x ⎛⎫∈ ⎪⎝⎭时，2(0,)x π∈，所以()f x 在π0,2⎛⎫
⎪⎝⎭
上先增后减，C 错误；
对于D ，
sin ()tan cos x f x x x
==，最小正周期ππ1T ==，满足()(2π)f x f x =+， 且由正且函数的单调性可知函数在ππ,22⎛⎫
- ⎪⎝⎭
单调递增，故D 满足题意．故选：BD ．
11．【答案】BC
【解析】直线AF 在平面11BB C C 内的投影为直线BF ， 显然直线BF 与直线1B C 不垂直，
由三垂线定理可知直线1B C 与直线AF 不垂直，A 错误；
因为11//AG D F ，1AG ⊄平面1D EF ，1D F ⊂平面1D EF ， 所以1
//AG 平面1D EF ，B 正确； 易证得1EF B C ⊥，EF CD ⊥，又1B C CD C = ， 所以EF ⊥平面11A B CD ，又EF ⊂平面1D EF ，
所以平面1D EF ⊥平面11A B CD ，C 正确；
若点C 和点1A 到平面1D EF 的距离相等，则有1
//AC 平面1D EF ， 或平面1D EF 过1AC 的中点，显然这两条都不满足，所以D 错误．
12．【答案】ABC 【解析】由2
212
x y +=
得1c ===，所以(1,0)F ，
联立22
12
y kx m x y =+⎧⎪⎨+=⎪⎩，消去y 并整理得222(12)4220k x mkx m +++-=， 2222164(12)(22)0m k k m ∆=-+->，即2212k m +>，
设11(,)A x y 、22(,)B x y ，则122412mk x x k +=-+，2122
22
12m x x k
-=+， 所以
AB ==
=，对于A ，当
m k =时，:(1)l y k x =+过椭圆的左焦点(1,0)-， 此时
AB =
=，若4FA FB +
=
，则由4AB FA FB a +==
+ ，得4AB =
-
4=-，解得21k =，k =，所以存在k =
，使得4FA FB +=
，故A 正确；
1
对于B ，当m k =时，2122
412k x x k +=-+，312122242()221212k k
y y k x x k k k k
+=++=-+=++，
所以1212(11,)x x y FA y FB =-+-+=+
== 令2
121k t +=≥，则2
21k t =-
，则FA F B ==+= ， 因为101t
<≤，所以当1
1t =，即1t =，0k =时，FA FB + 取最小值2，故B 正确；
对于C ，当1k =
时，4AB FA FB ==<=+ ，此时存在m ∈R 使得4FA FB +=
，故C 正确；
对于D ，当1k =时，12244123mk m x x k +=-
=-+，1212
42
2233
y y x x m m m m +=++=-=，
所以1212(11,)x x y FA y FB =-+-+=+
===，
因为2212k m +>且1k =，所以23m <
，所以m <<
，
所以当65m =-时，FA FB +
取最小值2<．故D 不正确．故选：ABC
13．【答案】1 【解析】()f x 的定义域为(,0)(0,)-∞+∞ ，因为()f x 为奇函数，所以()()0
f x f x -+=对任意非零实数恒成立，所以11
(1)(1)21022
f f a a a +-=+-+=-=，解得1a =．
14．【答案】1
2
##0.5 【解析】()2sin 2f x x '=-
，所以()()cos 22sin 22h x x x x ϕ=-=+
，其中
cos ϕ=
，sin ϕ=
，tan 2ϕ=，因为函数()h x 关于(,0)a 对称，所以π2π2a k ϕ+=+，π2π,2a k k ϕ=+-∈Z ，所以sin
cos 12tan 2tan tan 22sin 2cos
2a k πϕππϕπϕϕπϕϕ⎛⎫- ⎪
⎛⎫⎛⎫⎝⎭=+-=-=== ⎪ ⎪⎛⎫⎝⎭⎝⎭- ⎪⎝⎭
．
15．【答案】
【解析】因为90APB ∠︒≥，所以45APO ∠︒≥，
所以2sin sin 45O OA APO O P P ∠==︒=≥ 解得OP ≤P 的坐标为(x ，
，解得x ≤，所以动点P 的轨迹的长度为 16．【答案】32)##44
【解析】设球的半径为r ，由题意可知四面体A BCD -为正四面体，棱长为4r ，所以四面体A BCD -的
，所以2
311(4)3V r r ==，要使12V V 取得最大值，则使2V 取最小值，由题
意可知此时该三角垛与四面体1111A B C D -相切．
取11C D 的中点1E ，过点A 作平面BCD 的垂线，垂足为1O ，如图可得截面111A B E ，过A 作11AF A E ⊥于
F ，则AF r =，设1AO 平面BCD O =，则点1O 为111B C D △ 的中心，点O 为BCD △的中心，且
1OO r =，设正四面体1111A B C D -的棱长为a ，则11A B a =
，1111A E B E ==
，11B O =
，
11O E =
，11AO =，由1111113O E AF AA A E ==，可得133AA AF r ==
，又AO =，
所以11114AO AA AO OO r =++=
+，
所以
11
2AO AO ==-，
所以3
3
1112442)A V V O AO ⎛⎫== ⎪-=⎝⎭
．
17．【答案】（1）1
(2)n n a -=-；（2）1920526
T -=
【解析】（1）当1n =时，11321S a =+，故11a =，························································1分
当2n ≥时，1
1321321n n n n S a S a --=+⎧⎨=+⎩，两式相减得113()(21)(21)n n n n S a a S --+-+-=，即
1322(2)n n n a a a n -=-≥，进而得12(2)n n a a n -=-≥，·················································3分
所以数列{}n a 是以首项为1，公比为2-的等比数列，·······················································4分
所以 ．····················································································································5分 （2）当1n =时，111M m a ==，11b a ∴=，··································································6分
1(2)n n a -=- ，12n n a -∴=单调递增；当n 为奇数时，12n n a -=，且0n a >，当n 为偶数时，1
2n n a -=-，
且0n a <，因此当n 为大于1的奇数时，{}n a 的前n 项中的最大值为1(2)n n a -=-，最小值为2
1(2)n n a --=-，
因此当n 为偶数时，{}n a 的前n 项中的最大值为2
1(2)
n n a --=-，最小值为1
(2)
n n a -=-，
11,
1,22
n n n n b a a n -=⎧⎪
=+⎨⎪⎩≥，·····························································································7分
因此{}n b 的前20项和201351924620()()T b b b b b b b b =+++++++++
32541918341920121222222a a a a
a a a a a a a a a ++++++⎛⎫⎛⎫=++++++++ ⎪ ⎪⎝⎭⎝⎭
·····················8分
19
192019192011911(2)222222S S S S a a S +++-=+=+=++
····················································9分 191919
1(2)1(2)5212226
----=++=
+．·············································································10分 18．【答案】（1）1
2
；（2
【解析】（1）解法一：锐角ABC △中，M 是BC 的中点，且4AB =，2AC =，如图所示：
BM MC ∴=，sin sin(π)sin AMB AMC AMC ∠=-∠=∠，············································1分
在ABM △中，由正弦定理，有sin sin =∠∠AB BM
AMB BAM
，··············································3分
E 1B 11
在ACM △中，由正弦定理，有
sin sin AC MC
AMC MAC
=∠∠，··············································4分
则
sin sin 1sin sin 2
BM AMB BAM AC AB MC AMC MAC AB AC
∠∠===∠∠．(6分) 解法二：因为M 是BC 的中点，所以ABM ACM S S =△△，(2分)
所以sin s 2in 11
2BA AB AM M M M C AC A A ∠=⨯⨯⨯∠⨯⨯⨯，(4分) 即sin s 114n 222i BAM A M M C AM A ∠=⨯⨯⨯∠⨯⨯⨯，所以
sin 1
sin 2
BAM MAC ∠=∠．···················6分 （2）锐角ABC △
中，由cos MAC ∠=MAC ∠
为锐角，sin MAC ∴∠=，···········7分
由（1
）得1sin sin 2BAM MAC ∠=∠=，·······························································8分
又BAM ∠
为锐角，cos BAM ∴∠=，····································································9分
∴sin sin()sin cos cos sin BAC BAM MAC BAM MAC BAM MAC ∠=∠+∠=∠∠+∠∠········10分
=
=
，··············································································11分 所以ABC △
的面积为11sin 4222ABC S AB AC BAC =⋅⋅∠=⨯⨯=△．···················12分
19．【答案】（1）见解析；（2
．
【解析】（1）证明：因为AB 为半球M 的直径，C 为 AB 上一点，所以AC BC ⊥，··············1分 又因为AC PC ⊥，BC PC C = ，,BC PC ⊂平面PBC ，所以AC ⊥平面PBC ，·············2分 又因为PB ⊂平面PBC ，所以AC PB ⊥，····································································3分 又因为P 为半球面上一点，所以PA PB ⊥，···································································4分 PA AC A = ，,PA AC ⊂平面PAC ，所以PB ⊥平面PAC ，·········································5分 PC ⊂平面PAC ，所以PB PC ⊥．·············································································6分 （2）解法一：因为三角形ABC 为直角三角形，24AB AM ==，2AC =
，所以BC =，
又因为PB =PB ⊥平面PAC
，所以PC =，·····················································7分 又因为三角形PAB
也是直角三角形，所以PA ．·····················································8分
所以11
222PAC S AC PC =
⋅⋅=⨯=△·································································9分
11
22
PAB S PA PB =⋅⋅==△，··································································10分
设点C 到平面PAB 的距离为h ，则有C PAB B PAC V V --=，即11
33
PAB PAC S h S PB ⋅=⋅△△，
所以PAC PAB S PB h S ⋅===△△·····································································11分
设直线PC 与平面PAB 所成的角为θ
，则
sin h PC θ===
．······························12分 解法二：因为三角形ABC 为直角三角形，24AB AM ==，2AC =
，所以BC =，
又因为PB =
PB ⊥平面PAC
，所以PC =，·····················································7分
取BC 中点O ，连接PO ，MO ，因为PB PC =，所以PO BC ⊥，因为//MO AC ，且AC ⊥平面PBC ，所以MO ⊥平面PBC ，因为,PO BC ⊂平面PBC ， 所以MO PO ⊥，MO BC ⊥，
所以OB ，OM ，OP 两两垂直．(8分)
如图，分别以OB ，OM ，OP 为x ，y ，z 轴建立空间直角 坐标系，
则(2,0)A ，(3,0,0)B
，(C
，
P (9分)
设平面PAB 的一个法向量为(,,)n x y z = ，
因为2,0)AB =-
，PB =
，
所
以200n AB y y x z n PB ⎧⎧⋅=-==⎪⎪⇒⎨⎨=⎪⋅==⎪⎩⎩
，取1x =，
得n =
，··········································································································10分
又因为CP =
，所以cos ,n CP n CP n CP
⋅〈〉===⋅
···························11分
设直线PC 与平面PAB 所成的角为θ
，则sin cos ,n CP θ=〈〉=
．·······························12分 20．【答案】（1）0.61q <≤；（2）使用B 型号炮弹，理由见解析． 【解析】（1）因为发射B 型号炮弹3发至少命中一发的对立事件是一发都未命中，且每发炮弹击中与否相互独立，··················································································································1分 所以发射B 型号炮弹3发至少命中一发的概率为31(1)q --，·············································2分 所以31(1)0.936q --≥，所以3(1)0.064q -≤，10.4q -≤，0.6q ≥，·························3分 又因为01q <<，所以0.61q <≤．·············································································4分 （2）选用B 型号炮弹使得目标飞行物坠毁的概率更大，理由如下：·····································5分 记事件M =“发射A 型号炮弹目标飞行物坠毁”，i A =“发射A 型号炮弹命中i 发”(0,1,2,3)i =， 所以112233()()(|)()(|)()(|)P M P A P M A P A P M A P A P M A =⋅+⋅+⋅
122233223
333C (1)0.6C (1)1C 1.8(1)3(1)1p p p p p p p p p p =-+-⨯+=-⨯⨯-++
2233321.8(12)330.20.6 1.8p p p p p p p p p =-++-+=--+，·······································7分 同理记N =“发射B 型号炮弹目标飞行物坠毁”，i B =“发射B 型号炮弹命中i 发”(0,1,2,3)i =， 所以112233()()(|)()(|)()(|)P N P B P N B P B P N B P B P N B =⋅+⋅+⋅
122233223
3330.4C (1)0.8C (1)C 1.2(1) 2.4(1)q q q q q q q q q q =-+-+=-+-+
223331.2(12) 2.4 2.40.2 1.2q q q q q q q q =-++-+=-+···················································9分 因为1p q +=，所以1q p =-，
则323()()0.20.6 1.80.2(1) 1.2(1)P M P N p p p p p -=--++---
322330.20.6 1.80.2(133) 1.2 1.20.4 2.41p p p p p p p p p =--++-+--+=-+-，·············10分
令3()0.4 2.41(00.4)f p p p p =-+-<≤，则2
() 1.2 2.4f p p '=-+，
因为00.4p <≤，所以()0f p '>恒成立，所以()f p 在(0,0.4]上单调递增，
又4(0.4)0.4 2.40.410.02560.9610f =-+⨯-=-+-<，················································11分 则()(0.4)0f p f <≤，故()()0P M P N -<，即()()P M P N <，
所以使用B 型号炮弹使得目标飞行物坠毁的概率更大．·····················································12分
21．【答案】（1）2
213
x y -=，
（2）32 【解析】（1）解法一：由已知得2c =，222a b c +=，22
41
13a b -=，·································2分
解得a =
1b =，·································································································3分
O x y
z
所以双曲线的标准方程为2
213
x y -=．··········································································4分
解法二：2a ==
，a ∴=···················································2分 又因为2c =
，所以1b =
=，···········································································3分
所以双曲线的标准方程为2
213
x y -=．··········································································4分
（2）设切线l 的方程为：(0)x my n m =+≠，
联立22
13
x my n x y =+⎧⎪⎨-=⎪⎩，整理得222
(3)230m y mny n -++-=，·············································5分 由题知222244()3)0(3m n n m ∆=-⨯⨯-=-，化简得223m n +=，··································6分 设11(,)P x y ，22(,)Q x y ，则1F P l ⊥，2F Q l ⊥，
则1111
112x my n y x m
=+⎧⎪
⎨⋅=-⎪+⎩，解得2
121221(2)1n m x m m n y m ⎧-=⎪⎪+⎨+⎪=-⎪+⎩；·································································7分 同理2222
112x my n y x m
=+⎧⎪
⎨⋅=-⎪-⎩，解得2
222221(2)1n m x m m n y m ⎧+=⎪⎪+⎨-⎪=-⎪+⎩
．·····························································8分 点(0,0)到直线x my n =+
的距离==d 所以OPQ △
的面积1122
S d PQ d =⨯⨯=⨯
22421
2211n m mn m m ==⨯=++． （或由向量三角形的面积公式，可得OPQ △的面积
223122122222
(2)(2)(2)(2)221
22(1)2(12)1m n n m m n n m mn m n S x y x y mn m m m
++---+=-==++=+）·······10分 又2
2
3m n +=
，m ∴=
S =，
令24=-t n ，由203n <<，则14t <<，
所以====S ················11分 所以当158=t ，即85t =
时，max 33242
==⨯=S ．
所以OPQ △面积最大值为3
2
．····················································································12分
解法二：（5分至8分同解法一）2222(2),11n m m m P n m -+-++⎛⎫∴ ⎪⎝⎭，2222(2),11n m m n m Q m +⎛⎫
⎪⎝--+⎭
+，······9分。