山东省青岛市一模2024届高三数学答案

2024年高三年级第一次适应性检测
数学参考答案及评分标准
一、单项选择题：本题共8小题，每小题5分，共40分．1--8：A D B A C C B A
二、多项选择题：本题共3小题，每小题6分，共18分．9．AB 10．AC 11．BCD
三、填空题：本题共3个小题，每小题5分，共15分．12．0；
13．
51
2
；14
．1．
四、解答题：本题共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.（13分）解：（1）由题知：各组频率分别为：0.15，0.25，0.3，0.2，0.1······················3分日均阅读时间的平均数为：
300.15500.25700.3900.21100.167⨯+⨯+⨯+⨯+⨯=（分钟）·······················6分（2）由题意，在[60,80),[80,100),[100,120]三组分别抽取3,2,1人·······················7分
ξ的可能取值为：0,1,2·················································································8分
则30423
61
(0)5
C C P C ξ===················································································9分21423
63
(1)5C C P C ξ===···············································································10分12423
61
(2)5
C C P C ξ===··············································································11分所以ξ的分布列为：
ξ
012P
15351
5131
()0121555
E ξ=⨯+⨯+⨯=······································································13分
16．（15分）
解：（1）当1a =时，21
()x x f x x
-+'=，0()1f x '=解得01x =························3分
又因为1(1)2f =-，所以切线方程为：3
02
x y --=···········································5分
（2）()f x 的定义域为(0,)+∞，21
()x ax f x x
-+'=········································6分
当0a ≤时，得()0f x '>恒成立，()f x 在(0,)+∞单调递增·································8分
当0a >时，令22
()1,4g x x ax a =-+∆=-····················································9分（ⅰ）当0∆≤即02a <≤时，
()0f x '≥恒成立，()f x 在(0,)+∞单调递增··········································11分
（ⅱ）当0∆>即2a >时，(22()x x f x
x --'=···············12分
由()0f x '>得，02a x -<<或2a x +>，
由()0f
x '<
得，44
22
a a x +<
<
所以()f x
在(0,),(,)22
a a -+∞单调递增，
在
(,22
a
a -+单调递减···········································14分
综上：当2a ≤时，()f x 在(0,)+∞单调递增；
当2a >时，()f x 在(0,,)22
a a -+∞单调递增；
()f x 在44
(,)22
a a +单调递减·······················15分
17．（15分）
解：(1)取棱1A A
中点D ，连接BD ，因为1AB A B =，所以1BD AA ⊥·················1分因为三棱柱111ABC A B C -，所以11//
AA BB ······················································2分所以1BD BB ⊥，所以BD =
·····································································3分
因为2AB =，所以1AD =，12AA =；
因为2AC =，1A C =，所以2
2
2
11AC AA A C +=，所以AC ⊥1AA ，············4分同理AC ⊥AB ，·························································································5分因为1AA AB A = ,且1,AA AB ⊂平面11A ABB ，所以AC ⊥平面11A ABB ，因为AC ⊂平面ABC ，
所以平面11A ABB ⊥平面ABC ··············6分（2）取AB 中点O ，连接1A O ，取BC 中点P ，连接OP ，则//OP AC ，
由（1）知AC ⊥平面11A ABB ，
所以OP ⊥平面11A ABB ·······················7分因为1
AO ⊂平面11A ABB ，AB ⊂平面11A ABB ，所以OP ⊥1AO ，OP ⊥AB ，
因为11AB A A A B ==，则1A O AB ⊥······························································8分
1
B P
O 1
C C
M
B 1
A A x
y
z
D
以O 为坐标原点，OP ，OB ，1OA 所在的直线为x 轴、y 轴、z 轴，建立如图所示的空间直角坐标系O xyz -，则(0,1,0)A -
，1A
，1B ,(2,1,0)C -，
设点((02)N a a ≤≤，
···········································································9分11(0,2,0)A B =
，1(2,1,A C =-
，1(A N a =
，
设面11A B C 的法向量为(,,)n x y z = ，得11100
n A B n A C ⎧⋅=⎪⎨⋅=⎪⎩
，得20
20y x y =⎧⎪⎨-=⎪⎩，
取x =02y z ==，
，所以2)n =
·············································10分设直线AN 与平面11A B C 所成角为θ，
则||sin |cos ,|||||n AN n AN n AN θ⋅=<>=⋅
=
=
=
································11分
若0a =，则sin
7θ=
，·········································································12分若0a ≠
，则sin 7θ=≤=
，·······························13分
当且仅当4
a a
=，即2a =时，等号成立，·······················································14分
所以直线AN 与平面1A MB 所成角的正弦值的最大值7
·································15分18．（17分）
解:（1）设(,)M x y ，切点为N ，则2
2
2
2
||||||||MN MW OM OW ==+，
所以2
2
2
|2|4x x y +=++··············································································3分化简得24y x =，所以C 的方程为：2
4y x =····················································4分
（2）（ⅰ）因为12//l l ，所以可设,GA GA GB GB λλ''==
，
又因为1()()22
GE GA GB GA GB GF λλ''=+=+= ，
所以,,G E F 三点共线，同理，,,H E F 三点共线，
所以,,G E H 三点共线···················································································6分（ⅱ）设11223344(,),(,),(,),(,)A x y B x y A x y B x y ''，AB 中点为E ，A B ''中点为F ，
将x y m =+代入2
4y x =得：2
440y y m --=，所以12124,4y y y y m +==-，
所以12
22
E y y y +=
=，同理2F y =（,,,G E H F 均在定直线2y =上）················8分因为1//TT l '，所以EAT ∆与EAH ∆面积相等，EBT '∆与EBH ∆面积相等；
所以四边形GTET '面积等于四边形GAHB 面积···············································10分设(,2),(,2)G H G x H x ，
直线13
1113
:()y y AA y y x x x x -'-=--，即213112
231()444
y y y y y x y y --=--，整理得：直线13
13
4:x y y AA y y y +'=+，又因为2G y =，所以13132()4G y y y y x +-=，
同理，直线23
23
4:x y y BA y y y +'=+，2H y =，所以23232()4H y y y y x +-=·············12分
所以||||G H GH x x =-123()(2)||4y y y --=34
123()(
)
2||
4
y y y y y +--=1
234|()()|
8
y y y y --=····························14分所以四边形GAHB 面积2123412()||1
||||
216
y y y y S GH y y -⋅-=⋅-=
=
(1616)161616
m +=
=22(1)14[]2(22)162m n m m n +++≤=+++=············16分
当且仅当2
(1)1m n +=+，即22
226
m m n n m m ⎧+=⎪⎨++=⎪⎩，即13m n =⎧⎨=⎩时取等号，所以GAT ∆面积的最大值为16······································································17分
19．（17分）
解：（1）因为3
3
414233241()m x a b a b a b a b x =+++，
所以414233241m a b a b a b a b =+++···································································4分（2）因为({}{})({})({})n n n n f a b f a f b ⊗=⋅，
所以({})({})({})({}{})({})(({}{}){})n n n n n n n n n f a f b f c f a b f c f a b c ⋅⋅=⊗⋅=⊗⊗，又因为({})({})({})({})[({})({})]
n n n n n n f a f b f c f a f b f c ⋅⋅=⋅⋅({})({}{})n n n f a f b c =⋅⊗({}({}{}))
n n n f a b c =⊗⊗所以(({}{}){})({}({}{}))
n n n n n n f a b f c f a b f c ⊗⊗=⊗⊗所以({}{}){}{}({}{})n n n n n n a b c a b c ⊗⊗=⊗⊗···············································10分（3）对于{},{}n n a b S ∈，因为1
11121212()()n n n n n n a a x a x b b x b x d d x d x ---++++++++=++++
所以1
112111121()()()n n k n k n n n n k n k n n d x
a b x a x b x a x b x a x b ------+--=+++++ ，
所以1211121n n n k n k n n d a b a b a b a b a b -+--=++++++ ，所以11
{}{}{}{
}n
n n n k n k
k a b d a b
+-=⊗==∑····························································13分
2200
100
200
100100
2002012012012012
1
1
101
1
1
(1)1
(1)2
k k k k k k
k k k k k k k k k d a b a b a b a b k k ----+=====++==+
==+∑∑∑∑∑
·········14分所以100
2002
11
21(1)21k k d k k +==
+
-+∑100100
212
11111[]22(1)2k k k k k k k +++===+-⋅+⋅∑∑10211021210122
=-<⨯········································································17分。