山东省烟台德州东营2024年高考诊断性测试 数学答案

2025-05-16 0 0 220.37KB 4 页 10玖币
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数学参考答案(第 1 页,共 4 页)
2024 年高考诊断性测试
数学参考答案及评分标准
一、选择题
A C B C B D A A
二、选择题
9.ABD 10.BCD 11.BC
三、填空题
12.
4
13.
10
,
5
10 2
+
14.
95
[ , ]
42
四、解答题
15.解:1
x
axxf 2
12)(' +=
, ···································
2
直线
2 1 0xy+ + =
的斜率
2
1
=k
,
由题意知
···································
4
2114 =+a
,所以
2
1
=a
. ····································
5
2
)(xf
的定义域为
)0( +
. ···································
6
因为
( ) 0fx
,所以
xxxb ln2
2
12+
.
),0(,ln2
2
1
)( 2++= xxxxxg
,则
max
()b g x
. ························
8
x
xx
x
xx
x
xxg )2)(1(22
1)(' 2++
=
+
=+=
···················
9
)1,0(x
时,
0)(' xg
,所以
)(xg
)1,0(
单调递增,
),1( +x
时,
0)(' xg
,所以
)(xg
),1( +
单调递减, ··············· 11
所以
max 3
( ) (1) 2
g x g= = −
.
所以
2
3
b
. ······························· 13
16.解:1)因为
AB AC
33AB AC==
所以
60ACB=
13
2
OA BC==
. ············································ 1
因为
3AB =
2AD DB=
,所以
1DB =
.
DBO
中,
30DBO=
1DB =
3OB =
由余弦定理
2 2 2
1 ( 3) 2 1 3 cos30 1OD
= +   =
,所以
1OD =
. ········· 3
ADO
中,
1OD =
2AD =
3AO =
,由勾股定理
AO OD
. ····· 4
因为
1
AO
平面
ABC
OD
平面
ABC
所以
1
AO OD
. ····················································· 5
{#{QQABSYKEogCoAAAAAQgCEwEoCkGQkAGCAKoOAAAIMAAAyQFABCA=}#}
摘要:

数学参考答案(第1页,共4页)2024年高考诊断性测试数学参考答案及评分标准一、选择题ACBCBDAA二、选择题9.ABD10.BCD11.BC三、填空题12.13.,14.四、解答题15.解:(1),···································2分直线的斜率,由题意知,···································4分即,所以.····································5分(2)的定义域为.···································6分因为,所以.设,则.···············...

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