高三数学参考答案_2025年1月_250120湖南益阳市2024-2025学年高三上学期期末考试_数学

益阳市 2024 年下学期普通高中期末质量检测
高三数学（参考答案）
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．
题号 1 2 3 4 5 6 7 8
答案 C C B A A B D B
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．
题号 9 10 11
答案 BCD AC BD
三、填空题：本题共3小题，每小题5分，共15分．
5 1 1
12． 3i 13． 14．[- , ]
4 e4 e4
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．(本小题满分13分)
x2 y2 2 1
解：(Ⅰ)点P( 2,1)代入方程C:  1(a 0,b 0)得  1①
a2 b2 a2 b2
且2b2c②，a2 b2 c2③···································································3分
由①②③可解得：a2 4,b2 2，
x2 y2
所以椭圆C:  1··········································································5分
4 2
2
(Ⅱ)设直线l的方程： y  xm，点A(x ,y )、B(x ,y ).
2 1 1 2 2
x2 y2
直线方程代入椭圆C:  1得2x2 2 2mx2m2 40，
4 2
由 8m2 8(2m2 4)32(4m2)0，得2m2
x x  2m，x x m22·································································8分
1 2 1 2
则弦长 AB  1k2 (x x )2 4x x  3(4m2) ，
1 2 1 2
2
点P( 2,1)到直线l： y  xm的距离
2
2
 21m
2 m 2
d    m2 ······················································· 10分
2 3 3
1( )2
2 2
1 1 2 1
ABP面积S  AB d  3(4m2) m2  m2(4m2)  2········12分
2 2 3 2
当且仅当m2 4m2即m 2 时等号成立，
2
故ABP面积取得最大值 2 时直线l的方程为y  x 2························13分
2
高三数学参考答案 第 1 页 共 5 页
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b2cosA 2cosC
【详解】(Ⅰ)法一：由  可得bsinC 2cosAsinC2sinAcosC ,
sinA sinC
bsinC2sin(AC)2sinB····································································3分
由正弦定理可得：bc2b·········································································5分
所以c2·······························································································7分
b2cosA 2cosC b2cosA 2cosC
法二：由  可得  ·····································2分
sinA sinC a c
b2 c2 a2 a2 b2 c2
b
bc ab
由余弦定理得：  ···········································4分
a c
b2 c2 a2 a2 b2 c2
化简得：bc  ，即bc 2b，所以c2··················7分
b b
(Ⅱ)法一：tanC 2tanB，则sinCcosB2sinBcosC，
所以3sinCcosB2sinBcosC2sinCcosB2sinBC2sinA·····················9分
由正弦定理可得：3ccosB2a，又因为c2，所以3cosBa························ 11分
1 3 3
所以VABC面积为：S  acsinB3cosBsinB sin2B ···························13分
2 2 2
π 3
当sin2B1即B 时取等. 所以VABC面积的最大值为 ·····························15分
4 2
1 2sin Asin B 2sin Asin B
法二：S  absinC  
2 sinC sin(AB)
2sin AsinB 2tan AtanB
  ················································ 9分
sin AcosBcosAsinB tan AtanB
tanBtanC 3tanB
又tan Atan(BC)  ································.12分
tanBtanC1 2tan2B1
3tan2 B
2
2tan2 B1 6tanB 6tanB 3 
则S     ,当B  时取等··········15分
3tanB 2tan2 B2 4tanB 2 4
tanB
2tan2 B1
法三：tanC 2tanB，易知B,C都为锐角
如图，作边BC上的高AD，
则有BD2 AD2  AB2  4,
tanC 2tanB
AD 2AD
  ,即BD2CD·····································································11分
CD BD
1 1 3 3 3 BD2AD2 3
S  BCAD  BDAD BDAD  
ABC
2 2 2 4 4 2 2

当且仅当BD  AD ,即B  时取等·························································· 15分
4
法四：tanC 2tanB，则sinCcosB2sinBcosC，ccosB 2bcosC ,
a2 c2 b2 a2 b2 c2
余弦定理可得  ，即a23b2 3c2 12······················ 10分
2a a
由余弦定理可得：a2 b2c22bccosAb244bcosA,
高三数学参考答案 第 2 页 共 5 页
{#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#}则有12a23b2 4b244bcosA,
b22
化简可得b2bcosA2，即cosA ····················································12分
b
1 b2 2
所以S  bcsinAbsinAb 1( )2  b45b24
2 b
5 3
当b2  时，S  ············································································15分
2 max 2
法五：同法四可得a23b2 3c2 12··························································· 10分
如图过点C作底边AB的高CD，
不妨设AD 1d,BD 1d,CD h,
则有b2 (1d)2 h2,a2 (1d)2 h2,
则a23b2 3(1d)23h2(1d)2h2 12，
1 9
整理可得h2 (d  )2  ,······································································13分
2 4
9 3 1
所以h2  ，即h ，当且仅当d  时取等，
4 2 2
1 3
则有S  ch ·················································································15分
2 2
17．(本小题满分15分)
解：(Ⅰ)设ACBD O ，连接PO，过点A作PO的垂线，垂足即为点A在平面PBD内
的射影H. ·····························································································2分
下面证明:
AB  AD,CD  BC ，点A,C在线段BD的中垂线上，即有AC BD
PA平面ABCD,BD 平面ABCD ，PA BD
又PAAC=A，PA,AC 平面PAC ，BD平面PAC ·························4分
又BD平面PBD，平面PBD 平面PAC
又平面PBD平面PAC  AO，AH  PO，AH 平面PAC
AH  平面PBD ，故点H为点A在平面PBD内的射影·······························6分
(Ⅱ)由(Ⅰ)可知，可建立如图空间直角坐标系O-xyz， AB AD2,，又BAD120，
BCD60，PB PD且PB PD，易知OA1，OB  3，PB  2OB  6,
3
在RtPAB中，PA PB2AB2  2 ，在RtPAO中，PO 3，OH 
3
1 2
则O(0,0,0),A(0,1,0),B( 3,0,0), H(0, , )······································9分
3 3

设平面HAB的法向量为n
x,y,z
，
  2 2
AB ( 3,1,0)，AH (0, , )
3 3
   3x y 0 H
 nAB 0 
则  ， 2 2
nAH 0  y z  0
3 3

不妨取x1,则n(1, 3, 6)

平面DAB的法向量m 0,0,1  ································································12分
高三数学参考答案 第 3 页 共 5 页
{#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#} 
  mn 6
二面角H ABD的平面角为，|cos||cos m,n|    ··········14分
m n 10
10
又(0,)，sin 1cos2
5
10
二面角H ABD的正弦值为 ···························································15分
5
18．(本小题满分17分)
解：(Ⅰ)记事件A为抽到一件合格品，事件B为抽到另一件为不合格品，
C1 ×C1 160
P(AB)= 80 20 = ········································································2分
C2 495
100
C2 -C2 476
P(A)= 100 20 = ········································································4分
C2 495
100
P(AB) 160 40
P(B| A)= = = ······························································5分
P(A) 476 119
1
(Ⅱ)(i)由题：若
X ~B(100, )
，则EX50,DX25·······································7分
2
1
又 PX kCk ( )100 PX 100k,
100 2
所以 P0 X 25 1 P(0 X 25 或 75 X 100) 1 P  X 50 25 ··········· .9分
2 2
由切比雪夫不等式可知， P  X 50 25  25  1
252 25
1
所以 P0 X 25 ·········································································11分
50
(ii)设随机抽取100件产品中合格品的件数为X ，假设厂家关于产品合格率为95%的说法
成立，则X :B100,0.95，所以EX95,DX4.75·····························13分
由切比雪夫不等式知， PX 80P  X 95 15 
950.05
0.021·············15分
152
即在假设下100个元件中合格品为80个的概率不超过0.021，此概率极小，由小概率
原理可知，一般来说在一次试验中是不会发生的，据此我们有理由推断工厂的合格率
不可信.·······························································································17分
19．(本小题满分17分)
1
解：(Ⅰ) f(x)的定义域为(1,)， f(x) ··············································2分
x1
 f(0)1， f(0)0，
 f(x)在原点处的切线方程为： y01(x0)，即 y  x·························4分
(Ⅱ)将x看成变量，n看成常数，
1 1 1 1 x
设g(x) ( x)ln(x1)   ln(x1) ，x≥0，
x1 2n 2n x1 x1
高三数学参考答案 第 4 页 共 5 页
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1 x
所以g(x)
2n

1

2n
·····················································6分
(x1)2 x1 (x1)2
1 1
所以x[0, )时，g(x)＜0，x( ,)时，g(x)＞0，
2n 2n
1 1
所以g(x)在[0, )上单调递减，在( ,)上单调递增·······························8分
2n 2n
1 1
所以g(x)≥g( ) ln( 1)
2n 2n
1 1 1 1 1
又 f( )=ln( 1) ， f( )≤ ( x)ln(x1) ····························10分
2n 2n 2n x1 2n
 1 1 1 x
f( )≤ ( x) ln(x1)

21 x1 21 x1

(Ⅲ)由(2)可知：

1 1 1 x
f( )≤ ( x) ln(x1)
 2n x1 2n x1
对n累加，有
1 1 1 1 1 1 nx
f( ) f( )＜ (   ) nln(x1) ···················12分
21 2n x1 21 22 2n x1
1 1
(1 )
1 2 2n nx
    nln(x1)
x1 1 x1
1
2
1 1 nx 1nx
 (1 )  nln(x1)＜  nln(x1) ································15分
x1 2n x1 x1
1 1 1n(e1) n1
令xe1，则 f( ) f( )＜ n  ···························17分
21 2n e e
高三数学参考答案 第 5 页 共 5 页
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