山东省烟台市2023-2024学年高二上学期1月期末学业水平诊断数学试题（PDF版含答案）

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2023～2024 学年度第一学期期末学业水平诊断
高二数学参考答案
一、选择题
C A D B D C B C
二、选择题
9. BC 10. BCD 11. ACD 12. ABD
三、填空题
120 y = ±2x 11 a = 2n 113. 14. 15. 16. n +1 ， S
1 1
n = 2 2n
+1
四、解答题
17.解：（1）由2Sn = (n +1)an ，得2Sn 1 = nan 1 (n ≥ 2) ,
两式相减得2an = (n +1)an nan 1 . ································ 2 分
an n a 2 a 3 a n
即 = (n ≥ 2) 2 3 na n 1 ，所以
= , = , , = ，
n 1 a1 1 a2 2 an 1 n 1
以上各式相乘，得an = n(n ≥ 2)， ··································· 4分
经检验a1 =1 a = n(n∈N 也适合上式，故 n ) . ····························· 5 分
2
2 b =| n 9 | {n 9} n Q n 17n（ ）由题意 n ，数列 的前 项和 n = ， ················· 6 分 2
2
n ≤ 9 T = (a + a + a n +17n所以，当 时， n 1 2 n ) = Qn = ， ··············· 7 分 2
2
当 n > 9 时，Tn = (a1 + a2 + a9 ) + a10 + + an = Q 2Q
n 17n
n 9 = + 72 ，9 分 2
n2 +17n

, n ≤ 9
2
综上，Qn = . 2
n 17n
································ 10 分
+ 72, n > 9 2
x2 y2
18.解：（1）设双曲线的方程为 2 2 =1(a > 0,b > 0) ,a b
由已知c 3 b 2= ， = ， ·············································· 2 分
a 2
所以a = 2 ，b =1. ··················································· 4 分
x2
所以双曲线方程为 y2 =1. ·········································· 5 分
2
高二数学答案（第 1 页，共 5 页）
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x2
y2 =1
（2）联立方程 2 ,得 x2 + 4mx + 2m2 + 2 = 0，
y = x +m
x1 + x
2
1 = 4m, x1 x1 = 2m + 2， ····································· 7 分
所以 | AB |= 2 | x1 x2 |
2 (x x )2 4x x = + ································· 9 分 1 2 1 2
= 2 × ( 4m)2 4(2m2 + 2)
令 2 × ( 4m)2 4(2m2 + 2) = 4 2 ，解得m = ± 3 . ······················ 11 分
经检验 > 0符合题意，所以m = ± 3 . ·································· 12 分
19.（1）证明：抛物线的焦点坐标为 F (1,0)， ··································· 1 分
y2 = 4x 2
设直线 l 的方程为 x = my +1，联立 得， y 4my 4 = 0，
x = my +1
设点 P(x1, y )，Q(x . ·························· 3 分1 2 , y2 )，则 y1 + y2 = 4m, y1 y2 = 4
因为M ( 1, y ) y，所以 k = 1 = y ， ············································· 4 分 1 OM 1 1
k y
2
又 2
y2 y 4=
OQ = ， x2 = ， 1x 4 y
，
2 2
k k y 4 4所以 = y 2 = y = 0 ···································· 5OM OQ 1 分 x2 y
2 2
2 y2
所以O,Q, M 三点共线. ····································································· 6 分

（2）因为 PF = 3FQ，所以 y1 = 3y2 . ····················································· 8 分
2 y 2于是 y1 y2 = 3y2 = 4，即 2 = ± . ·············································· 9 分 3
m y1 + y2 3y2 + y2 y 3所以 = = = 2 = ± . ······································· 11 分
4 4 2 3
所以，直线 l 的方程为 3x ± y 3 = 0 . ············································· 12 分
高二数学答案（第 2 页，共 5 页）
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20.解：（1）A 品牌化妆品年销售利润构成首项为3.8、公差为0.5的等差数列.
B 品牌化妆品年销售利润构成首项为4 、公比为1.08的等比数列.
设销售 A、B 品牌化妆品前 n 年总利润分别为 Sn ,T ，则 n
S 10×910 =10×3.8+ ×0.5 = 60.5（万元） ··········································· 2 分2
T 4(1 1.08
10 )
10 = ≈ 57.95（万元），故 S10 > T10， ·································· 4 分 1 1.08
所以，A 品牌化妆品的前 10 年总利润更大. ············································· 5 分
n 1
（2）可得，an = 0.5n + 3.3，bn = 4× (1.08) ， ······································· 6 分
所以cn = an bn = 0.5n + 3.3 4× (1.08)
n 1
，
则 cn+1 = 0.5(n +1) + 3.3 4× (1.08)
n
，cn 1 = 0.5(n 1) + 3.3 4× (1.08)
n 2
令 cn ≥ c
n 1
n+1，得到1.08 ≥1.5625， ·················································· 8 分
令 cn ≥ cn 1，得到1.08
n 2 ≤1.5625 . ·················································· 10 分
由参考数据1.085 ≈1.469,1.086 ≈1.587,1.087 ≈1.714，
知 n 1> 5，n 2 < 6，所以n = 7 .
所以，第7 年时两种化妆品在同一年的利润差额cn = an bn 最大. ··········· 12 分
21.解：（1）由已知得，4Sn+1 = 3Sn 8，
当 n ≥ 2 时，4S ， n = 3Sn 1 8
an+1 3
两式相减得4an+1 = 3an ，即 = (n ≥ 2)a . ······························· 2 分 n 4
又 4S = 3S 8 ，即4(a + a ) = 3a 8，且a1 = 2
3
，所以a2 = ， 2 1 1 2 1 2
a2 3 a 3
所以 =
n+1
，亦满足 = (n =1) . ······························ 3 分a1 4 a 4
n
3 n 1
所以，数列{an}为等比数列，an = ( 2)( ) . ············ 4 分4
所以a 3 n 4 1n+1 = ( 2)( )4 ，
bn+1 = b +3 n (3
，
)n
4
即 (
3)n bn+1 (
3)n 1b =1
4 4 n ，
{(3)n 1 3 0所以，数列 b4 n
}是以 ( ) b1 = 1为首项、14 为公差的等差数列. ····· 6 分
高二数学答案（第 3 页，共 5 页）
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3 n 1 4
（2）由（1）可得， ( ) bn =1+ (n 1)×1= n，即bn = n( )
n 1
4 3 . ··········· 7 分
T 4 4所以 n =1× ( )
0 + 2× ( )1 + 3× (4)2 + + n (4× )n 1
3 3 3 3
4 Tn =1 (
4)1 2 4 4 4 4× + × ( )2 + 3× ( )3 + + (n 1)× ( )n 1 + n× ( )n
3 3 3 3 3 3 ，
两式相减得，
1
T = (4)0n + (
4)1 (4)2 4 4 4 4 4+ + ( )3 + n 1 n n n + ( ) n( ) = 3[( ) 1] n( )
3 3 3 3 3 3 3 3 3
所以T
4 n
n = (3n 9)( ) + 93 . ·············································· 9 分
由已知条件 (6nλ 54)(
4)n (n + 3)(Tn 9) ≤ 0，可得：3
(6nλ 4 4 54)( )n (n + 3)(3n 9)( )n ≤ 0 ，即2nλ ≤ n
2 + 9 . ···· 10 分
3 3
n2 + 9 n 9
即λ ≤ = +2n 2 2n ，
n 9 n 9
因为 + ≥ 2 × = 32 2n 2 2n ，当且仅当
n = 3时等号成立，
所以λ ≤ 3 . ·········································· 12 分
22.解：(1)设Q(x, y)，因为点 P 为 PD的中点，所以 P(x, 2y) . ·············· 1分
因为点 P 在圆 x2 + y2 = 4上，所以 x2 + (2y)2 = 4， ······························ 2 分
x2
化简得 +y2 =1,
4
x2
所以，求点Q的轨迹Γ的方程为 +y2 =1 (y ≠ 0) .
4 ·······················
3分
（2）（i）设过点T (0, 2) 的直线方程为 y = kx + 2，
x2
+y2 =1
联立 4 ，得： (1+ 4k 2 )x2 +16kx +12 = 0，
y = kx + 2
设点M (x1, y1), N (x2, y2 ) ,则 x x
16k
1 + 2 = 2 , x1 x
12
= . ············ 4 分
1+4k 2 1+4k 2
高二数学答案（第 4 页，共 5 页）
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BM : y 1 y1 +1 y2 1因为直线 + = x ，直线 AN : y 1= xx x ， 1 2
y 1 (y 1)x kx x + x
所以 = 2 1 = 1 2 1y +1 (y +1)x kx x ，+ 3x ··············································· 5 分 1 2 1 2 2
x x 3注意到 1 2 = (x1 + x2 )， 4k ····················································· 6 分
kx x + x k(
3
(x1 + x
1
2 )) + x1 x
3
x
1 2 1 = 4k
1 2 1
于是 4 43 = 9 3 = kx1x2 + 3x2 k( (x1 + x2 )) + 3x2 x2 x
3
4k 4 4 1
y 1 1 1 1
所以 = ，解得 y = ，所以点G 在直线 y = 上. y +1 3 2 2 ················· 8 分
BM : y y1 +1 x 1 1 x 3x1（ii）因为点G 在 = 上，令 y =x ，可得2 G
=
1 2y 2
.
1 +
y
H BN : y = 2
+1 1 3x
同理，点 直线 x 1上，且 yH = ， xH =
2
x . 2 2 2y + 2
······· 9 分
2
1 1
因为 S GAB = | AB | | xG |=| xG |， S 2 HAB
= | AB | | x
2 H
|=| xH |， ·········· 10 分
S 9 x1x2 9 x1 x2所以
GAB S =| x | | x |= | | = | | HAB G H 4 (y 1)(y 1) 2 , 1 + 2 + 4 k x1x2 + 3k(x1 + x2 ) + 9
16k 12
将 x1 + x2 = 2 , x1 x2 = 2 ，代入得： 1+4k 1+4k
9 12 9 12
S
GAB S HAB = | 2 2 |= × = 3. 4 k (12) + 3k( 16k) + 9(4k +1) 4 9
所以 S GAB S HAB 的面积之积为定值3 . ······································· 12 分
高二数学答案（第 5 页，共 5 页）
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