2023～2024 学年度第二学期期末学业水平诊断
高二数学参考答案及评分标准
一、选择题
C C A D B D C A
二、选择题
9. ABD 10.BCD 11.AC
三、填空题
1 4 2 3
12.−80 13.(−∞, ] 14.( )n−1L， L2
e 3 45
四、解答题
15.解：（1）根据已知条件，可得：
获奖 没有获奖 合计
选修阅读课程 8 12 20
不选阅读课程 2 28 30
合计 10 40 50
······················································ 3分
零假设为H ：创新作文比赛获奖与选修阅读课程无关联，
0
根据列联表中数据计算得到，
50×(8×28−2×12)2 25
χ2= = ≈8.333>7.879. ······························· 6分
20×30×10×40 3
根据小概率值α=0.005的独立性检验，推断H 不成立，即认为创新作文比赛获奖与
0
选修阅读课程有关联，此推断犯错误的概率不大于0.005. ···························· 7分
（2）由题意可知X 的可能取值为1,2,3,则 ··································· 8分
C1C2 1 C2C1 7
P(X =1)= 8 2 = ，P(X =2)= 8 2 = ，
C3 15 C3 15
10 10
C3 7
P(X =3)= 8 = ， ········································ 11分
C3 15
10
所以,随机变量X 的分布列为:
X 1 2 3
1 7 7
P
1 5 1 5 1 5
1 7 7 12
所以E(X)=1× +2× +3× = . ·························· 13分
15 15 15 5
16.解：（1）当a =−2时， f(x)=(x2 −2x+1)ex，所以 f′(x)=(x2 −1)ex. ········· 1分
设切点为(x ,y )，则y =(x 2 −2x +1)ex 0 ，k =(x2 −1)ex 0，
0 0 0 0 0 0
高二数学答案（第 1 页，共 4 页）
{#{QQABRYYQoggoAIBAAAhCAQXKCkOQkACACQgOBEAIMAAAARNABAA=}#}所以，切线方程为y−(x2 −2x +1)ex 0 =(x2 −1)ex 0(x−x ). ························ 3分
0 0 0 0
将(1,0)代入得(x −1)2x =0，解得x =0或x =1. ····························· 5分
0 0 0 0
故过(1,0)的切线方程为y =0或x+ y−1=0. ················································ 7分
（2） f ′(x) =(2x+a)ex +(x2 +ax+1)ex =(x+a+1)(x+1)ex. ····················· 8分
当a = 0时， f ′(x) =(x+1)2ex，恒有 f ′(x)≥ 0，函数 f(x)单调递增. ········· 10分
当a > 0时，−a−1< −1，当x∈(−∞,−a−1)，或x∈(−1,+∞)时， f ′(x) > 0，函
数 f(x)单调递增，当x∈(−a−1,−1)时， f ′(x)< 0，函数 f(x)单调递减. ···· 12分
当a <0时，−a−1> −1，当x∈(−∞,−1)，或x∈(−a−1,+∞)时， f ′(x) > 0，函数
f(x)单调递增，当x∈(−1,−a−1)时， f ′(x)< 0，函数 f(x)单调递减. ······· 14分
综上，当a = 0时，f(x)在R上单调递增，当a > 0时，f(x)在(−∞,−a−1)，(−1,+∞)
上单调递增，在(−a−1,−1)上单调递减，当a <0时，f(x)在(−∞,−1)，(−a−1,+∞)
上单调递增，在(−1,−a−1)上单调递减. ······························ 15分
17.解：（1）由题意可知，b −b =a ，即b −1=−1，故b =0. ························ 1分
2 1 2 2 2
由b −b =a ，可得a =1. ······················································ 2分
3 2 3 3
所以数列{a }的公差d =2，所以a =−1+2(n−2)=2n−5. ······················ 3分
n n
由b −b =a ，b −b =a ，，b −b =a ，
n n−1 n n−1 n−2 n−1 2 1 2
(n−1)(−1+2n−5)
叠加可得 b −b =a +a ++a = ，
n 1 2 3 n 2
整理可得 b =n2 −4n+4(n≥2)；当n=1时，满足上式，
n
所以b =n2 −4n+4 ················································································ 5分
n
(n−2)2 +5
（2）不妨设a =b (m,n∈N∗)，即2m−5=(n−2)2，可得m= ， ········ 6分
m n 2
9
当n=2k时，m=2k2 −4k+ ，不合题意，
2
当n=2k−1时，m=2k2 −6k+7=2k(k−3)+7∈N∗， ································ 7分
所以b 在数列{a }中均存在公共项，
2k−1 n
又因为b =b <b <b <，所以c = b =(2n−1)2. ································· 9分
1 3 5 7 n 2n+1
5
（3）当n=1时，T =1< ，结论成立， ············································ 10分
1 4
1 1 1 1 1 1
当n≥2时， = < = ( − )， ····················· 12分
c (2n−1)2 (2n−2)×2n 4 n−1 n
n
高二数学答案（第 2 页，共 4 页）
{#{QQABRYYQoggoAIBAAAhCAQXKCkOQkACACQgOBEAIMAAAARNABAA=}#}1 1 1 1 1 1
所以T <1+ (1− + − ++ − )
n 4 3 3 5 n−1 n
1 1 5 1 5
=1+ (1− ) = − < ，
4 n 4 4n 4
5
综上，T < . ·················································· 15分
n 4
18.解：（1）记事件A=“第2次取出的小球为黑球”；事件B=“第1次取出的小球为白球”，
3 3 3 3 11
则P(A)= × + × = ， ············································ 2分
6 6 6 5 20
3 3 3 P(AB) 6
P(AB)= × = ，所以P(B| A)= = ； ·································· 4分
6 5 10 P(A) 11
（2）由题意，X 的可能取值为0,1,2,3，则 ·············································· 5分
3 3 3 1
P(X =0)= × × = ，
6 6 6 8
3 3 3 3 3 3 3 3 3 91
P(X =1)= × × + × × + × × = ,
6 5 5 6 6 5 6 6 6 200
3 2 3 3 3 2 3 3 2 37
P(X =2)= × × + × × + × × = ,
6 5 4 6 5 5 6 6 5 100
3 2 1 1
P(X =3)= × × = ，
6 5 4 20
所以，随机变量X 的分布列为：
X 0 1 2 3
1 91 37 1
P
8 200 100 20
··········································· 10分
（3）由题意可知，前n−1次取了一个白球，第n次取了第二个白球，则：
3 3 3 3 3 3 3 2
P =[ ×( )n−2 + × ×( )n−3++( )n−2× ]× ··························· 12分
n 6 5 6 6 5 6 6 5
3 2 3 3 3 3 3 3
= × ×[( )n−2 + ×( )n−3+( )n−3× +( )n−2]
6 5 5 6 5 6 5 6
1 3 5 5 5
= ×( )n−2[1+ +( )2 +( )n−2]
5 5 6 6 6
5
1−( )n−1
= 1 ×( 3 )n−2 6 =2×[( 3 )n−1−( 1 )n−1](n≥2,n∈N*). ···················· 16分
5 5 5 5 2
1−
6
3 1
所以P =2×[( )n−1−( )n−1] (n≥2,n∈N*). ·································· 17分
n 5 2
19.解：（1）函数 f(x)定义域为(0,+∞)，
1 x+1
f′(x)=alnx+a(x+1)⋅ +1=a(lnx+ )+1， ···································· 1分
x x
高二数学答案（第 3 页，共 4 页）
{#{QQABRYYQoggoAIBAAAhCAQXKCkOQkACACQgOBEAIMAAAARNABAA=}#}x+1 1
显然a≠0，令 f ′(x) = 0，可得lnx+ = − ，
x a
x+1 1
令t(x) =lnx+ ，由 f(x)有两个不同极值点得t(x)=− 有两个不同的正根. ·· 3分
x a
1 1 x−1
因为t′(x) = − = . 当x∈(0,1)时，t′(x)<0，t(x)单减，x∈(1,+∞)时，
x x2 x2
t′(x) >0，t(x)单增. ················································································ 5分
所以t(x)的极小值即最小值t(1) = 2，又当x→0时，t(x)→+∞，且x→+∞时，
1 1
t(x)→+∞，所以− >2，即− <a<0. ··········································· 6分
a 2
（2）设x ,x 为函数 f(x)的极值点，由（1），不妨设x <1< x ，下证x +x >2.
1 2 1 2 1 2
要证：x >2−x >1，只要证t(x )>t(2−x ).
2 1 2 1
令g(x)=t(x)−t(2−x)(0< x<1). ···························· 8分
x−1 1−x −4(x−1)2
因为g′(x)=t′(x)+t′(2−x)= + = <0. ··········· 10分
x2 (2−x)2 x2(2−x)2
所以g(x)在(0,1)上单调递减，所以g(x)> g(1)=0，
故t(x )>t(2−x )，即x + x > 2. ························· 11分
2 1 1 2
1
由（1）可知，在(0,x )上， f′(x)=a(t(x)+ )<0， f(x)单调递减，在(x ,x )
1 1 2
a
上， f ′(x) > 0， f(x)单调递增，在(x ,+∞)上， f ′(x)< 0， f(x)单调递减，
2
又因为 f(1) =0，所以 f(x )< f(1) =0，
1
1 1 1
因为− < a <0，所以 < −2，所以ea <e−2 <1，
2 a
1 1 1 1 1
而 f(ea) = a(ea +1)lnea +ea −1= 2ea >0，
1
所以 f(x)在(ea,x )上存在点x ，使得 f(x ) =0， ····························· 13分
1 3 3
1 − 1
同理 f(x ) > f(1) = 0，又− > 2，e a >e2 >1，
2
a
1 1 1 1
− − − −
f(e a) = a(e a +1)lne a +e a −1= −2<0，
1
−
所以 f(x)在(x ,e a)上存在点x ，使得 f(x ) =0， ····························· 14分
2 4 4
故 f(x)存在3个零点x ,1,x ，
3 4
1 1 1 1 1 1
注意到 f( ) = a( +1)ln + −1= − (a(x+1)lnx+ x−1) = − f(x)， · 15分
x x x x x x
1 1
所以x = ，所以x + x = x + > 2. ··································· 16分
3 x 3 4 3 x
4 3
所以x + x + x + x +1>5，即m+n>5. ···································· 17分
1 2 3 4
高二数学答案（第 4 页，共 4 页）
{#{QQABRYYQoggoAIBAAAhCAQXKCkOQkACACQgOBEAIMAAAARNABAA=}#}
