四川省巴中市普通高中2023级“零诊”考试数学答案_2025年9月_250919四川省巴中市普通高中2023级“零诊”考试（巴中零诊）_四川省巴中市普通高中2023级“零诊”考试数学

巴中市普通高中 2023 级“零诊”数学试题
参考答案
一、单选题：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 A D B B C C D B
二、多选题：本题共 3 小题，每小题 6 分，共18 分。在每小题给出的四个选项中，
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
BC ACD AB
三、填空题：本题共 3 小题，每小题 5 分，共 15分.
题号 12 13 14
答案
1  1 e2
24  1, 
4  e2 4 
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）
3a  a 
解析：(1)在等差数列 {a } 中： S  1 3 3a 9 ，
n 3 2 2
则 a 3 ， a 2a 39·······························································3 分
2 5 2
公差 d  a 5 a 2  93 2，a a (n2)d 3(n2)22n1··················5 分
52 3 n 2
故数列
{a }
的通项公式为：a 2n1··············································6 分
n n
高2023级“零诊”数学参考答案 第 1 页（共 7 页）1 1 1 1 1 
（2）由(1)得： b       ···················8 分
n a a (2n1)(2n1) 22n1 2n1
n n1
则T b b b b
n 1 2 3 n
1 1 1 1 1 1 1 1 1 1 1 1 1 1
= （1 ）+ （  ）+ （  ）++（  ）+（  ）
2 3 2 3 5 2 5 7 2 2n3 2n1 2 2n1 2n1
1 1 1 1 1 1 1 1 1 1
= (1         )
2 3 3 5 5 7 2n3 2n1 2n1 2n1
1 1
= (1 )··································································12 分
2 2n1
1
由于 nN*，则： T  ························································13 分
n 2
16.（15 分）
解析：(1)在直三棱柱ABC  ABC 中，BB平面ABC，BC面ABC,则
1 1 1 1
BBBC
1
又ABC90，即ABBC ，又B
1
BABB
所以BC 平面ABB ··················································································2 分
1
又AB 平面ABB ，所以BC  AB
1 1 1
在直三棱柱ABC  ABC 中，AA  AB，则四边形ABB A 为正方形
1 1 1 1 1 1
所以AB  AB，又BCABB,所以AB 平面ABC ······································4 分
1 1 1 1 1
又AC 平面ABC，所以AC  AB ·····························································6 分
1 1 1 1
高2023级“零诊”数学参考答案 第 2 页（共 7 页）(2)由(1)知BA,BC,BB 两两互相垂直，
1
故以 所在直线分别为 轴建立如图所示的空间直角坐标系,
BA,BC,BB x,y,z
1
由于AC 2AB,设AB AA 1 2，则 BC 2 3 ，
B  0,0,0  ,A  2,0,0  ,C  0,2 3,0  ,B  0,0,2  ,A  2,0,2  ,E  1, 3,2  ·································9 分
1 1
设平面 EBC 的法向量为m   x,y,z ，  B  C    0,2 3,0  ，  B  E    1, 3,2 
 
mBC 2 3y0 
则：   ，取 z1,则y=0,x2 ，得m 2,0,1 ····················11分
mBE x 3y2z0

由(1)知AB 平面ABC ，则平面ABC 的一个法向量为AB 2,0,2 ··················12 分
1 1 1 1
由图可知二面角A BCE的平面角为锐角，记为.
1
 
2212
3 10
则： cos cos m,AB   ··········································· 14 分
1
41 44 10
3 10
即二面角A BCE的余弦值为 ························································15 分
1
10
17.（15 分）
解析：（1）表中数据入下：
语文成绩
数学成绩 合计
不优秀 优秀
优秀 48 36 84
不优秀 24 12 36
合计 72 48 120
··········································································································2 分
高2023级“零诊”数学参考答案 第 3 页（共 7 页）零假设为H ：数学成绩与语文成绩无关联；
0
120(48123624)2 20
根据表中数据，计算得到2   0.95246.635x ．
72488436 21 0.010
··········································································································4 分
依据小概率值0.010的独立性检验，没有充分证据推断出H 不成立，因此
0
可以认为H 成立，即认为数学成绩与语文成绩无关联；
0
··········································································································6 分
（2）由题意得分层随机抽样比为
4:3
，
则语文成绩不优秀、优秀的学生分别抽取4人，3人
CkC3k
X 的取值可能为 0，1，2，3， PX k 3 4 ,k 0,1,2,3·······················8 分
C3
7
则X的分布列为：
X 0 1 2 3
P 4 18 12 1
35 35 35 35
········································································································ 10 分
4 18 12 1 9
数学期望为： EX0 1 2 3  ································12 分
35 35 35 35 7
9 4 9 18 9 12 9 1 24
方差为： DX(0 )2 (1 )2 (2 )2 (3 )2 
7 35 7 35 7 35 7 35 49
········································································································ 15 分
18.（17 分）
解析：(1) f '(x)x3ax2bx2，
f '(1)0 a2
由已知：  ，得： ···························································3 分
f '(2)0 b1
高2023级“零诊”数学参考答案 第 4 页（共 7 页）此时， f '(x)x32x2x2(x1)(x1)(x2)，
令 f '(x)0，得：x1或1<x2； f '(x)0，得：1x1或x2；
f(x)在x1处取得极大值；在x2处取得极小值，符合题意，
故a2,b1····················································································5 分
1 2 1
(2)由(1)知： f(x) x4 x3 x22xc，则 f '(x)(x1)(x1)(x2)
4 3 2
易得 f(x)在[0,1]上单调递增，在[1,2]上单调递减，在[2,3]上单调递增.
····································································································7 分
f(0)c,f(1)13c,f(2)2c,f(3)15c,
12 3 4
f(x) maxf(1), f(3) f(3) 15cM,
max 4
f(x) minf(0), f(2) f(0)cm, ·····················································9 分
min
  19
又因为Mm 15c c ，即： 4c2+15c190 ，
4 4
又 c0，所以：c1·········································································11分
(3)因为g(x)在[0,3]上单调递减，则在[0,3]上g'(x) f '(x)k≤0,g'(x)0仅在端点处
成立.·································································································13 分
即k≥f '(x) x32x2 x2在[0,3]上恒成立·············································14 分
k≥f '(x) max  f '(0), f '(3) max  2,8 8.·············································16 分
max
即k的取值范围为：[8,).·······························································17 分
高2023级“零诊”数学参考答案 第 5 页（共 7 页）19.（17 分）
b 1


a 2

解析（1）由已知： 2c2 5 ·································································2 分
 a2 b2 c2


a2
 x2
解得： b1 ，则双曲线C的方程为：  y2 1····································4 分
 4
c 5
yxt
（2）联立方程 ，消x得： 3x2 8tx4t2 40
x2 4y2 40
64t212(4t24)16(t23)0 ，t 3或t  3 ······································ 6 分
 8t
设 Ax ,y ,Bx ,y ,则：    x 1 x 2  3 ················································7 分
1 1 2 2
 x x 
4t2 4
 1 2 3
则： AB  112 x x  2 x x 2 4x x
1 2 1 2 1 2
16(t2 3) 4 2
 2  ····························································9 分
3 3
解得：t 2满足0，所以t 2······················································10 分
（3）设 Ax ,y ,Bx ,y ,M x ,y 
1 1 2 2 0 0
设在 A 处的切线方程为：yk(xx ) y ，代入x2 4y2 4得
1 1 1 1
(14k2)x2 8k(y kx )x4(y kx )2 40，令0，整理得
1 1 1 1
(x2 4)k2 2x yk y2 10，解关于k的二次方程，其
1 1 1 1
高2023级“零诊”数学参考答案 第 6 页（共 7 页）2x y x y x
4x2y2k4(x2 4)(y2 1)16y2 4x2 160，则k  1 1  1 1  1 ，
1 1 1 1 1 1 2(x2 4) 4y2 4y
1 1 1
··································································································13 分
x x2
所以曲线 C在 A 处的切线方程为： y y  1  xx  ，又 1  y2 1 ，得：
1 1 1
4y 4
1
x x4y y4；同理，曲线C在 B 处的切线方程为：x x4y y4···················· 15 分
1 1 2 2
x x 4y y 4
又 Mx ,y 为两切线的交点，则：  1 0 1 0
0 0 x
2
x
0
4y
2
y
0
4
即A、B都在直线x x4y y4上，又A、B都在直线ykx3k 上，
0 0
4
而 ykx3k 过定点3,0，所以 3x 4 ， x  ········································16 分
0 0
3
4
综上，点M 的轨迹方程为： x ······················································17 分
3
x2 x x
注：另法1：对  y2 1求导得： 2yy'0，即y' ，则在A处的切线方
4 2 4y
x
程k  1
4y
1
x x
另法2：直接写出曲线C在 A 处的切线方程为： 1  y y1，即x x4y y4；
1 1 1
4
x x
曲线C在 B 处的切线方程为： 2  y y1，即x x4y y4请酌情考虑扣分。
2 2 2
4
高2023级“零诊”数学参考答案 第 7 页（共 7 页）