四川省绵阳市2026年高考适应性考试（绵阳三诊）数学+答案

2026年高考适应性考试

数学参考答案和评分标准

一、选择题：本题共8小题，每小题5分，共40分．

1．A2．C3．B4．D5．A6．B7．C8．D

二、选择题：本大题共3小题，每小题6分，共18分．全部选对的得6分，选对但不

全的得部分分，有选错的得0分．

9．ABD10．ACD11．ACD

三、填空题：本题共3个小题，每小题5分，共15分．

1

12．；13．−2；14．28，399+1

4

四、解答题：本题共5小题，第15题13分，第16、17小题15分，第18、19小题17

分，共77分．解答应写出文字说明、证明过程或演算步骤．

π

由正弦定理得：sinCsinB=sinBcos(C_)，又sinB≠0，·····························1分

6

∴sinC=cos············································································2分

∴sinC=cosCcossinCsin·····························································4分

∴tanC=·3，又C∈(0，π)，则C···················································6分

（2）方法一：在△BCD中，∵BD=CD，可得LBCD=B，LADC=2B，

又∵C·······················································································7分

∴LACDB，sinA=sin=sin····································8分

又在△ACD中，由正弦定理得：且CD=2AD，

·····································································9分

∴tanB则B由（1）知：C则A···························11分

······································································13分

数学答案第1页（共7页）

方法二：设AD=x，则BD=CD=2x，························································7分

π

在△ABC中，由余弦定理得：9x2=a2+b2_2abcos,

3

∴9x2=a2+b2_ab①,······································································8分

又············································································9分

平方得：4xab························································10分

ab②,································································11分

由①消去x2得：a2=2ab，又a≠0，所以·····································13分

16．解：（1）∵l过F1(−c，0)时，△ABF2的周长为8，则4a=8，a=2，············2分

设LAMO=θ,则cos2θ=1_2sin···············································4分

∴sin····················································································5分

又sin则c=2，·······················································6分

又a2=b2+c2，可得：a2=4，b2=2，··················································7分

∴E的方程为：································································8分

（2）已知直线l的方程为：y=kx+2，设A(x1，y1)，B(x2，y2)，················9分

联立，消y整理得：(2k2+1)x2+8kx+4=0，·······················10分

则：分

x1+xx1.x··················································11

y1y2(kx2+2)x1+(kx1+2)x1

∴k1+k2=+=··········································12分

x1x2x1x2

2kxx+2(x+x2)

=121································································13分

x1x2

=2k_4k=_2k，····································································14分

∴k1，_k，k2成等差数列．······························································15分

数学答案第2页（共7页）

．解：方法一：（）设，，，分

171AA1=aAB=bAC=c·································1

21

∵=1_1=.(b+c)_(_a)

32

111

=(b+c)_(b+c)+a=a_(b+c)，...................................3分

326

112

∴EF.BC=[a_(b+c)].(c_b)=a.c_a.b_(c_b2)

66

=1.2.cos60_1.2.cos60_0=0，......................5分

又..b=a.b

∴EF丄BC，EF丄AB，····································································7分

又∵ABBC=B，

∴EF⊥平面ABC；··············································································8分

（2）∵EF⊥平面ABC，以E为原点，建立如图所示的空间直角坐标系Exyz，

∴F(0，0，，A

C(_1，0，0)，A(0，3，0)，

设C1(x0，y0，z0)，由

可得

∴C

易知平面FA1E的一个法向量EB=(_1，0，0)，········································11分

设平面A1EC1的法向量为n=(x,y,z)，由

得可得一个法向量n=(_3，1，_·2)，··13分

∵cos<，n················································14分

∴平面FA1E与平面A1EC1的夹角的余弦值为··································15分

方法二：（1）连接A1B，A1C，易知△AA1B≌△AA1C，·······························1分

O

A1A=1，AB=AC=2，∠BAA1=∠CAA1=60，由余弦定理，

O

∴∠AA1B=∠AA1C=90，且A1B=A1C=3，··············································2分

由E为BC中点，则BC⊥A1E，

延长A1F交B1C1于点G，则A1G⊥B1C1，则A1G⊥BC，A1E∩A1G=A1，

数学答案第3页（共7页）

∴BC⊥平面A1GE，EF≤平面A1GE，

∴BC⊥EF，EF⊥B1C1，·······································································4分

在Rt△A1BE中，可得A1E=·2，···························································5分

222

在△A1EG中，EG=1，A1G，则A1G=EG+A1E，

∴A1E⊥EG，······················································································6分

又F为A1G上靠近点G的一个三等分点，FG，A1F

222

可得EG−GF=A1E−A1F∴EF⊥A1G，············································7分

又A1G∩B1C1=G，则EF⊥平面A1B1C1，

∴EF⊥平面ABC；··············································································8分

（2）由（1）知GE=AA1=1，A1E=2，A1G=3，

222

∴GE+A1E=A1G，则GE⊥A1E，··························································10分

又由（1）知BC⊥A1E，BC∩GE=E，BC≤平面GEC1，GE≤平面GEC1，

∴A1E⊥平面GEC1，

又GE≤平面GEC1，··········································································12分

∴C1E⊥A1E，····················································································13分

∴∠GEC1为平面A1EF与平面A1GE的夹角，···········································14分

在Rt△EGC1中，cos∠GEC········································15分

18．解：（1）若n=4，k=2时，X=0，1，2，···················································1分

P······································································2分

P································································3分

P·······································································4分

故X的数学分布列为：

X012

121

P

636

·······························5分

∴n(n_1)(n_2)=40(n_3)，································································6分

∴n2(n_6)+(n_6)(3n_20)=0，···························································7分

∴(n_6)(n2+3n_20)=0，···································································8分

又n≥3，n∈N*，且方程n2+3n_20=0无正整数根，

∴n=6；·····························································································9分

数学答案第4页（共7页）

第i个球同时在集合M，N中

（3）Ii，则XIi，

第i个球不同时在集合M，N中

由于两次抽取相互独立，且每个球被抽到的概率均为k，

n

因此E=E=n··········································11分

可得：E，其中E

2

又E(IiIj)=P(i，j∈M).P(i，j∈N)=[P(i，j∈M)]，

∵P(i，····································································12分

因此E，····································································13分

∵2

IiIj共有Cn项，

代入得：E·····························14分

22

2k(1)

∴D(X)=E(X2)_(E(X))=+=，·················15分

n(_1)_

①当n为偶数时，k最大，D····························16分

②当n为奇数时，k或kD(X)最大，D··17分

19．解：∵f-flnx3+a+x-2ln

由于a≥-，xÎ(-1，0)，则a·······························1分

令glnx2+x+1-2ln2，

要证xÎ(-1，0)，f(x)<f(1)，只需证：g(x)<0，··································2分

glnx2-3x+1，易知g¢(0)=0，

g=4lngⅱ(0)=0，（其中gⅱ(x)为函数g¢(x)的导函数）

数学答案第5页（共7页）

g可得gⅱ(x)>0，（其中gⅱ(x)为函数gⅱ(x)的导

函数）

∴gⅱ(x)在(-1，0)上单调递增，gⅱ(x)<gⅱ(0)=0，

∴g¢(x)在(-1，0)上单调递减，gⅱ(x)>g(0)=0，

∴g(x)在(-1，0)上单调递增，g(x)<g(0)=1-2ln2<0，··························3分

3

∴当a≥-时，xÎ(-1，0)，f(x)<f(1)；·············································4分

2

（2）∵flnx2+2ax，且f¢(0)=0，

∴f=4lnx+2a+2，fa，···························5分

（i）∵0为f(x)的极小值点，由于f¢(0)=f(0)=0，

3

∴必有fⅱ(0)=2a+3>0，即a>-，··················································6分

2

由于f

令f=0，则x····························································7分

ⅱ¢

∴存在-1<m<0<x0<n，使得在(-1，m)与(n，+∞)上满足f(x)<0，f(x)单

调递减；在(m，n)上fⅱ(x)>0，f¢(x)单调递增．··········································8分

∴存在-1<s<0<t，使得在(-1，s)与(0，t)上有f¢(x)>0，f(x)单调递增；在

(s，0)与(t，+∞)上有f¢(x)<0，f(x)单调递减．···········································9分

∴f(x)的极大值点为：x1=s，x2=t，

由于f¢(1)=5ln2+0.5+2a>5ln2+0.5-3>0，则x2>1，

f(x)在(1，x2)单调递增，则f(x2)>f(1)．·············································10分

3

由于_1<x1<0，a>_，

2

由（1）得：f(x1)<f(1)，

∴f(x1)<f(1)<f(x2)，则f(x1)<f(x2)；...........................................................11分

（ii）∵x1=0为f(x)的一个极大值点，

f¢(0)=0