烟台期末-数学答案_2.2025数学总复习_2023年新高考资料_3数学高考模拟题_新高考_山东烟台高三2022-2023学年度第一学期期末学业水平诊断数学

2022～2023 学年度第一学期期末学业水平诊断
高三数学参考答案及评分标准
一、选择题
D B B C A C D A
二、选择题
9.BC 10. ACD 11. ACD 12. ABD
三、填空题
3 1
13. 1 14. 15. 67 16.
2 2
四、解答题
17.解：（1）由正弦定理可得sinAcosC+sinAsinC =sinB， ······················· 1分
因为A+B+C =π，所以sinAcosC+sinAsinC =sin(A+C)，
即sin AcosC+sin AsinC =sinAcosC+cosAsinC， ···························· 2分
整理得：sin AsinC =cosAsinC，
因为0<C <π，所以sinC ≠0，所以tanA=1，
π
因为0< A<π，所以A= . ································································ 4分
4
（2）在∆ABD中，由余弦定理得：BD2 = AB2 + AD2 −2AB⋅ADcosA， ······ 5分
即9= AB2 + AD2 − 2AB⋅AD≥(2− 2)AB⋅AD， ································ 6分
9(2+ 2)
整理得
AB⋅AD≤
，当且仅当AB= AD时，等号成立.
2
1 π 2 9( 2+1)
所以S = AB⋅ADsin = AB⋅AD≤ ， ························ 8分
△ABD
2 4 4 4
  3 27( 2+1)
因为AD=2DC，所以S = S ≤ ，
△ABC
2
△ABD
8
27( 2+1)
所以△ABC面积的最大值为 . ··············································· 10分
8
18.解：（1）因为a a =2S (n∈N*)，所以a a =2S (n≥2)，
n n+1 n n−1 n n−1
两式相减得a (a −a )=2a (n≥2). ····································································· 1分
n n+1 n−1 n
高三数学答案（第 1 页，共 6 页）又因为a ≠0，所以a −a =2(n≥2)， ······························································ 2分
n n+1 n−1
所以数列
{a }和{a }都是以2为公差的等差数列.
2n−1 2n
因为a =1，所以在a a =2S 中，令n=1，得a =2，
1 n n+1 n 2
所以a =1+2(n−1)=2n−1, a =2+(n−1)×2=2n, ············································· 3分
2n−1 2n
所以a =n， ··················································································································· 4分
n
对于数列{b }，因为b =b ⋅b =2b ，且b ≠0，所以 b n+1 =2(n∈N*)， ··········· 6分
n n+1 1 n n n b
n
所以数列{b }是以2为首项，2为公比的等比数列，所以b = 2n. ························ 7分
n n
（2）因为T =1×2+2×22 +3×23 +...+n×2n
n
所以2T =1×22 +2×23+3×24 ++( n−1 )×2n +n×2n+1 ··································· 8分
n
两式相减得，−T =2+22 ++2n −n×2n+1 ··························································· 9分
n
2−2n+1
= −n×2n+1
1−2
=−2−(n−1)×2n+1 ························································································· 11分
所以T =(n−1)×2n+1+2. ····························································································· 12分
n
19. 解：（1）证明：取BC中点O，连接OA,OD，
因为∆ABC是以BC为斜边的等腰直角三角形，所以OA⊥ BC. ························· 1分
因为∆BCD是等边三角形，所以OD⊥ BC. ··························································· 2分
OAOD=O，OA⊂平面AOD，OD⊂平面AOD， ······································ 3分
所以BC ⊥平面AOD. ································································································· 4分
因为AD⊂平面AOD，故BC ⊥ AD. ···································································· 5分
z
（2）在∆AOD中，AO =1，OD= 3，AD= 7，由余弦定理可得， D
3
cos∠AOD=− ，故∠AOD=150. ·············· 6分
2
  C O B y
如图，以OA,OB及过O点垂直于平面ABC的方向为x,y,z轴
A
的正方向建立空间直角坐标系O−xyz， ·············· 7分
x
  
3 3 3 3
可得D(− ,0, )，所以BD=(− ,−1, )，CB =(0,2,0)，AB=(−1,1,0)，
2 2 2 2
高三数学答案（第 2 页，共 6 页）设n=(x ,y ,z )为平面ABD的一个法向量，则
1 1 1
 −x + y =0
 1 1
 3 3 ，令x= 3，可得n=( 3, 3,5)， ······················ 9分
− x − y + z =0
 2 1 1 2 1
设m =(x ,y ,z )为平面BCD的一个法向量，则
2 2 2
 2y =0
 2
 3 3 ，令x 2 = 3，可得m =( 3,0,3)， ···················· 11分
− x − y + z =0
 2 2 2 2 2
3+0+15 3 93
所以cos<n,m >= = ，
31× 12 31
3 93
故平面ABD与平面BCD夹角的余弦值为 . ······································ 12分
31
2
20.解：（1）设该容器的体积为V ，则V =πr2l+ πr3，
3
160 160 2
又V = π，所以l = − r， ·········································································· 2分
3 3r2 3
因为l ≥6r ，所以0<r ≤2. ························································································ 4分
9 160 2 9
所以建造费用y =2πrl× +3πr2m=2πr( − r)× +3πr2m，
4 3r2 3 4
240π
因此y =3π(m−1)r2 + ,0<r ≤2. ··································································· 5分
r
240π 6π(m−1) 40
（2）由（1）得y′=6π(m−1)r− = (r3 − ),0<r ≤2. ·········· 6分
r2 r2 m−1
9 40 40
由于m> ,所以m−1>0,令r3 − =0，得r = 3 . ··························· 7分
4 m−1 m−1
高三数学答案（第 3 页，共 6 页）40 40
若 3 <2 ，即m>6，当r∈(0,3 ) 时， y′<0 ， y(r) 为减函数，当
m−1 m−1
40 40
r∈(3 ,2)时， y′>0， y(r)为增函数，此时r = 3 为函数 y(r)的极小值点，
m−1 m−1
也是最小值点. ······················································································································ 9分
40 9
若3 ≥2，即 <m≤6， 当r∈(0,2]时，y′<0，y(r)为减函数，此时r =2
m−1 4
是y(r)的最小值点. ··········································································································· 11分
9
综上所述，当 <m≤6时，建造费用最小时r =2；当m>6时，建造费用最小时
4
40
r = 3 . ························································································································ 12分
m−1
21.解：（1）设A(−a,0)，B(a,0)，P(x ,y )，
1 1
y −0 y −0 y2 1
则k k = 1 × 1 = 1 = ， ·············································· 1分
AP BP x +a x −a x2 −a2 4
1 1 1
x2 y2
又因为点P(x ,y )在双曲线上，所以 1 − 1 =1. ····································· 2分
1 1 a2 b2
1 a2 b2
于是y2 = x2 − = x2 −b2，对任意x ≠0恒成立，
1 4 1 4 a2 1 1
b2 1
所以 = ，即a2 =4b2. ··································································· 3分
a2 4
又因为c= 5 ，c2 =a2 +b2，可得a2 =4，b2 =1，
x2
所以双曲线C的方程为 − y2 =1. ······················································ 5分
4
（2）设直线l的方程为：x=ty+ 5，M(x ,y ),N(x ,y )，由题意可知t ≠ ±2，
3 3 4 4
高三数学答案（第 4 页，共 6 页）x2
 − y2 =1
联立 4 ，消x可得，(t2 −4)y2 +2 5ty+1=0，

x=ty+ 5
−2 5t 1
则有y + y = ，y y = ， ··················································· 6分
3 4 t2 −4 3 4 t2 −4
假设存在定点D(m,0)，
 
则DMDN =(x −m)(x −m)+ y y
3 4 3 4
=(ty + 5−m)(ty + 5−m)+ y y ······················································ 7分
3 4 3 4
=(t2 +1)y y +( 5−m)t(y + y )+( 5−m)2
3 4 3 4
t2 +1 2 5( 5−m)t2
= − +( 5−m)2
t2 −4 t2 −4
(m2 −4)t2 −(4m2 −8 5m+19)
= ···························································· 8分
t2 −4
7 5
令4m2 −8 5m+19=4(m2 −4)，解得m= ， ·········································· 10分
8
 
245 11
此时DMDN =m2 −4= −4=− ， ························································ 11分
64 64
 
7 5 11
所以存在定点D( ,0)，使得DMDN为定值− . ··································· 12分
8 64
22.解：（1） f(x)= xex −ax2 −2ax，则 f′(x)=(x+1)(ex −2a)， ··················· 1分
当a>0时，方程ex −2a=0的根为x=ln(2a).
1
当ln(2a)>−1，即a> 时，当x∈(−∞,−1)和x∈(ln(2a),+∞)时， f′(x)>0，
2e
f(x)单调递增，当x∈(−1,ln(2a))时， f′(x)<0， f(x)单调递减. ············· 2分
1
当ln(2a)<−1，即0<a< ，当x∈(−∞,ln(2a))和x∈(−1,+∞)时， f′(x)>0，
2e
f(x)单调递增，当x∈(ln(2a),−1)时， f′(x)<0， f(x)单调递减. ············· 4分
1
当ln(2a)=−1，即a= 时，y′≥0恒成立，函数在R上单调递增， ·············· 5分
2e
高三数学答案（第 5 页，共 6 页）1
综上所述，当0<a< 时， f(x)在(−∞,ln(2a))，(−1,+∞)上单调递增，
2e
1 1
在(ln(2a),−1)上单调递减；当a= 时， f(x)在R上单调递增，当a> 时，
2e 2e
f(x)在(−∞,−1)，(ln(2a),+∞)上单调递增，在(−1,ln(2a))上单调递减. ········ 6分
（2）存在实数a使得 f′(x)≥b−2a对任意x恒成立，即b≤xex +ex −2ax恒成立.
令g(x)= xex +ex −2ax，则b≤ g(x) . ················································ 7分
min
因为 g′(x)=(x+2)ex −2a ，当 x≤−2 时， g′(x)<0 恒成立；当 x>−2 时，
g′′(x)=(x+3)ex >0，函数g′(x)在(−2,+∞)上单调递增，
且g′(−2)=−2a<0，g′(2a)=(2a+2)e2a −2a>0，
所以，存在x ∈(−2,2a)，使得g′(x )=0，且g(x)在(−2,x )上单调递减，
0 0 0
在(x ,+∞)上单调递增，所以g(x) = g(x )=(x +1)ex 0 −2ax . ··············· 9分
0 min 0 0 0
于是，原命题可转化为存在a使得b≤(x +1)ex 0 −2ax 在(−2,+∞)上成立，
0 0
又因为g′(x )=(x +2)ex 0 −2a =0，所以2a =(x +2)ex 0.
0 0 0
所以存在x ∈(−2,+∞)，使得b≤(x +1)ex 0 −(x2 +2x )ex 0 =ex 0(−x2 −x +1)成立.
0 0 0 0 0 0
···················································· 10分
令h(x)=ex(−x2 −x+1)，x∈(−2,+∞)，则h′(x)=ex(−x2 −3x)，所以当x∈(−2,0)
时，h′(x)>0，h(x)单调递增，当x∈(0,+∞)时，h′(x)<0，h(x)单调递减，所以
h(x) =h(0)=1，所以b≤1. ··························································· 12分
max
高三数学答案（第 6 页，共 6 页）