2021～2022 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
A A C B C D D C
二、选择题
9.ABD 10. AD 11.ABD 12.BCD
三、填空题
4 24 16
13. 2 14. 2 5 15. 16. ，
5 e 5 25
四、解答题
17.解：若选①:因为2a cos B = c，由正弦定理得2sin Acos B = sin C . ·············· 2 分
因为sin C = sin(A+ B) ，所以2sin Acos B = sin Acos B + cos Asin B ， ··········· 3 分
整理得sin(A B) = 0 .又 π < A B < π ，所以 A B = 0，即 A = B . ············ 5 分
2 2 2
所以b = a = 3 .所以cosC a + b c 1= = . ········································· 7 分
2ab 2
3 3 21
在 BCD中，由余弦定理 BD2 = ( 3)2 + ( )2 2× 3× cosC = ， ······· 9 分
2 2 4
所以 BD 21= . ···················································································· 10 分
2
若选②：因为m ⊥ n，所以m n = 0，即a(a b) + (b c)(c + b) = 0 ，
整理得a2 ab c2 + b2 = 0 . ····································································· 2 分
a2 + b2 2
所以cosC c 1= = . ··································································· 3 分
2ab 2
因为0 < C π< π ，所以C = . ····································································· 5 分
3
ABC 3 3 1在 中，由正弦定理 π = ，得sin A = ， sin sin A 2
3
因为0 < A < C π π= ，所以 A = . ······························································ 7 分
3 6
所以 B = π A π C = ，由勾股定理b = a2 + c2 = 2 3 ， ··························· 9 分
2
又 D为斜边 AC 中点，所以 BD b= = 3 ··················································· 10 分
2
高三数学答案（第 1 页，共 6页）
sin Acos B + cos Asin B 3 cosC
若选③：由已知 = ， ······························· 2 分
cos Acos B cos Acos B
所以sin(A+ B) = 3 cosC . ···································································· 3 分
又 sin C = sin(A+ B) ，所以 tan C 3 C 2π= ，所以 = ． ····························· 5 分
3
ABC 3 3 1在 中，由正弦定理 2π = ，得sin A = . sin sin A 2
3
0 2π因为 < A < C = ，所以 A π= . ····························································· 7 分
3 6
π
所以 B = π A C = ，故b = a = 3 .
6
在 BCD中，由余弦定理 BD2 = ( 3)2 + ( 3 )2 2 3 21× 3× cosC = ， ······· 9 分
2 2 4
所以 BD 21= . ···················································································· 10 分
2
18.解：（1）连接OP ，由题意O为 ABC 的中心，且 PO ⊥面 ABC ，又 AD 面 ABC ，
所以 PO ⊥ AD ，所以 POD 为直角三角形.
设半球与面 PBC 的切点为 E ，则 | OE |=1且OE ⊥ PD .
P
Rt ODE | OE |在 中， =| OD | 1 3= × | BC | ,
sinα 3 2 E C
A O D
| BC | 2 3所以 = . ················································2 分
sinα B
Rt POD | PD | | OD | 1在 中， = = . ················································ 4 分
cosα sinα cosα
（2）由题知， S = 3S PBC = 3
1 | BC | | PD | 3 2 3 1× × × = × × ，
2 2 sinα sinα cosα
S 3 3 π化简得 = ，α ∈ (0, )， ························································ 6 分
sin2α cosα 2
cosα t S(t) 3 3令 = ,则上述函数变形为 = 3 ， t∈ (0,1)， ································ 7 分 t t
2
所以 S ′(t) 3 3(3t 1)= 3 2 ， ········································································ 8 分 (t t )
高三数学答案（第 2 页，共 6页）
令 S ′(t) 3 3 3= 0 ，得 t = .当 t∈ (0, )时， S ′(t) < 0， S(t)单调递减，当 t∈ ( ,1) 时，
3 3 3
S ′(t) > 0 ， S(t)单调递增， ····································································· 10 分
3 3 27
所以当 t = 时，三棱锥的侧面积 S 的最小值为 S( ) = . ······················· 12 分
3 3 2
19.解：（1）取 AD 中点O，连接OM .因为在梯形 ABCD中，O, M 分别为 AD, BC 的
中点，所以OM // AB，又 AB ⊥ BC ，所以OM ⊥ BC . ······························ 2 分
因为 ADP 为等边三角形，故 PO ⊥ AD ，又面 ADP ⊥底面 ABCD ，面 ADP 面
ABCD = AD ， PO 面 ADP，
故 PO ⊥底面 ABCD . ············································································ 3 分
因为 BC 面 ABCD，所以 PO ⊥ BC .
又因为OP OM = O，所以 BC ⊥面 POM ， ··········································· 4 分
而 PM 面 POM ，故 PM ⊥ BC . ·························································· 5 分
（2）由（1）可知，以O为坐标原点，以向量 MB,OM ,OP的方向分别作为 x, y, z 轴的正
方向建立如图所示的空间直角坐标系O xyz ， ··········································· 6 分
则 B(1 , 3 ,0) C( 1 , 3 ,0) A(1 , 1 ,0) D( 1， ， ， , 1 ,0) ， P(0,0, 6 )，
2 2 2 2 2 2 2 2 2
所以 AD = ( 1,1,0)，OP 6= (0,0, ) ，
2
CB = (1,0,0) BP ( 1 , 3 6， = , ) ， ····················7 分
2 2 2
设m = (x1, y1, z1) 为平面 PAD 的一个法向量，
x + y = 0 m AD = 0 1 1
则 ,即 6 ，令 x1 =1，则m = (1,1,0) . ··························· 8 分
m OP = 0 z = 0 2 1
设n = (x2 , y2 , z2 ) 为平面 PBC 的一个法向量，则有
x = 0 n CB 0
= 2
则 ,即 ，令 z = 6 ，则n = (0, 2, 6) . ··· 10 分 1 3 6 2
n CP = 0 x2 y 2 2 2
+ z
2 2
= 0
cos m, n m n 2 5于是 < >= = = ，
| m || n | 2 10 5
所以面 PAD 与面 PBC 5所成的二面角的余弦值为 . ···································· 12 分
5
高三数学答案（第 3 页，共 6页）
20.解：（1）由题意2a = 4， ······································································ 1 分
2 6 8 1
又 P( ,1) 在椭圆上，所以 2 + 2 =1， ·········································· 2 分 3 3a b
解得a = 2，b = 3 . ············································································· 3 分
x2 y2
所以椭圆Γ的方程为 + =1． ··························································· 4 分
4 3
（2）由（1）可得 A( 2,0), B(2,0) ，设过点 (1,0) 的直线为 x = my +1，
C(x1, y1), D(x2 , y2 ) ， ············································································ 5 分
x2 y2
+ =1
联立 4 3 ，消 x 整理得 (3m2 + 4)y2 + 6my 9 = 0， ························· 6 分
x = my +1
∴ y1 + y
6m 9
2 = 2 , y1 y2 = 2 ． ····················································· 7 分 3m + 4 3m + 4
y1 y2
直线 AC 的方程为 y = (x + 2)x 2 ，直线+ BD的方程为
y = (x 2)
x 2 ， 1 2
y1 (x2 2) + y2 (x1 + 2)
联立两条直线方程，解得 x = 2 ( ) ( )．① ·························· 9 分 y2 x1 + 2 y1 x2 2
将 x1 = my1 +1， x2 = my2 +1代入①，
x 2 2my1 y2 + 3(y1 + y2 ) 4y1得 = 3(y y ) 2y ．② ··················································· 10 分 1 + 2 1
y y 6m , y y 9将 1 + 2 = 2 1 2 = 代入②， 3m + 4 3m2 + 4
2m 9 3 6m 9m× + × 4y
3m2 + 4 3m2 + 4 1 2
+ y
3m + 4 1
得 x = 2 = 4
3 6m 2y 9m
= 4．
×
3m2
+ y
+ 4 1 3m2 + 4 1
∴直线 AC, BD 的交点的横坐标为定值4 ． ·················································· 12 分
21.解：（1）由题知： 1 bn+1 = a2n+2 2 = a2n+1 + (2n +1) 22
1
= (a2n 2×2n) + (2n +1) 2 2
1
= a2n 1 ··························································· 2 分 2
高三数学答案（第 4 页，共 6页）
1
b a2n 1
所以 n+1 = 2 1= （ n∈N* ）. ························································· 3 分
bn a2n 2 2
1
又因为b1 = a2 2 = a1 +1 2 =1> 0， ······················································ 4 分 2
{b } 1 1所以数列 n 是以 为首项， 为公比的等比数列， ········································ 5 分 2
所以b = 21 nn ， n∈N
* . ············································································ 6 分
（2）由（1）知，a 1 n2n = bn + 2 = 2 + 2，（n∈N
* ）
所以a2 + a4 + + a2n = (b1 + b2 + bn ) + 2n
1 2 n
= 1 + 2n = 2 2
1 n 2n 2 2+ = n + 2n . ················································· 8 分
1 2
2
又 a2n = a
1
2n 1+1 = a2n 1 + 2n 1，所以a2n 1 = 2(a2n +1 2n)， ························ 9 分 2
所以a1 + a3 + + a2n 1 = 2(a2 + a4 + + a2n ) + 2n 4(1+ 2+ + n) ，
4 4 4n 2n 4n(n +1) 4= 2n + + = 4 n 2n + 4n .······································ 11 分 2 2 2
S 6所以 22n = 6 n 2n + 6n. ··································································· 12 分2
1
22．解：（1） f ′(x) = a， x∈ (0,+∞) . ··················································· 1 分
x
当 a ≤ 0 时， f ′(x) > 0恒成立， f (x) 在 (0,+∞)上单调递增． ··························· 2 分
a(x 1a 0 )
1
当 > 时， ′f ′(x) a ，当 x∈ (0, )时， f (x) > 0， f (x)= 单调递增；当
x a
x (1∈ ,+∞) 时， f ′(x) < 0， f (x) 单调递减． ············································ 3 分
a
综上，a ≤ 0 时， f (x) (0,+∞) 1在 上单调递增，当a > 0时 f (x) 在 (0, )上单调递增，在
a
(1 ,+∞)上单调递减. ··············································································· 4 分
a
（2）①注意到， f (1) = ln1 a + a = 0，
由（1）知，当a ≤ 0 时， f (x) 在[1,+∞) 上单调递增，对任意 x >1，恒有
f (x) > f (1) = 0，不合题意； ···································································· 5 分
高三数学答案（第 5 页，共 6页）
同理，当a ≥1 1 1时， f (x) 在 ( ,+∞)上单调递减，又 <1，所以对任意 x >1，恒有
a a
f (x) < f (1) = 0，不合题意； ·································································· 6 分
0 a 1 1 1 1当 < < 时， >1，由（1）知， f (x) 在[1, ) 上单调递增，在 ( ,+∞)上单调递减，
a a a
f (1所以 ) > f (1) = 0， 又当 x →+∞时， f (x) → ∞，由零点存在定理知，存在唯一
a
1
一点 x0 ∈ ( ,+∞)，使得 f (x0 ) = 0，满足题意. ··········································· 7 分 a
综上所述，a 的取值范围为{a | 0 < a <1}． ················································· 8 分
ln x
②由①知，当0 < a <1时， f (x0 ) = ln x0 ax
0
0 + a = 0，解得a = x ． 0 1
2 a 2(x0 1)
要证 x0 > ，只需证 ln xa 0
>
x ． 0 +1
g(x) ln x 2(x 1) x∈ (1,+∞) g′(x) 1 4 (x 1)
2
令 = ， ，则 = = > 0，
x +1 x (x +1)2 x(x +1)2
g(x) ln x 2(x 1)所以 = 在 (1,+∞)上单调递增，又 g(1) = 0，
x +1
所以 g(x) > 0 在 (1,+∞)上恒成立，
ln x 2(x0 1)> x 2 a即 0 x 1 ，即 0 > . ·························································· 10 分 0 + a
1 x 1
要证 x < ea ，只需证 ln x
1
< ln x < 00 ，即0 a 0 ln x
．
0
又因为 x0 >1 2，即证 (ln x0 ) < x0 1.
令 h(x) = (ln x)2 x +1， x∈ (1,+∞) h′(x) 2 ln x x，则 = ．
x
又 (2 ln x 2 x)′ = 1 2 x= ，所以函数 y = 2ln x x在 (1, 2)上单调递增，在 (2,+∞) 上
x x
单调递减，当 x = 2时，(2 ln x x) ，所以h′(x) < 0 在 (1,+∞)恒成立，max = 2ln 2 2 < 0
所以 h(x) 在 (1,+∞)上单调递减，又 x0 >1，所以 h(x0 ) < h(1) = 0，即 (ln x 20 ) < x0 1，
不等式得证. ························································································· 12 分
高三数学答案（第 6 页，共 6页）2021~2022学年度第一学期期末学业水平诊断
高三数学
注意事项
本试题满分150分,考试时间为120分钟
2.答卷前,务必将姓名和准考证号填涂在答题纸上。
3.使用答题纸时,必须使用0.5毫米的黑色签字笔书写,要字迹工整,笔迹清晰;超出答
题区书写的答案无效;在草稿纸、试题卷上答题无效
选择题: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
项符合题目要求。
已知集合A={-1,1,2},B={x(x+(x-2)<卟},则A∩B
A{0,}
B.{1,2}
C{-0,}D.{01,2
2命题“Yx∈R,2>0”的否定为
A.3x∈R,2≤0B.3x∈R,2<0c.x∈R,2≤0D.Wx∈R,22<0
3函数y-1n(x+1)
的定义域为
A.[-2,2]
B.(-1,2]
C.(-1,0)∪(0,2]D.(-1,1)∪(,2]
4在生活中,人们常用声强级y(单位:dB)来表示声强度Ⅰ(单位:W/m2)的相对
大小,具体关系式为y=10g(2),其中基准值=10+2W/m2若声强度为1时的声
强级为60dB,那么当声强度变为41时的声强级约为(参考数据:lg2≈03)
A 63 dB
B. 66dB-BC72 dB"i(i) D 76 dB
5若双曲线mx2-y2=1(m∈R)的一条渐近线方程为3x-4y=0,则其离心率为
4
7
6.已知|a|=1,|b=2,ab=、
,则cos
=
A
6
C
4
寡二些计甌:笆1而::A而1
若直线x-y+2=0将圆(x-a)2+(y-3)2=9分成的两段圆弧长度之比为1:3,则实
数a的值为
A.-4
B.-4或2
C.2
-2或4
8.若定义在R上的奇函数f(x)在(-∞,0)上单调递减,且f(2)=0,则满足
(2x-1)f(x+1)≥0的x的取值范用是
B.(-o,-3]U[,+∞)
C.[-3.-1]U[:,1
D.[3,JU[l,+∞)
选择题:本题共4小题,每小题5分,共20分。在每小题给出的选项中,有多项符
合要求。全部选对的得5分,部分选对的得2分,有选错的得0分。
9.已知a>0,b>0,则下列命题成立的有
A.若b=1,则a2+b2≥2B.若ab=1,则+≥2
C.若a+b=1,则a2+b2≤
D若a+b=1,则+≥4
10.函数f(x)=2sin(x+9)(>Q-2
,2下9<)的部分图象如图所示,则
A.O的值为2
()…,
(
B.φ的值为
,个一计个
C.(-,0)是函数f(x)的一个增区间
4
D.当x=+k(k∈Z)时,f(x)取最大
1.2已知抛物线C:x2=my的焦点为F(0,1),点A,B为C上两个相异的动点,则
A抛物线C的准线方程为y=-11)1,(
B设点P(2,3),则AP|+|AF|的最小值为4
C.若A,B,F三点共线,则AB|的最小值为21”共出
D.若∠AFB=60°,AB的中点M在C的准线上的投影为N,则MN|≤|AB