2020届深圳市高三二模理科数学试卷(含答案)

sin ADC 1 cos2 ADC 4 3 7
sin B sin( ADC 60 ) 4 3 1 1 3 = 3 3 7 2 7 2 14
ABC A1B1C1
CE FB1 AB
2 AA1
2
3 3
EB1
EF
CEB1
EF
CFB1
ABC
EF A1
F
AB AA1 C1
B1
A
C
E
B
5
19
AA1 2a 23 AB 2AA1 3 EB1 AB 2 2a EB1 6a BB1 2a E AB
x0 +
2k
(k Z) 2
x1+x2 + 2k
(k Z)
2
2
x2 x1 4k
2
2
1
2
x1
2k 3 2n
sin(2k 3 ) 2n
cos n
cos (0,1) n N* n 3 n
[1 ,1) 2
n2
n3 ||
cos 1 32
f (x) sin( x ) 2
f (x) ||
x
k (k Z)
2
2
[1 ,1)
2
2
1 [(1 5 )n (1 5 )n ] ( 2)11
52
2
5
an
1 [(1 5 )n (1 5 )n ]
52
2
{an}
an
( 2)11 5
an2
211 5
{an}
{an2 }
a7 13
an2
211 5
a8 21
n
a72
211 5
8
a82
211 5
log [(1 5)n (1 5)n ] 2n 11
6
3பைடு நூலகம்
26
g( )max
9
3
19
sin
6
t 2 cos
6
3
3
(2t t3 )max
2g( )max
46 9
ab 6 3
A EBCDF
V
23
23
2t t3 t( 2 t)( 2 t) 2t( 6 2 3 1t)( 6 2 3 1t)
2
2
2
2
t( 6 2 3 1t)( 6 2 3 1t)
2
2
2
2
t (6 2
CE
ABB1 A1
EF
ABB1 A1
CE EF
EC EB1 E
EF
CEB1
6
19
ABC A1B1C1
FAE EBB1 90
AB
AE AF
2 AA1
2
3 3
B1E
2 BB1 EB
FAE EBB1
EF
AB AA1
FEA BEB1 90
FEB1 =90
EF CE CE
CEB1 EB1
CEB1 CE EB1 E
n
2
6. D 12. D
8
C.
12.
A(x1, ) B(x2, ) C(x3, )
x3 x1 2
x0
=
x1 +x2 2
f (x)
AB
M (x0, )
x x0
1
19
AC nBC(n N*)
| AB | n 1(n N*) | AC | n
2(n 1)
2(n 1)
x2 x1
n
x2 x1
n
1
0
1 n N*
2 0 BAC 180 AD 0 DAC 90
DAC 60
ACD
BAC
10 3
4
19
ACD CD2 AD2 AC2 2 AC AD cos DAC 52 82 2 5 8 cos60 49
CD 7
ACD
CD = AD sin DAC sinC
sin C= AD sin DAC 5sin 60 5 3
[
2
3 1t) ( 6 2
2
2
3
3 1t) 2 ]3
26
9
2t t3 4 6 9
t6 3
ab 6 3
A EBCDF
V
9246 23
23
49
17
12
ABC D BC
CD sin B
AD
BAC AD 5 AC 8 ACD
10 3
A
B
D
C
AD 5 AC 8 1 5 8sin DAC 10 3 2 sin DAC 3
a2 b2
a2 b2 2ab
Vmax
9(1 ab) 2
ab 2ab
9 2 (2 ab ab ab) 4
t ab
t (0,1)
Vmax
9 2 (2t t3) 4
t (0,1)
f (t) 2t t3 t (0,1)
f (t) 2 3t 2
t6
f (t)
46
3
9
92 46
Vmax
4
23 9
ab 6 3
A EBCDF
f (x)
2
[
,]
11
2k (k Z)
16 2
=1
2
16 2
ncos 1(n 3) n
n 3 cos cos = 1
ncos 3 1(n 3)
n 32
n2
cos 1 0(n 3) nn
g(x) cos 1 (x 3)
g(x)
xx
n 3 g(n) g(3) 1 0 6
cos 1 0(n 3) nn
D.
EF
CEB1
CE
ABB1 A1
CE BB1
CE
ABC
ABC A1B1C1
A1B1
M
EB CE EM
EB EC EM
xyz
E
A1
C1
z
M
B1
F
A E
Bx
y C
E(0,0,0) C(0, 6a,0) F( 2a,0,a) B1( 2a,0,2a)
7
19
EF ( 2a,0,a) FC ( 2a, 6a, a) FB1 (2 2a,0, a)
EB 2a
EB12 EB2 BB12
EB BB1
ABC A1B1C1
ABB1 A1
ABB1 A1
F
AA1
FB12 A1F 2 A1B12 9a2 FE2 AF 2 AE2 3a2
FB12 EF 2 EB12
EF EB1
ABC CE AB
E AB
CE FB1
AB
ABB1A1 FB1
ABB1 A1
AB FB1
n cos 1(n 3) n
2
19
13. x y 1=0
14. 2
15. 14
16. 2 3
16.
| AE | 3a | AF | 3b a,b (0,1)
AEF
EF
h 3ab a2 b2
A EBCDF A EBCDF
S 9(1 ab) 2 AEF
EBCDF
1
ab ab
Vmax
Sh 9(1 3
) 2
CD
7
14
ACD AD AC
C
cosC= 1 sin2 C 11 14
DAC 60 BAC 2 DAC 120 B C 60 B 60 C
sin B sin(60 C) sin 60 cosC cos60 sin C 3 11 1 5 3 3 3 2 14 2 14 14
ACD
cos ADC 52 72 82 = 1 257 7
A
2020
1. B
2. D
3. C
4. A
5. A
7. A
8. C
9. D
10. B
11. C
11.
n
log [(1 5)x (1 5)x ] 2x 11
2
log [(1 5)n (1 5)n ] 2n 11 2
log [(1 5 )n (1 5 )n ] 11
22
2
(1 5 )n (1 5 )n ( 2)11
V
23
23
t 2 cos 2t t3 t(2 t2) 2 2 cos sin2
(,) 42
g( ) 2 cos sin2 [g( )]2 2cos2 sin4
(,) 42 (2 2sin2 ) sin2 sin2
[(2 2sin2 ) sin2 3
sin2 ]3 8 27
2 2sin2 sin2
sin