2020年高三理科数学考前大题强化练十一附答案详析

D
BB1Q
1 3
S BB1Q
CD
=
1 3
8
1 2
=
4 3
3D
,DA,DC, DD1
x ,y ,z ,
D 0, 0, 0 , A2,0,0 , B1 2, 2, 2 , Q 0, 2, , D1 0,0,2
AD1 2, 0, 2 , AQ 2, 2, , DB1 2, 2, 2
AQD1
,
1 3 3 2 1 k2 2 ,
k
2
1 3
,11
y
kx
3
x 2
y2 4
1
4 k 2 x2 2 3kx 1 0
9
M x1, y1 , N x2, y2 ,
x1
x2
2 3k 4 k2
x1x2
,
1 4 k2
MN 1 k 2 x1 x2 2 4x1x2 1 k 2
x
1
t 2
y
3t 2
t
t1 t1t2
t2
4
7
4 7
,
AB
t1 t2
t1
t2 2
4t1t2
82 7
23
fx = m − x + 4m > 0 fx − 2 ≥ 0
m
− 3, − 1
abc
1+ 1 + 1 =m
a 2b 3c
a + 2b + 3c ≥ 9.
I m=1
f(x − 2) = m − |x + 2| ≥ 0, |x + 2| ≤ m − m − 2 ≤ x ≤− 2 + m,
20 AB
O AB
r1 1 r2 2
OT
xy
l1 l2 l1 l2
P
1
OT O
2
l y kx 3
P
E
E MN
6
l
1
2
MN
P x, y ,OT x
x OA cos
,
y
OB
sin
x cos
y
2
sin
x2 y2 1
x2 y2 1
4 ,P
E
4
2
6
1
1 2, O
l
d
1 2
,
3 2
12k 2 4 k2
2
4 4 k2
4 1 k 2
4k2
4 1
4
3 k
2
MN
16 13
,
16 5
21
f(x) = ln(1 + x) − x(11++λxx).
x ≥ 0 f(x) ≤ 0 λ
{an}
an
=
1
+
1 2
+
1 3
+
+
1 n
a2n
−
an
+
1 4n
>
ln2.
f(0) = 0
f'(x)
=
(1−2λ)x−λx2 (1+x)2
f'(0) = 0.
λ<1
2
0 < x < 2(1 − 2λ) f'(x) > 0
f(x) > 0.
λ
≥
1 2
x > 0 f'(x) < 0
x > 0 f(x) < 0.
λ
12.
λ = 12.
x>0
f(x) < 0
x(2+x) 2+2x
>
ln(1
+
x).
x
=
1 k
53
3
CC1
Q
DB1
AQD1
9
Q
CC1
2
20 AB
O AB
r1 1 r2 2
OT
xy
l1 l2 l1 l2
P
1
OT O
2
l y kx 3
1
2
MN
P
E
E MN
6
l
21
f(x) = ln(1 + x) − x(11++λxx).
x ≥ 0 f(x) ≤ 0 λ
{an}
an
=
1
+
1 2
+
1 3
+
II
1:
1 a
+
1 2b
+
1 3c
=
1(a,b,c
>
0)
a
+
2b
+
3c
=
(a
+
2b
+
3c)(
1 a
+
1 2b
+
1 3c
)
=
3
+
(
a 2b
+
2b a
)
+
(
a 3c
+
3c a
)
+
(
2b 3c
+
3c 2b
)
≥
9
a = 2b = 3c,
a
=
3,b
=
3 2
,c
=
1
2:
1 a
+
1 2b
+
1 3c
=
1(a,b,c
2
B1 DBQ
3
CC1
Q
CC1
Q
DB1
AQD1
1
Q C1C
,
D1Q ,DC,AP
53 9
7
AP DC M,
D1M C1C
Q,
CP AD , MCP MDA ,
MC CP 1 MD AD 2
CQ D1D , CQ MC 1
MCQ MDD1 , DD1 MD 2
Q C1C
2V
B1 DBQ V
+
1 n
a2n
−
an
+
1 4n
>
ln2.
22
C
C E AB
1
x 2x
y
y
x
x
1
t 2
E
l
y
3t 2
t
3
1
C
M x, y x 3y
2
E
l
E
AB
23
fx = m − x + 4m > 0 fx − 2 ≥ 0
m
abc
1 a
+
1 2b
+
1 3c
=
m
− 3, − 1 a + 2b + 3c ≥ 9.
2020
17
an
n
Sn 1 an
1
an
2
S5
33 32
0
18
“
” 2018 4
“
”
200 P
P
100
600
40
20
1 40
2
40
“
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40
m
a 80 m a
M
M“
”
M“
”
22
22
99%
“
”
1
“
”
8
8
3
3
XX
.
19
ABCD A1B1C1D1
2 P BC
Q
CC1
1Q
D1Q DC AP
2
B1 DBQ
n x, y, z ,
nn
AD1 0 AQ 0
2x 2x
2z 2y
0
z
0
x2,z 2, y 2
n 2, 2 , 2
DB1
AQD1
sin cos n, DB1
n DB1
n DB1
பைடு நூலகம்
12 2
5 3
12 8 2 2 9
2 2 3 0
8
1
3 2
Q,
Q CC1
2k+1 2k(k+1)
>
ln(
k+1 k
).
a2n
−
an
+
1 4n
=
k2n=n−1
(
1 2k
+
1 2(k+1)
)
=
k2n=n−1
2k+1 2k(k+1)
>
k2n=n−1
ln
k+1 k
=
ln2n
−
lnn
=
ln2.
a2n
−
an
+
1 4n
>
ln2.
22
C
1
x
x
1
t 2
C
x 2x
y
y
E
l
y
3t 2
M
M“
”
M“
”
22
5
22
99%
“
”
“
”
8
8
3
3
XX
.
m 80 82 81
1
2
2
m 81 a 80
M 81
81 15
22
15
5
5
15
20
20
K 2 40 1515 55 10 6.635
20 20 20 20
P K 2 6.635 0.010
99%
“
”
81 5 20 20 40
17
an
n
Sn 1 an
1
an
0
2
S5
33 32
1 Sn 1 an , 0 , an 0
n 2 , Sn1 1 an1 ,
, an 1 an 1 an 1 an an 1 , 1 an an1 ,
0 , an 0
1 0
an 1, an1 1 , n 2 ,
4
an
q
,
1 ,
n 1
, S1 1 a1 a1 ,
a1
1 1
,
an
1 1
n1 1
2
S5
33 32
,
S5
1
1 1
4
1
33 32
,
1
5
33 32
1
1 32
,
1 1
1 2 ,
3
18
“
” 2018 4
“
”
200 P
P
100
600
40
20
1 40
2
40
“
”
40
m
a 80 m a
6
“
”
X
12 3
PX
1
C22C61 C83
6 56
3 28
PX
2
C21C62 C83
30 56
15 28
PX
3
C20C63 C83
20 56
5 14
X
2
6
X
1
2
3
3
15
5
P
28
28
14
X
19
E
X
1
3 28
2
15 28
3
5 14
9 4
ABCD A1B1C1D1
2 P BC
Q
CC1
1Q
D1Q DC AP
>
0)
3= a a + 2b + 3c ≥ 9
1 a
+
2b
1 2b
+
3c
1 3c
≤
a + 2b + 3c
a = 2b = 3c,
a
=
3,b
=
3 2
,c
=
1
1 a
+
1 2b
+
1 3c
.
11
t
E AB
1
C
M x, y x 3y
2
E
l
E
AB
1
2 x2 y2 ,
C x2 y2 1,
10
x cos
C
y
sin
,
x
3y cos
3 sin
2 sin
6
,
x 2 3y
2
2
x 2x
y
y
,
x
x 2
y y
x2 y2 1
C E2
,
l
,
E
7t2 4t 4 0 ,