全国高中数学联赛甘肃赛区预赛试题答案

.
g(x) + g(2 − x) = 0 g(x + 1) = −g(1 − x) g(x)
h(x) = f (x − 1) − g(x)
(1, 0)
h(x) 2019
(1, 0)
,
2018
2 × 1009 + 1 = 2019
2019.
8.
{an}
an+1
=
3n+1an an + 3n+1
,
a1
=
C= .
6
66
62
3
a2 + b2 − 2ab cos π = 7
sin B = 3 sin A
b = 3a
a = 1, b = 3
(
π)
3
(
π)
(
π)
2 g(x) = sin 2x + − 1
g(B) = sin 2B + − 1 = 0
sin 2B + = 1.
(
6)
6(
√)
(6
√
)
π π 13π 2B+ ∈ ,
O(x, y, z)
R
△AC D
Q
BCD
x, y, z
(xx2
+ −
y2 + z2 = R2 1)2 + y2 + z2
=
R2
xx22
+ +
(y y2
− 1)2 + z2 + (z − 1)2
= =
R2 R2
(
)
O′
111 ,,
333
.
Axy(0==, 0112, 0), C(1, 0, 0), D(0, 1, 0), B(0, 0, 1)
a1
an
)
1
−
=
an−1
(
11
+ 3
32
1 + 33
) 1 + · · · + 3n =
1 3
·
1
−
1 3n
1
−
1 3
=
3n − 1 2 · 3n
2 · 3n an = 3n − 1
2 · 3n
an
=
3n
−
. 1
9.
P
x + 2y = 4
P
( ) AB
答案
1 1, .
2
解析
P (x0, y0)
x0 + 2y0 = 4
3
{an}
h(x) = f (x−1)−g(x)
(1, 0)
(0, 0) (1, 0)
.
第4页
答案 解析
2 · 3n
an
=
3n
−
. 1
an+1
=
3n+1an an + 3n+1
1 an+1
=
an + 3n+1 3n+1an
=
1 an
+
1 3n+1 .
1 an
=
1 a1
( 1
+ a2
)
(
1
1
− +···+
×
2
×
2
34
=
8 27
8 .
27
7.
f (x) = x3 +sin x(x ∈ R)
g(x) g(x)+g(2−x) = 0(x ∈ R)
2019
.
答案 2019.
解析 f (−x) = (−x)3 + sin(−x) = −(x3 + sin x) = −f (x) f (x)
f (x − 1)
(1, 0)
=
√ 3
sin 2x
−
1 (cos2 x − sin2 x) −
1
=
√ 3
sin 2x −
1
cos 2x −
1
=
( sin 2x −
π) − 1.
(
π )2
2
(
π)
2
2
6
f (C) = sin 2C − − 1 = 0
sin 2C − = 1.
(6 )
6
π
π 11
ππ
π
2C − ∈ − ,
2C − =
2x + y = 4 x 1
−2 < a < 2
() 1
−2,2 .
7
70
z = ax + y
A(2, 0)
a
z = ax + y
x − 2y = 2 x
1
1
0 < −a <
0 ≥ −a > −2 − < a < 0
2
2
0≤a<2
2. △ABC
答案
7 −2.
解析 BC
AB = 4, AC = 3 P
8.
{an}
an+1
=
3n+1an an + 3n+1
,
a1
=
3
{an}
9.
P
x + 2y = 4
P
AB
P
x2 + 4y2 = 4
.
. h(x) = f (x−1)−g(x)
. AB
10.
x
lg(ax + 1) = lg(x − 1) + lg(2 − x)
a
.
6
80
√
11.
f (x) = 3 sin 2x − 1 (cos2 x − sin2 x) − 1, x ∈ R
1) .
f ′(t) = 0
1 t=
t=1
.
3
1 0<t<
f ′(t) < 0
1
1
<t<
f ′(t) > 0
1 t=
f (t)
3
3
2
3
()
1
f
= 4.
3
5.
AB C D
AB⊥AC, AB⊥AD, AC⊥AD AB = AC = AD = 1 Q
A − BCD 答案 解析
√ 23
. 3
AC, AD, AB
△BAE
14.
1 2ξ
5432 ,,,
6543 ξ
4321 ,,,.
5432
15. 1 2
m+n 3
QAB
x2 y2 C : a2 + b2 = 1
C
F
l
M (0, 2)
C
A, B
2 .
P
C
F (2, 0)
y
P
−→ −→ −−→ −−→ P A = mAF , P B = nBF .
Q
P
O
16. 1
12. 1
{an}
n
{an}
Sn
an+1 = 2Sn + 1(n ∈ N∗).
2
an an+1
n
n+2
dn
n
Tn
15
Tn <
. 8
1
13.
AB C D
AD
∥
BC, BA
=
AD
=
DC
=
BC 2
=
a
E
BC
AE
△B1AE
B1AE⊥ AECD F B1D
.
1
B1 − AECD
2
ADB1
E C B1
.
{} 1 dn
an+1 − an = 2(Sn − Sn−1) = 2an
an+1 = 3an(n ≥ 2).
{an}
a2 = 2a1 + 1
2
an+1 = an + (n + 1)dn
3a1 = 2a1 + 1
a1 = 1
1
n+1
n+1
dn = an+1 − an = 2 · 3n−1
an = 3n−1
34
n+1
z1, z2, z3
[0, 4]. −−→ −−→ OZ1 · OZ2 = 0
Z1, Z2, Z3
√ −−→ −−→ |z1| = |z2| = 2, OZ1 ·OZ2 = 0, |z1 +z2 −z3| = 2
.
−−→ −−→ OZ1⊥OZ2,
|z1
+
z2|2
=
|z1|2
+
|z2|2
=
4
|z1 + z2| = 2.
66
( π) sin A + ∈ (0, 1]
6
→−m · −→n ∈ (0, 1].
12. 1
{an}
n
{an}
Sn
an+1 = 2Sn + 1(n ∈ N∗).
2
an an+1
n
n+2
dn
n 解析
15
Tn
Tn <
. 8
1
an+1 = 2Sn + 1 an = 2Sn−1 + 1(n ≥ 2)
{} 1 dn
zR==
12 2√
3
.