资源简介

{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}
{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}
{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}
{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}
高二（下）期末考试参考答案
一．选择题
1 2 3 4 5 6 7 8 9 10 11
A D C B D B C B BCD AD ABD
二．填空题
13
12.135 13.1 14.
3
三．解答题
15.解：（1）由bcos A 3bsin A c a 0 及正弦定理得
sin B cos A 3 sin B sin A sinC sin A 0 ····················································· 1分
因为 sinC sin A B sin A B sin AcosB cos Asin B，
所以 3 sin B sin A sin AcosB sin A 0 ························································· 3 分
由于 sin A 0， 3 sin B cos B 1 0 ···························································4 分
所以 sin
1
B ··························································································5 分
6 2
又0 B ，故 B

··················································································· 6 分
3
（2）由题得 ABC S 1的面积 ac sinB 9 3 ，故 ac 9①································· 8分
2 4
而b2 a2 c2 2ac cosB，············································································ 9分
且b 2，故 a2 c2 18②，·········································································· 11 分
由①②得 a c 3 ························································································· 13分
16.（1）连接CA交 BD于点M ，连接MN .
{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}
因为 ABCD为正方形，M 为 AC中点，·····························································2分
又因为 N为 PC的中点，
所以MN为 CPA中位线，MN //AP .································································ 4分
又因为MN 平面 BND， AP 平面 BND，
所以AP//平面 BND．···················································································· 6分
（2）（方法一）因为 PD 平面 ABCD，AD，DC 平面 ABCD，所以 PD DA，PD DC，
在正方形 ABCD中，DA DC，

所以以 DA,DC ,DP 为正交基底建立空间直角坐标系O xyz，······························7分
因为 PD AD 3，
所以C 0,3,0 ， B 3,3,0 ， P 0,0,3 ，

所以 PB 3,3, 3 ,PC 0,3, 3 .·····································································8分

设平面 PBC 的一个法向量为m x, y, z ，

m PB 0, 3x 3y 3z 0,
所以 即 ·································································· 9分
m PC 0, 3y 3z 0,

解得 x 0，取 y 1，得 z 1，所以m 0,1,1 ，···············································11分

又平面 PAD的一个法向量为 n 0,1,0 ，·························································13分

cosm n
，
2 2
所以平面 PAD与平面 PBC 2夹角的余弦值为 ．···············································15分
2
（方法二）因为 PD 平面 ABCD， AD 平面 ABCD，CD 平面 ABCD，
所以 PD AD , PD CD .············································································ 8分
因为底面 ABCD为正方形，所以CD AD .
又因为DC 平面 PCD， PC 平面 PCD，DC PC C ,
所以 AD 平面 PCD .··················································································· 10分
在平面 PAD过 P作 l AD，有 l 平面 PCD ,
{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}
PD 平面 PCD , PC 平面 PCD ,所以 PD l , PC l , DPC为平面 PAD与平面 PBC 夹角
或其补角，································································································ 12分
因为底面 ABCD为正方形，所以 AD CD 3，又 PD 3，PD CD，故 DPC即为所求

又 DPC= ，··························································································· 14分
4
平面 PAD 2与平面 PBC 夹角的余弦值为 ．·····················································15分
2
n n 1
17.（1 ）设等差数列 an 的首项为 a1，公差为 d，则前 n项和为 Sn na1 d .2
a1 2d 7

所以 4 4 1 2 2 1
4a1 d 3 2a2 1
d
2
a1 2d 7
即 ，··············································································3分
4a1 6d 6a1 3d
解得 a1 3， d 2，·····················································································4分
所以 an 3 n 1 2 2n 1.
因此数列 an 的通项公式为 an 2n 1····························································· 6分
（2） cn anbn 2n 1 2n 1 ······································································7分
T 3 22n 5 2
3 7 24 2n 1 2n 1，
2T 3 23 4 5n 5 2 7 2 2n 1 2n 2 ··············································· 9分
所以Tn 2Tn 12 2 2
3 2 24 2 2n 1 2n 1 2n ·························· 11分
2412 2 2
n 2
2n 1 2 n 2 1 2n 2 n 2 4 ，··································· 13分
1 2
{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}
即 T 1 2n 2n 2n 4 ··············································································14分
T 2n 1 2n 2所以 n 4 ··········································································· 15分
18.（1）当 k 1时，赛制为三局两胜制，故 X的可能取值为 2,3，
1 2 2 2P X 2 5 ，
3 3 9
P X 1 2 4 3 C12 ,················································································· 2分3 3 9
所以 X的分布列为：
X 2 3
P 5/9 4/9
E X 2 5 3 4 22 ················································································ 4分
9 9 9
2
（2）①因为每局比赛中，机器人获胜的概率为 p ，
3
由题可知 P 1 为3局 2胜制时，机器人获胜的概率，机器人获胜的情形有两种：2 : 0或 2 :1，
2 1 2 2 2 2 20
所以 P 1 p C2 p 1 p p 3 2p （ ）2 3 2 ，·························· 6分3 3 27
P 2 为5局3胜制时，机器人获胜的概率，机器人获胜的情形有三种：3: 0或3 :1或3 : 2，
P 2 p3 C23 p3 1 p C24 p3 1 p
2 p3 6p2 15p 10
2 3 2 2 2 2
C2 1 2 C2 2 1 2 64 3 3 4
，·········································· 8分
3 3 3 3 3 3 81
所以 P 2 P 1 ，
所以 k 2时，5局3胜制对机器人更有利···························································9分
②随着 k的增大，机器人获胜的可能性越来越大.
证明如下：
2k 1
P k Ci i 2k 1 i由①可知， 2k 1 p 1 p （用全场打满的方法算）·························10分
i k 1
下面讨论 2k 3局与前 2k 1局的递推关系：（用先打 2k 1局，再打 2 局的方法）
（i)若前 2k 1局中机器人恰好赢了 k局，则后两场机器人都要赢才能获胜，
Ck pk 1 p k 1其概率为 p2，即Ck pk 2 1 p k 12k 1 2k 1 ··········································· 11分
（ii)若前 2k 1局中机器人恰好赢了 k 1局，则后两场机器人至少要赢一场才能获胜，
Ck 1 k 1 k 2 k其获胜概率为 k 1 k 22k 1p 1 p 1 1 p ，即C2k 1p 1 p 2 p ·················· 12分
（iii）若前 2k 1局中机器人至少赢了 k 2局，则后两场机器人无论输赢都获胜，
{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}
2k 1
Ci pi 1 p 2k 1 i其获胜概率为 2k 1 ································································ 13分
i k 2
2k 1
P k 1 C k2k 1 p k 2 1 p
k 1
C k 1 k 22k 1 p 1 p
k 2 2k 1 i p C i p i2k 1 1 p
i k 2
P(k k 1 1) P(k) Ck pk 2 1 p Ck 1 pk 22k 1 2k 1 1 p
k
2 p Ck 1 k 12k 1p 1 p
k
Ck k 12k 1p 1 p
k 1 2 p 1 ，·········································································15分
p 2 , 2 2 1 0 Ck2k 1p
k 1 1 p k 1 2 p 1 0，即 P(k 1) P(k).················ 17分
3 3
19. 解（1）已知 f (x) 2x ln x 1，对其求导可得 f '(x) 2(ln x 1)，令 f x 0，解得 x .
e
当 x变化时， f x ， f x 的变化情况如下表：

x 0,
1 1 1
,

e e e
f x - 0 ＋
f x 单调递减 极小值 单调递增
······················· 3分
x 0, 1 当 ， f x 0，又 f 1 0， f (e) 2e，
e
则不等式0 f (x) 2e的解集为 x1 x e ···················································· 5分
（2）由题意 g(x) 2x ln x 4x 2a的定义域为 0, ，且 g (x) 2(ln x 1) .
当 x 0,e 时， g x 0；当 x e, 时， g x 0 .
故 g x 在区间 0,e 上单调递减，在 e, 上单调递增.······································· 7分
a 0， g(x)min g(e) 2e 2a 0 ··························································· 8分
当 x 0,e2 时， x ln x 2x 0， a 0，故 g x 0；
当 x e2时， g(e2 a ) 2(2 a)e2 a 4e2 a 2a 2a(e2 a 1) 0
g x e2在 , 上单调递增， 当 a 0时， g x 有且仅有一个零点. ················ 10分
（其他方法请老师们酌情给分）
f x2 f x1 x2 ln x2 x1 ln x1
（3）由 f x0 ，得 ln x 1x2 x1 x2 x
0 ，
1
{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}
ln x x ln x x ln x则 0 2 2 1 1 1x x ，········································································ 11分2 1
2x1x2 x ln 2x1x2 ln x n 2x1x2 x2 ln x2 x1 lnx1要证 0 ，可证 0，即证 1 lx1 x2 x1 x2 x1 x
.
2 x2 x1
t x2 t 1 2tx1 tx1 ln t ln x1 x1 ln x令 ln 1 1x ，即证 ，1 t 1 (t 1)x1
即证 ln
2t t ln t
1 ···················································································· 13分
t 1 t 1
下证 t 1 ln 2t t ln t t 1 0 t 1 ，先证 ln x x 1 x 1 ，
t 1
p x x 1 ln x x 1 p x 1 1 x 1设 ， ， ，
x x
当 x 1， p x 0， p x 在 1, 上单调递增，
则 p x p 1 0，即 x 1 ln x ···································································· 15分
令 F x x 1 ln 2x x ln x x 1 x 1 ，则只需证明 F x 0，又 ln x x 1 x 1 ，
x 1
F x ln 2x x 1 1 1则 ln xx 1 x x 1
ln 2 x 1 2 x 1 1 x x 1 x 1 1 1 1
0
F x 1, x 1 x x 1 x 1 x x 1 x 1 x x 1 x 1 x ， 在
上单调递减，
2 1 2x1x
则 F x F 1 1 1 ln 1 ln1 1 1 0 2， xx x 0 ··································17分1 1 1 2
{#{QQABAYYghgiQkhSACBgqUUFWC0sYshMjJAgkRRAQuARDSANABAA=}#}

展开更多......

收起↑