四川省巴中市届高三第一次诊断性考试数学理试题PDF版.docx
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四川省巴中市届高三第一次诊断性考试数学理试题PDF版.docx
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四川省巴中市届高三第一次诊断性考试数学理试题PDF版
巴中市普通高中2017级“一诊”考试
理科数学参考答案与评分标准
一、选择题:
本大题共12小题,每小题5分,共60分.
题号123456789101112
答案DCBCADAACBDB
二、填空题:
本大题共4个小题,每小题5分,共20分.
13.1
2
;14.5
6
;15.2;16.9.
三、解答题:
共70分.
17.(12分)P
如图,在四棱锥PABCD中,底面ABCD是矩形,PAPD,PAAB,
N是棱AD的中点.
(1)求证:
PN平面ABCD;
(2)若APPD,且AB2,AD4,求二面角BPCN的余弦值.
解:
(1)证法一
N
D
C
由题意,知ABAD,ABPA
又PAADA,PA,AD平面PAD
AB
∴AB平面PAD···························································································2分
又PN平面PAD
∴ABPN···································································································3分
由PAPD,NAND得:
PNAD····································································4分
∵ABADA,AB,AD平面ABCD
∴PN平面ABCD·························································································6分证法二
由题意,知ABAD,ABPA
又PAADA,PA,AD平面PAD
∴AB平面PAD···························································································2分
又AB平面ABCD
∴平面ABCD平面PAD················································································3分
由PAPD,NAND得:
PNAD····································································4分∵平面ABCD平面PADAD,PN平面PAD
∴PN平面ABCD·························································································6分
(2)解法一:
向量方法
由APPD,NAND,AB2,AD4得:
NANDPNNE2
取BC的中点E,连结NE,则NE//AB,故NEAD
由
(1)知:
PNNA,PNNE,NENA···························································7分
以N为原点,NA,NE,NP分别为x,y,z轴建立空间直角坐标系,于是,有:
z
NA,B(2,2,0),C(2,2,0),P(0,0,2)
(0,0,0),(2,0,0)
P
∴NP(0,0,2),CB(4,0,0),CP(2,2,2)················································8分
设平面NPC的一个法向量为m(x,y,z)
mNP0,z0,
C
D则由得:
2x2y2z0.
mCP0.N
y
E
取y1得:
m(1,1,0)······················································································9分
A
B
x
一诊考试理科数学参考答案第1页共9页
设平面BPC的一个法向量为u(a,b,c)
uCB0,a0,
则由得:
2a2b2c0.uCP0.
取b1得:
u(0,1,1)·····················································································10分
∴cos,11
mu··································································11分
mu
|m||u|22
2
∴二面角BPCN的余弦值为1
2
解法二:
向量方法
.································································12分
如右图,连结CN,DE,由已知,得:
四边形NECD为正方形
∴DECN···································································································7分
由
(1)知:
PN平面ABCD,又DE平面ABCD
zP
∴PNDE
又PNCNN,PN,CN平面PNC
∴DE平面PNC,即DE为平面PNC的法向量··························Q·······················8分
连结PE,取PE的中点Q,连结NQ
由已知及
(1)可得:
PNBC,NEBC
∴BC平面PNE
N
D
E
C
y
∴BCNQ
由PN=NE得:
NQPE,又PEBCE
A
x
B
∴NQ平面PBC,即NQ为平面PBC的法向量·················································9分
以N为原点,NA,NE,NP分别为x,y,z轴建立空间直角坐标系
则有:
1()(0,1,1)
NQNENP,DEDNDC(2,2,0)································10分
2
NQDE
∴cos,21
NQDE······················································11分
|NQ||DE|222
2
∴二面角BPCN的余弦值为1
2
解法三:
综合法
.································································12分
取BC的中点E,连结NE,则NE//AB,NEAD,且NE2EC
取NC的中点F,连结EF,则EFNC
过F作FGPC,垂足为G,连结EG
由
(1)知:
PN平面ABCD,故PNEF
P
∴EF平面PNC
又PC平面PNC
G
∴EFPC···································································································7分
C
又EFFGF
D
F∴PC平面EFG
E
N
∴PCEG···································································································8分由EG平面PBC,FG平面PNC
AB
∴EGF为二面角BPCN的平面角······························································9分
在直角△ENC中,NE2EC
∴NC22,EFNEEC2,CFCE2EF22·································10分
NC
在直角△PNC中,由NC22,PN2得:
PC23
由△CFG∼△CPN得:
FGCF,故6
PNCP3
FG
∴2226
EGEFFG
3
∴cos1
EGF···················································································11分
FG
GE2
一诊考试理科数学参考答案第2页共9页
即二面角BPCN的余弦值为1
2
解法四:
综合法
.································································12分
取BC的中点E,连结NE
由
(1)知:
PNNA,PNNE,NENA
由已知,得:
PNNDNE2
以PN,ND,NE为棱将四棱锥PNECD补形为正方体NECDPGHM,如图············7分
连结NG,NC,DE
M
H
在正方体NECDPGHM中,由正方体的性质易知:
P
G
DE平面PNC,NG平面PEC·······································································9分
即DE,NG分别为平面PNC,平面PEC的法向量···············································10分
连结DH,EH,则由正方体的性质知:
NG//DH,△DEH为等边三角形D
C
∴HDE60·······························································································11分
N
E
∴DE与NG的夹角为60
A
∴二面角BPCN的余弦值为1B
.································································12分
218.(12分)
2
已知各项均为正数的数列{an}的前n项和Sn满足4Sn(an)(nN*).
1
(1)求数列{an}的通项公式;
(2)设2an
ba,求数列{bn}的前n项和Tn.
nn
解:
(1)由4
(1)2(*)
SnanN知:
n
2
当n1时,有4a1(a11),a10,解得a11····················································1分
22
由1
4Sn(an1),4Sn(an1)两式想减,得:
1
22
22220
4an1(an1)(an1),化简得:
a1a1aa····························2分
1nnnn
变形得:
(an1an)(an1an2)0·····································································3分
∵对nN*,有an0
∴an1an2,即an1an2········································································4分
故数列{an}是以1为首项,2为公差的等差数列··················································5分∴an2n1··································································································6分
(2)解法一
∵2an
ba,an2n1
nn
∴b2n122n1·························································································7分
n
∴Tn(132n1)(2232522n1)················································8分
nnn
nnn21n2
[1(21)]2(14)2422+32
2
21433
······························11分
∴
T
n
2n12
23n2
···················································································12分
3
解法二
由
(1)及已知,得:
Snn2··············································································7分
记2an
c,数列{cn}前n项和为An,则TnSnAn··············································8分
n
c
∵12,n14
c
c
n
∴数列{cn}是以2为首项,4为公比的等比数列···················································9分
∴
A
n
nn
2(14)2212
143
·············································································11分
∴
T
n
2n12
23n2
···················································································12分
3
一诊考试理科数学参考答案第3页共9页
19.(12分)“绿水青山就是金山银山”,“建设美丽中国”
已成为新时代中国特色社会主义生态文明建设的重要内容.某
班在一次研学旅行活动中,为了解某苗圃基地的柏树幼苗生长
0.20
频率
组距
情况,在这些树苗中随机抽取了120株测量高度(单位:
cm),经统计,树苗的高度均在区间[19,31]内,将其按[19,21),
0.10
[21,23),[23,25),[25,27),[27,29),[29,31]分成6组,制
成如图所示的频率分布直方图.据当地柏树苗生长规律,高度
2a
a
不低于27cm的为优质树苗.
019212325272931cm
(1)求图中a的值;
(2)已知所抽取的这120株树苗来自于A,B两个试验区,部分数据如下列联表:
A试验区B试验区合计
优质树苗20
非优质树苗60
合计
将列联表补充完整,并判断是否有99.9%的把握认为优质树苗与A,B两个试验区有关系,并说明理由;
(3)用样本估计总体,若从这批树苗中随机抽取4株,其中优质树苗的株数为X,求X的分布列和数学期望EX.附:
参考公式与参考数据:
2n(adbc)
2
K
(ab)(cd)(ac)(bd)
PK2…k0.0100.0050.001
()
0
其中nabcd
k6.6357.87910.828
0
解:
(1)根据频率直方图数据,有2(a22a0.1020.20)1,解得:
a0.025.················2分
(2)根据频率直方图可知,样本中优质树苗棵树有120(0.1020.0252)30··················3分
列联表如下:
······································································································5分
A试验区B试验区合计
优质树苗102030
非优质树苗603090
合计7050120
可得;
2
2120(10302060)7210.310.828
K················································6分
705030907
所以,没有99.9%的把握认为优质树苗与A,B两个试验区有关系······························7分
注:
也可由
2
2120(10302060)7210.28610.828
K得出结论
705030907
(3)用样本估计总体,由题意,这批树苗为优质树苗的概率为30
120
1
·······························8分
4
X的可能取值为0,1,2,3,4,由题意知:
X服从二项分布,即X∼(4,1)
B
413
∴PXk4k························································9分
()C()()(0,1,2,3,4)
kkk4
44
13811327即:
PXC004;113
(0)()()P(X1)C()();
44
442564464
1327133PX;331
(2)C()()P(X3)C()();
222
44
441284464
131440
P(X4)C()().
4
44256
∴X的分布列为:
··························································································11分
X0123
P
81
256
27
64
27
128
3
64
一诊考试理科数学参考答案第4页共9页
∴数学期望为()411
EX············································································12分
4
(或()08112722733411
EX).
2566412864256
评分说明:
13分布列的表示有解析法、列表法和图象法三种,本题若只写出PXk4k,
()C()()(0,
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