2014届执信中学高三年级十二月月考测试(理数答案)
理数参考答案
选择题1-8
BDCAB ACB
499) -84 10) 1 11)480 12) 13) 1006 14)1 15) 23
2216、解:(1)由与共线知„„2分 mna(a,c),b,c
2221a,c,bac222即a,c,b,ac,由余弦定理, „„„„„„„„5分 cosB,,,222acac
,又 „„„„„„„„„7分 (0,),B,?B,,3
(2), sinB,sin(A,C),sinAcosC,cosAsinC,cosAsinC
,,(0,)cos0,sinAcosC,0?又? „„„„„„„„„9分 AC,?C,C,,2
Rt中,为外接圆的直径, „„„„„„„10分 ,ABCAB?AB,2R,4
, „„„„„„„„„11分 ?B,?BC,2,AC,233
1?S,,2,23,23 „„„„„„„„„12分 ,ABC2
17. 解:(?)记“小球落入袋中”为事件,“小球落入袋中”为事件,则事件的对立事件为,AABBAB
33111,,,,PB(),,,,,,,224,,,,而小球落入袋中当且仅当小球一直向左落下或一直向右落下,故,从而B
13; PAPB()1()1,,,,,44333127,,,,3 B4,,,,,,,,PC(3),,,,4344464,,,,(?)显然,随机变量,故,( ,,,,E434
18、解:
CEGFGBG、解法一:(1) 证:取的中点,连结(
1CDGFDE//FGFDE,?为的中点,?且( 2
ACDACDAB,DE,?平面,平面,
1ABDE//GFAB//GFAB,ABDE,?,?(又,?( 2
GFABAFBG//?四边形为平行四边形,则( „„„„„„„„„3分 AF,BCEBG,BCE?平面,平面,
AF//BCE?平面( „„„„„„„5分
,ACDCDAFCD,F(2) 证:?为等边三角形,为的中点,? „„„„„„„6分
ACDACDDE,AF,DEAF,?平面,平面,?( „„„„„„„„7分
CDDED:,CDEAF, 又,故平面( „„„„„„„„„8分 BGAF//BG,CDE?,?平面( „„„„„„„„„9分
BG,BCEBCE,CDE ?平面,?平面平面( „„„„„„„„„10分
CDEFHCE,FBHH(3) 解:在平面内,过作于,连(
BCE,CDEBCEFH,?平面平面, ?平面(
BCE,FBHBF?为和平面所成的角( „„„„„„„„„12分
1
2ADDEABa,,,22设,则, FHCFa,:,sin452
FH22222BFABAFaaa,,,,,(3)2,在R t?中,( FHBsin,,,FBHBF4
2BCE?直线和平面所成角的正弦值为„„„„„„„„„14分 BF4
ADDEABa,,,22解法二:设,建立如图所示的坐标系,则 Axyz,
,,33(( ACaBaDaaEaaa000200,0,0,,,3,0,,3,2,,,,,Faa,,0,,,,,,,,,,,,,,22,,
,,,,,,,,,,,,,,33(1) 证:, AFaaBEaaaBCaa,,,,,,0,,3,,2,0,,,,,,,,,22,,
,,,,,,,,,,,,1AF,BCE?,平面, AFBEBC,,,,2
AF//BCE?平面(„„„„„„„5分
,,,,,,,,,,,,,,33(2) 证:?, AFaaCDaaEDa,,,,,,,0,,3,0,0,0,2,,,,,,,,22,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,?,?( AFCDAFED,,,AFCDAFED,,,,0,0,,,,
CDEAF//BCE?平面,又平面, AF,
BCE,CDE?平面平面( „„„„„„„„„10分 ,,,,,,,,,,,
BCE(3) 解:设平面的法向量为,由可得: nxyz,,,nBEnBC,,,,0,0,,
,,,,,,,33n,,1,3,2,取x=1,则( 又, BFaaa,,,,xyzxz,,,,,30,20,,,,,,22,,
|BF,n|2a2sin,,,,BCE,BF设和平面所成的角为,则„„„„14分 42a,22|BF|,|n|
1mxm,'fx0,,,19.解:(?)函数的定义域为,且 (——2分 fx,,,,,,,,,22xxx'fx,0xm,,令,得 ( ——————————————3分 ,,
xm,'fx0,,,m,0xm,,0fx,,0当时,,,函数在上是增函数;—5分 ,,,,,,2x
'0,,mfx,0fx0,,mm,0当时,在区间上,函数在上是减函数; ,,,,,,,,
',,,m,fx,0fx,,,m,在区间上,函数在上是增函数(———7分 ,,,,,,,,
xm,''1,efx,0fxm,,1fx,(?)由(?)知,(1)若,则在区间上,函数,,,,,,,,2x
1,efxf1fm13,,,在上是增函数,此时,取最小值,由,得,,,,,,,,
m,,,,,,31,;————————9分 ,,
2
'(2)若则在区间1,e上fx,0,函数fx在1,e上是减函数, m,,e,,,,,,,,,
mfxfem,,,,,,2e,e此时,取最小值,由,得;———11分 fe13,,,,,,,,,,,e
'1,,mfx,0fx1,,m(3)若,则在区间上,函数在上是减函数,,,,,e1m,,,,,,,,
',,,m,fx,0fx,,,m,在区间上,函数在上是增函数, ,,,,,,,,
2fxfm,fmm,,,,,ln13,,e,1此时,取最小值,由,得;——m,,e,,,,,,,,,,
fx1,e——13分. 综上所述,存在实数,使得在区间上取得最小值3( m,,2e,,,,
——————————14分
b20..解: (1)由焦点,渐进线方程为,根据对称性,取渐进线,则yx,,Fc(,0),bxay,,01a
,,bc091,又因为双曲线过点(.则,故双曲线的
标准
excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载
,,?,1,22a(3,1)22,,b222a8ba,
22xy22,,1 (或) 方程为xy,,888
22(2) 设动点, Pmnmn(,),8?,,
2222QPOmnn,,,,28由(1)可知, FF(4,0),(4,0),12
2222PFPFmnmn,,,,,,,(4)(4)12
22222,,,,,,,,mnmmnm168168PO而, 故=1,则为定值 2222222PFPF,12,,,,,,,,mnmnn166422464(8),,,,42222,,,,,,,43264(28)28nnnn
2PFPFPFPF+1212,,,=2POPFPF,,(3)由(2)可知,故 12POPO
22PFPFPFPFPFPF+4,,,,,,,32321212122,又因为 =448,,,,,,2228POPOPO
,,22故,综上:则实数 ,,(2,22]
21((本题满分14分)
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n,1解:(1)依题意,数列的通项公式为, „„„„„1分 {}bb,2nn
n由, abababababn,,,,,,,,,?(1)2111223311,,nnnn
n,1可得, n,2ababababn,,,,,,,,?(2)21,,11223311nn,,
n,1两式相减可得,即. „„„„„3分 an,abn,,,2nnn
,当,从而对一切,都有. „„„„„4分 nN,na,,11时,an,1n所以数列的通项公式是. „„„„„5分 {}aan,nn
d(2)设等差数列的首项为,公差为,则. {}aaaand,,,(1)n1n1
n,1n,2n,1由(1)得, „„„„„6分 n,2abnb,,,,2,即,,nnnand,,(1)1nn,,11n,22b,, „„„„„7分 nad,()adnd,,11,dn
bn,1是一个与无关的常数,当且仅当 „„„„„8分 要使ad,,0n1bn
n,12b,即:当等差数列的满足时,数列是等比数列,其通项公式是; {}aad,,0{}bn1nnd
当等差数列的满足时,数列不是等比数列( „„„„„9分 {}aad,{}b1nn
n,1(3)证法:由(1)知. abn,,2nnn111111,,,,,,? ,231,nabn,,,,,112232422,1iiin111111n,3,,,,,,? „12分 ,,,231,nab,,,,,1122222222,1iii
1n,2,1()11131112,,,, „„„„„14分 ,,,,11442148,12
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