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y=by=b一
2019年福中毕测数(科试解第卷选题本题12小题,每题5分共60分在小给的个选中只一项符题要的1.知集合Axx1
2
x20
,则
=(A.
1
B.
x1x
C.
1x
D.
x12.复数zz3i|z|(A.
C.2
D.23.
为弘扬中华民族传统文化,某中学生会对本校高一年级1000名生课余时间参加传统文化活动的情况,随机抽取50名学行调查,将数据分组整理后,列表如下:参加场数246参加人数占调查人数的百分比8%10%20%26%18%12%
4%2%估计该校高一学生参加传统文化动情况正确的是参活动次数是3场生约加活动次数是2场4场生约为人加活动次数不高于场的生为280参活动次数不低于场的生为360
.4.知双曲线
:
2ya2b2
0)
,直线与
的两条渐近线的交点分别为M
,
为坐标原点.若
为直角三角形,则C
的离心率为(A.
2
B.
3
C.
2
D.
55.知数列
a中n3
=2,7
数列}为数列,则a=(9nA.
12
B.
54
C.
45
D.
456.知
sin(
6
)
1,且,22
)
(第共23页
已知函数xf
为
f
C.1的导函数,则函数f
的部分图象大致为).ABD
在边长为
的等边中,点M满MA,则A
B.2
C.6
D.
如图,线段MN是半径为的圆O的条弦,且MN的为2在O内将线段MN绕点按逆时针方向转动,使点M移到圆上的新位置,继续将线段NM绕M点逆时针方向转动使移到圆上的新位置依继续转动点M的迹所围成的区域是图中阴影部分若在圆O随机取一点,则此点取自阴影部分内的概率为(
4
-63
C.
3310.已知函
x,fx2
,当
f
恒成立,则实数的值范围是()
.
C.
11.已知
FF1
为椭圆
4
的左焦点P是椭圆上异于顶点任意一点点是
1内切圆的圆心,过F作MPK1
于M,O
是坐标原点,则OM的取值范围为(A.
3
3.12.如图,长为1正体ABCDABD的块平面过点D且行于平面ACD,则木块在平面内正投影面积是(
32
C.
第题第页共23页
第卷本包必题选题部.~21题必题每试考都须答第、23题选题考根要作.二、填空题:本大题题,每小题5分,共分.把答案填在答题卡相应位置.13.若实数y
满足约束条件
xyxy
,则zy
的最小值等于.
xy14.知长方体ABCDCD的外接球体积为
A
直线A与面C所成的角为______.15.函数()xcos象,则.
Ra的象向左平移个位长度,得到一个偶函数图16.知数列
项为S,
a,Sn
(为常数数
n,且n
n
,则满足条件的的值合_.三解题解应出字明证过或算骤17.本小题满分分)在中=90点D,分在边,BC上,CE且的积为3.(1求边DE长;(2若AD,sin的.第页共23页
18.本小题满分分)峰谷电是目前在城市居民当中开展的一种电价类别.它是将一天小划分成两个时间段,把8—22共小时称为峰段,执行峰电价,即电价上调:—次:共10个时称为谷段,执行谷电价,即电价下调.为了进一步了解民众对峰谷电价的使用情况,从某市一小区机抽取了50户户进行夏季用电情况调查,各户月平均用电量以[100,300),,700),[700,度分组的率分布直方图如下:若将小区月平均用电量不低于700度住户称大户月均用电量低于度的住户称为一般用户.其中,使用峰谷电价的户数如下表:月平均用电度使用峰谷电价的户
9132数(1估计所抽取的户的月均用电量的众数和平均数(同一组中的数据用该组区间的中点值作代表(2i)“般用”和大户的数填入下面2的联表:一般用户
大用户使用峰谷电价的用户不使用峰谷电价的用户(ii根据(i)的列联表,能否有99%的握认为“用量的高低与使峰谷电有关?第页共23页
附:K
())(a)(b)
,
Dk
MFA
BE19.(本小题满分分)如图,四棱锥E,面平ABE,边为矩形,AD=6BE,F为CE上的点,且BF面.(1求证:BE;
,AB=5
,(2设M在段,且满足EMMD试在线段上定一点N,得MN
平面
,并求
的长20.本小题满分分)已知抛物线
1
:
py(
和圆
x+1+y
,斜角为
的直线
l
过
1
的焦点且与相.(1求
的值;(2)
M
在
C
的准线上,动点
A
在
1
上,
1
在A
点处的切线
l
交
轴于点B
,设MAMB求证:点N在定直线上,并求该定直线的方程.21.本小题满分分已知函数
f()ln
x
(R)
.第页共23页
(1求函数
fx)
的单调区间;()
e时,关于的程
f(
有两个不同的实数解
,求证:x11
(二选考题共分请生第、题中选题答如果做则所第个目分22.[选修4:标系与参数方程](10分在直角坐标系xOy
中,直线l参数方程为
1xtyat
(t为数,aR.坐标原点为极点,x
轴正半轴为极轴建立极坐标系的坐标方程为
交于
O,
两点,直线
l
与曲线C
相交于
AB
两点(1求直线l的通方程和曲线的直角坐标方程(2当ABOP时求a值.选修4
:不等式选讲]分已知不等式
2x2x
的解集为M(1求集合;(2设实数a,b,明:b.第页共23页
==b==b二
年班质测数(科试参答第卷选题本题12小题每题5分共60分在小给的个项,有项符题要的1.D5.C6.C7.A12.A13.已知集
A
B
=(
C.
满足(3+i),则【简解】B14.设复数z
,所以A
B
,故选D.
B.
C.2
2【简解一】因为
z
33+i
=
,所以z
,故选【简解二】因为(3+i)z,所以(3+i)(3+i)=,以z,选B.15.
为弘扬中华民族传统文化,某中学学生会对本校高一年级1000名生课余时间参加传统文化活动的情况,随机抽取名生进行调查,将数据分组整理后,列表如下:参加场数
13456参加人数占调查人数的百分比20%26%%
4%估计该校高一学生参加传统文化活动情况正确的是参活动次数是的学生约为人参活动次数是2场或场学生约为人C.参加活动次数不高于场的学生约为280人参活动次数不低于场的学生约为人
.【简解】估计该校高一学生参加活动次数不低于场学生约为:+0.12+0.04+0.02)=360
人,故选16.已知双线C:
x2ba2
,直线与C的条渐近线的交点分别M,
,第页共23页
,且33,,且33,O
为坐标原点.若
为直角三角形,则C的心率为(
2
C.
【简解依意得因直角三角形,所以双曲线的近线为
=
,即C是轴双曲线,所以的心率
e
2
,故选A17.已知数
{}中a3
=2,
.若数列{}为差数列,则=(
12
54
C.
45
45【简解】依题意得:
a
,因为数列{}为等数列,n所以
a17,所以978aa9
,所以
a
45
,故选C18.已知
,且cos()2C.
(
【简解一】由
10,62
π得,,代入cos3
得,
=
,故选【简解二】由
π122
得,
π
,所以
π
πcoscos6666
,故选.19.已知函f
为
f
的导函数,则函数f
的部分图象大致为).第页共23页
ABD【简解】依题意得:f
sincos为奇函数,排除D,设g()则g
x
,除
B
,故选A20.在边长3
的等边中点M满足BMMA,则CMA
32
B.2
C.
D.
【简解一】依题意得:2112CMCBCA)CA,选.333232【简解二】依意得:以C为原点,CA所在的直线为轴建平面角直角标,则53),),25所以CM,,故选D.22【简解三】依题意得:过M点MDAC于D,图所示,则
B
MCM(3cos60)
,故选D.
C
D
A21.如图,段MN是半径为2的的条弦,且MN的为2在O内将线段MN绕点按逆时针方向转动,使点M移到圆上的新位置,继续将线段NM绕M点逆时针方向转动使移到圆上的新位置依继续转动点M的迹所围成的区域是图中阴影部分若在圆O随机取一点,则此点取自阴影部分内的概率为(
4
-63
C.
33【简解一】依题意得:阴影部分的面积
1S(6
]=432P
3
,故选B.【简解二】依题意得:阴影部分的面积
S
3P
3
,故选B.第页共23页
12221212221222.已知函
f
,
,当
f
恒
成立,则实数的值范围是()
.
C.
【简解】依题意得:函数
f
在
上单调递减,
因为
f
,所以
,即
2x,以2(
,即
,故选B23.已知
FF1
为椭圆
4
的
左、右焦点,P是圆上异于顶点的任意一点点是
PF
内切圆的圆心过F作FMPK11
于M,
是坐标原点
OM的取值范围为(A.
3
3【简解】如图,延长
P,FM
相交于N
点,连接M
,因为K点是
PF
内切圆的圆心,所以PK
平分
PF12
,∵∴
FM,PNPF,M
F
中点,∵为FF中,为FN中,21∴
1PFPFPFFF32
,∴OM的值范围为24.如图棱长为正体ABD的块面过
点D且平行于平面
,则木块在平面内的正投影面积是
(
32
C.
第题第10页共23页
A2minA2min【简解长1正体BCD的块的三个面在平面内正投影是三个全的菱形(如图可以看成两个边长1222=3.22
的等边三角形,所以木块在内的正投影面积是第卷本包必题选题部.~21题必题每试考都须答第、23题选题考根要作.三、填空题:本大题题,每小题5分,共分.把答案填在答题卡相应位置.13.
14.
16.
13.若实数y
满足约束条件
xyxy
,则zy
的最小值等于______
xy【简解】依题意,可行域为如图所示的阴影部分的三
角形区域目函数化为:
x
则
的最小值即为
动直线在轴的截距的最大值.通平移可知在点
动直线在轴的截距最大:
xx
解得z
1,所以1722
x.
的最小值22.知长方体ABCDCD的外接球体积为
A
直线A与面C所成的角为_.第11页共23页
00【简解】设长方体ACD的接球半径为
,因为长方体ABCDBCD的接球体积为
3所即C=2BC2ABR,因为BC所以AB2
因为平面BBC,以C平面所成的角为,在eq\o\ac(△,Rt)ACB中,因为11
BC2,以B2,以ACB=11
.23.函数()xcos
Ra的象向左平移个单位长度,得到一个偶函数图象,则
______.【简解】因为
f(x)cosR
的图象向左平移
单位长度,得到偶函数图象,所以函数
f(x)cos
的对称轴为x
,b所以f()sin=f(0)=b因为,以.3a24.知数列
的前n项为,
a,且(为数.数列
满足n
n,b
,则满足条件的的值集合_.【简解】因为
,且S1n
a(常n所以
a
,解得
,所以
a,所以S
a
,所以
a
,所以
an
n
,因为
ab
,所以
2
,所以b+1
nn+28(nn2n
,解得
4
,又因为N
*
,所以或n.所以,当n=5或n=6时,
n
n
,即满足条件的的值集合为三解题解应出字明证过或算骤第12页共23页
25.本小题满分分)在中=90(1求边DE长;
点DE分在,BC上,,的积为.(2若AD
,求sin的值.(1析】如图,在△中
CEsinDCEDCE6所以DCE
25
,·········································································2分因为90
所以cos1
26=,··························································分55由
余
弦
定
理
得CECD2528,
DE.········································································7分(2因为ACB=90
,所以ACD
DCEcosDCE=,··········在ADC,正弦定理得3即15
ADCD,sinsin所以sinA.·············································································12分26.本小题满分分)峰谷电是目前在城市居民当中开展的一种电价类别.它是将一天小划分成两个时间段,把8—22共小时称为峰段,执行峰电价,即电价上调:—次:共10个时称为谷段,执行谷电价,即电价下调.为了进一步了解民众对峰谷电价的使用情况,从某市一小区机抽取了50户户进行夏季用电情况调查,各户月平均用电量以[100,300),,700),[700,度分组的率分布直方图如下:第13页共23页
若将小区月平均用电量不低于700度住户称大户月均用电量低于度的住户称为一般用户.其中,使用峰谷电价的户数如下表:月平均用电度使用峰谷电价的户
9132数(1估计所抽取的户的月均用电量的众数和平均数(同一组中的数据用该组区间的中点值作代表(2i)“般用”和大户的数填入下面2的联表:一般用户
大用户使用峰谷电价的用户不使用峰谷电价的用户(ii根据(i)的列联表,能否有
的把握认为“用量的高低与使峰谷电有关?附:K
())(a)(b)
,
第14页共23页
【】(1根据频率分布直方图的得到度到300度的频率为:0.001200200
,·············2分估计所抽取的50的月均用电量的众数为:
500+700
=600度·························3估
计
所
抽
取
的
50
户
的
月
均
用
电
量
的
平
均
数
为:0.00056008000.001210000.0006200640(度··············································································································分(2依题意,列表如下使用峰谷电价的用户不使用峰谷电价的用户
一般用户
大用户··················································································································8分K
的观测值
50400k3563
·····························11所以不能有
99%
的把握认为用电量的高与使峰谷电”有关.·······················12分27.(小题满分12分如图,四棱锥ABCD,面面ABE,边形ABCD为形,=6,AB=5,BE=3,F为上点,且平面(1求证:BE;(2设在线段DE,且满足EM,在线段AB上
D
确定一点N,使得//
平面BCE
,并求
的长
MF(1析】证明:
四边形ABCD为形BCAB
.
A
B平面与平面ABE平面ABCD
与平面=,
E
且
平面ABCD,
平面ABE.··························································································分又AE面ABE,BC
.···································································································2分第15页共23页
面ACE面ACE,BFAE.···································································································分又
BFB
,
平面
BCE
,BF面
BCE
,面
BCE
,···························································································4分平面,AEBE
.···································································································5分(2一中点作MG/AD
交于点ABE中作//交AB于
N
点,连
MN
(如图),·····································································6分EM,EGBN
.//,平面BCE,面BCE
,NG/
平面
BCE
.·························································································7分同理可证,
//
平面
BCE
.MG
,
平面
MGN//
平面BCE,···············································································分又
MN
平面
MGN
,MN/
平面BCE
,························································································分N
点为线段上靠近点的一个三等分点.················································10分,AB,=3MG
AD
,······························································11分MNMG
NG
.···························································分(2法二M点MG/CD交于G接在取点使得MG连MN(如图),·····························································································分
/CD
EMMD2MG//CD,CD3
AB/CD,BN,//BN,MGBN,················································································7分四边形
是平行四边形,第16页共23页
11//
,··································································································分又
MN面BCE,平面BCE,MN/
平面BCE
,························································································分N
点为线段上靠近点的一个三等分点,···············································10在△CBG,
AD
=CE63
cos
255
,··················11分BG25
255
,17
.·························································································12分28.本小题满分分)已知抛物线:1点且与C相.(1求的值;
pyp0)和圆C+12,倾斜角为的直线l过C的21(2)在C的线上,动点1
A在上,C在点的切线l交12
轴于点B,设MAMB求证:点在定直线上,并求该直线的方程.(1【解析题设直线
l
的方程为y
,···················································1分由已知得:圆C+1+
的心
(
,半径r2···························分因为直线l与圆相,12所以圆心到直线l:y的离
|1
2.···································3分即
p||22
,解得p或p舍去··············································分所以p.··································································································5(2法一依意设
(m,
(知物线
1
方程为
所y所y12
x,设()
,则以
为切点的切线l的斜率为k2
x1
,·················································6分所以切线l的程为y2
x).································································分第17页共23页
x12111x12111令,yx2y=,l交y
轴于点坐标)
,········分所以MAxmy3),·····················································································分MB3),···························································································分∴MBm
,··············································································分∴
OMMNx,3)
设N点坐标为
(,y)则,所以点N在直线上················································································12(2解法二:设(m,()知抛物线
C
方程为x
y,设A(x,),以为切点的切线l的方程为(x)②,联立①②得:x
]
,···································································6分因为k
kx
,所以k=,所以切线l的程为2
x(x).································································7分111令x,切l交y轴的点标为),·····················································分2所以MAxmy3),·····················································································分MB,··························································································10分∴MNMA=(xm
·················································································11∴
ONMN
,设点标为
(,y,则y,所以点N在直线上················································································1229.本小题满分分已知函数
f()ln
x
(R)
.(1求函数
fx)
的单调区间;()
e时,关于的程
f(
有两个不同的实数解
,求证:第18页共23页
xx212xx212x11
(1析f()的义域为(0,··································································1分f
ax)]x2x2x2
,···································分①当
a时即a
时,在)上f
在(1
,所以f()的调递增区间是)
上,单调递减区间是(1
;··························3分②当1
,即a
时,在0
,所以,函数f()单递减区间是(0,无递增区.··············································分(2证明:设
gxf(ax)+
a
=a+lnx············································5分所以
g
a(1)
x
,··············································································分当
时,
,数g)
在区间当x时
,数(x)
在区间
···································分所以
(x)
在x=1
处取得最大值.当
e时方程
f(
有两个不同的实数解
,2所以函数
(x)
的两个不同的零点
x,1
,一个零点比1小一个零点比1大·················8分不妨设
0x12
,由
(x),g()得xln(),且x=ln()122
,······································分则
=1
11e=ex=a2
xx
,···························································10分所以
xx1ex+1=x1212
,令
+=t1
,(t)=
et
,
eet(ttt2
.第19页共23页
2222t+x1212
,
所以
,·································································································11所以函数
在区间
上单调递增,
(t
,所以
xxe(+x)e1=xa224411
,又因为
x+x所xx12
.································································12分(二选考题共分请生第、题中选题答如果做则所第个目分22.[选修4
:坐标系与参数方程](分)在直角坐标系xOy
中,直线
l
1xt的参数方程为(yat
t
为参数,
aR
.坐标原点为极点,x
轴正半轴为极轴建立极坐标系
的极坐标方程为4cos
与曲线C交于P两,直线l与线相于AB两点(1求直线
l
的普通方程和曲线的直角坐标方程;(2当
时,求
的值【解析)直线l参数方程化为普通方程为
xy
··························2分由
4cos
得
,···································································从而
yx即曲线的角坐标方程为
··························(2解法一:由,
所以OP,·····························································································将直线l的数方程代入圆的方程
x
2xy
,t
23t
由
,得
2
··································································第20页共23页
33设A、两对应的参数为
t,
,则
t
tt43a22
·····································9分解得,a0
或
3
所以,所求的为0或43.·········································································10分解法二:将射线
化为普通方程为
,······················6分由()知,曲线C:
2
的心C
,半径为2,由点到直线距离公式,得C
到该射线的最短距离为:
33
,所以该射线与曲线C
相交所得的弦长为
OP2
.·························7分圆心C
到直线
l
的距离为:
233
22
,··············································由
,得
12
,即
3
,····················9分解得,a0
或
3所以,所求的为0或43.·········································································10分选修4
:不等式选讲]分已知不等式
x
的解集为M(1求集合;(2设实数a,b,明:b.【解析)法一:当
x
12
时,不等式化为:
,
,所以
12
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