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2020年天津一中高一(下)期中数学试卷

2020年天津一中高一(下)期中数学试卷
2020年天津一中高一(下)期中数学试卷

期中数学试卷

题号一二三总分

得分

一、选择题(本大题共10小题,共30.0分)

1.以下说法正确的有几个()

①四边形确定一个平面;②如果一条直线在平面外,那么这条直线与该平面没有公

共点;

③过直线外一点有且只有一条直线与已知直线平行;④如果两条直线垂直于同一条

直线,

那么这两条直线平行;

A. 0个

B. 1个

C. 2个

D. 3个

2.在△ABC中,角A,B,C的对边分别是a,b,c,且,则角

A的大小为()

A. B. C. D.

3.在△ABC中,若且∠BAC=30°,则△ABC的面积为()

A. B. C. D.

4.设α、β、γ为平面,为m、n、l直线,则下列判断正确的是()

A. 若α⊥β,α∩β=l,m⊥l,则m⊥β

B. 若α∩γ=m,α⊥γ,β⊥γ,则m⊥β

C. 若α⊥γ,β⊥γ,m⊥α,则m⊥β

D. 若n⊥α,n⊥β,m⊥α,则m⊥β

5.某三棱锥的三视图如图所示,其俯视图是一个等腰直角三角形,则此三棱锥的体积

为()

A. B. C. D.

6.点G为△ABC的重心,AB=2,BC=1,∠ABC=60°,则=()

A. B. C. D.

7.在正方体ABCD-A1B1C1D1中,点O是四边形ABCD的中心,关于直线A1O,下列

说法正确的是()

A. A1O∥D1C

B. A1O∥平面B1CD1

C. A1O⊥BC

D. A1O⊥平面AB1D1

8.一个圆锥SC的高和底面直径相等,且这个圆锥SC和圆柱OM的底面半径及体积

也都相等,则圆锥SC和圆柱OM的侧面积的比值为()

A. B. C. D.

9.平行六面体ABCD-A1B1C1D1的底面ABCD是菱形,且∠C1CB=∠C1CD=∠BCD=60°,

CD=2,,则二面角C1-BD-C的平面角的余弦值为()

A. B. C. D.

10.如图,在△ABC的边AB、AC上分别取点M、N,使,BN与CM

交于点P,若,,则的值为()

A. B. C. D. 6

二、填空题(本大题共6小题,共18.0分)

11.已知向量,满足,,,则=______.

12.如图,PA⊥平面ABC,∠ACB=90°且PA=AC,AC=2BC,

则异面直线PB与AC所成的角的正切值等于______.

13.如图,在直棱柱ABC-A1B1C1中,AB⊥AC,AB=AC=AA1=2,

则二面角A1-BC1-C的平面角的正弦值为______.

14.在△ABC中,角A、B、C的对边分别为a、b、c,2b(2b-c)cos A=a2+b2-c2,则内

角A的值为______.

15.已知正方体ABCD-A1B1C1D1的棱长为1,点E是棱BB1的中点,则点B1到平面ADE

的距离为______.

16.如图,在直角梯形ABCD中,,AB=AD=2.若M、

N分别是边AD、BC上的动点,满足,,

其中λ∈(0,1),若,则λ的值为______.

三、解答题(本大题共4小题,共48.0分)

17.设f(α)=,,其中向量=(),=(2sin,cos)

(1)若f(α)=-1,求cos的值.

(2)在△ABC中,角A,B,C的对边分别是a,b,若a cos B+b cos A+2c?cos C=0,求函数f(A)的取值范围.

18.如图,在几何体中,四边形ABCD是菱形,ADNM是矩形,平面ADNM⊥平面ABCD,

E为AB中点.

(1)求证:AN∥平面MEC;

(2)求证:AC⊥BN.

19.如图1所示,在矩形ABCD中,AB=2AD=4,E为CD的中点,沿AE将△AED折起,

如图2所示,O、H、M分别为AE、BD、AB的中点,且DM=2.

(1)求证OH∥平面DEC;

(2)求证平面ADE⊥平面ABCE;

(3)求三棱锥H-OMB的体积.

20.如图,四棱锥P-ABCD的底面是菱形,PO⊥底面ABCD,

O、E分别是AD、AB的中点,AB=6,AP=5,∠BAD=60°.

(1)求证:平面PAC⊥平面POE;

(2)求直线PB与平面POE所成角的正弦值;

(3)若F是边DC的中点,求异面直线BF与PA所成角的余弦值.

答案和解析

1.【答案】B

【解析】解:①错误,如空间四边形确定一个三棱锥.

②错误,直线可能和平面相交.

③正确,根据公理二可判断③正确.

④错误,在空间中,垂直于同一条直线的两条直线可能相交,也可能异面,也可能平行.故选:B.

利用反例判断①②④的正误;公理二判断③的正误即可.

本题考查科技中心与平面的位置关系的应用,命题的真假的判断,是基本知识的考查.2.【答案】B

【解析】解.由正弦定理得,即

,即,也即,故,

故选:B.

直接利用正弦定理和三角函数关系式的恒等变换求出结果.

本题考查的知识要点:正弦定理和三角函数的值的应用,主要考查学生的运算能力和转换能力及思维能力,属于基础题型.

3.【答案】C

【解析】解:因为在△ABC中,若且∠BAC=30°,

∴c?b?cos A=2?b?c×=2?bc=;

∴△ABC的面积为:bc sin A=××=;

故选:C.

根据数量积和角A的度数,求得bc=;再代入三角形的面积计算公式即可.

本题主要考查数量积的应用以及解三角形的有关知识,属于基础题目.

4.【答案】D

【解析】解:对于A,若α⊥β,α∩β=l,m⊥l,则当m?α时,m⊥β.当m?α时,m与β相交但不垂直,故A不正确;

对于B,α∩γ=m,α⊥γ,β⊥γ,则m与β,可能平行与可能相交,故B不正确;

对于C,α⊥γ,β⊥γ,m⊥α,则m∥β,或m?β,故C不正确;

对于D,若n⊥α,n⊥β,m⊥α,可得α∥β,并且m∥n,所以m⊥β,故D正确,

故选:D.

通过直线与平面垂直与平行,平面与平面的平行与垂直的判断与性质,推出正确选项即可.

本题考查的知识点是利用空间直线与平面之间的位置关系及平面与平面之间的位置关

系判断命题的真假,处理此类问题的关键是熟练掌握线面平行或垂直的判定方法和性质.

5.【答案】B

【解析】解:根据三视图可知,该几何体是一个底面等

腰直角三角形,高为2三棱锥,如图,该三棱锥为

D-ABC,

由三视图尺寸可知:AD=2,即AC=2,AB⊥BC,

则此三棱锥的体积为V=×=.

故选:B.

根据三视图,该几何体是一个底面为等腰直角三角形,

高为2的三棱锥,画出直观图.

本题考查了三视图的认识和尺寸关系,直观图以及几何体的体积的求法,属于基础题.6.【答案】A

【解析】解:以CA,CB为x,y轴,建立

直角坐标系,如图:

点G为△ABC的重心,AB=2,BC=1,

∠ABC=60°,

可得A(,0),B(0,1),则G(,),

=(,),=(-,)

则==-.

故选:A.

利用已知条件,建立直角坐标系,求出相关点的坐标,然后求解向量的数量积.

本题主要考查了平面向量数量积的定义的简单应用,善于利用平面向量的一些常见结论是求解本题的关键.

7.【答案】B

【解析】【分析】

本题考查命题真假的判断,考查空间中线线、线面、面面间的位置关系等基础知识,考查运算求解能力,考查数形结合思想,属于中档题.

推导出A1D∥B1C,OD∥B1D1,从而平面A1DO∥平面B1CD1,由此能得到A1O∥平面B1CD1.【解答】

解:∵在正方体ABCD-A1B1C1D1中,点O是四边形ABCD的中心,

∴A1D∥B1C,OD∥B1D1,

∵A1D∩DO=D,B1D1∩B1C=B1,

∴平面A1DO∥平面B1CD1,

∵A1O?平面A1DO,

∴A1O∥平面B1CD1.

故选:B.

8.【答案】C

【解析】解:如图,

设圆锥与圆柱的底面半径为r,则圆锥SC的高为2r,

设圆柱的高为h,由体积相等可得,可得h=.

∴,.

则圆锥SC和圆柱OM的侧面积的比值为.

故选:C.

设圆锥与圆柱的底面半径为r,则圆锥SC的高为2r,设圆柱的高为h,由体积相等可得h=,分别求出圆锥与圆柱的侧面积,作比得答案.

本题考查圆柱、圆锥的侧面积与体积,考查计算能力,是基础题.

9.【答案】D

【解析】解:取BD中点O,连结CO,C1O,

∵平行六面体ABCD-A1B1C1D1的底面ABCD是菱

形,

∠C1CB=∠C1CD=∠BCD=60°,CD=2,,

∴CO⊥BD,C1O⊥BD,

∴∠COC1是二面角C1-BD-C的平面角,

∵C1B=C1D==,

∴C1O==,CO==,

∴二面角C1-BD-C的平面角的余弦值为:

cos∠C1OC==.

故选:D.

取BD中点O,连结CO,C1O,推导出CO⊥BD,C1O⊥BD,∠COC1是二面角C1-BD-C 的平面角,求出C1B=C1D=,从而C1O=,CO=,由此利用余弦定理能求出二面角

C1-BD-C的平面角的余弦值.

本题考查二面角的余弦值的求法,考查空间中线线、线面、面面间的位置关系等基础知识,考查运算求解能力,是中档题.

10.【答案】D

【解析】【分析】

本题考查向量知识的运用,考查平面向量基本定理,考查学生的计算能力,属于中档题.用,作为基底分别表示,根据平面向量基本定理,求出λ,μ,即可得到结论. 【解答】

解:由题意,=

==.

=+

=+

=

根据平面向量基本定理,

可得,

∴,

∴=6.

故选D.

11.【答案】2

【解析】【分析】

本题考查了向量的数量积的运算和向量模的计算,属于基础题.

向量的数量积的运算和向量模即可求出答案.

【解答】

解:∵,,,

∴|+|2=||2+||2+2?=5,

∴2?=5-4-1=0,即?=0,

∴|2-|2=4||2+||2-4?

=4+4=8,

∴|2-|=2,

故答案为:.

12.【答案】

【解析】解:过A作AD∥BC,过B作BD∥AC,则四

边形ACBD为矩形,

即AC∥BD,

故直线PB和直线AC所成的角为∠PBD,

不妨设BC=1,则PA=AC=2,

由于PA⊥平面ABC,故PA⊥AC,PA⊥AD,PA⊥AB,

所以.

由于BD⊥AD,BD⊥PA,

所以BD⊥平面PAD,

所以BD⊥PD.

在直角三角形PBD中,

故答案为:.

先将三棱锥补成四棱锥,再由异面直线所成角的作法可得:直线PB和直线AC所成的角为∠PBD,再在直角三角形PBD中求解即可.

本题考查了异面直线所成角的求法,属中档题.

13.【答案】

【解析】解:在直棱柱ABC-A1B1C1中,AB⊥AC,

AB=AC=AA1=2,

以A为原点,AB为x轴,AC为y轴,AA1为z轴,建

立空间直角坐标系,

A1(0,0,2),B(2,0,0),C(0,2,0),C1

(0,2,2),

=(-2,0,2),=(-2,2,2),=(-2,2,0),

设平面A1BC1的法向量=(x,y,z),

则,取x=1,得=(1,0,

1),

设平面BCC1的法向量=(a,b,c),

则,取a=1,得=(1,1,0),

cos<>===,

∴二面角A1-BC1-C的平面角的正弦值为:=.

故答案为:.

以A为原点,AB为x轴,AC为y轴,AA1为z轴,建立空间直角坐标系,利用向量法能求出二面角A1-BC1-C的平面角的正弦值.

本题考查二面角的正弦值的求法,考查空间中线线、线面、面面间的位置关系等基础知识,考查运算求解能力,是中档题.

14.【答案】60°

【解析】解:因为2b(2b-c)cos A=a2+b2-c2,

由余弦定理可得:2b(2b-c)cos A=2ab cos C,

整理可得:(2b-c)cos A=a cos C,

由正弦定理可得2sin B cos A-sin C cos A=sin A cos C,

即2sin B cos A=sin C cos A+cos C sin A=sin B,B∈(0,π),

所以sin B≠0,

所以cos A=,A∈(0,π),

所以A=60°,

故答案为:60°.

由余弦定理可得a2+b2-c2=2ab cos C,再由正弦定理可得cos A=,再由A的取值范围可得

A的值.

本题考查正弦定理及余弦定理的应用,属于中档题.

15.【答案】

【解析】解:以D为原点,DA

为x轴,DC为y轴,DD1为z

轴,建立空间直角坐标系,

B1(1,1,1),A(1,0,0),

D(0,0,0),E(1,1,),

=(1,1,1),=(1,0,

0),=(1,1,),

设平面ADE的法向量=(x,y,

z),

则,取y=1,得=(0,1,-2),

∴点B1到平面ADE的距离为:

d===.

∴点B1到平面ADE的距离为.

故答案为:.

以D为原点,DA为x轴,DC为y轴,DD1为z轴,建立空间直角坐标系,利用向量法能求出点B1到平面ADE的距离.

本题考查点到平面的距离的求法,考查空间中线线、线面、面面间的位置关系等基础知识,考查运算求解能力,是中档题.

16.【答案】

【解析】解:由图可知:=+=+(1-λ),

==,

又,

所以[+(1-λ)]?()=-2,

所以2+λ(1-λ)=-2,

又=2,=||2=3,

可得:3λ2-5λ+2=0,

又0<λ<1,

所以,

故答案为:.

由平面向量的线性运算得:=+=+(1-λ),==,

由平面向量数量积的性质及其运算得:[+(1-λ)]?()=-2,所以2+λ(1-λ)=-2,又=2,=||2=3,可得:3λ2-5λ+2=0,又0<λ<1,所以,

得解.

本题考查了平面向量的线性运算、平面向量数量积的性质及其运算及向量投影的定义,属中档题.

17.【答案】(1)∵=(),=(2sin,cos-1),

由f(α)=,=-1,得,

∴,即,

∴;

(2)由题意,得f(A)==,

在△ABC中,由a cos B+b cos A+2c?cos C=0及正弦定理知,

sin A cos B+sin B cos A+2sin C cosC=0,

∴sin(A+B)+2sin(A+B)?cos C=0,

又∵sin(A+B)≠0,

∴cos C=-,∵C∈(0,π),∴C=,

∴0<A<,0<<,<<,

∴sin()∈().

∴函数f(A)=sin()-∈(0,).

即函数f(A)的取值范围是(0,).

【解析】(1)由数量积的坐标运算写出f(α),结合f(α)=-1,利用三角恒等变换求解cos的值;

(2)由(1)得到f(A),再由已知等式结合正弦定理求解C,进一步求得函数f(A)的取值范围.

本题考查平面向量的数量积运算,考查y=A sin(ωx+φ)型函数的图象与性质,是中档题.

18.【答案】证明:(1)连接BN,如图,使得BN∩CM=H,

则H是BN中点,

∵E为AB中点,∴EH∥AN,

∵AN?平面MEC,EH平面MEC,

∴AN∥平面MEC.

(2)∵ABCD是菱形,∴AC⊥BD,

平面ADNM⊥平面ABCD,

平面ADNM∩平面ABCD=AD,ADNM是矩形,ND⊥AD,

ND?平面ADNM,∴ND⊥平面ABCD,AC?平面ABCD,

∴AC⊥ND,BD∩ND=D,BD,ND?平面NDB,

∴AC⊥平面NDB,又BN?平面NDB,

∴AC⊥BN.

【解析】本题考查线面平行、线线垂直的证明,考查空间中线线、线面、面面间的位置关系等基础知识,考查运算求解能力,是中档题.

(1)连接BN,使得BN∩CM=H,推导出EH∥AN,由此能证明AN∥平面MEC.

(2)推导出ND⊥AD,从而ND⊥平面ABCD,进而AC⊥平面NDB,由此能证明AC⊥BN.19.【答案】证明:(1)取F为BC的中点,连OF、FH,

∵O、F分别为AE、BC的中点,

∴OF∥EC,

∵OF?面DEC,EC?面DEC,

∴OF∥面DEC,

同理可证,HF∥面DEC,

OF∩HF=F,

∴面HOF∥面DEC,又OH?面

HOF,

∴OH∥平面DEC;

(2)∵AD=DE=2,且点O是

AE的中点,

∴DO⊥AE,DO=,

∵M为AB的中点,∴OM=,

且AE⊥OM,

又∵DM=2,∴DO2+OM2=DM2,

∴DO⊥OM,

∵DO⊥AE,AE∩OM=O,∴DO⊥平面ABCE,

∵DO?平面ADE,∴平面ADE⊥平面ABCE;

解:(3)由(2)知,DO⊥平面ABCE,

∴点H到平面OMB的距离是DO=,

则V H-OMB==.

【解析】(1)根据条件取F为BC的中点,连OF、FH,由中位线性质和线面平行的判定证明面HOF∥面DEC,再证明

OH∥平面DEC;

(2)由E和O分别是中点得,DO⊥AE,再由长度和勾股定理证明DO⊥AE,根据线面垂直的判定证明DO⊥平面ABCE,再由面面垂直的判定得结论;

(3)由DO⊥平面ABCE和H是中点,求出三棱锥的高,代入体积公式求值.

本题考查了三棱锥的体积,线面平行的判定,线面垂直和面面垂直的性质、判定,熟练掌握空间直线与平面位置关系的定义、判定定理、性质定理是解答本题的关键,考查了空间想象能力、推理论证的能力.

20.【答案】(1)证明:∵四棱锥P-ABCD的底面是菱形,

∴AC⊥BD,

又∵O,E分别是AD、AB的中点,

∴OE∥BD,

∴OE⊥AC,

∵PO⊥底面ABCD,

AC?面ABCD,

∴PO⊥AC,

又∵OE∩OP=O,

∴AC⊥平面POE,

又∵AC?平面PAC,

∴平面PAC⊥平面POE.

(2)解:过点B作BM⊥OE于M,易证PO⊥BM,OE,OE∩OP=O,

∴BM⊥平面POE,

∴PM是PB在平面POE上的射影,∠BPM即为所求.

∵四边形ABCD是菱形,

∵E是AB中点,

∴BE=AE=,

又∵底面ABCD是菱形,∠BAD=60°,

∴△AOE是等边三角形,

∴∠AEO=∠BEM=60°,

∴BM=,

又∵AP=5,OA=3,

∴在直角三角形POA中得OP==4,

∵在直角三角形POB中,OB=,OP=4,

∴PB==,

∴在Rt△PMB中,,,

所以;

(3)解:取PB的中点T,AB的中点H,连BF,DH,TH.易证DH∥BF,TH∥PA

∴∠DHT即为异面直线BF与PA所成角或其补角,

在三角形PDB中,PD=5,DB=6,PB=,

在三角形PDT中,

所以,

在△DHT中,,,,

,.

【解析】(1)根据菱形的对角线互相垂直及线面垂直的性质得OE⊥AC,PO⊥AC,由面面垂直的判定定理可得平面PAC⊥平面POE.

(2)根据线面所成角的定义找到直线PB与平面POE所成角的,∠BPM即为所求再计算正弦值.

(3)根据中位线得DH∥BF,TH∥PA,∠DHT即为异面直线BF与PA所成角或其补角,结合余弦定理去计算DT,和计算角∠DHT的余弦值.

本题考查了空间中的垂直性质定理和判定定理,也考查了直线与平面所成的角,异面直线所成角,是中档题.

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