2018屆高考數(shù)學一模試卷(文科)

2018

12560

1U={1234}A={12}B={24}?UAUB=A{2}B{3}C{124}D{14}

2zzi2i=5z

ABCD

3py=fxqy=fxpq

AB

CD

4.1,2,3,4,5

A0.3B0.4C0.5D0.6

5.x,y1x-y-2<0z=2x+3y

A10B9C8D7

6{&}nSnS12S23S3{an}

A2B3CD

72

1

f

\

X

J

A.(1)nB.('■+2)nC.('■+3)nD.('■+4)n

8Cy2=2pxp0FACAF

AA

9.

y

1B

OAF

2C3D4

f

1O

x=Asin

f

f

p

jr

A03>0|

f0

~T

=1

'

A.1B.,C.

10.

7D.：

13

x=x3+x+1

x=1c=1d=0.1x

A0.6B0.69C0.7D0.71

11.fx=|2x2|+bX1X2X1X2

A.1vX1<2，X1+X2V2B.1vX1<2，X1+X2V1

C.X1>1，X1+X2V2D.X1>1，X1+X2V1

12.ABCDBCCDRtABCD1ABCD

AB

16n

ABCD

A'B.'C1D.-

333

4

13.=1x*=x1

14.fx=a,fa

4------

15.{an}dv0S10=20d=__

16.

abZ

ABAOB=120O

520

|；|?x=_________

a

nSn3S9

C

=1a0b0C

C

17.12

sinA

ABC

18.

ABC

33

A,BC

b

a,b,c3a=5csinAcosB=

12

1844

4646525455565758586366707375859094

51525558636365696970747877778383

89100

60

60

2X2

95%

P(K2

k

k

0.050

3.84i

0.010

6.635

0.001

10.828

n(ad-bc)2

@+b)(c+d)c)(b+d)

19.12

CA=CB

AiBC

ABC

AiB

ABCi

AiBiCi

ACi

ACCiAi

ABC

ABC

AiAC=60CA=2

20.i2E：J

J

2x=0

E

CD

ABCD

AB

AiBiBC=iab

ab

y=x+2,

0

AB

E

x2+y2

ABCD

21.

i2

fx

x0

fx=x+2

1

ax+1

a

4-4

10

[

22.

C:p-4pcos^=0

C

IC

]

xOy

I:

0

(t

I

x

OWaVn).

{y=tcos

[

23.

4-5

fx

x

]

=|x|

|x

fx

1|

|m

1|

m

M

1Mkaba2+b2=ka+b

2ab.

2018

12560

1U={1234}A={12}B={24}?UAUB=A{2}B{3}C{124}D{14}

AUB2

AUB={124}CUAUB={3}B

2zzi2i=5z

ABCD

(2i=5

zi

Zz

zi2i=5

32i(6+2i)(2+i)_10+1

z

22

A

3.py=fxqy=fxpq

AB

CD

q?p

R

P

q?p

pq

B

fx=x12

y=fx

q

fx=x12

y=fx

R

4.

1,2,3,4,5

A0.3B0.4C0.5D0.6

n=1=10

12345

n==10

12233445

m=104=6

p=.'-|-.

D

x+2y~5V0

5.

x,y

z=2x+3y

B

A10B9C8D7

x+2y5=0xy2=0x=0

z=2x+3y

B

31

::,

9

6.

{》}n

S

n

Si

2S2

3S3

{an}

A2B

x2S2

x2S2,

3CID

n

{an}

4aa2+a3=4ai+a2

{an}

???4ai+a?+a3=4aa2

qSi2S23S3Si+3S3=2

q???Si2S23S3Si+3S3=2

a3=3a2q=3

7.

21

kJ

A.(+1).(+2)nc.(+3)D.(-+4)n

S=nX12+2nX1X1X213+■)n

C

8.

y

A

C

1B2C

y=2px

3D4

p0

OAF

2

F

1

AC

O

AF

p

A

OAF

AC

1

Aab

AF

p

b2=2pa

A

1

y

=1a+2a

VI

p=2

B

IT

9.fx=AsineXA0e>0||

=1

(0

f

A1B.C

：

y=Asin(3X

Lj1'

TC

...f(

=Asin(+

=sin*=1

A

3，

f0

=Asin(ex?)

fx

(A>0，e>0，|?<—)

A

=Asin3x+.

fx

-->72

12i2

,n,n

3?.

37t7T

A?=1

=Acos

A='Uf0

y=Asin(ex?

e,

A

10.

13

f(x=x3+x+1

x=1c=1d=0.1x

r

A0.6B0.69C0.7D0.71

x

x=1f1=1v0cdx=1+0

x=0f0=10x=01c=0.dx=0.9

3x=0.9,f0.9)v0x=0.8

3x=0.8,f0.8)v0x=0.7

4x=0.7f0.7)v0x=0.6,

5x=0.6,f0.60x=0.7c=0.01vd

0.7

C

x

11.

fx=|2x2|+b

X

1

X

2

X

1

X

2

A.1vX1<2,X1+X2V2B.1vX1V2,X1+X2V1

C.X1>1,X1+X2V2D.X1>1,X1+X2V1

；

Fl，

fx=|2X2|+b

X1x2X1x2

y=|2X2|

y=|2X2|

y=

y=

b

b

fx=|2x XiX2y=|2X2|

？

2|+b

X1X2

y=b

I，?X1+X2<2-

y=|2x2|y=b

1xiv2

12.ABCDBCCDRtABCD1ABCDAB16nABCD

124

ABC1D..

ADBCDCBCDCAABCD

ABa=b=AC=2

ABCDV=ABC

ADBCDCBCDCA

J

ABCD

BC=aCD=b,....BD二

RtABCD

7-'iD

a0b0,.°..a=b=

a=b=

ABCD

ABC

AB

16n,AAB=4

ABCD

1

BCCDRtABCD

CECE=1

U；.：,BD==ab

=ab?f2ab2

--

=2

AC=2

2

1

V=

13

4

=1xg=x1

5

20

=|；|

x=：1

x

.1-I,-|L

x=1

1

X

O-|-

2x=x+1

'■I-

-1

J|

14.f

a

4

x

=

1

a,fa

-------

x=0

y=0

y=

Pa,

yy=-，

x=a

a

Y

='-(xa

a,

Iala=1

1

15.{an}dv0nSn3azS9

S10=20d=2

n

d

{an}

Sio=2O

(3^/5)(piai+SdJ

dv0n

n

S

n3

z

a

9

d<0

ai=11d=2

2

16.

C

=1a0b0

C

ab

AB

AOB=120

O

C

(2，+x).

tan60=abc

a

C

AB

AOB=120

y=±"

x

a

tan60=■，

a

b_a,b23a2

c2a23a2

c24a2

c2a

c

e=_2

a

：(2,+x).

17.122018?ABCABC

a,bc3a=5csinAcosB=

5

1sinA

2ABCb

1cosB=B

3sinA=5sinCsinA

sinA=sinB+C

sinA

s"B=&-80si

S

nC==

3a=5csinA

o

cosC=J—；

2

ABC

z

2

XXX

dsinbsinb

(1)tcosB=-

B

13'

sinB=,

3a=5csinA,

sinBsinC

3sinA=5sinCsinAsinA

19

B+C

sinA=sin

=sinBcosC+cosBsinC=_

13

2a___________cA

sinA_sinB_sinCaainBc“nB

1

33

5

=…―X

b265

ABC

b=10

0cosC=

d333

513565

C

331

X

.T1x/

1bsinAbsinC

■XsinB=X-

13

o

18.12

2018?

18

1

44464652545556575858636670737585

9094

51525558636365696970747877778383

89100

1

26060

P(K2k0.0500.0100.001

2n(ad^bc)

(a+b)(c+d)(a+c)(b+d)

k3.8416.63510.828K2=

2X295%

2

2X2

K2

1

887654

53

4

10

58

59?

877

51100

4494

5183

4666

4.0533.841

(22X2

18

18

36

10

4

14

8

14

22

=

95%

19.122018?ABCA1B1C1ACC1A

1

ABCABCCA=CBA1BAC

1

A1BCABC

1

A1AC=60CA=2A1B1BC

1ACBCBCACCiABCACi

AiB

2

ACi

UACi

AiBiBC

AiBC

-bjL

AiBC

ABCi

ivACCiA

CA=CB

???ACBC，vACCiAiA

ABC

ABC=AC

ABC

???BC

vAiBACi

vACi?ABCi

ACCiAvACi?BCAAiB=B

AiBC

ACCiAiACi

ABCi

BC

AiBC

ACi

：(2)vBC//BiCi，ACiAiBC，

Bi

AiBC

d=,.ACi

vABC

ACCiAi

=HW=

CA=CB=2AiAC=60

2d="=.

-J=2，

AiBiBC

ACi

AiBC

AiC=2

20.

x2+y2

E

ABCD

AB

2x=0

C

E

D

122018?

=1ab0

￡a

AB

E

ABCD

y=x+2,

1

C

a

b2=a2c2

2

I

2AB+d

AB

1)vE+7=1ab0

2a

AB

m

x2+y22x=0x2+y22x=0

C邁

Ia=b+c

E

I

2

1

a=T^

2

0

b=c=1

y=x+mAX1y1BX2y2

x223x2+4mx+2m22=0

丁+y=1

=16m24x32m22=2m2+30:vm<:

AB=

2

2

2',

2AB+d

0

I….=ABCDd==-'：：ABCD

3

UI:

21.122018?fx=x+2fx

x0

1

ax+1

a

1fx

2aa

1f'x=x+3

f'x0x3f'x)v0

f(x-x,-3)(-3,+x)

2av0

U■1exv0：L

x

aax+1

a0gx=x+1exax1

g'x=x+2exa

xv-3

i

ax+1

1gx0ax+1

x

g

a2

0

x

x+2ex2 [0+x)

1gx

0wagx

[0+

2

g

g'x

0=0

g

0

0=2av0

g'(a-2=aea210

-

X00vxvX0

gxvg0=0

a[02]

g

g

x=0

x)v0

[0+x)

gx0x°

1

x0X

[

22.

4-4

10

。

]

2018?

C

C

IC

xOy

2

p-4pCOS401=0

I

0

I

x

fy=4+1sind

r(t

y=tc

C

1

C2