2020年高考數(shù)學(xué)文科一模試卷 解析版

x22020高考數(shù)文科一模卷一選題在小給出四選中只一是合目求.x2{04}B{2x∈}∩A{04}B{4}C{24}

D{0.124}iAi

zBi

C1

DaAac

bBc

CcbDcaABCD3mnαβAm∥mβαβC?m⊥α⊥

Bm∥α∥mβD?αβmβ

n4131012212n4131012212A6B5CD3{}a++a20aA4B5CD8EABA4B3D

?

3[1+A

B

C

D10

bc

42×3

A0

B

C

D113a0bFFP4|+

|≥3|

3n84mn12223n84mn1222A1e

B≥

C1e

De123fxlnx+gxaxA

B

CDe二填題將案在題上133xy3xy1415{}a+2a+

163mxFFABx+xtyt

0三解題解應(yīng)出字明證過(guò)或算驟.171C2ABCCD2

18300

0

50100]100250]250

456

11121112

11120182018221000320001012319ABD⊥OAEF1DEF⊥2ABFOBD2HDλ

BAHC20

FCO∠

1CAxN4

1211222a1211222aaAExMNFMF21

1fx2lyaxbFxfxabRa[選修4-4：坐標(biāo)與數(shù)程22xOyCC

COx1l

l

2PPl

[選修4-5：不等選231+1|+|x5|2ababab

x202222x202222參考答案試題解析一選題在小給出四選中只一是合目求.{04}B{2x∈}∩A{04}B{4}C{24}D{0.124}B{1∴∩{4}i

z

AiBi

C1D1∵

∴z1DaAac

bBc

CcbDca0∵a

21loglogcloglog2∴cbC

0ABCD14125128∴A2628B∴CDD3mnαβAm∥mβαβC?m⊥α⊥

Bm∥α∥mβD?αβmβAαmββC

α⊥D⊥βmββA∥mAB∥α∥∥mCmαmβα⊥CDαα⊥mββCA6B5CD3S2n3n4

n4131012n461315110121n4131012n461315110121

an{}a++a20aA4B5CD8{}d∵+a+a+∴4+34da4+3420EAB∠ABC

A4B3

C

D2

2

3ABCD2∠ABC120°ABBDAD2E

2

2

4+2

22222212222222123[1+A

B

C

Dk+ykxykxxy42k∴[12]x4x+4C10bcA0B

42×3C

D

cos105°cos45°cos75°sin75°sin45°cos75°45°

C113a0bFFP4|+

|≥3|

133330000133330000A1e

B≥

C1e

Deeq\o\ac(△,F)eq\o\ac(△,)4|

+8|

|≥3|

|2≥6ceC123fxlnx+gxaxA

B

CDefgyyaf+xxy∴xx∴+xx+ax∴ax+∞yyyaxylnx∴yx∴aax∴xe∴a≤

ylnxyax133xy3xyx

zxyy+yx+A3z3×111114

n84mn1n86224m+1n8624mn1+n84mn1n86224m+1n8624mn1+

115{}a+2a+

{}a+2

aq+2aa

q

2

2m+n{}a+2∴

aq+2a∴q﹣2mn

++n

≥10+2

43∴4

2222AB2222AB2+t0x1+t1212163mxFFABx+xtyt

06pxp0l

B22214∴x+2x++2+m6三解題解應(yīng)出字明證過(guò)或算驟.17

1C2ABCCD2A+C2aABC∵BaA

cos

bB∴sinBBsin?sin

sincosA

∴

A

sin2BB2sin2Aa

2B

∴

+2B

22222222222222B∴C+B

22ADBD∠

2∠∵∠cos∠∴

6ba+b∵ca+bcos∴8a×ab4ABCSab×4×

18300

0

50100]100250]250

456

1112018201822

10003200010123[100201821012

m

[01000.3∴1020181121230.002×50×1010.004×50×1020.006×50×103n15m

6∴3000p19ABD⊥OAEF

ADOHAOFEAOFFADOBADOHAOFEAOFFADOB1DEF⊥2ABFOBD2HDλ

BAHC2V

F

﹣

﹣

FO∵⊥ABCD∴⊥AOABCDAO⊥∴⊥BDF∵EFAO∴EF⊥BDFDEF∴⊥2V

D

D﹣

2V

﹣

ADOBAHCD

F

﹣

﹣

H﹣

∵

∴

∴1λFO2∵ABFOBD∴

1112121211212211121212112122222221221211120

FCO∠

1CAxN4AExMNFMFa2ANkMBF|∴AF|+|BF2a4a2∠∴b

cb+ca4∴ab

c1∴

2ANx

3+4k32kxAxx+∵x

x

12101210∴Ey0x

+∴1∴MF0∴NF|MF021

1fx2lyaxbFxfxab∈a+2a+x

+01f0xf0xx01+2Fxfl+

1122212222212222211222122222122222hx01x01hxxhx+∞x1+[選修4-4：坐標(biāo)與數(shù)程22xOyCC

COx1l

l

2PPl

CααCy1xxρρ0l

2l

1CCαα+y1x+0ρρcosl

4+

02sin

ρcos12x+2+1

2aaabaab2aaabaab[1（]a[1bbaabbabaaba