2020年北京市高考數(shù)學試卷(潘裕老師審校)

2020北京市高考數(shù)學試卷一、選擇題：共小，小4分共。每題出的個項，出符題要的項A2}Bx0

B

){1}

{0

{2}

{12}(1,2)iz

)1i

1i

(x2)

x

()

()

6

3

123(3,4)()

f(x)f(x)()((((1OFlFQ()

OPPQl

OP

OP{}aaa({T}()

RkZ

(

sin()

Day)

f()f()f()f()n66邊2()

(sin3(sin

3030)ntan)606060tann)nnn二、填空題：共5小題每題5分共25分f

(x)lnxx

2:CABCDP

AP(ABAC)

PD|

PDf(x)x

)cosxtWft)，[a][tt]tt[0t][t][t][0t]三、解答題：共6小題共85分。解應出字明演步或明程ABCDACD//ADEAD

11

sinCABC

1cos7

19cosA8

3pp0

f()2yf(yf((f())(t)(t)

22:(abb

aiaiC(l

MAx，|Q||{

}

{}(ij){}aijaj

{}({}aa()l

al

a(){}a){}{}{}

2020北京市高考數(shù)學試卷參考答案與試題解析一、選擇題：共小，小4分共。每題出的個項，出符題要的項A2}Bx0A

){1}

{0

{2}

{12}A2}B{|3}A

D(1,2)iz

)i

1i

i

ii)

(x2)

x

()

r

xrrr5rx(C2()6

3

123

3

1111

133D22(3,4)()

OBAABOB4f(x)(

f(x)()(((12

y(0,1)f(x2

yxyf(x)(0)

DOFlFQ()

OPPQlO

OP2xPQAFP

xF(1,0)lP(1,2)Ql

x

A(1,0)QAFPFQP{

}

a({T}(n)

{a6

}

{

}

da

an2n

112

nN*{}T63T945T{}

(1)(((1)((kZ

()

kknn

sin

)

sin

n

nZ

n

Z

sin

Day)

n66邊2()

(sin3(sin

3030)ntan)606060tann)nnn6na

6nnanb

2sinn12nn

2

6ntan)n)2nnn

二填題共小，小5分，25。f

(x)lnxx

{

0x20x2xx{|0}{x0}

xyC:6

C

2y2:c

c(3,0)

36

y

ABCDPPD

APAB)

|5BC解

1AP(AC)2

D

CPD|2

PDPB(PCPCCD解

AB

1AP(AC)2

PDPBPDf(x)x

)cosx2

)cos

2

2

sin(x

sin

sinsin

cos

2

f(xsinxsinxsincossinx2

cossin

sincos

sinsin

f(t)f(t)tf(t)f(t)tg()()f(x)

cos

2

cos

sin

2

kkWtWf(t)

f()f()

[a][tt]tt[0t][tt][tt][0t]tWtWf(t)tt)[tt]tf(t)f()g()()f()ft)g()g(t)ttf(t)

t(t)tttt

[0

t]

[t

t]

[

t]

[

t]

三解題共小，85分。解應出字明演步或明程。ABCDCDBB//ADEAD

/ADAADABAAAAADEADEcosAAAA|||AAAB/CD//ADBCEADADEBC//ADAADx(00)

1A)D(a(0a)2AA(0,0,a)ADa,0,a)aa)

ADmx,,z)

mADmAE

x)()

77zxym2)

AAsinAD

cosAA

AAm|||

2|311

sinC

cos

19cosA8(a)()11

inCaa

a6bsins

a

bccos

)49b(a)(a49ba11baa49asinAcos

2

47

aAC

437sin3a

13SabC22ABCsinAsinCcoscosB

B)sinAcos

3B8

absinAABbB5aaABCC

B)Csin()sinAcosB

35718168411absin244

3ppB

13P),()3004

22

(A),()

P(C)

111313()))33

P

f(x)

yf(yf((f())(t)(t)

f(x)

(,)mn

xt(

f()f(m)myyf((f(t))

(12

))x

6ytt11()t)2t

S((S(t)

t

1(t)(t)(12)4t

3(4)(tt)tt

t2(t)S()(t)t(t)

22:(abC(l

CNNAxQ

|||

AN2y2AN2y21212AN1124

b

(4)(x)(xy)AMyy(2)y(2)xk)kyyx(a2bb4

b

22l

(x4)

(x4)

y

(1k

)

64k

12

My)

(y)

x2

k22

x

yy(x

(2)

yk(1k2)yyxx)xk2)(1)PBQBx|PB|kxkxxx)x|BQx4)](xkxkx)kk2k(2kxk)8(2

|

(2xk2)(xx)kx(2xk2)(xx)kk2)x

kx)k(2)

)x)x

||

ailaili{}{}aaij){}aijj

{}a(n3){}()l

(){}an{}