西宁市大通县一中2019-2020学年上学期高一数学期中考试卷附答案详析

fxx3x15ff=2

CA3A4A5A6A7A8A9A10A1112

B52

A1,292

A

U

D32

U1,2,3,4,5B2,3CB

D1,2(,2]

B3,4[2,)1x21ABA(,2]

)2a

()A{x|x20}

BB{x|xa}

C(D[2,)

fx13x2,0B,22,0C2,1D,22,1f(xf(x)x2

11)Bx22xf(x)x2

f(x)2x1Cf(x)x2

1Df(x)x2

2x1yx2(2a1)x13,2yx

B(,2]

Ca

3

,23

,2

D3

,2

By2x23

Cy2xDyx2,x[0,1]

(

)

Dfx2ax21(a0

a1)

C2,1cba

B2,1cab

1,1()

1,1acba0.993.3,b3.30.99,clog3.30.99,

BCabc

()Dyaxa(a0,a1)

1ABCD12fxm1x22mx3f1BDf2f

3AC3f2f1f2f3f1f

A2,3,B1,2,af

3f2f1f1f3f2a=____131415ABUZ

yf(x)

__________A{1,0,1,2},Bx|x2x0R

x0

A(UB)________.f(x)2x3

ab

f(2)

16fxax2bx3ab

Ax2x5a1   2aBxp1x2p117pABB

18UR

Ax|1x3Bx|2x4AB

a

UB.191,1fxfa1f2a2xx-1x0

20fx

1,fxx1x21fxfx.R22x0,1fxx226ax3a2

21A{}C.B{}.CNDNAC.B,D.CAC...2x1(x1)

fxx3x1

.5ff=2

CA12DB52.92D32x1,x1f(x)=x3,x1ff

515

f3

222

D5

=f2

12=13+1=223AU1,2,3,4,5A1,2B2,3CA

U

B

D1,2CB3,4U12U

BAB.U1,2,3,4,5B2,3U

B1,4,53A1,2AUB1.C..4AA{x|x20}

(,2]

BBB{x|xa}

CABA(,2]

Da

()[2,)[2,)

Ax|x2B

5Aa

2,.

fx13x1x2()2,0B,22,0C2,1D,22,1A13x0

x20

6Af(x)x2Cf(xf(x11)Bx22xf(x)x0

x2

f(x)2x1C2x0A.x2f(x)x21Df(x)x22x11)x22xf(x)f(x1)x22x=x+12

-1f(x)x21C7Ayx2(2a1)x13,2

BB(,2]

Ca

3

,23

,2

D3

,2



yx22a1x1

x

2

2a1

2

a

32

2a12

,2a

3,

2

B

48Ayx

BBy2x23

Cy2xDyx2,x[0,1]

ACf(x)xf(x)B.Bf(x)2x23f(x)

Df(x)f(x),f(x)f(x)

.x0,1f(x)

.f(x)9Afx2ax21(a0

a1)

C()

D2,1BB2,10,11,1fx1,1fx2ax21(a0

x20fxa1)

0

x2yf22a1211

2,1B10Aa0.993.3,b3.30.99,clog3.30.99,

()

Ccba

BBcababc

Dacba0.993.30,1clog3.30.990

B11b3.30.991cab

yaxa(a0,a1)

()5ABCDCa

a

.a1a1yaxa(a0,a1)

Ayax

(1,0)yax

(1,0)

CBx1yaa0

Dyaxa(a0,a1)

0a1

0a1x1

Cyaxa(a0,a1)

yaxa(a0,a1)

a

yaa0

.12fxm1x22mx3f1BDf2f

3AC3f2f1f2f3f1f

Bf

3f2f1f1f3f2m

.f(x)(m1)x22mx3m0

f(x)x23

(0,)f(1)f(1)

f(2)f(2)f(x)x23

f(1)f(2)f(3)B.f(1)f(2)f(3)6.133A2,3,B1,2,aABABa=____a3

A{1,0,1,2},Bx|x2x0A(UB)________.14UZ

{1,2}

BB.Bx|x2x00,1U

B0,1

A(UB){1,2}

.{1,2}

15-1yf(x)

__________Rx0

f(x)2x3f(2)

fx16f2f21

1.fxax2bx3ab

13a1   2aab

fxax2bx3ab

fxax23a

a1  2a

.b0

ab

1.3a

13.17pAx2x5p3p12p1

p12p1

Bxp1x2p1ABB

ABBp12p1

BA

p7pBBA

pABB

p12p1

BA

p2

BBAABB

p2

p12p1BA

p2

Bx3x33p2

Bxp1x2p13p3

ABB

p2

p12p1BAp2

2p15

p122p3ABB

pp3

.18B

Bx|2x4AB

UR

Ax|1x3UB.AB{x|2x3}

BACUB{x|x3x4}.AB

UB.Ax|1x3Bx|2x4CUB{x|x2x4}ACUB{x|x3x4}..AB{x|2x3}

191,10a12fxfa1f2aa

“11fx”81a11

12a1a12a

0a12a20fx

2xx-1

1,fx2

fx1fx2fx2

2x1

2x1x1x21

x1x21

2x2x122

fx1fx2

x11x21x11x21x1x21

x2x10

x110

x110

fxfx1fx20

fxfx.fx1fx21,.21Rx0

fxx1xx1x,x0

fx0,x0

x1x,x0

x0

fxRx0

fxf00

fxfxfxfxx0

x0x0

fxfxx1xx1xfxf00

fxx1xx0Rx1x,x0



fx0,x0

x1x,x0

9.22x0,1fxx226ax3a2

23a,

3a26a3,26a6a1,

132a

3a

fxmin12a33fxfxx226ax3a2x3a1

3a103a1103a11

.fxx226ax3a2

x3a1

3a10a

13fx0,1fminxf03a23a11

a

23fx0,1fminxf13a26a303a11

12a33

x3a1

0,1fminxf3a16a26a1



3a2,

3a26a3,26a6a1,

132a

3a

fxmin12a33

.10