苏州高二数学答案.docx

苏州高二数学答案.docx

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苏州高二数学答案

第一学期学业质量阳光指标调研卷

高二数学参考答案及评分建议2019.1

一、填空题

1．∀x∈R，x2-x+1≠0

2．（2,0）3．-1

2

4．k<1或k>2

5．2

6．x2+（y-1）2=5

7．（0,+∞）

8．必要不充分

9．3010．5-1

2

11．③12．2+ln2

13．[2

-2,42+2]

14.

（-∞,1-33）（1+33,+∞）

22

二、解答题

15．（本题满分14分）

解：

（1）因为等腰梯形ABCD，AB∥DC，AD=BC=4，AB=8，DC=6．

所以A（-4,0），B（4,0），C（3,15），D（3,-

15）．·································2分

所以CA=

=8，CB=

=4．

因为2a=CA-CB=4，所以a=2．···················································5分

又因为A，B为双曲线x

a2

y2

b21（a0,b0）的焦点，所以2cAB8，所以c4．

所以b=

=2．·······························································7分

x2y2

所以双曲线的方程为

-=1．······················································8分

412

（2）双曲线的离心率e=c=2．·····················································11分

a

双曲线的渐近线方程为y=±3x．···················································14分

16.（本题满分14分）证明：

（证法一）

（1）在CD上取点P，使DP=1PC．

2

在△DCF中，DN=1NF，DP=1PC．

EF

N

DPC

22M

AB

所以NP//FC，又因为NP⊄平面BCF，FC⊂平面BCF，

所以NP//平面BCF．·····································································2分

在△ACD中，AM=1MC，DP=1PC，

22

所以MP//DA，又四边形ABCD为正方形，所以DA//BC，

又因为MP⊄平面BCF，BC⊂平面BCF，所以MP//平面BCF．···········4分又因为MPNP=P，MP,NP⊂平面BCF，

所以平面MNP//平面BCF．·····························································6分

因为MN⊂平面MNP，

所以MN//平面BCF.······································································8分

（2）因为NP//FC，DC⊥CF，所以NP⊥CD．

因为MP//AD，DC⊥AD，所以MP⊥CD．又因为MPNP=P，且MP,NP⊂平面MNP，

所以CD⊥平面MNP．···································································12分又因为MN⊂平面MNP，

所以MN⊥CD.·············································································14分

（证法二）

（1）在CF上取点R，使CR=1RF，在CB上取点S，使BS=1CS．

22

在△DCF中，DN=1NF，CR=1RF，F

22

所以NR//CD且NR=2CD．·····················2分

3R

在△ABC中，AM=1MC，BS=1CS，C

22

所以MS//AB且MS=2AB．······················4分AB

3

因为四边形ABCD为正方形，所以AB//CD且AB=CD，所以NR//MS，且NR=MS，

所以四边形NRSM为平行四边形，所以MN//RS．································6分又因为MN⊄平面BCF，RS⊂平面BCF，

所以MN//平面BCF．···································································8分

（2）因为CD⊥CF，CD⊥BC，CFBC=C，CF,BC⊂平面BCF，

所以CD⊥平面BCF．···································································12分又RS⊂平面BCF，所以CD⊥RS．·················································13分又因为MN//RS，所以MN⊥CD.·····················································14分

17．（本题满分15分）

解：

（1）将圆C：

x2+y2+2x-4y+3=0化标准方程为（x+1）2+（y-2）2=2，

所以圆心C（-1,2），半径r=.························································2分又因为圆C的切线l在x轴和y轴上的截距相等，且截距不为零，

所以设切线l的方程为x+y-a=0.····················································3分

因为直线l与圆C相切，所以圆心C到直线l的距离等于半径，

即=．·······································································5分

解得：

a=-1或a=3．···································································6分所以切线l的方程为x+y+1=0或x+y-3=0.·····································7分

（2）因为PM为切线且M为切点，所以PM2=PC2-MC2.

又因为PM=2PO，所以PC2-MC2=2PO2.

又因为P（x1,y1），O（0,0），MC=r=

所以（x+1）2+（y

-2）2-2=2（x2+y2），

1111

化简可得：

x2+y2-2x+4y

-3=0

①；··········································11分

1111

因为点P在直线y=2x-6上，所以y1=2x1-6②.

⎧⎪x2+y2-2x+4y

联立①②可得：

⎨1111

⎪⎩y1=2x1-6

-3=0

，

消去y可得：

5x2-18x

+9=0，解得x=3或x=3．·························13分

111

151

将x=3代入②可得：

y=-24，所以点P的坐标为⎛3,-24⎫．

1515ç⎪

将x1=3代入②可得y1=0，所以点P的坐标为（3,0）．

综上可知，点P的坐标为⎛3,-24⎫或（3,0）.········································15分

⎝⎭

18．（本题满分15分）

解：

（1）因为物体P到光源A的距离为x，所以物体P到光源B的距离为10-x.

因为P在线段AB上且不与A,B重合，所以0

所以P点受A光源的照度为：

8k1k2，················································2分

x2

P点受B光源的照度为：

k1k2

（10-x）2

，·················································4分

所以物体P受到A,B两光源的总照度y=8k1k2+

x2

k1k2

（10-x）2

，x∈（0,10）.······6分

（2）因为f（x）=8k1k2+

x2

k1k2

（10-x）2

x∈（0,10）.

16kk

2kk

2kk

（3x-20）（3x2-60x+400）

所以f'（x）=-12+12=12

．···········9分

x3（10-x）3x3（10-x）3

令f'（x）=0，解得x=20.·······························································11分

3

当0

3ç3⎪

⎝⎭

当200，所以在⎛20,10⎫上单调递增．····················13分

3ç3⎪

⎝⎭

因此，当x=20时，f（x）取得极小值，且是最小值.···························14分

3

答：

在连接两光源的线段AB上，距光源A为20处，物体P受到光源A,B的总照度

3

最小.··························································································15分

19．（本题满分16分）

解：

（1）因为椭圆的离心率为3，右准线方程为x=43，

23

⎧c=

⎪a2

所以⎨

⎪a2=43

，·········································································2分

⎪⎩c

⎧⎪a=2

解得⎨

⎪⎩c=

3

．···············································································3分

又因为b=

=1．

x22

所以椭圆C的标准方程为+y

4

=1．···············································4分

（2）设A（x1,y1），B（x2,y2），M为椭圆的上顶点，则M（0,1）．

①解法一：

因为直线l经过原点，由椭圆对称性可知B（-x1,-y1）．

x2x2

因为点A（x,y）在椭圆上，所以1+y2=1，即y2-1=-1．

因为k1

11

=y1-1，k

x1

4114

=y2-1=y1+1．

x2x1

y-1y+1y2-11

所以kk=1⨯1=1=-．·············································6分

122

111

⎧k-k=5

⎧k=1⎧1

⎪12

所以⎨

4⎪1

，解得⎨

⎪k=

1⎨4

．····································8分

⎪kk

⎪⎩12

=-1

4

⎪⎩k2=-4

⎪⎩k2

=-1

因为点A在第三象限内，所以k>1，所以k=1，则直线MA的方程为y=x+1．

⎧x2

121

⎧x=-8

⎪+y2=1⎧x1=0⎪2583

联结方程组⎨4

，解得⎨y

=1或⎨

，所以A（-,-）．······10分

355

⎪y=x+1⎩1⎪y=-

⎩⎪⎩25

（解出k=1，k=-1，也可根据k

=y1-1=1，k

=y1+1=-1，求出点A的坐标）

124

1x14

解法二：

因为k1

=y1-1，k

x1

=y2-1=y1+1，

x2x1

所以k-k=y1-1-y1+1=-2．·····················································6分

x1x1x1

所以-2=5，解得：

x=-8．························································8分

x14

15

x223

将其代入椭圆方程+y

4

=1，解得y1=±5．

因为点A在第三象限内，所以点A的坐标为（-8,-3）．························10分

55

②直线l过点（−2,−1），设其方程为y+1=k（x+2）．

⎧x2

⎨

联列方程组⎪4

+y2=1

，消去y可得（4k

2+1）x2

+8k（2k-1）x+16k（k-1）=0．

⎪⎩y=kx+2k-1

当△>0时，由韦达定理可知x+x

=-8k（2k-1），xx

=16k（k-1）．·····12分

124k2+1124k2+1

又因为k+k

=y1-1+y2-1=x1y2+x2y1-（x1+x2）=2k+2（k-1）（x1+x2）．14分

x1x2x1x2x1x2

=2k+2（k-1）⨯[-8k（2k-1）]=2k+（1-2k）=1．

16k（k-1）

所以k1+k2为定值1．·····································································16分

20．（本题满分16分）

解：

（1）当b=1时，因为f（x）=alnx+（x-1）（x-2），所以f'（x）=a+2x-3．····2分

x

因为f（x）在x=2处取得极小值，所以f'

（2）=0，解得：

a=-2．·············3分

此时，f'（x）=-2+2x-3=（2x+1）（x-2），

xx

当x∈（0,2）时，f'（x）<0，f（x）单调递减，当x∈（2,+∞）时，f'（x）>0，f（x）单调递增．所以f（x）在x=2处取得极小值．

所以a=-2符合题意．······································································4分

（2）当a=1时，因为f（x）=lnx+b（x-1）（x-2），

'12bx2-3bx+1

所以f（x）=+b（2x-3）=．

xx

令g（x）=2bx2-3bx+1.

①因为f（x）在（1,2）上单调递增，所以f'（x）≥0在（1,2）上恒成立，

即g（x）≥0在（1,2）上恒成立.······························································5分

1︒当b=0时，则g（x）=1，满足题意.·················································6分

2︒当b≠0时，因为g（x）的对称轴为x=3<1，

4

⎩

所以⎧g

（1）≥0，解得-1≤b<0或0

⎨g

（2）≥02

综上，实数b的取值范围为⎡-1,1⎤.······················································8分

⎣⎢2⎥⎦

②1︒当b=0时，f（x）=lnx，与题意不符.·········································9分

2︒当b<0时，取x=3-1，则x>1.

0b0

令h（x）=lnx-x+1，则h'（x）=1-1，

x

当x∈（0,1）时，h'（x）>0，h（x）单调递增，当x∈（1，+∞）时，h'（x）<0，h（x）单调递减，所以h（x）≤h

（1）=0，即lnx≤x-1.

所以f（x0）=lnx0+b（x0-1）（x0-2）≤（x0-1）+b（x0-1）（x0-2）=2b-1<0，

所以b<0符合题意.·······································································11分

3︒当0

因为g（x）=2bx2-3bx+1在（1,+∞）递增且g

（1）=1-b≥0

所以f'（x）=g（x）≥0在（1,+∞）上恒成立，所以f（x）在（1,+∞）上单调递增，

x

所以f（x）≥f

（1）=0恒成立，与题意不符.··········································13分

4︒当b>1时，

因为g

（1）=1-b<0，g

（2）=2b+1>0，

由零点存在性原理可知，存在x1∈（1,2），使得g（x1）=0，所以当x∈（1,x1）时，f'（x）<0，f（x）单调递减，

取x0=x1>1，则f（x0）

（1）=0，符合题意.······································15分

（注：

将g（x）=0大于1的根求出，并赋值给x0同等赋分）

综上可知，实数b的取值范围为（-∞,0）（1,+∞）.···································16分