苏州高二数学答案.docx
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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20 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
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