2016年河南省普通高等招生全国统一考试猜题卷（二）数学（理）

绝密★启用前普通高等招生全国统一考试猜题卷（二）理科数学本试卷分第1卷（选择题）和第Ⅱ卷（非选择题）两部分．共150分，考试时间120分钟．第Ⅰ卷（选择题共60分）注意事项：1．答题前，考生在答题纸上务必用直径0.5毫米黑色签字笔将自己的姓名、准考证号填写清楚，并贴好条形码．请认真核准条形码的准考证号、姓名和科目，2．每小题选出答案后，用2B铅笔把答题纸上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号．在试题卷上作答无效．一、选择题（本大题共12小题，每小题5分，共60分．在每小题给出的四个选项中，只有一项是符合题目要求的）1．复数z=QUOTE
=··········································································()A.-2+iB．iC．2-iD．-i：2．已知集合M=QUOTE
，N=QUOTE
，则MUN=·································()A.[-2，4）B．（-2，4）C．(o，2)D．(o，2]3．采用系统抽样方法从1000人中抽取50人做问卷调查，为此将他们随机编号1，…，1000，适当分组后在第一组采用简单随机抽样的方法抽到的号码为8，抽到的50入中，编号落人区间[1，400]的人做问卷A，编号落入区间[401，750]的人做问卷B，其余的人做问卷C，则抽到的人中，做问卷C的人数为·······()A．12B．13C．14D．154．已知命题p:函数y=ln(x2+3)+QUOTE
的最小值是2；命题q：x>2是x>l的充分不必要条件．则下列命题为真命题的是··············································································()A.pQUOTE
qB．￢pQUOTE
qC．￢pQUOTE
qD．pQUOTE
q5．已知点A是抛物线C1:y2=2px(p>0)与双曲线C2:QUOTE
(a>0，b>0)的一条渐近线的交点，若点A到抛物线C1，的焦点的距离为p，则双曲线C2的离心率等于·····································()A．QUOTE
B．QUOTE
C．QUOTE
D．QUOTE
6．QUOTE
的展开式中含x的正整数指数幂的项数是········································()A1B．2C．3D．4数学试题（二）第1页（共8页）7．设{an}是等差数列，下列结论中正确的是··················································()A.若a2+a5>0,则a1+a2>0B.若a1+a3<0,则a1+a2<0C.若0<a1<a2，则a3>QUOTE
-D.若a1<0，则（a2-a1）（a4-a2）>08．如图，正四棱锥P-ABCD庇面的四个顶点A，B，C，D在球O的同一个大圆上，点P在球面上，如果VP-ABCD=QUOTE
，则球O的表面积是····································()A．4QUOTE
B．8QUOTE
C．12QUOTE
D．16QUOTE
3r+y-2≤0，9．变量x，y满足线性约束条件QUOTE
，目标函数z=kx-y仅在点(O，2)取得最小值，则k的取值范围是·········································································()A．k<-3B．k>lC．-l<k<1D．-3<k<l10．某几何体的三视图如图所示，当a+b取最大值时，这个几何体的体积为·······················()A．QUOTE
B．QUOTE
C．QUOTE
D．QUOTE
11.已知M是△ABC内一点（不含边界），且QUOTE
·QUOTE
=2QUOTE
，∠BAC=300，若△MBC，△MCA，△MAB的面积分别为x，y，z，记f(x，y，z)=QUOTE
，则f(x，y，z)的最小值为··································()A.26B．32C．36D．4812.已知集合M=QUOTE
，若对于任意（x1，y1）∈M，存在(x2，y2)∈M，使得x1x2+y1y2=0成立，则称集合M是“商高线”．给出下列四个集合：①M=QUOTE
；②M=QUOTE
；③M=QUOTE
；④M=QUOTE
．其中是“商高线”的序号是······························································()A.①②B.②③C.①④D.②④数学试题（二）第2页（共8页）绝密★启用前普通高等招生全国统一考试猜题卷（二）理科数学第Ⅱ卷（非选择题共90分）注意事项：1．答题前，考生先在答题纸上用直径0.5毫米黑色签字笔将自己的姓名、准考证号填写清楚，然后贴好条形码．请认真核准条形码上的准考证号、姓名和科目．2．第Ⅱ卷共6页，请用直径0.5毫米黑色签字笔在答题纸上各题的答题区域内作答，在试题卷上作答无效．本卷包括必考题和选考题两部分．第13～21题为必考题，每个试题考生都必须作答，第22～24题为选考题，考生根据要求作答．二、填空题（本大题共4小题，每小题5分，共20分）13.执行如图所示的程序框图，若输入x=0.1，则输出的m的值是__________________.14.已知f(x)是定义在R上的奇函数，当x≥0时，f(x)=3x+m（m为常数），则f(-log35)的值为________________.15.关于函数f(x)=2(sinx—cosx)cosx的四个结论：P1:最大值为QUOTE
；P2:把函数f(x)=QUOTE
的图象向右平移QUOTE
个单位后可得到函数f(x)=2(sinx一cosx)cosx的图象；P3:单调递增区间为QUOTE
，k∈Z；P4:图象的对称中心为QUOTE
，k∈z．其中正确的结论有_________个．16.已知数列{an}满足a1=QUOTE
，an-1-an=QUOTE
(n≥2)，则该数列的通项公式为_________________.三、解答题（本大题共6小题，共70分，解答应写出文字说明、证明过程或演算步骤）17.（本小题满分12分）在△ABC中，角A，B，C的对边分别为a，b，c，已知角A=QUOTE
，sinB=3sinc．（1）求tanC的值；（2）若a=QUOTE
，求△ABC的面积．数学试题（二）第3页（共8页）18．（本小题满分12分）某青少年研究中心为了统计某市青少年（18岁以下）2015年春节所收压岁钱的情况进而研究青少年的消费去向，随机抽查了该市60名青少年所收压岁钱的情况，得到如下数据统计表（图①）：已知“超过2千元的青少年”与“不超过2千元的青少年”人数比恰好为2：3．（1）试确定x，y，p，q的值，并补全频率分布直方图（图②）．（2）该机构为了进一步了解这60名青少年压岁钱的消费去向，从“压岁钱超过2千元的青少年”、“压岁钱不超过2千元的青少年”中用分层抽样的方法确定10人，若需从这10人中随机选取3人进行问卷调查．设S为选取的3人中“压岁钱超过2千元的青少年”的人数，求ɛ的分布列和数学期望．（3）若以颇率估计概率，从该市青少年中随机抽取15人进行座谈，若15人中“压岁钱超过2千元的青少年”的人数为η，求η的期望．数学试题（二）第4页（共8页）19.（本小题满分12分）如图所示的多面体中，四边形ABCD是菱形，ED∥FB，ED⊥平面ABCD，AD=BD=2，BF=2DE=2QUOTE
(1)求证：AE⊥CF；(2)求二面角A-FC-E的余弦值．20．（本小题满分12分）已知椭圆C的中心在原点，焦点在x轴上，长轴长为4，且点QUOTE
，譬在椭圆C上．(1)求椭圆C的方程5(2)设P是椭圆C长轴上的一个动点，过P作斜率为QUOTE
的直线L交椭圆C于A,B两点，求证：QUOTE
为定值，数学试题（二）第5页（共8页）21.（本小题满分12分）已知函数f(x)=-x3+x2(x∈R)，g(x)满足gˊ(x)=QUOTE
（a∈R，x>0），且g(e)=a，e为自然对数的底数．(1)已知h(x)=e1--xf(x)，求h(x)在(1，h(1))处的切线方程；(2)若存在x∈[1，e]，使得g(x)≥一x2+（a+2）x成立，求a的取值范围；(3)设函数F(x)=QUOTE
，O为坐标原点，若对于y=F(x)在x≤一1时的图象上的任一点P，在曲线y=F(x)(x∈R)上总存在一点Q，使得QUOTE
·QUOTE
QUOTE
，且PQ的中焦在y轴上，求a的取值范围，数学试题（二）第6页（共8页）请考生在第22、23、24三题中任选一题作答，如果多做，则按所做的第一题计分，作答时请写清题号．22.（本小题满分10分）选修4-1：几何证明选讲如图所示，在△ABC中，AB=AC，过点A的直线与其外接圆交于点P，交BC的延长线于点D．(1)求证：QUOTE
=QUOTE
；(2)若AC=3，求AP·AD的值．23.（本小题满分10分）选修4-4：坐标系与参数方程在直角坐标系中，以原点为极点，z轴的正半轴为极轴建立极坐标系，已知曲线C：QUOTE
(a>0)，过点P（-4，-2）的直线L的参数方程为QUOTE
（t为参数），L与C分别交于点M，N.（1）写出C的直角坐标方程和L的普通方程；（2）若QUOTE
，QUOTE
，QUOTE
成等比数列，求a的值，数学试题（二）第7页（共8页）24.（本小题满分10分）选修4-5：不等式选讲已知函数f(x)=lx--1l+lx+1l.（1）求不等式，f(x)≥3的解集；（2）若关于x的不等式，f（x）˃a2-x2+2x在R上恒成立，求实数a的取值范围．数学试题（二）第8页（共8页）答案1.B解析：解法一：z=QUOTE
=QUOTE
=i解法二：z=QUOTE
=QUOTE
2.A解析：M=QUOTE
，N=QUOTE
，QUOTE
MQUOTE
N=[-2，4）3.A解析：若采用系统抽样方法从1000人中抽取50人做问卷调查，则需要分为50组，每组20人，若第一组抽到的号码为8，则以后每组抽取的号码分别为28，48，68，88，108，…，所以编号落入区间[1，400]的有20人，编号落入区间[401，750]的有18人，所烈做问卷C的有12人．4.C解析：命题p为假命题，命题q为真命题，所以￢pQUOTE
q为真命题．5.C解析：因为点A到抛物线C,的焦点的距离为p，所以点A到抛物线准线的距离为p，从而点A的坐标可以为（QUOTE
，±p），所以双曲线的渐近线方程为了y=±2x．所以QUOTE
=2，所以b2=4a2．又b2=c2-a2，所以c2=5a2．所以双曲线的离心率为QUOTE
6．B解析：QUOTE
的展开式中第r+1项为QUOTE
．（-QUOTE
）=(-1)rQUOTE
，当QUOTE
为正整数时，r=0或2，所以QUOTE
的展开式中含x的正整数指数幂的项数是2．7．C解析：设等差数列{an}的公差为d，若a2+a5>0，则a1十a2=(a2-d)+(a5-3d)=(a2+a5）-4d，由于d的正负不确定，因而a1+a2的符号不确定，故选项若a1+a3<0，则a1+a2=（a1+a3）-d，由于d的正负不确定，因而a1+a2的符号不确定，故选项B错误；若0<a1<a2，则a1>0,d>0,a2>0,a3>0,a4>0,所以QUOTE
=QUOTE
，a3>QUOTE
，故选项C正确；（a2–a1）（a4-a2）=d(2d)=2d2，由于d有可能等于0，故选项D错误．8．D解析：连接PO，由题意知，PO⊥底面ABCD，PO=R，S正方彤ABCD==2R2．因为VP-ABC=QUOTE
，所以QUOTE
，解得R=2，所以球0的表面积是16QUOTE
．9．D解析：如图，作出不等式组表示的平面区域，由z=kx-y得y=kx-z，要使目标函数z=kr-y仅在点A(O，2)处取得最小值，则阴影部分区域在直线y=kx+2的下方，所以目标函数的斜率A满足-3<k<l．10．D解析：由该几何体的三视图可得其直观图为如图所示的三棱锥，其中从点A出发的三条棱两两垂直，且AB=1，PC=QUOTE
，PB=a，BC=b，则PA2+AC2=a2—1+b2-1=6，即a2+b2=8.所以(a+b)2=8+2ab≤8+2QUOTE
，即a+b≤4，当且仅当a=b=2时取等号，此时，PA=QUOTE
，AC=QUOTE
.所以该几何体的体积V=QUOTE
．11.C解析：由QUOTE
·QUOTE
=2QUOTE
．∠BAC=300得S△ABC=1，即x+y+z=1，f(xyz)=QUOTE
=14+QUOTE
QUOTE
=14+4+6+12=36.当且仅当y=2x，z=3x，2z=3y，即x=QUOTE
，y=QUOTE
，z=QUOTE
时取等号12．D解析：如果对于函数图象上的任意一点M(x1，y1)，在其图象上都存在点N(x2，y2)使OM⊥ON，则函数图象上的点的集合为商高线．对于①，若取M(1，1)，则不存在这样的点；对于③，若取M(1，O)，则不存在这样的点．故选D．13.0解析：若输入x=0.1，则m=lg0.1=-1，因为m<0,QUOTE
m=-1+1=0.所以输出的仇的值为0．14.-4解析：因为f(x)是定义在R上的奇函数，QUOTE
f(0)=1+m=0，QUOTE
m=-1．QUOTE
f(-log35)=-f(log35)=-(3log35-1)=-4.15.2解析：因为f(x)=2sinx·cosxQUOTE
2cos2x=sin2xQUOTE
cos2xQUOTE
1=QUOTE
，所以其最大值为QUOTE
QUOTE
1，QUOTE
P1错误f(x)=QUOTE
的图像向右平移QUOTE
个单位后得到函数为f(x)=QUOTE
=QUOTE
，QUOTE
P2错误由QUOTE
，kQUOTE
Z得函数f(x)的单调递增区间为QUOTE
，kQUOTE
Z，QUOTE
P3正确由QUOTE
=kQUOTE
，kQUOTE
Z，得x=QUOTE
，kQUOTE
Z，QUOTE
P4正确16.QUOTE
解析：因为an-1-an=QUOTE
(nQUOTE
)，QUOTE
QUOTE
，QUOTE
QUOTE
.QUOTE
QUOTE
QUOTE
QUOTE
.所以QUOTE
，QUOTE
QUOTE
，经检验，n=1时也适合此公式QUOTE
QUOTE
17.解(1)因为A=QUOTE
，QUOTE
B+C=QUOTE
QUOTE
sin(QUOTE
-C)=3sinC···································································2分QUOTE
QUOTE
cosC+QUOTE
sinC=3sinC······························································4分QUOTE
QUOTE
cosC=QUOTE
sinC，QUOTE
tanC=QUOTE
·························································6分(2)由QUOTE
，sinB=3sinC得b=3c····················································8分ΔABC中，由余弦定理得a2=b2+c2-2bccosA=9c2+c2-2×(3c)×c×QUOTE
=7c2························10分因为a=QUOTE
，QUOTE
c=1，b=3QUOTE
ΔABC的面积为S=bsinA=QUOTE
·························································12分18.解：(1)根据题意，有QUOTE
，解得QUOTE
·····························2分p=0.15，q=0.10(2)用分层抽样的方法，从中选取10人，则其中“压岁钱超过2千元的青少年”有10QUOTE
=4人，“压岁钱不超过2千元的青少年”有10QUOTE
=6人···············································5分故QUOTE
的可能取值为0，1，2，3P(QUOTE
=0)=QUOTE
，P(QUOTE
=1)=QUOTE
，P(QUOTE
=2)=QUOTE
，P(QUOTE
=3)=QUOTE
·····················7分所以QUOTE
的分布列为：QUOTE
0123P

QUOTE
··············································10分(3)以频率估计概率，从该市青少年中随机抽取1人为“压岁钱超过2千元的青少年”的概率为QUOTE
，则η~B(15，QUOTE
)，所以随机变量η的期望为E(η)=15×QUOTE
=6······································12分19.(1)证明：方法一：由题意知，在ΔAEF，AE=QUOTE
，EF=QUOTE
，AF=QUOTE
······························1分QUOTE
AE2+EF2=AF2，QUOTE
AE⊥EF.································································2分在ΔAEC中，AE=QUOTE
， EC=QUOTE
，AF=QUOTE
···················································3分QUOTE
AE2+EC2=AC2，QUOTE
AE⊥EC································································4分又EFQUOTE
EC=E,QUOTE
AE⊥平面ECF····························································5分又平面ECFᴝFC，QUOTE
AE⊥FC.······························································6分方法二：因为四边形ABCD是菱形，AD=BD=2，QUOTE
AC⊥BD,AC=QUOTE
······························1分因为ED⊥平面ABCD,BD=2，BF=2，DE=2QUOTE
，故可以O为坐标原点，以OA，OB所在直线为x轴、y轴，建立如图所示的空间直角坐标系。························································2分

则A(QUOTE
，0,0)，E(0，-1，)，C(-QUOTE
，0,0)，F(0，1，2QUOTE
)·································2分QUOTE
QUOTE
=(-QUOTE
，-1，)，QUOTE
=(QUOTE
，1，2QUOTE
)，·················································4分QUOTE
QUOTE
·QUOTE
=(-QUOTE
，-1，)·(QUOTE
，1，2QUOTE
)=-3-1+4=0·········································5分所以AE⊥CF···········································································6分（2）解：由（1）方法二知A(QUOTE
，0,0)，E(0，-1，)，C(-QUOTE
，0,0)，F(0，1，2QUOTE
)则QUOTE
=(-，1，2QUOTE
)，QUOTE
=(-2QUOTE
，0，0)，QUOTE
=(0，2,QUOTE
)，QUOTE
=(-QUOTE
,1，-QUOTE
)················7分设平面AFC的一个法向量为n1=（x1，y1，z1），由QUOTE
·n1=0，QUOTE
·n1=0，得-QUOTE
x1+y1+2QUOTE
z1=0且-2QUOTE
x1=0··································8分令z1=1，得n1=（0,-2QUOTE
，1）·························································9分设平面EFC的一个法向量为n2=（x2，y2，z2），由QUOTE
·n2=0，QUOTE
·n2=0，得2y2+QUOTE
z2=0，且-QUOTE
x2+y2-QUOTE
z2=0·······························10分令y2=-1，得n2=（-QUOTE
，-1，QUOTE
）······················································11分设二面角A-FC-E的大小为QUOTE
，则cosQUOTE
=QUOTE
····························12分20.（1）解：因为2a=4，所以a=2·····························································1分又焦点在x轴上，所以设椭圆方程为QUOTE
········································2分将点（1，QUOTE
）代入椭圆方程得b2=1···················································3分所以椭圆方程为QUOTE
+y2=1······························································4分（2）证明：p（m，0）（-2QUOTE
mQUOTE
2），则直线L的方程是y=QUOTE
·································5分由方程组QUOTE
消去y得2x2-2mx+m2-4=0（*）···································7分设A(x1，y1)，B（x2，y2），则x1，x2是方程（*）的两个根所以有x1+x2=m，x1x2=QUOTE
···························································8分所以QUOTE
=QUOTE
=QUOTE
=QUOTE
=QUOTE
=QUOTE
=5··································11分所以QUOTE
为定值····························································12分21.解：（1）因为h(x)=(-x3+x2)e1-x，QUOTE
h′(x)=(x3-4x2+2x)e1-x····························································1分QUOTE
h(1)=0，h′(1)=-1······························································2分QUOTE
h(x)在（1，h(1)）处的切线方程为y=-(x-1)，即y=-x+1······························3分（2）因为g′(x)=QUOTE
QUOTE
g(x)=alnx+c(c为常数)QUOTE
g(e)=alne+c=a+c=aQUOTE
c=0QUOTE
g(x)=alnx······································································4分由g(x)QUOTE
-x2+(a+2)x，得(x-lnx)aQUOTE
x2-2x.由于xQUOTE
时，lnxQUOTE
1QUOTE
x，且等号不能同时成立，所以lnxQUOTE
x，x-lnxQUOTE
0从而aQUOTE
QUOTE
，为满足题意，必须aQUOTE
QUOTE
···································5分设t(x)=QUOTE
，xQUOTE
，则t′(x)=QUOTE
因为xQUOTE
，所以x-1QUOTE
0，lnxQUOTE
1，x+2-2lnxQUOTE
0，从而t′(x)QUOTE
0所以t(x)在QUOTE
上为增函数························································6分从而t(x)max=t(e)=QUOTE
,从而aQUOTE
QUOTE
··················································7分（3）设P(t,F(t))为y=F(X)在xQUOTE
1时的图象上的任意一点，则tQUOTE
1因为PQ的中点在y轴上，所以Q得坐标为(-t,F(-t))···································8分因为tQUOTE
1所以-tQUOTE
1，所以P(t，-t3+t2)，Q(-t，aln(-t))因为QUOTE
·QUOTE
=-t2-at2(t-1)ln(-t)QUOTE
0所以a(1-t)ln(-t)QUOTE
1····························································9分当t=-1时，a(1-t)ln(-t)QUOTE
1恒成立，所以aQUOTE
R····································10分当tQUOTE
-1时，aQUOTE
QUOTE
，令QUOTE
(t)=QUOTE
(tQUOTE
-1)则QUOTE
′(t)=QUOTE
因为tQUOTE
-1，所以t-1QUOTE
0，tln(-t)QUOTE
0，所以QUOTE
′(t)QUOTE
0从而QUOTE
(t)=QUOTE
在(QUOTE
)上为增函数，由于tQUOTE
时，QUOTE
(t)=QUOTE
QUOTE
0所以QUOTE
(t)QUOTE
0，所以aQUOTE
0综上可知a的取值范围是QUOTE
··················································12分22.（1）证明：因为∠CPD=∠ABC,∠PDC=∠PDC,所以ΔDPC~ΔDBA··································2分所以QUOTE
·································································3分又AB=AC，所以QUOTE
························································5分（2）解：因为∠ABC+∠APC=1800，∠ACB+∠ACD=1800，∠ABC=∠ACB,所以∠APC=∠ACD································································7分又∠CAP=∠DAC,所以ΔAPC~ΔACD