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(三)
寿县一中2017届高三数学(文)第五次月考试卷
一、选择题(共12个小题,每小题5分,共60分.)
1.设集合A{x|1
2
x2},B{x|x21},则AB( ) A.{x|-1x2} B.{x|1
2
x1} C.{x|x2} D.{x|1x2}
2.复数zi
2017
,则z的虚部为( )
A.i B.i C.1 D.1
3.已知直线 l1:ax(a2)y10,l2:xay20,则“l1//l2”是“a1”的 ( ) A.充分不必要条件 B.必要不充分条件 C.充要条件 D.既不充分也不必要条件 4.若函数fxxalnx不是单调函数,则实数a的取值范围是( ). A.0, B.,0
C.,0 D.0, 5
.直线x2y50被圆x2
y2
2x4y0截得的弦长为( ) A.1 B.2 C.4 D
. 6
.函数f(x)(1x)cosx的最小正周期为( ) A.2 B.
32 C. D.2
7.已知程序框图如图所示,则该程序框图的功能是( )
A.求数列{1的前10项和(nΝ) B.求数列{1
n
n的前10项和(nΝ2) C.求数列{1的前11项和(nΝ) D.求数列{1
的前11项和(nΝ)
8
.已知OAuurn
1,uuOBuruuOArg
uuOBur2n0,点C在AOB内,且AOC
4
,设mn, 则n
m
等于( )
A.
12 B
.2
C
.2 9.若实数x,y满足|x3|y1,则z
2xy
xy
的最小值为( ) A.
53 B.2 C.315 D.2
.........................................................................................................................................................................................................................................................
10.已知偶函数yfx对于任意的x0,
2
满足fxcosxfxsinx0,
(其中fx是函数fx的导函数),则下列不等式中成立的是( ) A
3f4 B
3f4 C. f
0
D
.
f4
43
11.设Fx2y2
1、F2是双曲线Ca2b
21(a0,b0)的两个焦点,点P是C上一点,若|PF1||PF2|6a,
且△PF1F2最小内角的大小为30,则双曲线C的离心率是( ) A.
3
2
12.已知函数f(x)
log5(1x),
x1(x2)2
2,
x1
,则方程f(x
1
x
2)a的实根个数不可能为( ) A.8个
B.7个 C.6个 D.5个
二、填空题(本大题共4小题,每题5分,满分20分.)
13.现有5根竹竿,它们的长度(单位:m)分别为2.5,2.6,2.7,2.8,2.9,若从中一次随机抽取2根竹竿,则它们的长度恰好相差0.3m的概率为 .
14.若抛物线y22px(p0)上的点A(x0,2)到其焦点的距离是A到y轴距离的3倍,则p_____.
15.已知fxsin
x30,f6f3,且fx在区间63
有最小值,无最大值,则 .16.课本中介绍了应用祖暅原理推导几何体体积公式的做法.请在研究和理解球的体积公式求法的基础上,解答
x2y2
以下问题:已知椭圆的
标准
excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载
方程为a2b
21(ab0),将此椭圆绕y轴旋转一周后得到椭球体,则它的体积
等于 .
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所对的边分别为a、b、c,且c2,C60. (Ⅰ)求
ab
sinAsinB
的值;
(Ⅱ)若abab,求ABC的面积.
.........................................................................................................................................................................................................................................................
18.(本小题满分12分)
已知等比数列{an}的公比为q(q1),等差数列{bn}的公差也为q,且a12a23a3. (Ⅰ)求q的值;
(Ⅱ)若数列{bn}的首项为2,其前n项和为Tn, 当n2时,试比较bn与Tn的大小.
19.(本小题满分12分)
一个几何体的三视图如下图所示,其中主视图与左视图都是腰长为6的等腰直角三角形,俯视图是正方形. (Ⅰ)请画出该几何体的直观图,并求出它的体积;
(Ⅱ)用多少个这样的几何体可以拼成一个棱长为6的正方体ABCDA 1BC11D1? 如何组拼?试证明你的结论;
21.(本小题满分12分) 已知函数f(x)(2a)lnx
1
2ax. x
(Ⅰ)当a0时,讨论f(x)的单调性;
(Ⅱ)若对任意的a(3,2),x1,x21,3恒有(mln3)a2ln3|f(x1)f(x2)|成立,求实数m的取 值范围.
请考生从22、23两题任选1个小题作答,如果多做,则按所做的第一题计分. 22. (本小题满分10分) 选修4-4:坐标系与参数方程选讲
H是AC(Ⅲ)在(Ⅱ)的正方体ABCDA1BC11D1中,求证:B1D与平面AC11B的交点11B的重心.
20.(本小题满分12分) 已知右焦点为F(c,0)的椭圆M:(Ⅰ)求椭圆M的方程;
1
x1t2
在直角坐标系xoy中,直线l
的参数方程为(t为参数).以原点O为极点,以x轴的正半轴为极轴建
y立极坐标系,曲线C的极坐标方程为6cos. (Ⅰ)写出直线l的普通方程和曲线C的直角坐标方程;
(Ⅱ)若点P的直角坐标为1,0,曲线C与直线l交于A,B两点,求PAPB的值. 23. (本小题满分10分) 选修4-5:不等式选讲 已知函数fxx
(Ⅰ)解关于x的不等式fxx10;
2
xy
1(a>0)关于直线xc对称的图形过坐标原点. a23
22
(Ⅱ)过点(4,0)且不垂直于y轴的直线与椭圆M交于P,Q两点,点Q关于x轴的对称点为E,证明:直线PE
(Ⅱ)若gxx4m,fxgx的解集非空,求实数m的取值范围. 与x轴的交点为F.
寿县一中2017届高三第五次月考数学 (文数)
答案
八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案
一、选择题(每小题5分,共60分)
二、填空题
(每小题5分,共20分) 13.
1425 14. 2 15. 14
3 16. 3
ab
三、解答题(共6大题,共70分)
17、()原式=c2sinCsin60
3
, „„„„„„„„„„„„„„„„„„„6分 (II)由c2
a2
b2
2abcosC可得(ab)2
3ab4, „„„„„„„„„„8分
又abab,所以ab2
3ab40,解得ab4或ab1 (舍去),
所以S1ABC
2absinC1242
„„„„„„„„„„„„„„„12分 18.()由已知可得a12a1q3a21q, „„„„„„„„„„„„„„„„„„„1分 ∵{an}是等比数列,a10∴3q22q10. „„„„„„„„„„„„„„2分 解得q1或q
13. ∵q1, ∴ q1
3
. „„„„„„„„„„„„„„4分 (II)由()知等差数列{bn}的公差为
13, ∴ b1)(7nn2(n3)3
,„5Tnn13n2(n1)(3)n2
n26, „„„7分
Tnbn
(n1)(n14)
6
, „„„„„„9分
当n14
时,Tnbn
;当n14
时,Tnbn
;当2n1时
4,Tnbn故当2n14时,Tnbn;当n14时,Tnbn;当n14时,Tnbn.„12分 19.()该几何体的直观图如图1所示,易知V
13
62
672. „„„„„„„„4分 (II)依题意,正方体体积是原四棱锥体积的3倍,故用三个这样的四棱锥可以拼出一个
棱长为6的正方体, 即由四棱锥D1ABCD,D1BB1C1C,D1BB1A1A组成, 其拼法如图2所示. „„„„„„„„„„„„„„„„„„„„„„„„„8分 (Ⅲ)连接A1H,BH,C1H,易证B1H面A1BC1,又VA
1BC1为等边三角形, 所以易得A1HBHC1H,故H为外心,故H为重心. „„„„„„„„„„12分
20.()由题意得椭圆M的焦点在x轴上,. „„„„„„„„„„„„„„„„„1分 由椭圆M关于直线xc对称的图形过(0,0),故a2c. „„„„„„„„„„3分 由a23c2,得a2
4 , „„„„„„„„„„„„„„„„„„„„„„„„4分
x2 故椭圆M的方程为y2
43
1. „„„„„„„„„„„„„„„„„„„„5分 (II)易知直线PQ的斜率必然存在,设PQ的方程为yk(x4)(k0),代入
x2y2
1得(34k21143
)x232k2x64k2120,由0,得k(2,2).„„7分 设P(x(x32k21,y1),Q(x2,y2),Q2,y2),则x1x234k2
,x64k212
1x234k2
,„„„„8分 则直线PE的方程为yyy1y2
1
x(xx1). „„„„„„„„„„„„„„„„9分
1x2
令y0,得xyx1x2xx1y2x2y1xgk(x24)x2gk(x14)
11=1
y 分 1y2y1y2k(x1x28)
64k21232k22 =2x1x24(x1x2)xx2421 ,28
32k2„„„„„„„„„„„„„11分
134k2
8
故直线PE过定点(1,0),又M的右焦点为(1,0),故直线PE与x轴的交点为F.„12分
.
21.()f(x)
2ax1x2a(2x1)(ax1)
2x2
, 令f(x)0,得x11
1=2,x2a
, „„„„„„„„„„„„„„„„„„„„2分
当a2时,f(x)0,函数f(x)在(0,)上单调递减;
当2a0时,f(x)在(0,12
)和(
1a,)上单调递减,在(11
2,a
)上单调递增; 当a2时,f(x)在(0,1111
a)和(2,)上单调递减,在(a,2
)上单调递增.„6分
(II)由(1)知当a(3,2)时,函数f(x)在区间1,3单调递减; 所以当x1,3时,f(x)maxf(1)12a,
f(x)1
minf(3)(2a)ln33
6a.„„„„„„„„„„„„„„„„„„„„9分 即对任意的a(3,2), 恒有(mln3)a2ln312a(2a)ln31
3
6a成立,
即am22
34a,因为a0,所以m(3a
4)min,
∴实数m的取值范围为(,13
3
].„„„„„„„„„„„„„„„„„„„„„12分22.()直线l
y0; „„„„„„„„„„„„„„2分
曲线C的直角坐标方程为: x32
y2
9.„„„„„„„„„„„„„„„„5分
1
(Ⅱ)把直线的参数方程
x12t(t为参数)代入曲线C的方程化简得:
y t2
2t50. „„„„„„„„„„„„„„„„„„„„„„„„„„„„8分 ∴t1t2
2,t1t250,
∴∣PA∣+∣PB∣=
t1t2=t1
t2
. „„„„„„„10分
23. ()由题意原不等式可化为:
x1x2,
即:x11x2
或x1x2
1,由x11x2
得x1或x2 ,
由x1x2
1得x1或x0, „„„„„„„„„„„„„„„„„„4分 综上原不等式的解为
x|x1或x0. „„„„„„„„„„„„„„„„„5分
(Ⅱ)原不等式等价于x1x4m的解集非空,
令h
xx1x4,即hxx1x4minm,
∴即h
xmin5, „„„„„„„„„„„„„„„„„„„„„„„„„„9分
∴m5. „„„„„„„„„„„„„„„„„„„„„„„„„„„„„10分