南京金陵中学2011年高考数学预测卷2
(满分160分,考试时间120分钟)
一、填空题:本大题共14小题,每小题5分,共计70分.
1.命题“若一个数是负数,则它的平方数正数”的逆命题是 .
2.设全集U={1,3,5,7},集合M={1,a-5},M
U,
={5,7},则实数a= .
3.某工厂生产了某种产品3000件,它们来自甲、乙、丙三条生产线.为检查这批产品的质量,决定采用分层抽样的方法进行抽样.若从甲、乙、丙三条生产线抽取的个数分别为a,b,c,且a,b,c构成等差数列,则乙生产线生产了 件产品.
4.若
=
+
是偶函数,则实数a= .
5.从分别写有0,1,2,3,4五张卡片中取出一张卡片,记下数字后放回,再从中取出一张卡片.两次取出的卡片上的数字之和恰好等于4的概率是 .
6.如右图,函数y=
的图象在点P处的切线方程,y=-x+5,在
-
= .
7.定义某种新运算
:S=a
b的运算原理如图所示,则5
4-3
6= .
8.如图,四边形ABCD中,若AC=
,BD=1,则
= .
9.有三个球和一个正方体,第一个球与正方体的各个面相切,第二个球与正方体的各条棱相切,第三个球过正方体的各个顶点,则这三个球的表面积之比为 .
10.若A,B,C为△ABC的三个内角,则
+
的最小值为 .
11.双曲线
=1(a>0,b>0)的左、右焦点分别是
,
,过
作倾斜角
的直线交双曲线右支于M点,若
垂直于x轴,则双曲线的离心率e= .
12.在平面直角坐标系中,点集A={( x,y) |
+
≤1},B={( x,y) | x≤4,y≥0,3x-4y≥0},则点集Q={( x,y) |x=
+
,y=
+
,(
,
)∈A,(
,
)∈B}所表示的区域的面积为 .
13.已知函数
=
+
+3x+b的图象与x轴有三个不同交点,且交点的横坐标分别可作为抛物线、双曲线、椭圆的离心率,则实数a的取值范围是 .
14.定义函数
=
,其中
表示不超过x的最大整数, 如:
=1,
=-2.当x∈
,
(n∈
)时,设函数
的值域为A,记集合A中的元素个数为
,则式子
的最小值为 .
二、填空题:本大题共6小题,共计70分.请在指定区域内作答,解答时应写出文字说明、证明过程或演算步骤.
15.(本小题满分14分)
在△ABC中,角A,B,C的对边分别是a,b,c,且A,B,C成等差数列.
(1)若
=
,b=
,求a+c的值;
(2)求
的取值范围.
16.(本小题满分14分)
如图,四面体ABCD中,O,E分别为BD,BC的中点,CA=CB=CD=BD=2,AB=AD=
.
(1)求证:AO⊥平面BCD;
(2)求点E到平面ACD的距离.
17.(本小题满分14分)
如图,某市拟在道路的一侧修建一条运动赛道,赛道的前一部分为曲线段ABC,该曲线段为函数y=
(A>0,
>0,
<
<
),x∈[-3,0]的图象,且图象的最高点为B(-1,
);赛道的中间部分为
千米的水平跑到CD;赛道的后一部分为以O圆心的一段圆弧
.
(1)求
,
的值和∠DOE的值;
(2)若要在圆弧赛道所对应的扇形区域内建一个“矩形草坪”,如图所示,矩形的一边在道路AE上,一个顶点在扇形半径OD上.记∠POE=
,求当“矩形草坪”的面积最大时
的值.
18.(本小题满分16分)
在直角坐标系xOy中,直线l与x轴正半轴和y轴正半轴分别相交于A,B两点,△AOB的内切圆为圆M.
(1)如果圆M的半径为1,l与圆M切于点C (
,1+
),求直线l的方程;
(2)如果圆M的半径为1,证明:当△AOB的面积、周长最小时,此时△AOB为同一个三角形;
(3)如果l的方程为x+y-2-
=0,P为圆M上任一点,求
+
+
的最值.
19.(本小题满分16分)
已知数列
满足
=0,
=2,且对任意m,n∈
都有
+
=
+
(1)求
,
;
(2)设
=
-
( n∈
),证明:
是等差数列;
(3)设
=(
-
)
( q≠0,n∈
),求数列的前n项的和
.
20.(本小题满分16分)
对于函数y=
,x∈(0,
,如果a,b,c是一个三角形的三边长,那么
,
,
也是一个三角形的三边长, 则称函数
为“保三角形函数”.
对于函数y=
,x∈
,
,如果a,b,c是任意的非负实数,都有
,
,
是一个三角形的三边长,则称函数
为“恒三角形函数”.
(1)判断三个函数“
=x,
=
,
=
(定义域均为x∈(0,
)”中,那些是“保三角形函数”?请说明理由;
(2)若函数
=
,x∈
,
是“恒三角形函数”,试求实数k的取值范围;
(3)如果函数
是定义在(0,
上的周期函数,且值域也为(0,
,试证明:
既不是“恒三角形函数”,也不是“保三角形函数”.
参考
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
1.若一个数的平方是正数,则它是负数.解析:因为一个命题的逆命题是将原命题的条件与结论进行交换,因此逆命题为:“若一个数的平方是正数,则它是负数”.
2.8.解析:由a-5=3,得a=8.
3.1000.解析:因为a,b,c构成等差数列,根据分层抽样的原理,所以甲、乙、丙三条生产线生产的产品数也成等差数列,其和为3000件,所以乙生产线生产了1000件产品.
4.-3.解析:由
是偶函数可知,
=
对任意的x∈R恒成立,即
+
=
+
,化简得2a=-6,a=-3.
5.
.解析:从0,1,2,3,4五张卡片中取出两张卡片的结果有5×5=25种,数字之和恰好等于4的结果有(0,4),(1,3),(2,2),(3,1),(4,0),所以数字和恰好等于4的概率是P=
.
6.3.解析:函数y=
的解析式未知,但可以由切线y=-x+5的方程求出
=2,而
=
=-1,故
-
=3.
7.1.解析:由题意知5
4=5×(4+1)=25,3
6=6×(3+1)=24,所以5
4-3
6=1.
8.2.解析:
=
=
=
=
=2.
9.1︰2︰3.解析:不妨设正方体的棱长为1,则这三个球的半径依次为
,
,
,从而它们的表面积之比为1︰2︰3.
10.
.解析:因为A+B+C=
,且(A+B+C)·(
+
)=5+4·
+
≥5+
=9,因此
+
≥
,当且仅当4·
=
,即A=2(B+C)时等号成立.
11.
.解析:如图,在Rt△
中,∠
=
,
=2c,所以
=
=
,
=
=
.所以2a=
-
=
-
=
,故e=
=
.
12.18+
.解析:如图所示,点集Q是由三段圆弧以及连接它们的三条切线围成的区域,其面积为:
+
+
+
+
=
×4×3+(3+4+5)×1+
=18+
.
13.(-3,-2).解析:由题意知,三个交点分别为(1,0),(
,0),(
,0),且0<
<1<
.
由
=0可知b=-a-3,所以
=
+
+3x+b=(x-1)(
+ax+a+3),故
+ax+a+3=0的两根分别在(0,1),(1,
)内.
令
=
+ax+a+3,则
得-3<a<-2.
14.13.解析:当x∈
,
时,
=
=
=0;
当x∈
,
时,
=
=
=
=1;
当x∈
,
时,再将
,
等分成两段,x∈
,
时,
=
=
=
=4;x∈
,
时,
=
=
=
=5.
类似地,当x∈
,
时,还要将
,
等分成三段,又得3个函数值;将
,
等分成四段,得4个函数值,如此下去.当x∈
,
(n∈
)时,函数
的值域中的元素个数为
=1+1+2+3+4+…+(n-1)=1+
,于是
=
+
-
=
-
,所以当n=13或n=14时,
的最小值为13.
15.解析:(1)因为A,B,C成等差数列,所以B=
.
因为
=
,所以
=
,所以
=
,即ac=3.
因为b=
,
,所以
=3,即
=3.
所以
=12,所以a+c=
.
(2)
=
=
=
.
因为0<C<
,所以
∈
.
所以
的取值范围是
.
16.解析:(1)连结OC.因为BO=DO,AB=AD,所以AO⊥BD.因为BO=DO,CB=CD,所以CO⊥BD.
在△AOC中,由已知可得AO=1,CO=
.而AC=2,所以
=
,所以∠AOC=
,即AO⊥OC.因为BD
OC=O,所以AO⊥平面BCD.
(2)设点E到平面ACD的距离为h.因为
=
,所以
=
EMBED Equation.DSMT4 .
在△ACD中,CA=CD=2,AD=
,所以
=
=
.
而AO=1,
=
=
,所以h=
=
=
.
所以点E到平面ACD的距离为
.
17.解析:(1)依题意,得A=
,
=2,因为T=
,所以
=
,所以y=
.
当x=-1时,
=
,由
<
<
,得
=
,所以
=
.
又x=0时,y=OC=3,因为CD=
,所以∠COD=
,从而∠DOE=
.
(2)由(1)可知OD=OP=
,“矩形草坪”的面积
S=
EMBED Equation.DSMT4 =
=
=
,
其中0<
<
,所以当
=
,即
=
时,S最大.
18.解析:(1)由题可得
=
,
=
.所以l:y=
+
+1.
(2)设A(a,0),B(0,b) (a>2,b>2),则l:bx+ay-ab=0.由题可得M (1,1).
所以点M到直线l的距离d=
=1,整理得(a-2)(b-2)=2,即ab-2(a+b)+2=0.于是ab+2=2(a+b)≥
,
≥2+
,ab≥6+
.当且仅当a=b=2;所以面积S=
≥3+
,此时△AOB为直角边长为2+
的等腰直角三角形.
周长L=a+b+
≥
+
=(2+
)·
≥
=6+
,此时△AOB为直角边长为2+
的等腰直角三角形.
所以此时的△AOB为同一个三角形.
(3)l的方程为x+y-2-
=0,得A(2+
,0),B(0,2+
),
:
+
=1,设P(m,n)为圆上任一点,则
+
=1,
+
=2(m+n)-1,
+
=1≥
,2-
≤m+n≤2+
.
+
+
=
+
-(4+
)(m+n)+
=(9+
)-(
-2)(m+n).
当m+n=2-
时,
=(9+
)-(
-2)( 2-
)=17+
.此时,m=n=1-
.
当m+n=2+
时,
=(9+
)-(
-2)( 2+
)=9+
.此时,m=n=1+
.
19.解析:(1)由题意,令m=2,n=1,可得
=
-
+2=6,再令m=3,n=1,可得
=
-
+8=20.
(2)当n∈
时,由已知(以n+2代替m)可得
+
=
+8,于是[
-
]-(
-
)=8,即
-
=8.所以
是公差为8的等差数列.
(3)由(1)(2)可知
是首项
=
-
=6,公差为8的等差数列,则
=8n-2,即
-
=8n-2.另由已知(令m=1)可得,
=
-
.那么
-
=
-2n+1=
-2n+1=2n,于是
=
.
当q=1时,
=2+4+6+…+2n=n (n+1).
当q≠1时,
=2·
+4·
+6·
+…+2n·
,两边同乘以q,可得
=2·
+4·
+6·
+…+2n·
.上述两式相减,得
=
-2n
=
-2n
=
,
所以
=
.
综上所述,
=
20.解析:(1)对于
=x,它在(0,
上是增函数,不妨设a≤b≤c,则
≤
≤
,因为a+b>c,所以
+
=a+b>c=
,故
是“保三角形函数”.
对于
=
,它在(0,
上是增函数,,不妨设a≤b≤c,则
≤
≤
,因为a+b>c,所以
+
=
+
=
>
>
=
,故
是“保三角形函数”.
对于
=
,取a=3,b=3,c=5,显然a,b,c是一个三角形的三边长,但因为
+
=
<
=
,所以
,
,
不是三角形的三边长,故
不是“保三角形函数”.
(2)解法1:因为
=1+
,所以当x=0时,
=1;当x>0时,
=1+
.
①当k=-1时,因为
=1,适合题意.
②当k>-1时,因为
=1+
≤1+
=k+2,所以
∈
,
.从而当k>
-1时,
∈
,
.由1+1>k+2,得k<0,所以-1<k<0.
③当k<-1时,因为
=1+
≥1+
=k+2,所以
∈
,
,从而当k>-1时,所以
∈
,
.由
得,k>
,所以
<k<-1.
综上所述,所求k的取值范围是(
,0).
解法2:因为
=
=
,
①当k=-1时,因为
=1,适合题意.
②当k>-1时,可知
在
,
上单调递增,在
,
上单调递减,而
=1,
=k+2,且当x>1时,
>1,所以此时
∈
,
.
③当k<-1时,可知
在
,
上单调递减,在
,
上单调递增,而
=1,
=k+2,且当x>1时,
<1,所以此时
∈
,
.
(以下同解法1)
(3)①因为
的值域是(0,
,所以存在正实数a,b,c,使得
=1,
=1,
=2,显然这样的
,
,
不是一个三角形的三边长.
故
不是“恒三角形函数”.
②因为
的最小正周期为T(T>0),令a=b=m+kT,c=n,其中k∈
,且k>
,则a+b>c,又显然b+c>a,c+a>b,所以a,b,c是一个三角形的三边长.
但因为
=
=
=1,
=
=2,所以
,
,
不是一个三角形的三边长.
故
也不是“保三角形函数”.
(说明:也可以先证
不是“保三角形函数”,然后根据此知
也不是“恒三角形函数”.)
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