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高三第一次周练

高三第一次周练
高三第一次周练

高三第一次周练

考试范围:全部考试时间:2020.8.7 命题人:侯竹春满分:100分

一、选择题(每小题7分,

共56分)

1.设,则

A. B. C. D.

2.某地区经过一年的新农村建设,农村的经济收入增加了一倍.实现翻番.为更好地了解该地区农村的经济收入变化情况,统计了该地区新农村建设前后农村的经济收入构成比例.得到如下饼图:

则下面结论中不正确的是

A.新农村建设后,种植收入减少

B.新农村建设后,其他收入增加了一倍以上

C.新农村建设后,养殖收入增加了一倍

D.新农村建设后,养殖收入与第三产业收入的总和超过了经济收入的一半

3.设为等差数列的前项和,若,,则

A. B. C. D.

4. 函数的图像大致为 ( )

A. B. C. D.

5.设函数,若为奇函数,则曲线在点处的切线方程为

A. B. C. D.

6.某圆柱的高为2,底面周长为16,其三视图如图所示,圆柱表面上的点在正视图上的对应点为,圆柱表面上的点在左视图上的对应点为,则在此圆柱侧面上,从到的路径中,最短路径的长度为()

A. B. C. D.2

7.设抛物线C:y2=4x的焦点为F,过点(–2,0)且斜率为的直线与C交于M,N两点,则=

A.5

B.6

C.7

D.8

8.已知函数.若g(x)存在2个零点,则a的取值

范围是

A.[–1,0)

B.[0,+∞)

C.[–1,+∞)

D.[1,+∞)

二、填空题(每小题7分,

共14分)

9.的展开式中的系数为_________.

10.从2位女生,4位男生中选3人参加科技比赛,且至少有1位女生入选,则不同的选法共有_____________种.(用数字填写答案)

三、解答题(每小题15分,

共30分)

11.在平面四边形中,,,,.

(1)求;

(2)若,求.

12.如图,四边形为正方形,分别为的中点,以为折痕把折起,使点到达点的位置,且.

(1)证明:平面平面;

(2)求与平面所成角的正弦值.

参考答案

一、选择题

1.答案:C

解析:首先根据复数的运算法则,将其化简得到,根据复数模的公式,得到,从而选出正确结果.

解:因为,

所以,故选C.

2.答案:A

解析:首先设出新农村建设前的经济收入为M,根据题意,得到新农村建设后的经济收入为

2M,之后从图中各项收入所占的比例,得到其对应的收入是多少,从而可以比较其大小,并且得到其相应的关系,从而得出正确的选项.

解:设新农村建设前的收入为M,而新农村建设后的收入为2M,

则新农村建设前种植收入为0.6M,而新农村建设后的种植收入为0.74M,所以种植收入增加了,所以A项不正确;

新农村建设前其他收入我0.04M,新农村建设后其他收入为0.1M,故增加了一倍以上,所以B 项正确;

新农村建设前,养殖收入为0.3M,新农村建设后为0.6M,所以增加了一倍,所以C项正确;新农村建设后,养殖收入与第三产业收入的综合占经济收入的,所以超过了经济收入的一半,所以D正确;

故选A.

3.答案:B

解析:首先设出等差数列的公差为,利用等差数列的求和公式,得到公差所满足的等量关系式,从而求得结果,之后应用等差数列的通项公式求得,从而求得正确结果.

解:设该等差数列的公差为,

根据题中的条件可得,

整理解得,所以,故选B.

4.答案:B

解析:通过研究函数奇偶性以及单调性,确定函数图像.

解:为奇函数,舍去A,

舍去D;

所以舍去C;因此选B.

5.答案:D解析:利用奇函数偶此项系数为零求得,进而得到的解析式,再对求导得出切线的斜率,进而求得切线方程.

解:因为函数是奇函数,所以,解得,

所以,,

所以,

所以曲线在点处的切线方程为,

化简可得,故选D.

6.答案:B解析:首先根据题中所给的三视图,得到点M和点N在圆柱上所处的位置,点M在上底面上,点N在下底面上,并且将圆柱的侧面展开图平铺,点M、N在其四分之一的矩形的对角线的端点处,根据平面上两点间直线段最短,利用勾股定理,求得结果.

解:根据圆柱的三视图以及其本身的特征,

可以确定点M和点N分别在以圆柱的高为长方形的宽,圆柱底面圆周长的四分之一为长的长方形的对角线的端点处,

所以所求的最短路径的长度为,故选B.

7.答案:D解析:首先根据题中的条件,利用点斜式写出直线的方程,涉及到直线与抛物线相交,联立方程组,消元化简,求得两点,再利用所给的抛物线的方程,写出其焦点坐标,之后应用向量坐标公式,求得,最后应用向量数量积坐标公式求得结果.

解:根据题意,过点(–2,0)且斜率为的直线方程为,

与抛物线方程联立,消元整理得:,

解得,又,

所以,

从而可以求得,故选D.

8.答案:C解析:首先根据g(x)存在2个零点,得到方程有两个解,将其转化为有两个解,即直线与曲线有两个交点,根据题中所给的函数解析式,画出函数的图像(将去掉),再画出直线,并将其上下移动,从图中可以发现,当时,满足与曲线有两个交点,从而求得结果.

解:画出函数的图像,在y轴右侧的去掉,

再画出直线,之后上下移动,

可以发现当直线过点A时,直线与函数图像有两个交点,

并且向下可以无限移动,都可以保证直线与函数的图像有两个交点,

即方程有两个解,

也就是函数有两个零点,

此时满足,即,故选C.

9. 解析:写出,然后可得结果

解:由题可得

令,则

所以

10.答案:16

解析:首先想到所选的人中没有女生,有多少种选法,再者需要确定从6人中任选3人总共有多少种选法,之后应用减法运算,求得结果.

解:根据题意,没有女生入选有种选法,

从6名学生中任意选3人有种选法,

故至少有1位女生入选,则不同的选法共有种,故答案是16.

11.解析:分析:(1)根据正弦定理可以得到,根据题设条件,求得,结合角的范围,利用同角三角函数关系式,求得;

(2)根据题设条件以及第一问的结论可以求得,之后在中,用余弦定理得到所满足的关系,从而求得结果.

解:(1)在中,由正弦定理得.

由题设知,,所以.

由题设知,,所以.

(2)由题设及(1)知,.

在中,由余弦定理得

.所以.

12.答案:(1).由已知可得,BF⊥PF,BF⊥EF,又,所以BF⊥平面PEF.

又平面ABFD,所以平面PEF⊥平面ABFD.(2)..

解析:分析:(1)首先从题的条件中确定相应的垂直关系,即BF⊥PF,BF⊥EF,又因为,利用线面垂直的判定定理可以得出BF⊥平面PEF,又平面ABFD,利用面面垂直的判定定理证得平面PEF⊥平面ABFD.

(2)结合题意,建立相应的空间直角坐标系,正确写出相应的点的坐标,求得平面ABFD的法向量,设DP与平面ABFD所成角为,利用线面角的定义,可以求得,得到结果.

解:(1)由已知可得,BF⊥PF,BF⊥EF,又,所以BF⊥平面PEF.

又平面ABFD,所以平面PEF⊥平面ABFD.

(2)作PH⊥EF,垂足为H.由(1)得,PH⊥平面ABFD.

以H为坐标原点,的方向为y轴正方向,为单位长,建立如图所示的空间直角坐标系H?xyz.

由(1)可得,DE⊥PE.又DP=2,DE=1,所以PE=.又PF=1,EF=2,故PE⊥PF.

可得.

则为平面ABFD的法向量.

设DP与平面ABFD所成角为,则.

所以DP与平面ABFD所成角的正弦值为.

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