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2017年高考山东理科数学试题及答案(word版)2017年高考山东理科数学试题及答案(word版)2017年高考山东理科数学试题及答案(word版)#2017年普通高等学校招生全国统一考试(山东卷)数学(理科)第Ⅰ卷(共50分)一、选择题:本大题共10小题,每小题5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)【2017年山东,理1,5分】设函数EMBEDEquation.3的定义域为EMBEDEquation.DSMT4,函数EMBEDEquation.3的定义域为EMBEDEquation.DSMT4,则EMBEDEquation.DSMT4()(A)EMBEDEquation.DSMT4(B)EMBEDEquation.DSMT4(C)EMBEDEquation.DSMT4(D)EMBEDEquation.DSMT4【答案】D【解析】由EMBEDEquation.DSMT4得EMBEDEquation.DSMT4,由EMBEDEquation.DSMT4得EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,故选D.(2)【2017年山东,理2,5分】已知EMBEDEquation.3,EMBEDEquation.DSMT4是虚数单位,若EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,则EMBEDEquation.DSMT4()(A)1或EMBEDEquation.DSMT4(B)EMBEDEquation.DSMT4或EMBEDEquation.DSMT4(C)EMBEDEquation.DSMT4(D)EMBEDEquation.DSMT4【答案】A【解析】由EMBEDEquation.DSMT4得EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4,故选A.(3)【2017年山东,理3,5分】已知命题EMBEDEquation.DSMT4:EMBEDEquation.DSMT4,EMBEDEquation.DSMT4;命题EMBEDEquation.DSMT4:若EMBEDEquation.DSMT4,则EMBEDEquation.DSMT4,下列命题为真命题的是()(A)EMBEDEquation.DSMT4(B)EMBEDEquation.DSMT4(C)EMBEDEquation.DSMT4(D)EMBEDEquation.DSMT4【答案】B【解析】由EMBEDEquation.DSMT4时EMBEDEquation.DSMT4有意义,知EMBEDEquation.DSMT4是真命题,由EMBEDEquation.DSMT4可知EMBEDEquation.DSMT4是假命题,即EMBEDEquation.DSMT4,EMBEDEquation.DSMT4均是真命题,故选B.(4)【2017年山东,理4,5分】已知EMBEDEquation.DSMT4、EMBEDEquation.DSMT4满足约束条件EMBEDEquation.DSMT4,则EMBEDEquation.DSMT4的最大值是() (A)0(B)2 (C)5(D)6【答案】C【解析】由EMBEDEquation.DSMT4画出可行域及直线EMBEDEquation.DSMT4如图所示,平移EMBEDEquation.DSMT4发现,当其经过直线EMBEDEquation.DSMT4与EMBEDEquation.DSMT4的交点EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4最大为EMBEDEquation.DSMT4,故选C.(5)【2017年山东,理5,5分】为了研究某班学生的脚长EMBEDEquation.DSMT4(单位:厘米)和身高EMBEDEquation.DSMT4(单位:厘米)的关系,从该班随机抽取10名学生,根据测量数据的散点图可以看出EMBEDEquation.DSMT4与EMBEDEquation.DSMT4之间有线性相关关系,设其回归直线方程为EMBEDEquation.DSMT4,已知EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,该班某学生的脚长为24,据此估计其身高为() (A)160 (B)163(C)166(D)170【答案】C【解析】EMBEDEquation.DSMT4,故选C.(6)【2017年山东,理6,5分】执行两次如图所示的程序框图,若第一次输入的EMBEDEquation.DSMT4值为7,第二次输入的EMBEDEquation.DSMT4值为9,则第一次、第二次输出的EMBEDEquation.DSMT4值分别为()(A)0,0(B)1,1(C)0,1(D)1,0【答案】D【解析】第一次EMBEDEquation.DSMT4;第二次EMBEDEquation.DSMT4,故选D.(7)【2017年山东,理7,5分】若EMBEDEquation.DSMT4,且EMBEDEquation.DSMT4,则下列不等式成立的是()(A)EMBEDEquation.DSMT4(B)EMBEDEquation.DSMT4(C)EMBEDEquation.DSMT4(D)EMBEDEquation.DSMT4【答案】B【解析】EMBEDEquation.DSMT4EMBEDEquation.DSMT4,故选B.(8)【2017年山东,理8,5分】从分别标有1,2,…,9的9张卡片中不放回地随机抽取2次,每次抽取1张,则抽到在2张卡片上的数奇偶性不同的概率是()(A)EMBEDEquation.DSMT4(B)EMBEDEquation.DSMT4(C)EMBEDEquation.DSMT4(D)EMBEDEquation.DSMT4【答案】C【解析】EMBEDEquation.DSMT4,故选C.(9)【2017年山东,理9,5分】在EMBEDEquation.DSMT4中,角EMBEDEquation.DSMT4、EMBEDEquation.DSMT4、EMBEDEquation.DSMT4的对边分别为EMBEDEquation.DSMT4、EMBEDEquation.DSMT4、EMBEDEquation.DSMT4,若EMBEDEquation.DSMT4为锐角三角形,且满足EMBEDEquation.DSMT4,则下列等式成立的是()(A)EMBEDEquation.DSMT4(B)EMBEDEquation.DSMT4(C)EMBEDEquation.DSMT4(D)EMBEDEquation.DSMT4【答案】A【解析】EMBEDEquation.DSMT4所以EMBEDEquation.DSMT4,故选A.(10)【2017年山东,理10,5分】已知当EMBEDEquation.DSMT4时,函数EMBEDEquation.DSMT4的图象与EMBEDEquation.DSMT4的图象有且只有一个交点,则正实数EMBEDEquation.DSMT4的取值范围是()(A)EMBEDEquation.DSMT4(B)EMBEDEquation.DSMT4(C)EMBEDEquation.DSMT4(D)EMBEDEquation.DSMT4【答案】B【解析】当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4单调递减,且EMBEDEquation.DSMT4,EMBEDEquation.DSMT4单调递增,且EMBEDEquation.DSMT4,此时有且仅有一个交点;当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上单调递增,所以要有且仅有一个交点,需EMBEDEquation.DSMT4,故选B.第II卷(共100分)二、填空题:本大题共5小题,每小题5分(11)【2017年山东,理11,5分】已知EMBEDEquation.DSMT4的展开式中含有EMBEDEquation.DSMT4的系数是54,则EMBEDEquation.DSMT4.【答案】4【解析】EMBEDEquation.DSMT4,令EMBEDEquation.DSMT4得:EMBEDEquation.DSMT4,解得EMBEDEquation.DSMT4.(12)【2017年山东,理12,5分】已知EMBEDEquation.DSMT4、EMBEDEquation.DSMT4是互相垂直的单位向量,若EMBEDEquation.DSMT4与EMBEDEquation.DSMT4的夹角为EMBEDEquation.DSMT4,则实数EMBEDEquation.DSMT4的值是.【答案】EMBEDEquation.DSMT4【解析】EMBEDEquation.DSMT4,EMBEDEquation.DSMT4EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4EMBEDEquation.DSMT4,解得:EMBEDEquation.DSMT4.(13)【2017年山东,理13,5分】由一个长方体和两个EMBEDEquation.DSMT4圆柱体构成的几何体的三视图如图,则该几何体的体积为.【答案】EMBEDEquation.DSMT4【解析】该几何体的体积为EMBEDEquation.DSMT4.(14)【2017年山东,理14,5分】在平面直角坐标系EMBEDEquation.DSMT4中,双曲线EMBEDEquation.DSMT4(EMBEDEquation.DSMT4,EMBEDEquation.DSMT4)的右支与焦点为EMBEDEquation.DSMT4的抛物线EMBEDEquation.DSMT4(EMBEDEquation.DSMT4)交于EMBEDEquation.DSMT4、EMBEDEquation.DSMT4两点,若EMBEDEquation.DSMT4,则该双曲线的渐近线方程为.【答案】EMBEDEquation.DSMT4【解析】EMBEDEquation.DSMT4,因为EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4渐近线方程为EMBEDEquation.DSMT4.(15)【2017年山东,理15,5分】若函数EMBEDEquation.DSMT4(EMBEDEquation.DSMT4是自然对数的底数)在EMBEDEquation.DSMT4的定义域上单调递增,则称函数EMBEDEquation.DSMT4具有M性质。下列函数中所有具有M性质的函数的序号为.①EMBEDEquation.DSMT4②EMBEDEquation.DSMT4③EMBEDEquation.DSMT4④EMBEDEquation.DSMT4【答案】=1\*GB3①=4\*GB3④【解析】=1\*GB3①EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上单调递增,故EMBEDEquation.DSMT4具有EMBEDEquation.DSMT4性质;=2\*GB3②EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上单调递减,故EMBEDEquation.DSMT4不具有EMBEDEquation.DSMT4性质;=3\*GB3③EMBEDEquation.DSMT4,令EMBEDEquation.DSMT4,则EMBEDEquation.DSMT4,EMBEDEquation.DSMT4当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上单调递减,在EMBEDEquation.DSMT4上单调递增,故EMBEDEquation.DSMT4不具有EMBEDEquation.DSMT4性质;=4\*GB3④EMBEDEquation.DSMT4,令EMBEDEquation.DSMT4,则EMBEDEquation.DSMT4,EMBEDEquation.DSMT4EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上单调递增,故EMBEDEquation.DSMT4具有EMBEDEquation.DSMT4性质.三、解答题:本大题共6题,共75分.(16)【2017年山东,理16,12分】设函数EMBEDEquation.DSMT4,其中EMBEDEquation.DSMT4,已知EMBEDEquation.DSMT4.QUOTE(1)求EMBEDEquation.DSMT4;(2)将函数EMBEDEquation.DSMT4的图象上各点的横坐标伸长为原来的2倍(纵坐标不变),再将得到的图象向左平移EMBEDEquation.DSMT4个单位,得到函数EMBEDEquation.DSMT4的图象,求EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上的最小值.解:(1)因为EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4,由题设知EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4,EMBEDEquation.DSMT4.故EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,又EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4.(2)由(1)得EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4.因为EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4,当EMBEDEquation.DSMT4,即EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4取得最小值EMBEDEquation.DSMT4.(17)【2017年山东,理17,12分】如图,几何体是圆柱的一部分,它是由矩形EMBEDEquation.DSMT4(及其内部)以EMBEDEquation.DSMT4边所在直线为旋转轴旋转EMBEDEquation.DSMT4得到的,EMBEDEquation.DSMT4是EMBEDEquation.DSMT4的中点.(1)设EMBEDEquation.DSMT4是EMBEDEquation.DSMT4上的一点,且EMBEDEquation.DSMT4,求EMBEDEquation.DSMT4的大小;(2)当EMBEDEquation.DSMT4,EMBEDEquation.DSMT4时,求二面角EMBEDEquation.DSMT4的大小.解:(1)因为EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4EMBEDEquation.DSMT4平面EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4平面EMBEDEquation.DSMT4,又EMBEDEquation.DSMT4平面EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4,又EMBEDEquation.DSMT4,因此EMBEDEquation.DSMT4.(2)解法一:取EMBEDEquation.DSMT4的中点EMBEDEquation.DSMT4,连接EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4.因为EMBEDEquation.DSMT4,所以四边形EMBEDEquation.DSMT4为菱形,所以EMBEDEquation.DSMT4.取EMBEDEquation.DSMT4中点EMBEDEquation.DSMT4,连接EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4.EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4为所求二面角的平面角.EMBEDEquation.DSMT4,EMBEDEquation.DSMT4.在EMBEDEquation.DSMT4中,EMBEDEquation.DSMT4,由余弦定理EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4,因此EMBEDEquation.DSMT4为等边三角形,故所求的角为EMBEDEquation.DSMT4.解法二:以EMBEDEquation.DSMT4为坐标原点,分别以EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4所在的直线为EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4轴,建立如图所示的空间直角坐标系.由题意得EMBEDEquation.DSMT4EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,故EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,设EMBEDEquation.DSMT4是平面EMBEDEquation.DSMT4的一个法向量.由EMBEDEquation.DSMT4可得EMBEDEquation.DSMT4取EMBEDEquation.DSMT4,得平面EMBEDEquation.DSMT4的一个法向量EMBEDEquation.DSMT4.设EMBEDEquation.DSMT4是平面EMBEDEquation.DSMT4的一个法向量.由EMBEDEquation.DSMT4可得EMBEDEquation.DSMT4取EMBEDEquation.DSMT4,可得平面EMBEDEquation.DSMT4的一个法向量EMBEDEquation.DSMT4.所以EMBEDEquation.DSMT4.因此所求的角为EMBEDEquation.DSMT4.(18)【2017年山东,理18,12分】在心理学研究中,常采用对比试验的方法评价不同心理暗示对人的影响,具体方法如下:将参加试验的志愿者随机分成两组,一组接受甲种心理暗示,另一组接受乙种心理暗示,通过对比这两组志愿者接受心理暗示后的结果来评价两种心理暗示的作用,现有6名男志愿者A1,A2,A3,A4,A5,A6和4名女志愿者B1,B2,B3,B4,从中随机抽取5人接受甲种心理暗示,另5人接受乙种心理暗示.(1)求接受甲种心理暗示的志愿者中包含A1但不包含B1的概率;(2)用EMBEDEquation.DSMT4表示接受乙种心理暗示的女志愿者人数,求EMBEDEquation.DSMT4的分布列与数学期望EMBEDEquation.DSMT4.解:(1)记接受甲种心理暗示的志愿者中包含EMBEDEquation.DSMT4但不包含EMBEDEquation.DSMT4的事件为M,则EMBEDEquation.DSMT4.(2)由题意知X可取的值为:0,1,2,3,4.则EMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4因此X的分布列为X01234PEMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4X的数学期望是EMBEDEquation.DSMT4=EMBEDEquation.DSMT4.(19)【2017年山东,理19,12分】已知EMBEDEquation.DSMT4是各项均为正数的等比数列,且EMBEDEquation.DSMT4,EMBEDEquation.DSMT4.(1)求数列EMBEDEquation.DSMT4的通项公式;(2)如图,在平面直角坐标系EMBEDEquation.DSMT4中,依次连接点EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,…,EMBEDEquation.DSMT4得到折线EMBEDEquation.DSMT4,求由该折线与直线EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4所围成的区域的面积EMBEDEquation.3.解:(1)设数列EMBEDEquation.DSMT4的公比为EMBEDEquation.DSMT4,由已知EMBEDEquation.DSMT4.由题意得EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4,因为EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4,因此数列EMBEDEquation.DSMT4的通项公式为EMBEDEquation.DSMT4(2)过EMBEDEquation.DSMT4……EMBEDEquation.DSMT4向EMBEDEquation.DSMT4轴作垂线,垂足分别为EMBEDEquation.DSMT4……EMBEDEquation.DSMT4,由(1)得EMBEDEquation.DSMT4记梯形EMBEDEquation.DSMT4的面积为EMBEDEquation.DSMT4.由题意EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4……+EMBEDEquation.DSMT4EMBEDEquation.DSMT4……+EMBEDEquation.DSMT4=1\*GB3①又EMBEDEquation.DSMT4……+EMBEDEquation.DSMT4=2\*GB3②=1\*GB3①—=2\*GB3②得EMBEDEquation.DSMT4=EMBEDEquation.DSMT4,EMBEDEquation.DSMT4.(20)【2017年山东,理20,13分】已知函数EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,其中EMBEDEquation.DSMT4是自然对数的底数.(1)求曲线EMBEDEquation.DSMT4在点EMBEDEquation.DSMT4处的切线方程;(2)令EMBEDEquation.DSMT4(EMBEDEquation.DSMT4),讨论EMBEDEquation.DSMT4的单调性并判断有无极值,有极值时求出极值.解:(1)由题意EMBEDEquation.DSMT4,又EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4,因此曲线EMBEDEquation.DSMT4在点EMBEDEquation.DSMT4处的切线方程为EMBEDEquation.DSMT4,即EMBEDEquation.DSMT4.(2)由题意得EMBEDEquation.DSMT4,因为EMBEDEquation.DSMT4EMBEDEquation.DSMT4EMBEDEquation.DSMT4,令EMBEDEquation.DSMT4,则EMBEDEquation.DSMT4,所以EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上单调递增.所以当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4单调递减,当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4EMBEDEquation.DSMT4,当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4单调递减,当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4单调递增,所以当EMBEDEquation.DSMT4时EMBEDEquation.DSMT4取得极小值,极小值是EMBEDEquation.DSMT4;当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,由EMBEDEquation.DSMT4,得EMBEDEquation.DSMT4,EMBEDEquation.DSMT4,①当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4单调递增;当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4单调递减;当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4单调递增.所以当EMBEDEquation.DSMT4时EMBEDEquation.DSMT4取得极大值.极大值为EMBEDEquation.DSMT4,当EMBEDEquation.DSMT4时EMBEDEquation.DSMT4取到极小值,极小值是EMBEDEquation.DSMT4;②当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,函数EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上单调递增,无极值;③当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4所以当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4单调递增;当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,EMBEDEquation.DSMT4单调递减;当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4,、EMBEDEquation.DSMT4单调递增;所以当EMBEDEquation.DSMT4时EMBEDEquation.DSMT4取得极大值,极大值是EMBEDEquation.DSMT4;当EMBEDEquation.DSMT4时EMBEDEquation.DSMT4取得极小值.极小值是EMBEDEquation.DSMT4.综上所述:当EMBEDEquation.DSMT4时,EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上单调递减,在EMBEDEquation.DSMT4上单调递增,函数EMBEDEquation.DSMT4有极小值,极小值是EMBEDEquation.DSMT4;当EMBEDEquation.DSMT4时,函数EMBEDEquation.DSMT4在EMBEDEquation.DSMT4和EMBEDEquation.DSMT4和EMBEDEquation.DSMT4上单调递增,在EMBEDEquation.DSMT4上单调递减,函数EMBEDEquation.DSMT4有极大值,也有极小值,极大值是EMBEDEquation.DSMT4,极小值是EMBEDEquation.DSMT4;当EMBEDEquation.DSMT4时,函数EMBEDEquation.DSMT4在EMBEDEquation.DSMT4上单调递增,无极值;当EMBEDEquation.DSMT4时,函数EMBEDEquation.DSMT4在EMBEDEquation.DSMT4和EMBEDEquation.DSMT4上单调递增,在EMBEDEquation.DSMT4上单调递减,函数E
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