2019年龙泉驿区高2017级统一模拟试题数学（理）参考答案

12019年龙泉驿区高2017级统一模拟试题数学（理）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 一、选择题二、填空题13.02yx,14.2,15.4,16.4.三、填空题17.解：（1）由 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 中数据得2K的观测值828.105.1220302030)661424(5022K，所以根据统计有99.9%的把握认为视觉和空间能力与性别有关…………5分（2）由题可知在选择做立体几何题的6名女生中任意抽取三人，抽取 方法 快递客服问题件处理详细方法山木方法pdf计算方法pdf华与华方法下载八字理论方法下载 有2036C种，其中甲、乙、丙三人都没被抽到有133C种；恰有一人被抽到有92313CC种；恰两人被抽到有91323CC种，三人都被抽到有331C∴X可能取值为0，1，2，3201)0(XP，209)1(XP，209)2(XP，201)0(XPX的分布列为：∴232013209220912010)(XE…………12分18.（1） 证明 住所证明下载场所使用证明下载诊断证明下载住所证明下载爱问住所证明下载爱问 ：令PD中点为F，连接EF，点,EF分别是PCD的中点，EF//12CD,EF//AB.四边形FABE为平行四边形.…………4分题号123456789101112答案CBDCCDACCCAAX0123P2012092092012//BEAF,AF平面PAD,EF平面PAD(三个条件少写一个不得该步骤分)PADBE面//…………6分（2）作PCBH于H，在等腰梯形PABC中，PCADPCAB,//，62PCADAB，，,2CHDHPDADABABCDPAD面侧面,ADPDPD面ABCD…………………7分以D为原点，,,DADCDP所在直线为,,xyz轴建立空间直角坐标系．则)0,2,2(),0,4,0(),2,0,0(),0,0,0(BCPD.PC中点E的坐标为（0,2,1）易知）（2,0,0DP即平面BDC的法向量．…………………8分设面BDE的法向量为,,mxyz则00DBmDEm．即002yxzy令1x，得2,1zy，得)2,1,1(m．…………………10分3621-124,cos222）（DPm∴二面角QBDP的余弦值为36．…………12分（采用其它方法求出的参照相应步骤给分）19.解：（1）由mn得0coscos2AcCba，…………1分则由正弦定理得0cossincossin2sinACCBA…………2分得CBCAcossin2)sin(，即CBBcossin2sin由于sin0B，得21cosC，…………4分又因2,0C，因此60C.…………………5分3（2）∵c=6，sinC=32∴由正弦定理得：622sinsinBsinC32abcA,……………6分即a=22sinA，b=22sinB，又A+B=π-C=23，即B=23-A∴a+b+c=22（sinA+sinB）+6=22[sinA+sin（23-A）]+6=22（sinA+sin23cosA-cos23sinA）+6=26（sinAcos6+cosAsin6）+6=26sin（A+6）+6，……………9分∵△ABC是锐角三角形，∴6＜A＜2，∴32＜sin（A+6）≤1，……………11分则△ABC周长的取值范围是（6+32，36]．……………12分20.解：（1）由已知可得222123cbabace，∴2a，1b，∴椭圆的方程2214xy.……………4分（2）当0k时，直线和椭圆有两交点只需11m；……………5分当0k时，设点P、Q的坐标分别为2211,,,yxyx，弦PQ的中点为00,yxR，由1422yxmkxy，得014814222mkmxxk,由于直线与椭圆有两个不同的交点，所以40，即1422km①因14422210kmkxxx，从而14200kmmkxy，因0,0mk时显然不合题意，则00x，∴kmkmxykMR4141200又∵MQMP∴1431414,22kmkkmkmPQMR，即则②……10分将②代入①得mm32，解得30m，由②得01342mk得31m，故所求的m取值范围是3,31.……………11分综上知，0k时，m的取值范围是3,31；0k时，m的取值范围是11（－，）……………12分21.解：（1）由题，2(2)exfxx，所以当2x时，0fx，fx在(,2)上单调递增．当2x时，0fx，fx在(2,)上单调递减，fx有极大值(2)1fa.···············································3分且当x时，ya；x时，y．所以，当0a或1a时，fx恰有一个零点；10a时，fx有两个零点；1a时，fx没有零点．··········································································5分（2）由（1）可知，2(3)e1xx．①当1x时，不等式22(43)eln20xxxxxx可化为2(3)eln21xxxxxx．·····························································6分5记ln21xxxgxx，得22ln1xxgxx．设2lnnxxx，则1110xnxxx，∴nx在1，是单调增函数．又31ln30,42ln40nn，nx在3,4上图象是不间断的，∴存在唯一的实数034x,，使得00nx．∴当01xx时，00nxgxgx，，在01x,上递减，当0xx时，0,0nxgxgx，在0,x上递增．∴当0xx时，gx有极小值，即为最小值，0000012lnxxxgxx．·································································································9分又0002ln0nxxx，所以00ln2xx，所以002gxx．又03,4x，002(1,2)gxx，∴0()1gxgx．所以，2ln21(3)e1xxxxxx，即22(43)eln20xxxxxx．·································································································10分②当01x时，设()ln2hxxxx，则()ln0hxx．()hx在(0,1]单调递减，()(1)10hxh，所以222(43)eln2(1)(3)eln20xxxxxxxxxxxx．综上所述，22(43)eln20xxxxxx．···································12分22.解：（1）由12112xyxtttxyyttt，，，两式相乘得，224xy．·····························································1分因为cos,sin,xy所以曲线C的极坐标方程为2222cossin4，即2cos24．··········································································3分因为cos6，所以cos62，则曲线2C的直角坐标方程为0622xyx．·································5分6（2）联立2cos246，，得22A，·············································7分联立6cos6得33B，··················································9分故2233BAAB．··················································10分23.解：（1）由题意，可得01)3(xxf，即||1x1011xx，或，···················································2分又因为解集为(,1][1,),所以211．································································5分（2）不等式130)(axxaxxf，··························6分axx3表示数轴上到点3x和ax的距离之和，····················8分则2a或4a．于是，当关于x的不等式()||0fxxa≥对xR恒成立时，实数a的取值范围是，42,．······································································10分（其它解法的参照相应步骤给分）