黑龙江省齐齐哈尔市2021-2022学年高二上学期期末考试数学试题答案

高二数学试卷参考答案及评分标准一.单选题：本大题共8小题，每小题5分，共40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.题号12345678答案BCDCAACD二.多选题：本大题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求，全部选对的得5分，有选错的得0分，部分选对得2分.题号9101112答案ABABACDABD三.填空题：本大题共4小题，每小题5分，共20分.把正确答案写在答题卡相应题的横线上.322n13、14、8415、116、x（注：第一空2分；第二空3分）2n1四.解答题：共70分，解答应写出文字说明、解答过程或演算步骤.17.（10分）解：（1）【解法一】2直线l的斜率为k··················································································································································1分l32l//m直线m的斜率为k·························································································································2分m32直线m的方程为y3(x2)······················································································································3分3即2x3y130····························································································································································4分【解法二】l//m直线m的方程为2x3yC0············································································································1分直线m过点(2,3)2233C0C13························································································3分直线m的方程为2x3y130··························································································································4分（2）易知圆心C(1,2)，半径r1······························································································································5分|21326|2213圆心C到直线l的距离d······································································7分22321313224613|AB|2rd21··················································································································9分1313613线段AB的长为···············································································································································10分13高二数学试卷参考答案及评分标准第1页（共4页） 18.（12分）a22a5123a19d12a11解：（1）设等差数列{an}的公差为d·····································3分3a42a110a19d10d1a1(n1)1n等差数列{a}的通项公式为an···········································································5分nnnn（2）bn2··································································································································································6分n123n1nS122232(n1)2n2①·····················································································7分n234nn12S122232(n1)2n2②·····················································································8分n23nn1①－②得：S2222n2·······································································································9分nn2(12)n1Sn2············································································································································10分n12n1数列{b}的前n项和S(n1)22············································································································12分nn19.（12分）解：（1）3csinBbcos(C)，由正弦定理可得：3sinCsinBsinBcos(C)····················1分33130BsinB03sinCcosCsinC·····················································································3分223tanC··········································································································································································5分30CC角C的大小为·····················································································································6分6622222（2）cab2abcosCc12b6b·······························································································7分222222223c3b2ac8b8b12b6bb3b20b1，或b2···················9分13①当b1时，c7SabsinC·····························································································10分ABC221②当b2时，c2SabsinC3································································································11分ABC23ABC的面积为3或·········································································································································12分2高二数学试卷参考答案及评分标准第2页（共4页） 20.（12分）证明：（1）四边形ABCD是等腰梯形，AD//BCOD1,OB2··················1分PEDO1连接EO，EO//PD··································································2分PBDB3EO平面AEC，PD平面AEC·····································································3分PD//平面AEC·············································································································4分（2）PO平面ABCD，AC平面ABCDPOACOPtanPAC22OP2·························································································································5分OAOBOC以O为坐标原点，分别以OB，OC，OP所在直线为x，y，z轴，建立空间直角坐标系OxyzO(0,0,0)，P(0,0,2)，A(0,1,0)，C(0,2,0)，B(2,0,0)······························································7分PA(0,1,2)，PC(0,2,2)，PB(2,0,2)，···························································································8分nPC2y2z0设平面PBC的法向量为n(x,y,z)nPB2x2z0令x1y1，z1n(1,1,1)···························································································································10分|PAn|315设PA与平面PBC所成角为sin|cosPA,n|··························11分|PA||n|53515PA与平面PBC所成角的正弦值为···············································································································12分521.（12分）b3解：（1）由题，可知a2········································································································································2分2b23a2解得·······································································································································································3分b322xy双曲线C的方程为1.·······························································································································4分4322xy122（2）43,消y得:(34k)x8kx160·······················································································6分ykx1高二数学试卷参考答案及评分标准第3页（共4页） 234k031k1，且k··············································································7分222(8k)64(34k)08k16设A(x,y)，B(x,y)xx，xx··············································································8分112212212234k34k2OAOBxxyyxx(kx1)(kx1)(1k)xxk(xx)11·······························9分12121212121222216(1k)8k1016k23320034k0k，或k··················11分22234k34k34k2233k的取值范围为1k，或k1································································································12分2222.（12分）log3(xn1)解：由题可知y(x0，nN)，·········································································································1分nnx1n*xn1（1）n2，且nN,x3x2x13(x1)，3(常数)··································2分nn1nn1x1n1x13{x1}是首项为3，公比为3的等比数列································································································3分1nn1nnnx1333x31数列{x}的通项公式为x31···························································4分nnnnnlog3nyn13n1（2）y01yy数列{y}单调递减··········································5分nnnn1nn33y3nn12y最大值为y·9t18mt0恒成立···········································································································6分n1329t18t0m[1,1]t2，或t229t18t0t的取值范围为(，2)(2，)···························································································································7分(yy)(xx)4n1nn1n1n（3）四边形PQQP的面积是T································································8分nnn1n1n2333113()·························································································································10分n(4n1)n(n1)nn1111111111113[(1)()()()]3(1)·······················11分T2TnT22334nn1n112n*13nN3(1)33n1n11113.············································································································································12分T2TnT12n如有其他解法，保证各问分值不变的前提下，酌情给分.高二数学试卷参考答案及评分标准第4页（共4页）