中学生标准学术能力诊断性测试2024年1月测试数学参考答案一、单项选择题：本题共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．12345678ADBACCAC二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对但不全的得2分，有错选的得0分．9101112ABBCDCDACD三、填空题：本题共4小题，每小题5分，共20分．5．14．134131．，15．216−2四、解答题：本题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤．17．（10分）21（1）当n2时，数列an的前n项和为Sn，满足SaSnnn=−，222111即SSSSSSSnnnnnnn=−−=−−+(nnSS−−−111)，222整理可得2SSSSnnnn−−11=−········································································1分1S=1，则2SSSS=−，即21SS=−，可得S=·······························2分121122223211由2SSSS=−，即SS=−，可得S=,,2323333335*以此类推可知，对任意的nSN,0n，11在等式两边同时除以SSnn−1可得−=2·······················4分SSnn−1第1页共9页11所以数列为等差数列，且其首项为=1，公差为2·································5分SSn111=+−=−12121nn()，因此，Sn=············································6分Sn21n−nS2111111（）解：b==+=+−n1，2n2142121482121nnnnn+−+−+()()n11=+−Tn1············································································8分4821n+2*22不等式Tnm−3m+n对所有的nN恒成立，则mm−+30，39+579−57即m或m·····································································9分66因此，满足条件的正整数m的最小值为3······················································10分18．（12分）1（1）证明：由==BACBADDAC,，知==BADDAC,，2263111SSSAB=++=ADADACAB,sinsinAC，ABCABDACD26232即ADADACAC+=32，21两边同除以ADAC，得−=3······················································5分ADAC（2）设=BAD，则=DAC2，ABBDABD中，由正弦定理，得=①，sinsinBDAACDCACD中，由正弦定理，得=②，sinCDAsin22②①，结合sinsin,4==BDACDADCBD，得AC=····················7分cos1sin33sin−4sin3S=ABACsin3===3tan−4tansin2ABC2coscos第2页共9页tan3tantan23−=−=3tan4tan····································9分1tan1tan++223tt−3设tan0,3=t，即求函数yt=,0,3的最大值，()1+t2()(3−3t2)(1+t2)−(3t−t3)2t(23−3−tt22)(23+3+)y=22=，(11++tt22)()t2−(0,233)时，y0，函数单调递增；t2−(233,3)时，y0，函数单调递减，当2时，函数有最大值，，t=−233ymax=−639ABC面积的最大值为639−····························································12分19．（12分）（1）记蚂蚁爬行n次在底面ABCD的概率为Pn，212由题意可得，PPPP==+−,1()····················································3分11333nnn+111111PPPnnn+1−=−−−,是等比数列，首项为，公比为−，232263111111nn−−11························································5分PPnn−=−=+−,263263（2）X=0,1,2，X=2时，蚂蚁第3次、第5次都在C处，1121211212211111PX(==2222)++++=66363663633666618··············································································································7分X=1时，蚂蚁第3次在C处或第5次在C处，设蚂蚁第3次在C处的概率为P1，1121211211515211P1=++222++=66363663666663318··············································································································8分设蚂蚁第5次在C处的概率为P2，第3页共9页设蚂蚁不过点C且第3次在D1的概率为P3，设蚂蚁不过点C且第3次在B1的概率为P4，设蚂蚁不过点C且第3次在A的概率为P5，由对称性知，PP34=，11121213P=+=43，36663635412122211P=+=6，56363332712117得PPP=+=222···················································11分2356366545==+=PXPP(1)，122741PXPXPX(==−=−==0112)()()，54X的分布列为：X0124151P5427188X的数学期望E(XP)=XP001122=+XP(=+X)==()()············12分2720．（12分）（1）过点E作AM的平行线交AD于点F，过点N作AB的平行线交AC于点G，连接FG．因1为点E是线段DM的中点，BNNC=3，==EFNGAM，且EFNG，四边2形EFGN是平行四边形．由NEFGNE,平面DAC，FG平面DAC，NE平面DAC······················································································5分（2）解法1：以点A为原点，AB，AC所在的直线为x轴、y轴，过点A垂直于平面ABC的直线为z轴，建立空间直角坐标系·····································································6分13设ABAC==2，则AMN(0,0,0,),(1,0,0),,,0，设D(x,,,yz)，22第4页共9页因为平面DMC⊥平面ABC，所以点D在平面ABC上的射影落在直线CM上，y+x=1①，232由题意可知，DMDNxyz==−++=1,2,11()22②，2221329x−+y−+z=③，2228221182211由①②③解得，x=,,,,,y=−z=D−··························8分77777782211816211ADCD=−=−,,,,,，777777设平面ACD的法向量为nx=yz(,,)，AD=n04x−y+11z=0，即，取xyz===11,0,4−······················11分CD=n04x−8y+11z=0取平面ABC的法向量m=(0,0,1)．设二面角D−−ACB的平面角为，mn43则coscos,===mn，mn943所以，二面角的余弦值为···················································12分9解法2：如图，过点B作直线MN的垂线交于点I，交直线CM于点H．由题意知，点D在底面ABC上的射影在直线BI上且在直线MC上，所以点H即点D在底面上的射影，即DH⊥平面ABC·····················································································6分310设AB=2，则BM=1,BN=2,MBN=，由余弦定理，得MN=，2421031010cosBMN=−,sinBMN=,MI=BMcos(−BMN)=，101010第5页共9页1053102572coscos=−IMHIMBHMB=+=()，10510510MI5MH==．cos7IMH过点H作AC的垂线交于点O，连接DO，由三垂线定理知，DOA⊥C，DOH是二面角D−−ACB的平面角········································································9分AMCM8211由=，解得HODHDMMH==−=,22，HOCH77DH1143tan==DOH，得cos=DOH，HO4943所以，二面角的余弦值为··················································12分921．（12分）2（1）设点CxyDxy(1122,,,)()，设直线l的方程为ykxk=+10()，代入抛物线yx=−1，得xkx2−−=20（*），CMDMkxk11++x222172+k2===−1221,2··········4分CDk222+8218+−+kxxk12221（2）CxxD(1122,1xxQ,,1−−−),,0()，设Tm(n,)，k由（*）式，知xxkx1212+==x−,2······························································5分直线AC的方程为yxx=−+(111)()，直线BD的方程为yxx=+−(211)()，xx+21(xx23+x−xx−−)x−()解得xy===12,121212，x2−xx12+−222xx12+−1x+xx+23(xx12−−)所以点P的坐标为12,···············································7分xxxx2121−+−+22xx12+23(xx12−−)1TPmn=−−=TQm−−n−,,,，x2−xx12+−22xk1+xx12+123(x