普通高中教学检测数学参考答案及评分建议一、选择题（每小题5分，共40分）12345678DCCBBABD二、选择题（每小题6分，共18分。在每小题给出的选项中有三个正确，选对一个给2分；有二个选项正确，选对一个给3分。出现错误选项，该小题0分）91011BCABDACD三、填空题（每小题5分，共15分）1212．13．14．2或4(2分，答案不完整不给分）；11（3分）33四、解答题15.解：（1）由已知，切点坐标为(0,1).····························································1分又f'(x)=2cos2x+2ax−1，······························································3分则切线斜率k=f'(0)=1，所以切线方程为y=x+1；···································································5分（2）由（1）知f'(x)=2cos2x+2ax−1，则f(x)=−4sin2x+2a，·······6分2因为x[−,]，2x[−,]，····················································7分6333数学参考答案及评分建议第1页（共7页）学科网（北京）股份有限公司3所以−≤sin2x≤1，······································································8分2因为f'(x)在[−,]上具有单调性，63则fx()0≥或fx()0≤恒成立，··························································9分即x[−,]时，有2ax≥4sin2，或2ax≤4sin2，63即24a≥，或22a≤3−，有a≥2，或a≤−3.所以a的取值范围是(−,−3][2,+).·············································13分16．解：(1)零假设H0:喜欢户外运动与性别无关.··················································1分由列联表可知400(10080−100120)22=4.040··················5分(100+100)(120+80)(100+120)(100+80)由于4.0403.841＞，···········································································6分故零假设不成立，所以喜欢户外运动与性别有关.··································7分（2）不喜欢户外运动的员工按分层抽样抽取9人，其中女员工4人.··············8分X可能的取值为0，1，2，3.····································································9分03C4C55P(X=0)=3=，C94212C4C520P(X=1)=3=，C942数学参考答案及评分建议第2页（共7页）学科网（北京）股份有限公司21C4C515P(X=2)=3=，C94230C4C52P(X=3)=3=.·····································································13分C942X的分布列为0123520152P42424242的期望为520152564E（X）=0+1+2+3==.·······························15分4242424242317．解：（1）由已知c=2，············································································1分因为△F1PQ的周长为4a，··································································2分即4a=82，a=22，····································································3分所以b2=a2−c2=4.x2y2椭圆E方程为+=1；·································································5分84（2）设l:x=my+2(m0)，P（x1,y1），Q（x2,y2）,M（t,y1），N（t,y2）.········································································································6分xmy=+2联立xy22，得(m2+2)y2+4my−4=0，····································7分+=184−4m−4所以y+y=，yy=，··············································8分12m2+212m2+2y1−y2−y2(x1−t)直线PN方程为y−y2=(x−t).令y=0，得x=+t.x1−ty1−y2数学参考答案及评分建议第3页（共7页）学科网（北京）股份有限公司y2−y1−y1(x2−t)直线QM方程为y−y1=(x−t).令y=0，得x=+t，···10分x2−ty2−y1−y2(x1−t)−y1(x2−t)由已知，有=，·················································11分y1−y2y2−y1即y2x1−y2t=−y1x2+y1t，即y2x1+y1x2−t(y1+y2)=0，······················12分化简得2my1y2+(2−t)(y1+y2)=0，····················································13分−4−4m则2m+(2−t)=0，m2+2m2+2有2+2−t=0，t=4.········································································15分18．解：(1)取线段BC中点E，连接EA，ED.·················································1分因为△ABC是正三角形，有BC⊥AE.··················································2分又BC⊥AD,所以BC⊥平面ADE，·······································································3分有BC⊥DE，···············································································4分又E为线段中点，则BD=CD.···················································································5分（2）（i）连接EP，AP由（1）可知，CE⊥EP，···································································6分有CP2=CE2+EP2=1+EP2.···························································7分3EP≥AE−=AP（当且仅当点P在线段AE的中点处，等号成立），······8分27则CP=+1EP2≥.27所以线段CP的最小值为；····························································10分2数学参考答案及评分建议第4页（共7页）学科网（北京）股份有限公司（ii）由（1）及已知，平面ADE⊥平面BCD.······································································································11分以E为坐标原点，以EC，ED分别作为x轴，y轴正方向建立空间直角坐标系如图所示.····························································································12分33则A(0,,)，C(1,0,0)，D(0,3,0)，2233于是CD=(−1,3,0)，AD=(0,,−).·········································13分22−x+3y=0,设平面ACD的法向量为m=(x,y,z)，有33y−z=0.22令y=3，有x=3，z=1，即m=(3,3,1).······································14分设直线l的方向向量n=(0,m,n)，直线与平面ACD所成角为，|3m+n|3m2+n2+23mn于是.·································15分sin==2213m2+n213(m+n)因为23233mnmnmn=+（）≤22（当且仅当m=3n时取“=”），（42mn1322+）所以sin≤=.······················································16分1313（mn22+）当n=(0,1,−3)，sin=0.213故0≤sin≤.13数学参考答案及评分建议第5页（共7页）学科网（北京）股份有限公司213即直线l与平面ACD所成角的正弦值的取值范围是[0,].·················17分1319．解：1sinan+11（1）已知tanan+1=，即=，··································1分cosancosan+1cosan22sinan+111−cosan+11即2=2，2=2，··································3分cosan+1cosancosan+1cosan111化简得2−2=1，又2=2cosan+1cosancosa11所以数列2是首项为2公差为1的等差数列；································5分cosan1（2）由（1）可知2=2+(n−1)=n+1，cosan11n所以cos2a=，sin2a=1−=.nn+1nn+1n+1又a(0,)，n21n所以cosa=，sina=.················································6分nn+1nn+1n1b=log(sina)=log=[logn−log(n+1)].n2n2n+12221所以S=[log1−log2+log2−log3++logn−log(n+1)]n22222221=−log(n+1).·······························