2023年福州市普通高中毕业班质量检测数学试题参考答案与评分细则一、单项选择题：1-8DBCABDCD8【解析】因为f(x)+g(x−32)=，所以f(x+32)+g(x)=，又f(12−x)+g(x)=，则有f(x+31)=f(−x)，因为fx(+1)是奇函数，所以f(x+11)=−f(−x)，可得f(x+31)=−f(x+)，即有f(x+2)=−f(x)与f(x+42)=−f(x+)，即f(x+=4)f(x)，所以fx()是周期为4的周期函数，故gx()也是周期为的周期函数.因为−f(−x)=f(x+2)，所以f(−=x)f(x)，所以fx()为偶函数.故A错误；由fx(+1)是奇函数，则f(10)=，所以f(30)=，又f(2)+f(4)=f(2)+f(0)=0，20所以f(k)=5f(1)+f(2)+f(3)+f(4)=0，所以C选项错误；k=1由f(10)=得g(02)=，所以B选项错误；因为g(2)=2−f(5)=2−f(1)=2，g(1)32+g()=−f(4)+2−f(64)=−f(4)+f(24)=，20所以g(0)+g(1)+g(2)+g(3)=8，所以g(k)=5g(0)+g(1)+g(2)+g(3)=40，k=1所以D选项正确.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，部分选对的得2分，有选错的得0分．9.BCD10.BD11.BC12.AD12xy+=3cos,1322x=−3cossin,12【解析一】由x+y+y=3，令可得243y=2sin,y=3sin,2故，故正确；错误；2xy+=23cos−2sin+2sin=23cos−23,23AB22x+−=−=−yxy32xy32(3cos−sin)2sin=−54sin2+1,9，故C错误；D6正确，故选AD.12【解析二】令2x+=yt，则y=−t2x代入x22+xy+y=3，整理可得，3x22−3tx+t−3=0，因为xR，所以=9t2−12(t2−3)=−3(t2−12)0，解得−23t23，故A正确；B错数学参考答案第1页共12页误；由x2+xy+y2=33x2+y2=−xy，因为x22+y2xy，所以32−xyxy，所以−31xy，又x22+xy+y=32−xy，从而13−2xy9，故C错误；D正确，故选AD.v212【解析三】令x=+uv，y=−uv，代入x22+xy+y=3，得u2+=1，3令uv==cos,3sin（0,2π）.π2x+y=3u+v=3cos+3sin=23sin+−23,23，故A正确，B错误；3x2+y2−xy=u2+3v2=cos2+9sin2=1+8sin21,9，故D正确，C错误；故选AD.三、填空题：本大题共4小题，每小题5分，共20分．2413.−14.0.6415.1116.xy−+40=715【解析一】：令f(x)=x32−3x+6x+2，则f(x)=3x2−6x+6，设Pm(,1)，Q(n,f(n))，依题意f(m)=f(n)，所以3m22−6m+6=3n−6n+6，则m22−n=2(m−n)，显然mn，则mn+=2，因为f(x)=(x−1)3+3(x−1)+6，所以fx()的图象关于点(1,6)中心对称，所fn()+1以点P与点Q关于点(1,6)对称，所以=6，则fn()=11.2215【解析二】：令，因为f(x)=3(x−1)+30，故fx()在R上单调3232递增，令x−3x+6x+2=1，设其根为xP，得xPPP−3x+6x=−1.由于fx()在点P处的切线与在点Q处的切线平行，得f(x)=k存在两实根，其中一个为xP，设另一个为xQ.即23x−6x+6=k两根为xP，xQ，由韦达定理得xxPQ+=2，则xxQP=−2，从而3232f(xQQQQPPP)=−++=−x3x6x2(2x)−−3(2x)+−+6(2x)232232=−+xPPPPPPPPP6x−12x+−83x+12x−+−12126x+=−−2(x3x+6)1011x+=.16【解析一】：由条件得B点为线段MN中点，设B点坐标为y(,)xy00，得Mx(20,0)、Ny(0,20)，由|MA|:|AB|=1:2得A坐标N225xy00xy为A(,)，将A、B坐标分别代入+=1中，得B33126A22MxOxy00+=1,126x0=−2,解得则M、N坐标分别为(−4,0)、2225xyy0=2,00+=1,12969(0,4)，直线l的方程为.数学参考答案第2页共12页m16【解析二】：设直线l方程为y=+kxm（km,0），可得M−,0，Nm(0,)，根据k5mmmm|MA|:|AB|:|BN|=1:2:3得A,B坐标分别为−,,−,.将,坐标代入C方66k22k程中，可得25mm22+=1,22221236k263625m+2mk=1236k,即两式相减得mk=4，222222mm+=1,m+2mk=124k,124k264将其代入m2+2m2k2=124k2，得k=1，从而直线的方程为xy−+40=.四、解答题：本大题共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.（本小题满分10分）【命题意图】本小题主要考查正弦定理、余弦定理、三角恒等变换、基本不等式等基础知识；考查运算求解能力；考查化归与转化思想、函数与方程思想等；导向对发展逻辑推理、数学运算等核心素养的关注；体现基础性和综合性.满分10分.【解析一】（1）由余弦定理可得b2=c2+a2−2accosB，代入b2−=a22c2，得到(c2+a2−2accosB)−a2=2c2，······································1分化简得c2+=2accosB0，即c+=2acosB0，····················································································2分由正弦定理可得sinCAB+=2sincos0，即sin(ABAB+)+2sincos=0，····································································3分sinABABABcos+cossin+2sincos=0，3sinABABcos=−cossin，·········································································4分tanB可得=−3.·························································································5分tanA（2）由（1）可知tanBA=−3tan，所以tanA与tanB异号，又b2−a2=20c2，故ba，则有BA，因为AB,(0,π)，故A为锐角，B为钝角，则C为锐角，···································6分tanABAAA+−tantan3tan2tan所以tanCAB=−tan(+)===······················7分tanABAAtan−1−3tan22−13tan+1223==，·························································8分13tanA+233tanA3π等号当且仅当tanA=，即A=时成立，····················································9分36π所以C的最大值为.················································································10分6数学参考答案第3页共12页sinBb【解析二】（1）由正弦定理可得：=，··················································1分sinAab2+−c2a2cosA由余弦定理可得：=2bc，·························································2分cosBa2+−c2b22acb2+−c2a2tanBsinBcosAbb2+−c2a2则==2bc=，········································3分tanAsinAcosBaa2+−c2b2a2+−c2b22actanBc22+2c因为b2−=a22c2，所以==−3.···················································5分tanAc22−2ca2+−b2c2（2）cosC=·············································································6分2abba22−ab22+−=2······································································7分2ab3ab22+3ab==+4ab44ba32163=.······················································································8分2π所以C，等号当且仅当ba22=3时，即ba=3时成立，·································9分6π所以C的最大值为·················································································10分6【解析三】（1）由b2−=a22c2，由正弦定理可得，sin2BAC−=sin22sin2，··········1分因为sin2BABBABAA−sin2=sin2−sin2sin2+sin2sin2−sin2=sin2BAABBAAB(1−sin2)−sin2(1−sin2)=sin2cos2−sin2cos2=(sinBABABABAABBAcos+cossin)(sincos−cossin)=sin(+)sin(−)，···············3分又因为ABC++=π，所以sinCAB=+sin()，所以sin(ABBAAB+)sin(−)=2sin2(+)，因为0+ABπ，所以sin(AB+)0，所以sin(BAAB−)=2sin(+)，化简得−=3sinABBAcossincos，·································································4分tanB可得=−3.·························································································5分tanA（2）同解析一.························································································10分数学参考答案第4页共12页18.（本小题满分12分）【命题意图】本小题