数学(质检-三明卷)福建2026届高中毕业班高三年级4月适应性练习(4.8-4.10)

三明市2026届高中毕业班适应性练习（四月）评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则。2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分。3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。4．只给整数分数。选择题和填空题不给中间分。一、单项选择题题号12345678答案ACDCCBCD二、多项选择题题号9答案ADBCDACD三、填空题四、解答题15.本小题主要考查函数的奇偶性、函数的零点、三角恒等变换、等差数列求和等基础知识，考查运算求解能力、逻辑推理能力等，考查函数与方程思想、分类与整合思想等，考查逻辑推理、数学运算等核心素养，体现基础性.满分13分.解法一1）因为f(x)为奇函数，所以f(_x)=_f(x)，····························································1分即sin(_2x)_sin(_x+φ)=_[sin(2x)_sin(x+φ)]恒成立.所以sinxcosφ+sinφcosx+sinφcosx_sinxcosφ=0恒成立，················································3分所以sinφcosx=0恒成立，·····························································································4分所以sinφ=0，············································································································5分数学参考答案及评分细则第1页（共15页）（2）因为φ=，所以f=sin2x_sin解得xk1k1∈Z或xk2k2∈Z，·································································11分k*，20234所以S．························································13分所以sinφ=0，············································································································3分经检验，f(x)=sin2x_sin(x+kπ),k∈Z是奇函数，（2）因为所以f(x)=sin2x_cosx，·····································································7分令f(x)=0，则sin2x_cosx=0，····················································································9分所以cosx=0或sinx解得xk1k1∈Z或xk2k2∈Z或xk3k3∈Z，······································11分n2025)25)，所以S···························································13分（2）因为所以f(x)=sin2x-sin(x+)，因为f(x+2π)=f(x)，所以2π是f(x)的一个周期，··························································································7分当0<x2π时，令f(x)=0，则sin2x=sin(x+π)，····························································9分2所以f(x)在区间(0,2π]的零点之和为．················································11分令an=x4n-3+x4n-2+x4n-1+x4n，2023452216.本小题主要考查导数的几何意义、导数的应用等基础知识，考查逻辑推理能力、运算求解能力等，考查函数与方程思想、化归与转化思想、分类与整合思想等，考查逻辑推理、数学运算等核心素养，体现基础性．满分15分．解法一1）函数f(x)的定义域为(0,+∞)，f,=x···················································2分当a=2时，因为f,=x所以f,(1)=-1，···························································3分又f·······································································································4分所以曲线y=f(x)在点(1,f(1))处的切线方程为y即2x+2y_3=0．·······················································（2）(i)当a<0时，falne不符合题意，舍去；························9分(ii)当a=0时，fx2>0显然成立；··································································11分解法二1）同解法一．······························································································7分（2）由已知，得x2_alnx>0．222lnx2因为0<x<1，所以x<0，·············2lnx又因为x→0时22lnx 12时，g,(x)<0，g(x)单调递减；117.本小题主要考查椭圆的定义、直线与椭圆的位置关系、三点共线等基础知识，考查逻辑推理能力、直观想象能力、运算求解能力等，考查函数与方程思想、数形结合思想、分类与整合思想、转化与化归思想等，考查逻辑推理、直观想象、数学运算等核心素养，体现基础性与综合性．满分15分．解法一1）当MF2丄x轴时，|M所以|M································································1分故C的方程为·····················································································6分所以=(_3,yP)，=(x0_1,y0)，.······························10分所以0yP=0，F2M.F2Q=3(x0_1)+y0yQ=0，················又因为=(x0+4,y0_yP)，QM=(x0_4,y0_yQ)，(x0y0解法二1）当MF2丄x轴时，|M所以······························································································2分又a2_b2=1②,··············································································故C的方程为·····················································································6分（2）设M(x0,y0)(y0≠0)，则，即y．··········································7分(i)当直线MF1，MF2斜率均存在时，kMkMF所以直线Py_1，Qy+1，···················································9分所以(ii)当直线MF1或MF2斜率不存在时，根据对称性，不妨设MF2斜率不存在，且y0>0，此时点MQ(4,0)，kM故直线PFy_1，从而P(_4,4)，则kMQkMP18.本小题主要考查随机变量的分布列、数学期望、条件概率与全概率公式等基础知识，考查数学建模能力、运算求解能力等，考查分类与整合思想、概率与统计思想等，考查数学运算、逻辑推理、数据分析、数学建模等核心素养等.体现基础性，应用性.满分17分.化简可得k，···················································································2分当时，k，即顾客一次性购买文创盲盒数量的平均值为16.····························································4分9（2i）设事件=“一次性购买i个文创盲盒”（i=0,1,2,3事件B=“顾客为幸运客户”，则P(A0)=k2，P(A1)=kα,P(A2)=k，P(A3)=k(1_α).因为每个盲盒是否为封面款相互独立，又由题意知，B=A0BA1BA2BA3B，且A0B,A1B,A2B,A3B两两互斥，···························9分由（1）得，k，代入化简可得P（ii）设事件C=“一次性购买的文创盲盒全部是封面款”，且C=A1CA2CA3C，A1C,A2C,A3C两两互斥，由（i）得，P所以幸运客户中，一次性购买的文创盲盒全部是封面款的概率为P·······································································16分由题意P可得解得“，1719.本小题主要考查空间点、线、面位置关系，直线与平面所成角，二面角，平面轨迹方程等基础知识；考查运算求解因为AB丄AD，所以BD2=a2+b2，PB2=a2+c2，PD2=由余弦定理，得cosLPBD···········································3分（2i）因为PA丄Y，Q在CP上，且LCQB=LCQD，由对称性知B,D在同一个轨迹上，且轨迹关于AC对称，故以A为原点，,分别为x轴和z轴的正方向建立如图所示的空间直角坐标系A-xyz．设B(x1,y1,0)，D(x2,y2,0)，因为AP=AC=3，所以P因为Q是线段CP上靠近C的三等分点，-2,y1,-1)，依题意得化简得x_y=3，··················································6分且x1_1>0，即x1>1，故x1≥/3，又点B不在直线AC上，故x1>/3，同理，x_y=3，且x2>3，················································································7分故在坐标平面xAy中，B,D是双曲线x2_y2=3右支上的动点，且B,D在x轴的两侧，如图．2因为x2_y2=3的两条渐近线分别为y=x和y=_x，它们的夹角为π,22因为平面PAB平面PAD=PA，PA丄AB，PA丄AD，所以LBAD是二面角B_AP_D的平面角，所以二面角B_AP_D为锐角．··························9分（ii）因为△PBD不是任何一个长方体的截面，所以△PBD是直角三角形或钝角三角形．······10分证明如下：若△PBD为锐角三角形，有PB2+PD2_BD2>0，PB2+BD2_PD2>0，BD2+PD2_PB2>0，可令a由（1）知，若△PBD是长方体的截面，则△PBD是锐角三角形，2222因为PA丄Y，所以LPBA,LPDA分别是直线PB,PD与Y所成的角，即LPBA=α,LPDA=β,且LPBD≥π>LPDB．2作QM丄BD于M，因为平面QBD丄Y，平面QBDY=BD，QMC平面QBD，因为Q是线段CP上靠近C的三等分点，所以M是线段AC上靠近C的三等分点，所以M(2,0,0)，即直线BD过M(2,0,0)，···································································14分所以LPBD=LPBM所以.=(2_x1,_y1,0).(_x1,_y1,3)=x12_2x1+y12≤0，··············15分这样，问题等价于在平面直角坐标系xAy中，B(x1,y1),D(x2,y2)在双曲线x2_y2=3的右支上，直线BD过点M(2,0)，AB<AD，x12_2x1+y12≤0，求tan的最小值．如图，不妨设点B在第四象限，则y1<0，x1<2．因为B,D都在双曲线的右支，故kBD2，又x12_2x1+y12≤0，x_y=3，所以tan当x即LPBD时，等号成立．故tan2θ的最小值为．·················································································17分又因为AB丄AD，故可以A为原点，,,分别为x轴，y轴和z轴的正方向，建立如图所示的空间直角坐标系A_xyz．························································································2分.=(_a,b,0).(_a,0,c)=a2>0，所以LPBD为锐角，.=(a,_b,0).(0,_b,c)=b2>0，所以LPDB为锐角，.=(a,0,_c).(0,b,_c)=c2>0，所以LBPD为锐角，所以△PBD是锐角三角形．·····················································································4分（2i）同解法一．·····························································································9分（ii）因为△PBD不是任何一个长方体的截面，所以△PBD是直角三角形或钝角三角形．······10分证明如下：若△PBD为锐角三角形，有PB2+PD2_BD2>0，PB2+BD2_PD2>0，BD2+PD2_PB2>0，PB2+BD2_PD22,,PD2+BD2_PB22PB2PB2+PD2_BD22,则存在以A,B=a,,A,D=b,,A,P=c,为共点棱的长方体，△PBD为该长方体的截面．由（1）知，若△PBD是长方体的截面，则△PBD是锐角三角形，作QM丄BD于M，因为平面QBD丄Y，平面QBDY=BD，QMC平面QBD，因为Q是线段CP上靠近C的三等分点，所以M是线段AC上靠近C的三等分点，所以M(2,0,0)，即直线BD过M(2,0,0)．···································································12分在平面直角坐标系xAy中，设直线BD的方程为x=ty+2，联立2_1≠0,|y1+y2=_t2_1因为y1y2<0，所以t2<1．(x2_x1,y2_y1,0)，y2(x1_x2)+y1(y1_y2)=x+y_(x1x2+y1y2)，不妨设x+y≤x+y，则必有=x+y_(x1x2+y1y2)≤0．3t23t2因为x1=ty1+2且y1≠0，所以t，代入上式得到又因为x1>3，所以x·····································································15分因为PA丄Y，所以LPBA,LPDA分别是直线PB,PD与Y所成的角，即LPBA=α,LPDA=β,当x即LPBD时，等号成立．故tan2θ的最小值为·················································································17分又因为AB丄AD，所以在△PBD中，所以LPBD为锐角，···········································2分所以LPDB为锐角，··········································3分2所以△PBD是锐角三角形．·····················································································4分（2i）同解法一．·······································································