2026届福建省福州市高三4月适应性练习数学试题含答案

数学友情提示:请将所有答案填写到答题卡上！请不要错位、越界答题！4．设α,β是两个不重合的平面，则α∥β的充要条件是A．存在无数条直线与α,β都平行B．存在无数个平面与α,β都垂直C．对任意的直线lα,都存在直线mβ,使得l∥mD．对任意的直线lα,都存在直线mβ,使得l丄m43若该三棱锥的四个顶点都在球O的球面上，则球O的表面积为A．24τB．48πm(x,x2,x3，且x1A．m为奇数39．已知抛物线C:y2=2px的焦点为F(1,0)，准线为l，圆M过点F．下列说法正确C．若圆心M在C上，则圆M与l相切D．若圆M与l相切，则圆心M在C上示，点A(0,−)，B(,0)在f(x)的图象上.下列说法正确的是A．f(x)的最小正周期是B．f(x)在区间单调递增C．f(x)的一个对称中心是(,0)D．f(x)的图象可以由g(x)=tan2x的图象向左平移个单位长度得到{an}的前n项和为Sn，公比为q的等比数列{bn}的前n项和时，d=0当上ADB最大时，四边形ABCD的面积为.（1）若f(x)是奇函数，求φ;项和S20.已知函数x2−alnx.（1）当a=2时，求曲线y=f(x)在点(1,f(1))处的切线方程；（2）若f(x)>0，求a的取值范围.的动点，且M不在x轴上．当M丄x轴时，|M.P,M,Q三点共线.某盲盒商店调查数据显示，顾客一次性购买某种文创盲盒数量X的分布列为X0123Pk(1−α)2kαkk(1−α)（2）已知该种文创盲盒分为封面款与非封面款两类，且每个盲盒为封面款的概选取一人.（i）求该顾客为幸运客户的概率f(α)；求α的取值范围.已知PA丄平面Y，垂足为A，直线ACY，B,D是Y内的动点，且B,D始终在AC的两侧.（2）若PA=AC=3，Q是线段CP上靠近C的三等分点，上CQB=上CQD=且△PBD不是任何一个长方体的截面，求tan2θ的最小值.PγCADγCADB2026届高中毕业班适应性练习（四月）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12345678ACDCCBCD9ACD 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，考查运算求解法一1）因为f(x)为奇函数，所以f(−x)=−f(x)，···························································1分所以sinxcosφ+sinφcosx+sinφcosx−sinxcosφ=0恒成立，···············································3分所以sinφcosx=0恒成立，······················································································所以sinφ=0，······················································································解得φ=kτ,k∈Z.····································（2）因为φ=，所以f(x)=sin2x−sin(x+)，令f(x)=0，则sin2x=sin(x+)，········································································k*，45)解法二1）因为f(x)为R上的奇函数，所以f(0)=0，·························································2分所以sinφ=0，······················································································解得φ=kτ,k∈Z，··················经检验，f(x)=sin2x−sin(x+kτ),k∈Z是奇函数，所以φ=kτ,k∈Z.···································（2）因为φ=，所以f(x)=sin2x−cosx，·····························令f(x)=0，则sin2x−cosx=0，······························································所以cosx(2sinx−1)=0，···············2k*，25)，所以S20=··························解法三1）同解法一.·································································因为φ=所以f(x)=sin2x−sin(x+)，因为f(x+2τ所以2τ是f(x)的一个周期，·························································································7分当0<x„2τ时，令f(x)=0，则sin2x=sin(x+)，··············································解得x=τ,τ,5τ,3τ,···························所以f(x)在区间(0,2τ]的零点之和为···············································11分则{an}是以3τ为首项，8τ为公差的等差数列，································································12分245······························16.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，考解法一1）函数f(x)的定义域为.··················································2分当a=2时，因为f,，所以f,=−1，··························································3分又，·····································································································4分所以曲线y=f(x)在点(1,f(1))处的切线方程为即2x+2y−3=0.····································································（2）(i)当a<0时，f(e)=1(e)2−alne=1e−1<0不符合题意，舍去；······················9分22(ii)当a=0时，f(x)=x2>0显然成立；····················································所以f(x)在(0,a)单调递减，在(a,+∞)单调递增.··················································13分所以f(x)min=f()=a−aln>0，解得0<a<e.···············································14分解法二1）同解法一.····························································································7分（2）由已知，得x2−alnx>0.(i)当0<x<1时，可得a>.···········································································8分因为0<x<1，所以<0，···············································································9分又因为x→0时，所以a…0；······································································································10分(ii)当x=1时，x2−alnx>0恒成立，所以a∈R；·····················································11分令g(x)=x2(x>1)，g2.=x(2lnx−1)，···················2lnx2(lnx)22(lnx)211当x>e2时，g’(x)>0，g(x)单调递增；·································1所以g(x)min=g(e2)=e，所以a<e.·······································································14分综上所述，a的取值范围为[0,e).··········································································15分17.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． \42=+4=， \42|=5+3=4，···············································从而a=2，b2=3，····························································································5分故C的方程为=1.····················································································6分（2）设M(x0,y0)(y0≠0)，P(−4,yP)，Q(4,yQ)，························································7分则x02+y0202.····································································8分43因为PF1两式相加、减，得yP+yQ=，yP−yQ又因为PM=(x0+4,y0−yP)，QM=(x0−4,y0−yQ)，(x0+4)(y00所以PM∥QM，故P,M,Q三点共线.··········································································15分PyMQF1OF2x解法二当M丄x轴时，|M，所以M(1,)或M(1,−)，·····················································································1分所以=1①,·····························································································2分2=1②,································································································4分由①②,解得a2=4，b2=3，···············································································5分故C的方程为=1.····················································································6分设M，则即y02=3−.·········································7分(i)当直线MF1，MF2斜率均存在时所以直线:x=−y−1，Q:x=−y+1，····················由y+1,得Q，························································所以PM∥QM，故P,M,Q三点共线.·····································································12分(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，所以P,M,Q三点共线.························································································14分综上，P,M,Q三点共线.····················································································15分18.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，考查数学建模能解1）由题可知，k(1−α)2+kα+k+k(1−α)=1，··············································化简可得，·················································································2分即顾客一次性购买文创盲盒数量的平均值为.··························································4分（2i）设事件Ai=“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”，······················································································································5分则P(A0)=k2，P(A1)=kα,P(A2)=k，P(A3)=k(1−α).依题意=0，P，···································································6分所以，·········································8分又由题意知，B=A0BA1BA2BA3B，且A0B,A1B,A2B,A3B两两互斥，··························9分所以.····································································12分（ii）设事件C=“一次性购买的文创盲盒全部是封面款”，依题意i,i=1,2,3，······························且C=A1CA2CA3C，A1C,A2C,A3C两两互斥，所以，·············································14分所以幸运客户中，一次性购买的文创盲盒全部是封面款的P(BC)P(C)2(1+2α)P(B)P(P(BC)P(C)2(1+2α)P(B)P(B)5+α,由题意P(C|B)„，可得解得α„,又因为0<α<1，所以α∈(0,].············································································17分解法一1）因为PA丄平面Y，AB,ADY,所以PA丄AB，PA丄AD.··································1分因为AB丄AD，所以BD2=a2+b2，PB2=a2+c2，PD2=b2+c2，所以BD≤PB≤PD，所以上PBD为△PBD的最大内角.·················································2分所以上PBD∈(0,)，所以△PBD是锐角三角形.·························································4分（2i）因为PA丄Y，Q在CP上，且上CQB=上由对称性知B,D在同一个轨迹上，且轨迹关于AC对称，故以A为原点，AC,AP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.zPyγxCQADγxCQADB设B(x1,y1,0)，D(x2,y2,0)，因为AP=AC=3，所以P(0,0,3)，C(3,0,0).因为Q是线段CP上靠近C的三等分点，故AQ=AC+AP=(2,0,1)，即Q(2,0,1)，······························································5分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，·,···············································································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.yDxAxB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，所以0<上BAD<.····························································································8分因为平面PAB平面PAD=PA，PA丄AB，PA丄AD，所以上BAD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分若△PBD为锐角三角形，有PB2+PD2−BD2>0，PB2+BD2−PD2>0，BD2+PD2−PB2>0，PB2+BD2−PD22PDPD2+BD2−PB22PB2PB2+PD2−BD22,则存在以AB=a,AD=b,AP=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△P所以△PBD不是任何一个长方体的截面等价于△PBD是直角三角形或钝角三角形.···············11分由（i）知，0<上BAD<，所以AB.AD>0，又因为PA丄AB，PA丄AD，因为PA丄Y，所以上PBA,上PDA分别是直线PB,PD与Y所成的角，即上PBA=α,上PDA=β,不妨设AB≤AD，则α≥β,且PB≤PD，所以θ=α,tan2θ=，····················13分作QM丄BD于M，因为平面QBD丄Y，平面QBDY=BD，QM平面QBD，所以QM丄Y，又PA丄Y，所以PA∥QM.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，AB<AD，x12−2x1+y12≤0，求tan2θ=的最小值.yDxAMBxA故tan2θ的最小值为.················································································17分解法二1）因为PA丄平面Y，AB,ADY，所以PA丄AB，PA丄AD.·································1分又因为AB丄AD，故可以A为原点，AB,AD,AP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zPCQyDAxBγCQyDAxBγ设AB=a,AD=b,AP=c，所以2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分若△PBD为锐角三角形，有PB2+PD2−BD2>0，PB2+BD2−PD2>0，BD2+PD2−PB2>0，PB2PB2+BD2−PD22则存在以AB=a,AD=b,AP=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△P所以△PBD不是任何一个长方体的截面等价于△PBD是直角三角形或钝角三角形.···············11分zPyγxCQMADγxCQMADB作QM丄BD于M，因为平面QBD丄Y，平面QBDY=BD，QM平面QBD，所以QM丄Y，又PA丄Y，所以PA∥QM.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+2，2联立{2联立{因为y1y2<0，所以t2<1.因为PB=(x1,y1,−3),PD=(x2,y2,−3),BD=(x2−x1,y2−y1,0)，y1y2+2t+13=>0，······················································不妨设x+y≤x+y，则必有BPBD=x+y−(x1x2+y1y2)≤0.22222223t2+32622222223t2+326···························································14分 因为x+y≤x+y，所以AB≤AD，所以α≥β,所以θ=α,·····································16分故tan2θ的最小值为.················································································17分解法三1）因为PA丄平面Y，AB,ADY，所以PA丄AB，PA丄AD.·································1分又因为AB丄AD，所以在△PBD中，>0，所以上PBD为锐角，···