福州市2026届高中毕业班4月适应性练习数学+答案

（在此卷上答题无效）

福州市2026届高中毕业班4月适应性练习

数学

（完卷时间：120分钟；满分：150分）

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！

一、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。

1．设全集U={x|0<x<6,x∈Z}，集合A={1,2,3}，则UA=

A．{4,5}B．{4,5,6}C．{1,2,3}D．{x|3<x<6}

2．已知复数(1+i)(a+bi)(a,b∈R)在复平面内对应的点位于第二象限，则下列结论一定

成立的是

A．a>0B．a<0C．b>0D．b<0

3．某次测试中，某10人的成绩（单位：分）分别为：48，75，58，66，78，82，84，

78，86，91，则这组数据的第80百分位数是

A．78B．82C．84D．85

4．设“,β是两个不重合的平面，则“∥β的充要条件是

A．存在无数条直线与“,β都平行

B．存在无数个平面与“,β都垂直

C．对任意的直线lc“，都存在直线mcβ,使得l∥m

D．对任意的直线lc“，都存在直线mcβ,使得l丄m

5．已知函数f(x为增函数，则a的最小值是

A．4B．2C．4D．5

3

6．已知三棱锥P一ABC的体积为93，LBAC=90o，AB=AC=32，PB=PC=6．

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．24πB．48πC．96πD．108π

高三数学—1—（共4页）

n

7．已知数列{an}的前n项和为Sn，若an+1+(一1)Sn=2n+1，则a6=

A．16B．18C．20D．22

．已知函数mn（*，，）有且仅有个极值点，

8f(x)=(x一a)(x一b)m,n∈Nm<na≠b3x1,x2,x3

且x1<x2<x3，则

A．m为奇数B．n为奇数

C．若a<b，则2x2>x1+x3D．若a>b，则2x2>x1+x3

二、选择题：本题共3小题，每小题6分，共18分。在每小题给出的选项中，有多

项符合题目要求。全部选对的得6分，部分选对的得部分分，有选错的得0分。

9．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

的是

A．p=1

B．l的方程为x=一1

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

π

10．已知函数f(x)=tan(⑴x+φ)(⑴>0，|φ|<)的部分图象如图所

2

π

示，点A(0,一3)，B(,0)在f(x)的图象上.下列说法正确的是

6

π

A．f(x)的最小正周期是

2

B．f(x)在区间单调递增

C．f(x)的一个对称中心是

π

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

3

11．已知公差为d的等差数列{an}的前n项和为Sn，公比为q的等比数列{bn}的前n项和

为Tn，且a1=b1>0，a10=b10．下列命题正确的是

A．当d>0时，S10>T10

B．当S10=T10时，d=0

C．当一1<q<0时，S10>T10

D．当q<一1时，集合{n|an=bn}可能有三个元素

高三数学—2—（共4页）

三、填空题：本题共3小题，每小题5分，共15分。

12．已知单位向量a,b满足a丄(a一2b)，则a,b=

13．为了应对新能源产业爆发式增长带来的挑战，某研究所设立了资源组、电芯组、

基建组三个攻关小组.现安排甲、乙等5名工作人员到这三个小组协助工作，

且每个小组至少安排一人，每人只能去一个小组，同时，要求安排到电芯组的人

数比资源组的人数多，甲、乙两人不能被安排到资源组，则不同的安排方案种数

是.（用数字作答）

14．在平面凸四边形ABCD中，LBAC=60o，AB=2，BC=23，△BCD的面积为33.

当LADB最大时，四边形ABCD的面积为

四、解答题：本题共5小题，共77分。解答应写出文字说明、证明过程或演算步骤。

15．（13分）

已知函数f(x)=sin2x一sin(x+φ).

（1）若f(x)是奇函数，求φ;

（2）当时，f(x)的所有正零点从小到大排列构成数列{xn}，求{xn}的前20

项和S20.

16．（15分）

已知函数fx2一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．

17．（15分）

已知椭圆C的左、右焦点分别为，，M是C上

12

的动点，且M不在x轴上．当MF2丄x轴时，|MF(__1,0)F(1,0)

（1）求C的方程；

（2）点P,Q分别在直线l1:x=一4与l2:x=4上，且PF1丄MF1，QF2丄MF2.证明：

P,M,Q三点共线．

高三数学—3—（共4页）

18．（17分）

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

X0123

P

2

其中k>0，0<“<k1.(1__“)k“kk(1__“)

（1）当时，求顾客一次性购买该种文创盲盒数量的平均值；

（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

率为1，每个盲盒是否为封面款相互独立.若顾客一次性购买的盲盒中，封

3

面款的数量大于非封面款的数量，则称此顾客为幸运客户.现从顾客中随机

选取一人.

（i）求该顾客为幸运客户的概率f(“)；

（ii）若该顾客是幸运客户，他购买的盲盒全部是封面款的概率不超过1，

2

求“的取值范围.

19．（17分）

已知PA丄平面Y，垂足为A，直线ACCY，B,D是Y内的动点，且B,D始终在AC

的两侧．

（1）若AB丄AD，证明：△PBD是锐角三角形；

（2）若PA=AC=3，Q是线段CP上靠近C的三等分点，LCQB=LCQD

（i）证明：二面角B_AP_D为锐角；

（ii）直线PB,PD与Y所成的角分别为“,β,记θ=max{“,β}．若平面QBD丄Y，

且△PBD不是任何一个长方体的截面，求tan2θ的最小值．

P

Q

D

AC

卩B

高三数学—4—（共4页）

2026届高中毕业班适应性练习（四月）

数学参考答案及评分细则

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91011

答案BCDADACD

三、填空题

°

12．6013．3014．4·3

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.

得sin(x+φ)+sin(_x+φ)=0恒成立，···············································································2分

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

解得φ=kπ,k∈Z.········································································································6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

令f=0，则sin2x=sin··················································································8分

所以2x=xk1k1∈Z或2x+xk2k2∈Z，··················································10分

解得xk1k1∈Z或xk2k2∈Z，·································································11分

*

令an=(2n_)π,bn=(n_)π,则b3k__2<ak<b3k__1,k∈N，

所以S20=x1+x2++x20=(a1+a2+a3+a4+a5)+(b1+b2++b15)，··································12分

所以S．························································13分

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，············································································································3分

解得φ=kπ,k∈Z，·····································································································4分

经检验，f(x)=sin2x_sin(x+kπ),k∈Z是奇函数，

所以φ=kπ,k∈Z.········································································································6分

（2）因为所以f(x)=sin2x_cosx，·····································································7分

令f(x)=0，则sin2x_cosx=0，····················································································9分

所以cosx(2sinx_1)=0，·····························································································10分

所以cosx=0或sinx

解得xk1k1∈Z或xk2k2∈Z或xk3k3∈Z，······································11分

令an=bn=cn=

数学参考答案及评分细则第2页（共15页）

*

则(2k-2)π<bk<a2k-1<ck<a2k<2kπ,k∈N，

所以S20=x1+x2++x20=(a1+a2++a10)+(b1+b2++b5)+(c1+c2++c5)，

所以S···························································13分

解法三：（1）同解法一．·····································································································6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

π

当0<x.2π时，令f(x)=0，则sin2x=sin(x+)，····························································9分

2

解得x·································································································10分

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，

则{an}是以3π为首项，8π为公差的等差数列，································································12分

所以S20=x1+x2++x20=a1+a2+a3+a4+a5

5(a+a5)5x(3π+35π)

=1==95π13分

22

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

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

数学参考答案及评分细则第3页（共15页）

即2x+2y_3=0．································································································7分

（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

(iii)当a>0时，令f,(x)<0，得0<x<a·,令f,(x)>0，得x>；

所以在单调递减在单调递增分

f(x)(0,a)，(·a,+∞)．···················································13

所以fmin=fa_alna>0，解得0<a<e．················································14分

综上所述，a的取值范围为[0,e)．················································································15分

解法二：（1）同解法一．······························································································7分

1

（2）由已知，得x2_alnx>0．

2

x2

(i)当0<x<1时，可得a>．·············································································8分

2lnx

x2

因为0<x<1，所以<0，·················································································9分

2lnx

又因为x→0时

所以a开0；·······································································································10分

当x=1时x2_alnx>0恒成立，所以a∈R；······················································11分

x2

(iii)当x>1时，可得a<．

2lnx

令g············································12分

1

当1<x<e2时，g,(x)<0，g(x)单调递减；

1

当x>e2时，g,(x)>0，g(x)单调递增；···································································13分

所以gmin=ge，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，考查函数与方程思想、数形结合思想、分类与整合思想、转化与化归

数学参考答案及评分细则第4页（共15页）

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

所以53，分

2a=|MF1|+|MF2|=+=4···········································································3

22

从而a=2，b2=3，·····························································································5分

故C的方程为·····················································································6分

（2）设M(x0,y0)(y0≠0)，P(_4,yP)，Q(4,yQ)，·························································7分

22

则，即3x0+4y0_12=0．·····································································8分

又F1(_1,0),F2(1,0)，

所以1=(x0+1,y0)，1=(_3,yP)，=(x0_1,y0)，.······························10分

因为PF1丄MF1，QF2丄MF2，

所以1.1=_3(x0+1)+y0yP=0，F2M.F2Q=3(x0_1)+y0yQ=0，··································12分

两式相加、减，得yP+yQyP_yQ·····························································13分

又因为=(x0+4,y0_yP)，QM=(x0_4,y0_yQ)，

6x2+8y2_24

00，分

(x0+4)(y0_yQ)_(x0_4)(y0_yP)=x0(yP_yQ)+8y0_4(yP+yQ)==0·········14

y0

所以∥,故P,M,Q三点共线.···········································································15分

解法二：（1）当MF2丄x轴时，|M

33

所以M(1,)或M(1,_)，······················································································1分

22

数学参考答案及评分细则第5页（共15页）

9

所以······························································································2分

又a2_b2=1②,··································································································4分

由①②,解得a2=4，b2=3，················································································5分

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

y_1,

由得P·····································································10分

y+1,

由得Q···································································11分

所以

因为

所以∥,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，

此时点MQ(4,0)，kM故直线PFy_1，从而P(_4,4)，则kMQkMP

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

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.

数学参考答案及评分细则第6页（共15页）

解：（1）由题可知，k(1_α)2+kα+k+k(1_α)=1，··························································1分

化简可得k，···················································································2分

当时，k，

则E=kα+2k+3k=k

即顾客一次性购买文创盲盒数量的平均值为16.····························································4分

9

（2）（i）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

2

则P(A0)=k（1_α)，P(A1)=kα,P(A2)=k，P(A3)=k(1_α).

依题意，得P=0，P····································································6分

因为每个盲盒是否为封面款相互独立，

312

所以PP(B|A3)=()+C3xx()=，··········································8分

又由题意知，B=A0BA1BA2BA3B，且A0B,A1B,A2B,A3B两两互斥，···························9分

331172k(5+)

所以==.=+++_=α，分

P(B)ΣP(AiB)Σ[P(Ai)P(B|Ai)]0kαkk(1α)·············11

i=0i=0392727

由（1）得，k，代入化简可得P

所以fα∈(0，1).·····································································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，得Pi,i=1,2,3，········································································13分

且C=A1CA2CA3C，A1C,A2C,A3C两两互斥，

所以P·············································14分

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

数学参考答案及评分细则第7页（共15页）

P·······································································16分

由题意P可得解得“.，

1

又因为0<“<1，所以“∈(0,].············································································17分

7

19.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

解法一：（1）因为PA丄平面Y，AB,ADCY,所以PA丄AB，PA丄AD．···································1分

不妨设AB=a,AD=b,AP=c，且a≤b≤c，

因为AB丄AD，所以BD2=a2+b2，PB2=a2+c2，PD2=b2+c2，

所以BD≤PB≤PD，所以LPBD为PBD的最大内角．··················································2分

△

由余弦定理，得cosLPBD···········································3分

所以LPBD所以PBD是锐角三角形．···························································4分

△

（2）（i）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(0,0,3)，C(3,0,0)．

因为Q是线段CP上靠近C的三等分点，

故即Q(2,0,1)，·······························································5分

故=(1,0,-1)，=(x1-2,y1,-1)，

数学参考答案及评分细则第8页（共15页）

22

依题意得化简得x1_y1=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，

22

同理，x2_y2=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．

π

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,

2

π

所以0<LBAD<.······························································································8分

2

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

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，

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

π

由（△i）知，0<LBAD<,所以.>0△，又因为PA丄AB，PA丄AD，

2

2π

所以.=(+).(+)=+.>0，故0<LBPD<.···························12分

2

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,

不妨设AB≤AD，则α≥β,且PB≤PD，所以θ=α,tan····················13分

数学参考答案及评分细则第9页（共15页）

π

且LPBD≥>LPDB．

2

作QM丄BD于M，因为平面QBD丄Y，平面QBDY=BD，QMC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

22

所以LPBD=LPBM所以.=(2_x1,_y1,0).(_x1,_y1,3)=x1_2x1+y1≤0，··············15分

22

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x_y=3的右支上，直线BD

22≤tan

过点M(2,0)，AB<AD，x1_2x1+y10，求的最小值．

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kBD=kBM

2（222≤22

即_y1>2_x1>0，所以y1>2_x1），又x1_2x1+y10，x1_y1=3，

故解得·····································16分

所以tan

当x即LPBD时，等号成立．

故tan2θ的最小值为．·················································································17分

解法二：（1）因为PA丄平面Y，AB,ADCY，所以PA丄AB，PA丄AD．··································1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

数学参考答案及评分细则第10页（共15页）

设AB=a,AD=b,AP=c，所以B(a,0,0),D(0,b,0),P(0,0,c)，在△PBD中，

.=(_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

（2）（i）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_PD2PD2+BD2_PB2PB2+PD2_BD2

,,

可令a,=,b,=,c=

222

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，

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

△△

作QM丄BD于M，因为平面QBD丄Y，平面QBDY=BD，QMC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，

数学参考答案及评分细则第11页（共15页）

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

(4t

2y+y=_,

(|t_1≠0,|12t2_1

依题意，有且

〈22〈

Δ=(4t)_4(t_1)>0,

|ly1y2=.

|l

2

因为y1y2<0，所以t<1．

因为=(x1,y1,_3),=(x2,y2,_3),=(x2_x1,y2_y1,0)，

所以=x1x2+y1y2+9=(ty1+2)(ty2+2)+y1y2+9

分

y1y2+2t····························································13

22

=x1(x1_x2)+y1(y1_y2)=x1+y1_(x1x2+y1y2)，

22

同理=x2+y2_(x1x2+y1y2)，

222222

不妨设x1+y1≤x2+y2，则必有=x1+y1_(x1x2+y1y2)≤0．