福建厦门市 2026 届高中毕业班第二次质量检测数学+答案

第1页（共8页）123456789CABBDCBCACDABDACD解析：因为A≤cUB，所以AB=g，所以B≤CUA，故Bn(UA)=B，故选B．,解得k舍去故选B．解析：由题意得f是f的最大值，所以f，解得a=3．所以f故选D．6．答案：C解析：由f(x)=f(_x)可知f(x)是偶函数，当x>0时，f,(x)=ex_e_x>0，故f(x)在2，所以c=f>fb，故选C．解析：方法一：依题意，△ABD的面积为1，设AD=m，则AB=2m，所以Sm.2msinA=1，即m，在△ABD中，由余弦定理可得，BD222_2.2m.mcosA，所以BD，设则ksinA+4cosA=5≤k2+16，解得k≥3，当且仅当sinA=方法二：以BC的中点为坐标原点，建立平面直角坐标系，不妨设B(_m,0)，C(m,0)，A，所以D，所以BD，当且仅当m2=时解析：设△ABC与△BCD的外接圆半径分别为r1，r2，所以π(r12OM2+BM2=R2，所以R2_r12+R2_r22第2页（共8页）解析：对于选项A：由表格可知，y随xyxOBAyxOBA22对于选项B：设(x,y)在C上，因为(x,y)关于y=_x的对称点(_y,_x)也在C上，所以C关于直线y=_x对称，C为轴对称图形，故选项B正确；△PAB面积的最大值为，因为，所以不存在满足条件的点P.因为曲线C:x2_y2=满足题意的点P恰有两个，故选项D正确；故选ABD．8)7)所以ED选项正确；故选ACD．第3页（共8页）），15．(13分)记数列{an}的前n项和为Sn，已知a1=3，Sn=nan_n(n_1)．（1）证明{an}是等差数列，并求an；解1）当n≥2时，Sn_1=(n_1)an_1_(n_1)(n_2)，·········································1分两式相减得，an=Sn_Sn_1=nan_(n_1)an_1_(n_1)n+(n_1)(n_2)，·················2分即(n_1)(an_an_1_2)=0，········································································3分由于n_1≥1，所以an_an_1=2(n≥2)，······················································4分所以{an}是首项为3，公差为2的等差数列．····································（2）由（1）知Sn=n2+2n，···············1615分）设函数f(x)=e2x_2ax．（1）讨论f(x)的单调性；（2）证明：当a>0时，f≥lna解：解法一1）f,(x)=2(e2x_a)．······························································1分当a≤0时，e2x_a>0，所以f,(x)>0，f(x)在R上单调递增；·····················3分当a>0时，令f,(x)<0，解得x令f,，解得x>，···············5分所以f单调递减，在单调递增．综上，当a≤0时，f(x)在R上单调递增；当a>0时，f单调递减，在单调递增．····················6分（2）由（1）知，当a>0时，fmin=fa_alna．····························7分下证a_alna≥lna，即证lna8分令glna则g,=lna_a9分所以h(a)在(0,+∞)上单调递减．·······································即lna所以f≥lna······················································所以f≥lna（2）由（1）知，当a>0时，fmin=f（2）由（1）知，当a>0时，fmin=fa_alna．,即证lna令g=lna···············10分则g,···············10分,所以f≥lna································15分,所以f≥lna································15分（1）证明：BD丄平面PAC；P（2）已知PEEAB（ii）求平面PBD与平面PEC夹角的余弦值．ABDC解：解法一1）因为ABCD为菱形，所以BD丄AC．·······································1分因为DC所以BD丄平面PAC．··············································································4分2OBEF平面PEC=EF，又因为PA∥EB，所以OF∥EB，即O,F,E,B四点共面．·······························5分因为OB∥平面PEC，OBEF平面PEC=EF，PFOCAE所以OB∥EF，······················································································7分PFOCAE,即B故OF,即B故OF=BEAP··································································9分（ii）由（1）可知，BD丄平面PAC，因为BDC平面ABCD，DDABCD平面PAC=AC，所以平面ABCD平面PAC=AC，所以在平面PAC内作Oz垂直于AC， D23,0,0 B D23,0,0 BP0,_1,3EEADPC=(0,3,_3)CxOBPz第5页（共8页）0,3,1z2所以cos<m,n4分所以平面PBD与平面PEC夹角的余弦值为3．··········································15分解法二1）同解法一；（2i）同解法一；（ii）过P作BD的平行线l，因为l∥BD，EF∥BD，所以l∥EF，所以l为平面PBD与平面PEC的交线．······················································11分由（1）可得，BD丄PO，BD丄平面PAC，PCC平面PAC，所以BD丄PC， lP所以LOPC为平面PBD与平面PEC夹角.····················································13分 lPE3所以LAOP=60o，LOPCE3所以LAOP=60o，LOPC=30o，所以cosLOPC=所以平面PBD与平面PEC夹角的余弦值为3．DC15分A2B··········································OF21817分）已知椭圆C的左、右顶点分别为A，B，AB=4，直线（1）求C的方程；（2）点M在线段PQ上，直线AM，BM分别交C于D，E两点，直线AE，BD交于点N．（i）证明：MN丄AB；（ii）判断y轴上是否存在定点T，使得NT+NM为定值，若存在，求出T的坐标；若···························4分,yNT2)，DEPMQAOBx解：解法一1）依题意，AB=2a=4，所以a=2，··································易知点在C上，···········································································2分···························4分,yNT2)，DEPMQAOBx(2626),y1)，E(2626)设直线AM:x=(x0+2)y_2，BM:x=(x02+4y2_12(x0+2)y=0所以y············································································6分第6页（共8页）联立直线BM和C可得，y7分所以直线BD的斜率为所以直线BD:y，···························································8分同理直线AE的斜率为所以直线AE:y，···························································9分由可得，N所以MN丄AB．··················································（ii）假设存在点T(0,t)，使得NT+NM为定值m，NTEPDQAOBxMy即NTNMNTEPDQAOBxMy所以存在T，使得NTNM为定值．解法二1）同解法一；（2i）同解法一；（ii）由（i）可知，点N(xN,yN)在抛物线上，··············12分假设NT+NM=m，当M(0,1)时，N(0,3)，此时MN=2，则m≥2，如图，M到直线y=m+1的距离m，又1<yN≤3≤则N点到直线y=m+1的距离为m_NM=NT，所以N应在以T为焦点，y=m+1为准线的抛物线上，所以T只能为Γ的焦点，y=m+1为Γ的准线，············································14分可求得Γ的焦点为，准线为y=，···················································16分所以当T为(l(0,|时，NT=_yN，NM=yN_1，所以NTNMyN+yN所以存在T，使得NTNM为定值．···········································17分解法三1）同解法一；（2i）设S(xs,ys)为C上一点，则kSAxkSB········6分因为kMBxkAEkMAxkBD所以kAE，kBD，····8分第7页（共8页）所以直线BD:y，AE:y，······················9分由可得，N所以MN丄AB．·················································解法四1）同解法一；（2i）设D(x1,y1)，E(x2,y2)设直线lAM:x=t1y_2，令y=1，则M(t1_2,1)，+4)y2_12t1y=0，由韦达定理得y，代入直线lAM得x，所以D，所以kBD所以lBD:y①·······································································7分+12t2y=0，由韦达定理得y，代入直线lBM得E，得kAE所以lAE:y②······································································9分抽取一张，重复上述操作，直至n号从n_1号手中的三张卡片中随机抽取一张；（3）在一轮游戏结束后，将手持两张同色卡片的学生淘汰，余下的戏终止．求经过两轮游戏后只剩一个学生未被淘汰的概所以P所以一轮游戏结束后，1号学生恰有两张红卡的6=“从i号手中取出的卡为红卡”，所以PAi第8