2024年普通高等学校招生全国统一考试（模拟）数学试题参考答案一、单项选择题：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。题号12345678答案ADCDCCBB二、多项选择题：本题共3小题，每小题6分，共18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6分，部分选对的得部分分，有选错的得0分。题号91011答案CDBCDBCD三、填空题：本题共3小题，每小题5分，共15分。12．0；13．32，，答案不唯一；14．4．四、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．【解析】（1）此次测试的平均成绩为：0.2650.3750.4850.19579；···································5分（2）由题意可知，录取率为0.3，能进入第一梯队的概率为0.1；···········7分设录取分数为x，因为分数落在[90，100]的概率为0.1，分数落在[809，0)的概率为0.4，所以x[8090)，，令0.1(90)0.040.3x，解得x85，··········10分所以录取分数大概为85分，进入第一梯队的分数大概为90分，所以学生甲能被录取，但不能进入第一梯队．·····························13分16．【解析】若选择①caC2cos()（1）因为，bBcos由正弦定理得sinBCBCABcoscossin2sincos0,··················2分所以sin()2sincos0BCAB，即sinAB(2cos1)0，-1-1从而cosB，································································5分22因为B0，，所以B．··············································7分3ADc（2）在△ABD中，，sinsinBADBcBsin2所以sinADB，···············································10分AD2所以ADB，所以BADDAC，412所以ACBBAC，·····················································13分6所以△ABC是等腰三角形，且ac，所以ba2cos23．·······················································15分6若选择②sinsinACbc（1）因为，sinsinBCa由正弦定理得bacac222，···············································2分又由余弦定理bacacB2222cos，1从而cosB，··································································5分2B0，，所以．······················································7分（2）同①中第二问．若选择③B（1）因为2abAsin32sin，所以aBbA1cos3sin，2由正弦定理得sinABBA1cos3sinsin，·····························2分1整理得3sinBBcos1，所以sinB···························5分627因为，所以B，，666-2-52所以B，所以B.·················································7分663（2）同①中第二问．17．【解析】（1）取线段的中点为，连接，，AB1HEHFH1因为F为线段AC的中点，所以FHBC，且FHBC；·············2分121又E是AD的中点，所以ED，且EDBC；2所以EDFH，且EDFH，故四边形EDFH为平行四边形；所以DFEH，·······································································5分因为DF平面A1BE，EH平面，所以直线平面；························································7分DFA1BEzA1HAEFB（2）因为是的中点，DExy所以，所以；BEADBEAE1C因为平面平面，AB1EBCDE平面平面，A1BEBCDEBE所以平面.···························································8分AE1以为原点，分别为x轴，y轴，轴建立空间直角坐标系，EEB，ED，EA1z设，则，，，，，，AB2E(000)A1(0，0，1)B(300)，，C(320)，，则，，，··················9分EA1(0，0，1)BA1(3，0，1)BC(020)，，-3-nBA030xz设平面的法向量为，，，则1，即，A1BCn()xyznBC020y取x1，则n(1，03)，，·····················································11分设直线与平面所成角为，AE1A1BCnEA3则1，分sin|cos|n，EA1··································13||||nEA12所以直线AE与平面所成角为.·······································15分1318．【解析】（1）由题意可知，42p，所以p2，·········································2分所以抛物线E的方程为yx24.···············································4分2（2）（i）设Ax1，y1，Bx2，y2，将直线AB的方程代入yx4得：42kmm2kxkmxm222(24)0，所以xxxx，，········6分1212kk22因为直线PA与PB倾斜角互补，yykxmkxm21212222所以kkPAPB0，xxxx2121111111xx2即2(2)()kkmkkm2(2)012，xxxx212111(1)(1)422kmk2所以2(2)0kkm，(2)(2)kmkm422kmk2即20k，所以k1；·····································10分km2222（ii）由（i）可知x(2m4)xm0，所以xxmx1212xm42，，2则ABxxxxm1144212112，因为(2mm4)2240，所以m1，即13m，|3|m又点P到直线AB的距离为，21|3m|所以S421m2(3m)2(m1)，························13分221因为(3m)2(m1)(3m)(3m)(2m2)，2-4-1332232mmm()3，2327861所以S，当且仅当322mm，即m时，等号成立，9386所以△PAB面积最大值为.···············································17分9119.(1)解：因为XB(2,)，2所以X的分布列为：X01211P42(3分)1111113所以HX()(logloglog).(4分)4422442222(2)（i）解：记发出信号0和1分别为事件Ai,收到信号0和1分别为事件Bi,i=0,1,则PApPAp(),()1011，(5分)P(|)(|),BAP00111001BAq(|)(|)1PBAPBAq，(6分)所以P()()(|)()(|)BP0000101APBAPAPBApqpqpqpq(1)(1)12.(7分)P()A0(|)PBA00pq所以P(|).AB00(9分)P()12Bpqpq0(ii)证明：由(i)知，PBpqpq()120，所以P()1()2BP10Bpqpq，(10分)pp1所以KL(X||Y)pplog(1)log，(11分)22122pqpqpqpq11x设f(x)1lnx，则fx()，xx2当x∈(0,1)时，fx()0,fx()单调递增;当x(1,)时,fx()0,单调递减.-5-1所以fx(f)(1)0，即ln1x，xln11x所以log(1)x.(13分)2ln2ln2x11212pqpqpqpq所以KL(||)(1)(1)(1)0XYpp，(15ln2ln21pp分)pp11当且仅当1，即pq,01时等号成立.122pqpqpqpq2即KL(X||Y)≥0得证.(17分)【评分细则】1.第一问没有交待X的分布列，直接得到H(X)的值，得1分;若交待111PXPXPX(0),(1),(2)没有列表，不扣分;4242.第二问(i)直接得到PBpqpq()120没有交待过程，扣1分，第二问(ii)没有交待等号成立条件，扣1分。-6--7-