山东省烟台德州东营2024年高考诊断性测试 数学答案

2024年高考诊断性测试数学参考答案及评分标准一、选择题ACBCBDAA二、选择题9.ABD10.BCD11.BC三、填空题59512.413.10,10+14.[,]242四、解答题215.解：（1）f'(x)=2ax+1−,···································2分x1直线xy+2+1=0的斜率k=−,2由题意知f'(2)=2，···································4分1即4a+1−1=2，所以a=.····································5分2（2）f(x)的定义域为(0，+).···································6分1因为fx()0，所以b−x2−x+2lnx.21设g(x)=−x2−x+2lnx,x(0,+)，则bg()x.························8分2max2−x2−x+2(−x+1)(x+2)g'(x)=−x−1+==···················9分xxx当x(0,1)时，g'(x)0，所以g(x)在(0,1)单调递增，当x(1,+)时，g'(x)0，所以g(x)在(1,+)单调递减，···············11分3所以g(x)=g(1)=−.max23所以b−.·······························13分216.解：（1）因为AB⊥AC，AB==33AC，1所以=ACB60，OA==BC3.············································1分2因为AB=3，AD=2DB，所以DB=1.在DBO中，=DBO30，，OB=3，由余弦定理OD2=12+(3)2−213cos30=1，所以OD=1.·········3分在ADO中，，AD=2，AO=3，由勾股定理，AO⊥OD.·····4分因为AO1⊥平面ABC，OD平面，所以AO1⊥OD.·····················································5分数学参考答案（第1页，共4页）{#{QQABSYKEogCoAAAAAQgCEwEoCkGQkAGCAKoOAAAIMAAAyQFABCA=}#}因为AOAO1=O，所以OD⊥平面AOA1.······································6分因为AA1平面，所以AA1⊥OD；·····································7分（2）由（1）可知，OA,,ODOA1两两垂直，以O为坐标原点，方向分别为x,,yz轴正方向，建立如图所示的空间直角坐标系O−xyz.······8分因为AA1=23，AO=3，所以AO1=3.·············9分33则A(3,0,0)，A(0,0,3)，B(−,,0).···········10分12233333可得BA=−(,,3)，BA=−(,,0)，12222设m=(,,)xyz为平面ABA1的一个法向量，33x−y+30z=22则，取x=3，则y=3，z=1，333xy−=022故m=(3,3,1)，·····························12分由题意可知，n=(0,1,0)为平面AOA1的一个法向量，·······················13分mn3313因为cosmn,===，|mn|||1313313所以二面角B−−AAO的余弦值为.·······························15分11317.解：（1）两人得分之和大于100分可分为甲得40分、乙得70分，甲得分、乙得40分，甲得分、乙得分三种情况，所以得分大于100分的概率1121411141217p=++=.··························4分5332533253324515111（2）抢答环节任意一题甲得15分的概率p=+=.············7分212243（3）X的可能取值为2,3,4,5.12因为甲任意一题得分的概率为，所以任意一题乙得分的概率为.·····8分33111214PX(=2)=()2=，PXC(=3)=1=，39233327121228PXC(=4)=1()2+()4=，333338112121232PXCC(=5)=1()3+3()3=.···················12分4433333381所以的分布列为2345142832P9278181··································13分数学参考答案（第2页，共4页）{#{QQABSYKEogCoAAAAAQgCEwEoCkGQkAGCAKoOAAAIMAAAyQFABCA=}#}142832326所以EX()=2+3+4+5=.·····················15分927818181c18.解：（1）由题意知，a=2，=5，又因为c2=+a2b2，···················2分a解得b=4.xy22所以，双曲线C的方程为−=1.·············································3分416设直线l的方程为x=+my3，xy22−=122联立416，消x可得，(4m−1)y+24my+20=0.···············4分x=+my3不妨设P(x1,y1),Q(x2,y2)，1−24m20则m，且yy+=，yy=.·························5分21241m2−1241m2−yyyy所以12127分kkAPAQ==2·····················x1+2x2+2my1y2+5m(y1+y2)+254=−.·····························9分51（2）设直线AP的方程为y=+k(x2)，则直线DM:y=−(x−3)，ky=+k(x2)5k联立1，解得y=，·····································11分yx=−(−3)Mk2+1k4−100k用−替换上式中的k可得y=.·······························13分5kN25k2+16253125k2故SS=||yy=·································15分124MN(kk22++1)(2516)3125=.1625k2++41k2161625因为25kk22+225=40，当且仅当k=时，“=”成立，kk22531253125所以SS，故SS的最大值为.·························17分1281128119.解：（1）由题意可得yt=−1cos，||OB==BMt，所以x=|OB|−sint=t−sint，································2分数学参考答案（第3页，共4页）{#{QQABSYKEogCoAAAAAQgCEwEoCkGQkAGCAKoOAAAIMAAAyQFABCA=}#}所以x=−tsint，yt=−1cos.································4分（2）证明：由复合函数求导公式yt=yxxt，yxxtytsint所以yx===.··········································7分xttx1−costsint所以tan=，1−cost2cos22因为1+cos2=2cos2==sin2++cos2tan2122(1−cost)2===1−costy=，sint0()2+12−2cost1−cost1+cos2所以为定值1.·········································10分y0t（3）由题意，F(t)=(1−cost)22+sint=2−2cost=2|sin|.··········13分2tt因为0，sin022t所以Ft()=2sin，2t所以F(t)=−4cos+c（c为常数），······································15分2F(2)−F(0)=(−4cos+c)−(−4cos0+c)=8,所以OE的长度为8.·································17分数学参考答案（第4页，共4页）{#{QQABSYKEogCoAAAAAQgCEwEoCkGQkAGCAKoOAAAIMAAAyQFABCA=}#}