山东省烟台德州东营2024年高考诊断性测试数学答案(1)_2024年3月_013月合集_2024届山东省“烟台一模”2024年3月高考诊断性测试

2024 年高考诊断性测试 数学参考答案及评分标准 一、选择题 A C B C B D A A 二、选择题 9.ABD 10.BCD 11.BC 三、填空题 5 9 5 12.4 13.10, 10+ 14.[ , ] 2 4 2 四、解答题 2 15.解：（1） f'(x)=2ax+1− , ··································· 2分 x 1 直线x+2y+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 f(x)0，所以b− x2 −x+2lnx. 2 1 设g(x)=− x2 −x+2lnx,x(0,+)，则b g(x) . ························ 8分 2 max 2 −x2 −x+2 (−x+1)(x+2) g'(x)=−x−1+ = = ··················· 9分 x x x 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)=− . max 2 3 所以b− . ······························· 13分 2 16.解：（1）因为AB⊥ AC，AB= 3AC =3， 1 所以ACB=60 ，OA= BC = 3. ············································ 1分 2 因为AB=3，AD=2DB，所以DB=1. 在 DBO中，DBO=30 ，DB=1，OB = 3， 由余弦定理OD2 =12 +( 3)2 −21 3cos30 =1，所以OD=1. ········· 3分 在 ADO中，OD=1，AD=2，AO= 3，由勾股定理，AO⊥OD. ····· 4分 因为AO⊥平面ABC，OD平面ABC， 1 所以AO ⊥OD. ····················································· 5分 1 数学参考答案（第 1 页，共 4 页） {#{QQABSYKEogCoAAAAAQgCEwEoCkGQkAGCAKoOAAAIMAAAyQFABCA=}#}因为AO AO=O，所以OD⊥平面AOA. ······································ 6分 1 1 因为AA 平面AOA，所以AA ⊥OD； ····································· 7分 1 1 1 （2）由（1）可知，OA,OD,OA 两两垂直，以O为坐标原点，OA,OD,OA 方向分别为x,y,z 1 1 轴正方向，建立如图所示的空间直角坐标系O−xyz. ······ 8分 因为AA =2 3，AO= 3，所以AO=3. ············· 9分 1 1 3 3 则A( 3,0,0)， A(0,0,3)，B(− , ,0). ··········· 10分 1 2 2 3 3 3 3 3 可得BA =( ,− ,3)，BA=( ,− ,0)， 1 2 2 2 2 设m =(x,y,z)为平面ABA 的一个法向量， 1  3 3  x− y+3z =0  2 2 则 ，取x= 3，则y =3，z =1，  3 3 3 x− y =0   2 2 故m =( 3,3,1)， ····························· 12分 由题意可知，n=(0,1,0)为平面AOA的一个法向量， ······················· 13分 1 m n 3 3 13 因为cosm,n= = = ， |m||n| 13 13 3 13 所以二面角B− AA −O的余弦值为 . ······························· 15分 1 13 17.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70分、乙得40 分，甲得70分、乙得70分三种情况，所以得分大于100分的概率 1 1 2 1 4 1 1 1 4 1 2 1 7 p=    +    +    = . ·························· 4分 5 3 3 2 5 3 3 2 5 3 3 2 45 1 5 1 1 1 （2）抢答环节任意一题甲得15分的概率 p=  +  = . ············ 7分 2 12 2 4 3 （3）X 的可能取值为2,3,4,5. 1 2 因为甲任意一题得15分的概率为 ，所以任意一题乙得15分的概率为 . ····· 8分 3 3 1 1 1 2 1 4 P(X =2)=( )2 = ， P(X =3)=C1   = ， 3 9 2 3 3 3 27 1 2 1 2 28 P(X =4)=C1 ( )2 +( )4 = ， 3 3 3 3 3 81 1 2 1 2 1 2 32 P(X =5)=C1 ( )3 +C3( )3  = . ··················· 12分 4 3 3 3 4 3 3 3 81 所以X 的分布列为 X 2 3 4 5 1 4 28 32 P 9 27 81 81 ·································· 13分 数学参考答案（第 2 页，共 4 页） {#{QQABSYKEogCoAAAAAQgCEwEoCkGQkAGCAKoOAAAIMAAAyQFABCA=}#}1 4 28 32 326 所以E(X)=2 +3 +4 +5 = . ····················· 15分 9 27 81 81 81 c 18.解：（1）由题意知，a =2， = 5， 又因为c2 =a2 +b2， ··················· 2分 a 解得b=4. x2 y2 所以，双曲线C的方程为 − =1. ············································· 3分 4 16 设直线l的方程为x=my+3， x2 y2  − =1 联立 4 16 ，消x可得，(4m2 −1)y2 +24my+20=0. ··············· 4分   x =my+3 不妨设P(x ,y ),Q(x ,y )， 1 1 2 2 1 −24m 20 则m ，且y + y = ，y y = . ························· 5分 2 1 2 4m2 −1 1 2 4m2 −1 y y y y 所以k k = 1  2 = 1 2 ····················· 7分 AP AQ x +2 x +2 m2y y +5m(y + y )+25 1 2 1 2 1 2 4 =− . ····························· 9分 5 1 （2）设直线AP的方程为y =k(x+2)，则直线DM :y =− (x−3)， k  y =k(x+2)  5k 联立 1 ，解得y = ， ····································· 11分 y =− (x−3) M k2 +1   k 4 −100k 用− 替换上式中的k 可得y = . ······························· 13分 5k N 25k2 +16 25 3125k2 故S S = | y y |= ································· 15分 1 2 4 M N (k2 +1)(25k2 +16) 3125 = . 16 25k2 + +41 k2 16 16 2 5 因为25k2 + 2 25k2 =40，当且仅当k = 时，“=”成立， k2 k2 5 3125 3125 所以S S  ， 故S S 的最大值为 . ························· 17分 1 2 81 1 2 81 19.解：（1）由题意可得y=1−cost，|OB|= BM =t ， 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=t−sint ，y=1−cost. ································ 4分 （2）证明：由复合函数求导公式y= yx， t x t yx y sint 所以y = x t = t = . ·········································· 7分 x x x 1−cost t t sint 所以tan= ， 1−cost 2cos2 2 因为1+cos2=2cos2= = sin2+cos2 tan2+1 2 2(1−cost)2 = = =1−cost = y ， sint 2−2cost 0 ( )2 +1 1−cost 1+cos2 所以 为定值1. ········································· 10分 y 0 t （3）由题意，F(t)= (1−cost)2 +sin2t = 2−2cost =2|sin |. ·········· 13分 2 t t 因为0 ，sin 0 2 2 t 所以F(t)=2sin ， 2 t 所以F(t)=−4cos +c（c为常数）， ······································ 15分 2 F(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=}#}