贵州省毕节市2025届高三上学期第一次诊断考试数学答案_2025年1月_250125贵州省毕节市2025届高三上学期第一次诊断考试（毕节一诊）（全科）

保密★启用前 毕节市 届高三年级第一次诊断性考试 2025 数学参考答案及评分建议 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 有一项符合题目要求. 题号 1 2 3 4 5 6 7 8 答案 C B C D A D A C 二、多项选择题：本题共3小题，每小题6分，共18分.全部选对的得6分，部分选对的 得部分分，有选错的得0分.(备注：第9题选对一个得2分，第10，11题选对一个得3分) 9 10 11 ABC AD AC 三、填空题：本题共3小题，每小题5分，共15分． 1 12. (1,3) 13. 14. 2 1,1(第一空2分，第二空3分) 2 四、解答题：本题共5小题，共77分. 15. 解：（Ⅰ）m n，2asinB2 3bcosA0································（2分） 又结合正弦定理可得：sin AsinB 3sinBcosA0······························ （4分） sinB 0，sin A 3cosA0，tan A 3·······························（5分）   A(0,)， A ·······································································（6分） 3  1 （Ⅱ）由（Ⅰ）可知A ，S  bcsinA  3··································（8分） 3 ABC 2 1  3 bcsin  bc 3， bc4····················································（10分） 2 3 4 由a2 b2 c2 2bccosAb2 c2 bc2bcbc bc 4 （当且仅当bc2时取等）······························································（12分）  a2，即a的最小值为2································································（13分） 数学答案 第 1 页 （共 4 页） {#{QQABQYwAggCoAAJAABhCQQGyCgOQkACAAYgOAAAQoAAACQFABAA=}#}16. 解：（Ⅰ）∵ f(x)ex[ax2 (a5)x1] ∴ f(x)ex[ax2 (3a5)xa4]····················································· （2分） ∵直线l的斜率为4e,···································································（3分） 由题意知 f(1)(5a9)e4e，解得a 1·····································（5分） ∴ f(x)ex(x2 4x1) ∴ f(1)2e，即P(1,2e)····························································（6分） ∴曲线 y  f(x)在点P(1,2e)处的切线方程为 y2e4e(x1) 即4ex y2e0········································································（7分） （Ⅱ）由（Ⅰ）知 f(x)e（x x2 2x3） 由 f(x)0得x1或x3 由 f(x)0得1 x3·································································（10分） ∴ f(x)的单调递增区间为（，1）（, 3，）， f(x)的单调递减区间为（1，3）·························································（13分） 6 ∴x1时, f(x)有极大值，极大值为 f(1) ································（14分） e x3时, f(x)有极小值，极小值为 f(3)2e3···········（15分） 17.（Ⅰ）证明:连接BD交AC 于点F ,连接EF ··················（2分） ∵底面ABCD是菱形 ∴F 是BD的中点,又E是PD的中点 ∴EF //PB·······························································（4分） ∵EF 平面ACE,PB平面ACE 所以PB//平面ACE···················································（6分） 解：（Ⅱ）记AD中点为O,连接EO,OC 则EO//PA 又∵PA 底面ABCD ∴EO 底面ABCD ∵AD底面ABCD ∴EO  AD 又∵EC  AD, ECEO  E 所以AD 平面COE,CO平面COE ，∴AD CO 所以ACD是等边三角形....................（8分） ∵E是PD的中点，且AE PD，∴PA AD. 以O为原点,OA,OC ,OE分别为x轴,y轴,z 轴建立如图所示的空间直角 坐标系oxyz. 不妨设PA  AD 2,则A(1,0,0), C(0, 3,0), E(0,0,1),B(2, 3,0)·········（ 10 分 ） AC (1, 3,0) ,AE (1,0,1),BE (2, 3,1)·································（11分） 设平面ACE的法向量n (x,y,z) n AC x 3y 0 n AE xz 0 可取n=( 3,1, 3)············································································ （13分） 2 3 3 3  42 cosn,BE   ············································（14分） 7 8 14 数学答案 第 2 页 （共 4 页） {#{QQABQYwAggCoAAJAABhCQQGyCgOQkACAAYgOAAAQoAAACQFABAA=}#}42 记BE与平面ACE所成角为,则sin|cos n，BE |  14 42 即BE与平面ACE所成角的正弦值为 ············································（15分） 14 18.解：（Ⅰ）设事件A=“甲第i次射击命中目标”，设事件B =“乙第i次射击命中目标”， i i 2 1 设事件C=“第三次射击就结束训练”，则P(A ) ，P(B ) ··············（1分） i 3 i 3 所以P(C) P(A )P(A )P(B )P(A )P(B )P(B )································（3分） 1 2 1 1 1 2 1 2 1 2 2 1       ···························································（5分） 3 3 3 3 3 3 2  9 2 所以第三次射击就结束训练的概率为 ····················································（6分） 9 （Ⅱ）①设事件D=“甲射击一次就结束训练”··········································（7分） 则P(D) P(A 1 )P(A 1 )P(B 1 )·····························································（9分） 2 2 1 7  （1 ）  3 3 3 9 7 所以甲射击运动员射击一次的概率 ·····················································（11分） 9 ②设结束训练时，甲射击运动员射击次数为X，则X的可能取值为1，2，，k, ······································································································（12分） 2 2 1 7 P(X1) P(A)P(A )P(B ) (1 )  1 1 1 3 3 3 9 1 2 2 1 2 1 1 14 P(X2) P(A)P(B )P(A )P(A)P(B )P(A )P(B )        1 1 2 1 1 2 2 3 3 3 3 3 3 3 81  P(Xk)[P(A)P(A )P(A )][P(B )P(B )P(B )]P(A ) 1 2 k1 1 2 k1 k [P(A)P(A )P(A )][P(B )P(B )P(B )]P(A )P(B ) 1 2 k1 1 2 k1 k k 1 2 2 1 2 1 1 7 2 ( )k1( )k1 ( )k1( )k1   ( )k1 3 3 3 3 3 3 3 9 9  故甲射击运动员射击次数X 的分布列为： X 1 2 3  k  7 7 2 7 2 7 2 P ( )1 ( )2  ( )k1  9 9 9 9 9 9 9 ······································································································（17分） 数学答案 第 3 页 （共 4 页） {#{QQABQYwAggCoAAJAABhCQQGyCgOQkACAAYgOAAAQoAAACQFABAA=}#}19.解：（Ⅰ）设P(x,y)，由题意得k k 4········································· （1分） PF PF 1 2 y y 即  4···············································································（3分） x1 x1 y2 化简整理得x2  1 4 y2 所以曲线C的方程为x2  （1 x 1） ················································ （5分） 4 （Ⅱ）证明：由题意可知P (a ,b )，P (a ,b )都在第一象限， n n n n1 n1 n1 4a2 n1 b2 n1 4 (b b )(b b )  作差化简整理得a a  n1 n n1 n ·············（7分）  4a2 n b2 n 4 n1 n （4 a n1 a n ） b b 而b b 4，所以a a  n1 n 0········································（8分） n1 n n1 n a a n1 n y 设P P 的中点为Q ,所以a a  Q n k ······································（9分） n n1 n n1 n x OQ n Q n 因为曲线C的渐近线方程为y 2x， 所以k （0，2），所以0a a 2···············································（11分） OQ n n1 n （Ⅲ）由题意可知直线l的斜率存在，设直线l的方程为y1k(x1) A(x ,y ),B(x ,y ),D(x,y) 1 1 2 2 y1k(x1)  联立方程组 y2 整理得（4k2)x2 (2k2 2k)x(k2 2k5)0 x2  1  4 2k2 2k k2 2k5  0,x x  ,x x  ········································ （14分） 1 2 k2 4 1 2 k2 4 因为 MA  1k2 x 1  1k2(x 1)，同理由 MA DB  AD  MB 得 1 1 (x 1)(x x)(xx )(x 1)化简整理得2x x (x x )(x x 2)x 1 2 1 2 1 2 1 2 1 2 ······································································································（15分） k2 2k5 2k2 2k 2k2 2k 所以2  ( 2)x k2 4 k2 4 k2 4 4x5 化简整理得k  ········································································（16分） x1 代入 y1k(x1).化简整理得4x y40. 所以点D在定直线4x y40上······················································（17分） 数学答案 第 4 页 （共 4 页） {#{QQABQYwAggCoAAJAABhCQQGyCgOQkACAAYgOAAAQoAAACQFABAA=}#}