贵州省2025年基础教育质量提升普通高中教学检测数学答案_2025年5月_0502贵州省2025年基础教育质量提升普通高中教学检测（毕节三诊、遵义贵阳4月）（全）

普通高中教学检测 数学参考答案及评分建议 一、选择题（每小题5分，共40分） 二、选择题（每小题6分，共18分。在每小题给出的选项中有三个正确，选对一个给2 分；有二个选项正确，选对一个给3分。出现错误选项，该小题0分） 三、填空题（每小题5分，共15分） 12． 13． 14． 或 ( 分，答案不完整不给分）； （3分 ） 四、解答题 15. 解： （1）由已知，切点坐标为 . ···························································· 1分 又 ， ······························································ 3分 则切线斜率 ， 所以切线方程为 ； ··································································· 5分 （2）由（1）知 ，则 ， ······· 6分 因为 ， ， ···················································· 7分 数学参考答案及评分建议 第 1 页（共7 页） 学科网（北京）股份有限公司 1 3 2 3  f ' ( x ) = 2 k c o = s 2 f x ' ( + 0 ) 2 = a x 1 − ( 0 1 , 1 ) 2 4 2 1 1 y = x+1 f ' ( x ) = 2 c o s 2 x + 2 a x − 1 f"(x)=−4sin2x+2a [ 6 , 3 ]   x  − 1 2 3 4 5 6 7 8 D C C B B A B D 9 10 11 BC ABD ACD  2 2x[− , ] 3 33 所以− ≤sin2x≤1， ······································································ 8分 2 因为 在 上具有单调性， 则 数学参考答案及评分建议 第 2 页（共7 页） 学科网（北京）股份有限公司 f ( x ≥) 0 或 f ( x ) ≤ 0 恒成立， ·························································· 9分 即 时，有2a≥4sin2x，或2a≤4sin2x， 即2a≥4，或 2 a ≤ − 2 3 ， 有 a ≥ 2 ，或a≤− 3. 所以 的取值范围是 . ············································· 13分 16．解： (1)零假设 :喜欢户外运动与性别无关. ·················································· 1分 由列联表可知 ·················· 5分 由于4.040＞3.841， ··········································································· 6分 故零假设 不成立，所以喜欢户外运动与性别有关. ·································· 7分 （2）不喜欢户外运动的员工按分层抽样抽取9人，其中女员工 人. ·············· 8分 可能的取值为 . ···································································· 9分 ， ， f ' ( x ) [ 6 , 3 ]   [ 6 , 3 − ]   x  a − ( −  , − 3 ]  [ 2 , +  ) H 0 2 = ( 1 0 0 + 1 0 0 4 ) 0  0  (1 2 ( 1 0 0 + 0 8  0 8 ) 0  − (1 1 0 0 0 0 +  1 1 2 2 0 0 ) 2 )  ( 1 0 0 + 8 0 )  4 . 0 4 0  H 0 4 X P P ( ( X X = = 0 1 ) ) = = C C 0 0 3 C 4 5 3 C 9 1 2 C 4 5 3 C 9 ， 1 = = ， ， 2 5 4 2 2 0 4 2 3， . ····································································· 13分 的分布列为 的期望为 . ······························· 15分 17．解： （1）由已知 ， ············································································ 1分 因为△ 的周长为 ， ·································································· 2分 即 ， ， ···································································· 3分 所以 . 椭圆 方程为 ； ································································· 5分 （2）设l:x=my+2(m0)， ， , ， . ········································································································ 6分 x=my+2  联立x2 y2 ，得 ， ···································· 7分 + =1   8 4 所以 ， ， ·············································· 8分 直线 方程为 .令 ，得 . 数学参考答案及评分建议 第 3 页（共7 页） 学科网（北京）股份有限公司 P P X ( ( X X = = 2 3 ) ) = = C C 2 C 4 3 C 9 3 C 4 3 C 9 X 15 05 = = 1 4 4 5 2 2 2 0 1 2 3 P 5 4 2 20 42 1 4 5 2 2 42 X E （ 4 X a b E ） = 2 = F 1 8 = 0 P a  Q 2 2 c − 5 + 4 2 = 2 a 2 c 2 x 8 = = + 1  2 4 y 4 2 4 2 0 2 4 2 = a + 1 2  1 4 5 2 + 3  4 2 2 = 5 4 6 2 = 4 3 P（x ,y） Q（x ,y） 1 1 2 2 y 1 + y 2 = m − 4 2 m + 2 ( m y 2 1 +  y 2 2 ) y = 2 m + 4 − 2 m 4 + y 2 − 4 = 0 M （ t , y ）1 N （ t , y ）2 PN y − y 2 = y 1 x − 1 − y t 2 ( x − t ) y = 0 x = − y y 2 1 ( x − 1 y − 2 t ) + t直线 方程为 .令 ，得 ， ··· 10分 由已知，有 ， ················································· 11分 即 ，即 ， ······················ 12分 化简得 ， ···················································· 13分 则 ， 有 ， . ········································································ 15分 18．解： (1)取线段 中点 ，连接 ， . ················································· 1分 因为△ 是正三角形，有 . ·················································· 2分 又 , 所以 平面 ， ······································································· 3分 有 ， ··············································································· 4分 又 为线段 中点， 则 . ··················································································· 5分 （2）（i）连接 ， 由（1）可知， ， ··································································· 6分 有 . ··························································· 7分 3 EP≥ AE−AP = （当且仅当点 在线段 的中点处，等号成立）， ······ 8分 2 7 则CP= 1+EP2≥ . 2 所以线段 的最小值为 ； ···························································· 10分 数学参考答案及评分建议 第 4 页（共7 页） 学科网（北京）股份有限公司 y 2 Q x M 1 − y 2 − t = y y 2 1 − y − ( x 1 − y y x 1 − 2 2 y 1 ) t + = = y 1 t y − y 2 x − t 2 ( − y x 1 y − 2 1 2 ( x − y 1 − t ) t ) y = 0 x = − y y ( 1 2 x − 2 − y 1 t ) + t ( ) y x + y x −t y + y =0 2 1 1 2 1 2 2 2 m +  2 2 m − 2 m − t y 1 4 + = y 2 2 0 + + ( 2 ( 2 − − ) t ) t ( m y − 1 4 2 + m + y 2 2 ) = = 0 0 t =4 BC E EA B B E B C C C D P B A ⊥ C ⊥ = 2 = B A ⊥ D C C C D E D E B C 2 E C + A P E E D P ⊥ E 2 A E = P P 1 + E P B 2 C ⊥ E D A E P AE CP 2 7（ii）由（1）及已知，平面 平面 . ······································································································ 11分 以 数学参考答案及评分建议 第 5 页（共7 页） 学科网（北京）股份有限公司 E 为坐标原点，以 E C ， E D 分别作为 x 轴， 轴正方向建立空间直角坐标系如 图所示. ···························································································· 12分 则 A ( 0 , 2 3 , 3 2 ) ，C(1,0,0)，D(0, 3,0)， 于是CD =(−1, 3,0)， A D = ( 0 , 2 3 , − 3 2 ) . ········································· 13分 设平面 的法向量为 m = ( x , y , z ) ，有 令 ，有 ， ，即m=(3, 3,1). ······································ 14分 设直线l的方向向量 n = ( 0 , m , n ) ，直线 l 与平面 所成角为 ， 于是 . ································· 15分 因为 2 3 m n = 2 （m 3 n ） ≤ m 2 + 3 n 2 （当且仅当 时取“ ”）， 所以 s i n 1 4 3 m m 2 2 n n 2 2 2 1 1 3 3  ≤ （ （ + + ） ） = . ······················································ 16分 当n=(0,1,− 3)， s i n = 0  . 2 13 故0≤sin≤ . 13 A C D A D E ⊥ B C D y −x+ 3y =0,   3 3  y− z =0.  2 2 y = 3 x = 3 z = 1 ACD  s i n = | 1 3 3 m m + 2 n + | n 2 = 3 m 2 1 + 3 ( n m 2 2 + + 2 n 2 3 ) m n  m= 3n =即直线l与平面 所成角的正弦值的取值范围是 . ················· 17分 19．解： （1）已知 ，即 ， ·································· 1分 即 ， ， ·································· 3分 化简得 ，又 所以数列 是首项为 公差为 的等差数列； ································ 5分 （2）由（1）可知 ， 所以 ， . 又 ， 所以 ， . ················································ 6分 . 所以 . ················································································ 7分 于是 ， ···························· 8分 数学参考答案及评分建议 第 6 页（共7 页） 学科网（北京）股份有限公司 t a n a n A + 1 C = D c o 1 s a n [ 0 , 2 1 1 3 3 ] sina 1 n+1 = cosa cosa n+1 n s c i o n s 2 2 a a c n + 1 n + 1 1 2 o s = a n c o + 1 1 2 s − c a o n s 1 2 a 1 n − c = c o 1 o s s 2 2 a a n + n 1 + 1 = c o 1 2 s a n 1 =2 cos2a 1  c o 1 2 s a n  2 1 c o s ( 2 0 a , n 2 = ) n 1 + c 1 o s 1 2 s a i n n 2 = a 2 n + = ( 1 n − − n 1 ) 1 + = 1 n = + n 1 n + 1  b n a = n  c o l o S n s g a 2 = n ( s 1 2 = i n [ l a o n n g ) 2 1 + = 1 1 − l o l g o s 2 g i 2 n 2 a n n n + + = 1 l o = g 2 n 1 2 2 n + 1 [ l o g − l o 2 g n 2 − 3 l + o  g ( 2 + n l + o g 1 ) 2 ] n − l o g 2 ( n + 1 ) ] 1 =− log (n+1) 2 2 1 log 1 1 1 4S i +S i+1 =4 4(i+1)(i+2) = = − (i+1)(i+2) i+1 i+2, 9分 因为n≥1，所以 ≤ ， 即 ≤ ； ······································································· 10分 （3）证明：定义 ，原不等式即 （n≥1） ·························································································· 12分 下面证明 ， 即 ， ···································································· 13分 即证 （✱）. ·················································· 14分 设 ， ·························································· 15分 则 ， 于是 在区间 上是增函数. ····················································· 16分 因为 ，有 ，不等式（✱）成立. 故原不等式成立. ················································································· 17分 数学参考答案及评分建议 第 7 页（共7 页） 学科网（北京）股份有限公司 n i= 1 c o 4 1 6 s f f S + S = i i+1 n  4 i= 1 a c o s 0 c o 1 n n + n + 1 n ( x ) = x ( x ) = 1 f ( x ) n + 1 n n 1 1    − =   i + 1 i + 2   i= 1 1 1 1 − − 2 3 2 n 1 S + S  i i+1 2 c o s a = 1 0 a + c o s a c o s a + 1 1 2 n + s a c o s a  l n n − 1 n n n + 1  l n n 1 n n + 1 −  l n n + 1 n 1 − − 2 l n x ( x  1 ) x 2 1 2 ( x − 1 ) + − = 2 2 x x x ， (1 +  ) n + 1 f ( )  1 n 1 2 1 +  1   − 2 0 f 1 3  + (1 + 1 2 c o ) = 1 3 s 0 − a n 1 4 − 1 + c  o s + a n n 1 +  1 l n − 2 1 n + 1 + l 2 n = 3 2 1 2 + −  n + 1 + l 2 n n + n 1