山东省枣庄市滕州市2023-2024学年高三上学期期中考试数学答案(1)_2023年11月_0211月合集_2024届山东省枣庄市滕州市高三上学期期中考试_山东省枣庄市滕州市2024届高三上学期期中考试数学

2024 届高三定时训练 ( 3 − 6 2 9 , a 0 ] = 4 A = { x | 5  x  7 } 3 2 e B={x|x3 x5} A B = { x | 5  x  7 } a2 2a−1a+1 A =  A B a  2 2a−1a+1 A 2a−13 A B  a2  a a +  1 2  5 a=2 a4 {#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}a f ( x ) = ( x − 3 ) ( x − a )  a | a  2 a  4  f ( x a a a ) =  =  ( 3 3 3 x − 3 ) ( x − a ) f f f ( ( ( x x x ) ) )    0 0 0 (  ( a 3 , , 3 a ) ) x  [ 4 , +  ) f ( x )  − 9 a  x + x 9 − 3 9 9 9 x+ = x−3+ +32 (x−3) +3=9 x−3 x−3 x−3 x − 3 = x 9 − 3 x = 6 a9 a {a|a9} f(x)= 3sin2x+cos2x+1 = 2 ( 2 3 s i n 2 x + 1 2 c o s 2 x ) + 1 π =2sin(2x+ )+1 6 {#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}2 k π k π + + π 2 π 6   2 x x +  π 6 k π  + 2 2 k π π 3 + 3 π 2 k  Z k  Z π 2π [kπ+ ,kπ+ ] kZ 6 3 f(A)=2， A a  = ( 2 0 , π 3 ) s i n ( 2 2 A A + + π 6 π 6 = ) = 5 π 6 1 2 A = π 3 a2 =b2+c2−2bccosA b2+c2−bc=12 b c = b 2 + c 2 − 1 2  2 b c − 1 2 b c  1 2 b=c=2 3 S △ = A 1 2 B b C c s i n A = 4 3 b c  3 3 3 3 f ( x ) R x0 f ( x ) = x 2 e x x0时 −x0 f ( f x ( ) x = ) =  − x − x 2 e f ( − x ) 2 − x e , x x , x =   − [ 0 0 ( − x ) 2 e − x ] = − x 2 e − x f(x) R f(ax2 −3x−1)+ax2 −3x−1− f(5−ax)+ax−5 {#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}xR f ( a x 2 − 3 x − 1 ) + a x 2 − 3 x − 1  f ( a x − 5 ) + a x − 5 x  R h(x)= f(x)+x h(x) R a a x  2 − 0 ( 3 + a ) x + 4  0 a x 2 − 3 x − x 1   R a x − 5 x  R a0 ( 3 + a ) 2 − 1 6 a  0 1a9 a n = 1 (1 , 9 ) 2a2 =a2 +2 1 1 n  2 a 1 = 2 2a S =a2 +2n， n n n 可得 2 S n ( S n − S n − 1 ) = ( S n − S n − 1 ) 2 + 2 n S 2n − S 2n − 1 = 2 n S2 =S2 +(S2 −S2)+(S2 −S2)+ +(S2 −S2 )=n2 +n n 1 2 1 3 2 n n+1 S = n2 +n = n(n+1) n a =S −S = n(n+1)− (n−1)n ............................................ 5分 n n n−1 a = 2 1 {#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}a n + 1 a = n = ( n n + ( 1 n ) ( + n 1 + ) − 2 ) − ( n − n 1 ( n ) n + 1 ) 故 a n + 1 − a n = 2 n + n 2 + + 1 n − n + 2 1 + n n − 1 = 2 n + 1 ( ( n n + + 1 2 + + n n − ) ( 1 ) n − + 2 1 + n ( n n − + 1 2 ) + n ) 2(n+1)+2 n2 −1−2 n2 +2n−2n = ( n+2+ n)( n+1+ n−1) 2(1+ n2 −1)−2 n2 +2n = ............................................... 8分 ( n+2+ n)( n+1+ n−1) 1 + ( 1 + n 2 n − 2 1 −  1 ) 2 n = 2 n + 2 2 + n 2 n 2 − 1  n 2 + 2 n = ( n 2 + 2 n ) 2 a −a 0 n+1 n a a n 1   2 2 2 2 − − m m  a n  m f  ( x ) 2 1 (0，+) f(x)= +m x m  0 f ( x )  0 f(x) 1 m0 x(0,− ) f(x)0 f(x) m {#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}m x = = − − 1 1 m x  ( − 1 m f , + ( x )  ) f ( x ) l n  ( − 0 1 m ) = 0 f ( x ) x m ( + x 1 )  e e x x − l n l n x x x x + f 1 1 ( x ) (  0 ， g + (  x ) )  = − + x2ex +lnx (x)= x2 q ( x ) = x 2 e x + l n x 1 q(x)=(x2+2x)ex + 0 x q(x) q x ( 0 x e ) x0 = − l n x x 0 0 = x − 0 l n x 0 x  e x 2 e 0 0 − ln x0 + l q n ( x 1 2 0 ) =  0 0 q (1 )  0 h(x)= xex h(x )=h(−lnx ) 0 0 h ( x ) = x e x ( 0 ， +  ) x =−lnx 0 0 ( x )  ( x ) ( ( x 0 0 , ) x 0 ) e x0 l n x x 0 0 1 1 x ( 0 x 0 , +  x 0 x 0 ) 1 1   m  = − + = + − = (−,0] {#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}