山东省安丘市2023届高三下学期4月二模考前适应性练习（二）数学试题（PDF版含答案）

高三二模考前适应性练习（二） 数学参考答案 1.A 2.D 3 C 4 B 5 C 6 C 7 A c 8 D c = 2ln + a c 2ln c = a 2ln a f (x) = x 2ln x(x 0) a 2 x 2 f (x) =1 = f (x) 0 x 2 f (x) 0 0 x 2 x x f (x) (0,2) f (x) (2,+ ) f (x) = f (2) = 2(1 ln 2) 0 min b 1 0 a 2 c 2 2a +1= 5 x x x x 2 1 2 1 g (x) = + g (x) = + (0,+ ) 5 5 5 5 a a 2 1 g (2) =1 g (a) = + g (2) =1 5 5 b a 5 = 2a +1 5 b a b 5 = 2a +1 22 +1= 5 b 2 c 2 b a a c b 2 x x b 2 1 2 0 c 2 a 2 2a +1= 5 g (x) = + 5 5 x x 2 1 g (x) = + (0,+ ) g (2) =1 5 5 a a 2 1 b a g (a) = + g (2) =1 5 = 2a +1 5 a b 5 5 b 5 = 2a +1 22 +1= 5 b 2 a b 2 c a c b 2 a c b 2 . D 9 BD 10 BCD 11 BD ABCD 1 A , , , 4 B ; 2, O ABCD BO E, OE . 3 , 1 ABCD M △BCD O ABCD BM BO AM O AM . 2 2 4 3 4 3 4 6 AB = 4 , BM = 42 22 = , AM = 42 = . 3 3 3 3 2 2 BO2 = BM 2 +OM 2 = (AM OM )2 BO2 4 3 4 6 = +OM 2 = OM 3 3 4 3 4 3 BO = 6 , ABCD R = ( 6) = 8 6 . 3 3 ABCD , C . BE = AB = 4 , OE = 4 6 , D . 2 4 5 20 b a 13 1 14.10 15 2 2 16. L = + , 0, 4 2m sin cos 2 6 70 . 17. 1 sinB + sinC = 2sinAcosB sinB = 2sinAcosB sin (A+ B) = sin (A B) B = A B, A = 2B 2 B = A = 4 12 6 (b+ c)2 a2 (b+ c) 2 (b2 + c2 2bccosA) 2 = ac ac 2b (1+ cosA) 2sinB (1+ cosA) 2sinB (1+ cos2B) = = = = 2cosB 6 a sinA sin2B A+ B = 3B B 0, 8 3 (b + c + a)(b + c a) 2cosB (1,2) (1,2) . 10 ac 18 1 2 2 2 2 2 2 4a 2 bn+1 bn = = = = n = 2 2an+1 1 2an 1 1 2an 1 1 2an 1 2a 11 n 2 2 1 1 4a 2a n n 2 b1 = = 2 2a1 1 ∴ bn 2 2 4 2 bn = 2+ (n 1) 2 = 2n 5 3 b c nn = 6 + ( 1) n 1 2 n = 6n ( 4)n cn+1 = 6 n+1 + ( 1)n 22n+2 = 6 6n + 4 ( 4)n 5 2 c n+1 cn = 5 6 n + ( 4)n = 1+ ( ) n 6n 7 3 5 2 n *N cn+1 cn cn+1 cn = 1+ ( ) n 0 6n 3 2 1+ ( )n 0 3 n 1 9 n * 2 4 N n 0, = 9 3 9 2 4( )n 3 max n 2 2 1 3 n ,0 = . 11 3 3 2 2( )n 3 min 9 3 , * n N cn+1 cn . 12 4 2 19. 1 P PA = PB = PC D, E F . AB M PM ⊥ AB AC2 + BC2 = AB2 , CA⊥CB . 2 AM =CM PAM PCM PMA = PMC = 90 PM ⊥ MC PM ⊥ ABC 4 PM ABED ABED⊥ ABC . 5 2 ABC DEF h, DEF S1 ABC S2 VF BCD =VDEF ABC VB DEF VD ABC = 1 1 1 2 h(S1 + S1S2 + S2 ) h (S1 + S ) = h S . 6 2 1S2 = h h = 3 3 3 3 3 F FH ⊥DE H FH ⊥ ADE H HG ⊥ BD G FG FGH E BD F . 7 2 2 5 4 5 3 5 9 FH = , DH = , BD = 9+ = 8 5 5 2 2 4 3 2 sin HDG = sin DBA = = 9 3 9 2 8 5 FH 3 4 HG = , tan FGH = = cos FGH = 11 15 GH 4 5 4 E BD F . 12 5 1 1 1 V D BCF = VD BCP = VA BCP = VP ABC = 2 PM = 6 2 4 4 C CA CB x y 1 3 P (2,1,6) , B (0,2,0) , D 3, ,3 , F 1, ,3 6 2 2 AB = ( 4, 2,0) AP = ( 2,1,6) EBD n1 = (1, 2,0) 8 3 BF = 1, ,3 2 3 BD = 3, ,3 2 FBD n2 = (0,2,1) 10 4 cos n1,n2 = 11 5 4 E BD F . 12 5 20 1 12 2 C 2 12 2 2 11 11 11 1 P = = = 4 C212 66 6 2 10 (u1 v1) (u (u v )2 v2 ) 10 10 10 12 u v = x y 10 21 10 22 = 2965 430 = 2535 5 i i i i i=1 i=1 1 12 1 12 u = ( xi 20) =10.8,v = ( yi 43) = 22.7 10uv =10 10.8 22.7 = 2451.6 10 i=1 10 i=1 10 12 2 2 u2i = x 2 2 i 2 10 =1394 200 =1194 10u =10 10.8 =1166.4 6 i=1 i=1 5 10 uivi 10uv b = i=1 2535 2451.6 = 3.0 7 10 1194 1166.4 u2 2i nu i=1 a = v b u = 22.7 3 10.8 = 9.7 10 8 y = 3x 10 9 3 x =10 y = 3 10 10 = 20 10 21 20 =1 2 22 20 = 2 2 12 a 21 1 B (0,b) , P ,0 , BPO = 45 2 a = b a = 2b 1 2 ... ...