山东省泰安市2019-2020高二数学上学期期末试题(PDF版附答案)
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书书书 !"#$%&'()*+,-./ 2020 1 !、"#$%&: !  " 1 2 3 4 5 6 7 8 #  $ C D A B D C D B '、(#$%&: !  " 9 10 11 12 #  $ BD AB BCD ABC )、*+&: 13. y2 12 - x2 24 = 1      14. 2 3 - 2n      15. 4,槡8 3      16. 槡8 85 85 ,、-.&: 17. (10 %) &':①(%): ∵ m2 - n2 = 1 ∴ m2 = n2 + 1 1 %************************ ∴ m4 - n4 = (m2 + n2 )(m2 - n2 ) 3 %**************** = m2 + n2 = n2 + 1 + n2 = 2n2 + 1 4 %********************* ∴ m4 - n4 = 2n2 + 1 +, 5 %******************** ②-.): ∵ m4 - n4 = 2n2 + 1 ∴ m4 = n4 + 2n2 + 1 = (n2 + 1)2 8 %********************** ∴ m2 = n2 + 1 / m2 - n2 = 1 9 %************************ ∴ m2 - n2 = 1 +,. 01,m4 - n4 = 2n2 + 1 +,2(.345 m2 - n2 = 1. 10 %****** 18. (12 %) 6:(1)①7 m = 0 8,mx2 - mx - 1 = - 1 < 0,9:;+,. 1 %********* ②7 m≠0 8,mx2 - mx - 1 < 0 < x∈R ;+,,=>?@ m < 0 Δ = m2 + 4m{ < 0 6A - 4 < m < 0 4 %*********************** 01,BC m 2DEFG5(- 4,0]. 5 %**************(2)H f(x)= mx2 - mx - 1, ①7 m = 0 8,f(x)= - 1 < 0,9:;+,. 6 %************ ②7 m > 0 8,f(x)2IJKLM1,?@ f(3)< 0. / m < 1 6 , ∴ 0 < m < 1 6 9 %************************* ③7 m < 0 8,f(x)2IJKLM[,?@ f(1)< 0,/ m - m - 1 = - 1 < 0, 9:+, ∴ m < 0. 11 %************************** 01,BC m 2DEFG5(- ∞ ,1 6 ). 12 %************* 19. (12 %) 6:\$]:^34①(1)∵ a3 = 5,a2 + a5 = 6b2 ,a1 = b1 ,d = q,d > 1 ∴ a1 + 2d = 5 2a1 + 5d = 6a1{ d 2 %********************** 6A a1 = 1 d{ = 2 _ a1 = 25 6 d = 5{ 12 (`a) 4 %***************** ∴ b1 = 1 q{ = 2 ∴ an = a1 + (n - 1)d = 2n - 1 bn = b1 qn - 1 = 2n - 1 6 %*********************** (2)∵ cn = an bn ∴ cn = 2n - 1 2n - 1 = (2n - 1)× (1 2 )n - 1 7 %**************** ∴ Tn = 1 + 3 × 1 2 + 5 × (1 2 )2 + …+ (2n - 3)× (1 2 )n - 2 + (2n - 1)× (1 2 )n - 1 ∴ 1 2 Tn = 1 2 + 3 × (1 2 )2 + 5 × (1 2 )3 + …+ (2n - 3)× (1 2 )n - 1 + (2n - 1)× (1 2 )n 8 %********************** ∴ 1 2 Tn = 1 + 2[1 2 + (1 2 )2 + …+ (1 2 )n - 1 ]- (2n - 1)× (1 2 )n 9 %*** = 1 + 2 × 1 2 [1 - (1 2 )n - 1 ] 1 - 1 2 - (2n - 1)× (1 2 )n = 3 - (2n + 3)× (1 2 )n ∴ Tn = 6 - (2n + 3)× (1 2 )n - 1 12 %***************** QRCST!UV#$ W 2 X(Y 6 X)\$R:^34②(1)∵ b2 = 2,a3 + a4 = 3b3 ,a1 = b1 ,d = q,d > 1 ∴ a1 d = 2 2a1 + 5d = 3a1 d{ 2 ∴ a1 d = 2 2a1 + 5d = 6{ d 2 %*********************** 6A a1 = 1 d{ = 2 _ a1 = - 1 d{ = - 2 (`a) 4 %***************** ∴ b1 = 1 q{ = 2 ∴ an = a1 + (n - 1)d = 2n - 1 bn = b1 qn - 1 = 2n - 1 6 %*********************** (2)∵ cn = an bn ∴ cn = 2n - 1 2n - 1 = (2n - 1)× (1 2 )n - 1 7 %**************** ∴ Tn = 1 + 3 × 1 2 + 5 × (1 2 )2 + …+ (2n - 3)× (1 2 )n - 2 + (2n - 1)× (1 2 )n - 1 ∴ 1 2 Tn = 1 2 + 3 × (1 2 )2 + 5 × (1 2 )3 + …+ (2n - 3)× (1 2 )n - 1 + (2n - 1)× (1 2 )n 8 %********************** ∴ 1 2 Tn = 1 + 2[1 2 + (1 2 )2 + …+ (1 2 )n - 1 ]- (2n - 1)× (1 2 )n 9 %*** = 1 + 2 × 1 2 [1 - (1 2 )n - 1 ] 1 - 1 2 - (2n - 1)× (1 2 )n = 3 - (2n + 3)× (1 2 )n ∴ Tn = 6 - (2n + 3)× (1 2 )n - 1 12 %***************** \$b:^34③ ∵ S3 = 9,a4 + a5 = 8b2 ,a1 = b1 ,d = q,d > 1 ∴ a1 + d = 3 2a1 + 7d = 8a1{ d 2 %************************ 6A a1 = 1 d{ = 2 _ a1 = 21 8 d ={ 3 8 (`a) 4 %******************* ∴ b1 = 1 q{ = 2 ∴ an = a1 + (n - 1)d = 2n - 1 QRCST!UV#$ W 3 X(Y 6 X)bn = b1 qn - 1 = 2n - 1 6 %**************************** (2)∵ cn = an bn ∴ cn = 2n - 1 2n - 1 = (2n - 1)× (1 2 )n - 1 7 %**************** ∴ Tn = 1 + 3 × 1 2 + 5 × (1 2 )2 + …+ (2n - 3)× (1 2 )n - 2 + (2n - 1)× (1 2 )n - 1 ∴ 1 2 Tn = 1 2 + 3 × (1 2 )2 + 5 × (1 2 )3 + …+ (2n - 3)× (1 2 )n - 1 + (2n - 1)× (1 2 )n 8 %********************** ∴ 1 2 Tn = 1 + 2[1 2 + (1 2 )2 + …+ (1 2 )n - 1 ]- (2n - 1)× (1 2 )n 9 %*** = 1 + 2 × 1 2 [1 - (1 2 )n - 1 ] 1 - 1 2 - (2n - 1)× (1 2 )n = 3 - (2n + 3)× (1 2 )n ∴ Tn = 6 - (2n + 3)× (1 2 )n - 1 12 %***************** 20. (12 %) 6:(1)∵ cde ABEF Pfe ∴ BF∥AE g BF hi ADE,AE hi ADE ∴ BF∥hi ADE 2 %*********************** g BC∥AD, jklA:BC∥hi ADE 4 %******************* g BF∩BC = B,BF,BC hi BCF ∴ hi BCF∥hi ADE g CF hi BCF ∴ CF∥hi ADE 6 %***********************(2)mI,n A Popqr,s,tuvwopx A - xyz,y C(2,2,0),D(0,4,0),F(2,0,4) ∴ →AD = (0,4,0 ),→CD = (- 2,2,0 ),→CF = (0,- 2,4) 8 %********** z n = (x,y,z)5hi CDF 2]{|M},y n·→CD = 0 n·→CF{ = 0/ x - y = 0 y - 2z{ = 0 H y = 2,6A x = 2 z{ = 1 ∴ n = (2,2,1) 10 %********** g →AD 5hi AEFB 2]{|M}, QRCST!UV#$ W 4 X(Y 6 X)∴ cos〈n,→AD〉= n·→AD | n| ·| →AD| = 2 3 ∴ hi CDF ~hi AEFB +€Riw2‚EP 2 3 . 12 %****** 21. (12 %) 6:z c = kx2 (k≠0) 1 %************************* 7ƒ„P 300 …8,z†‡4CP x1 ,y - x1 + b = 300 kx2 1{ = 1800             ① 2 %****************** 7ƒ„P 320 …8,z†‡4CP x2 ,y - x2 + b = 320 kx2 2{ = 200 ② 3 %****************** ˆ,①②6A k = 2,b = 330 ∴ c = 2x2 ,p = - x + 330 5 %********************** (1)z‰Š‹Œ† x 4†‡2Ž+P y …,h+P s …,y y = 800 + 2x2 6 %************************* ∴ s = y x = 800 + 2x2 x = 800 x + 2x≥2 800 x ·2槡x = 80 7 %**************** 7‘’7800 x = 2x,/ x = 20 8,s DA“”E,“”EP 80. ∴ •–Œ†‰†‡ 20 48,h+“—,“—P 80 …. 8 %****(2)z‰Š‹Œ† x 4†‡82˜™P t …,y t = px - 800 - c = (- x + 330)x - 800 - 2x2 = - 3x2 + 330x - 800 9 %********************* H - 3x2 + 330x - 800≥8200 6A 50≤x≤60 11 %*********************** ∴ Z‰Š‹–˜™š—› 8200 …,y•–Œ†2†‡Cœš”› 50 4, š› 60 4. 12 %*********************** 22. (12 %) 6:(1)ž!Ÿ  2a = 4 1 2 × 2c × b 槡= 3 a2 = b2 + c{ 2 ,6A b = 1 _ b 槡= 3 ∵ b > c, ∴ b 槡= 3 ∴ ¡¢ C 2\£5x2 4 + y2 3 = 1. 4 %***************** (2)ž y = kx + m x2 4 + y2 3{ = 1,¤kA(4k2 + 3)x2 + 8kmx + 4m2 - 12 = 0. 6 %****** QRCST!UV#$ W 5 X(Y 6 X)∵ v¥ l ~¡¢ C ¦‘=¦]{§Yr M, ∴ m≠0 ‘ Δ = 0, ž Δ = 64k2 m2 - 4(4k2 + 3)(4m2 - 12)= 0,¨©A 4k2 - m2 + 3 = 0. z M(x0 ,y0 ),y x0 = - 4km 4k2 + 3 = - 4k m ,y0 = kx0 + m = 3 m , ∴ M - 4k m ,3( )m . ž x = 4, y = kx +{ m6A N(4,4k + m). 8 %**************** ªz«¬­r P ?@!Ÿ,žIe2

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