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# Key pat1 1-53

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• 1. F F F F PAT 1 ( . .53)1. ก Fp q F F F F 1. (p ⇒ q) ∨ p 2. (∼ p ∧ p) ⇒ q 3. [(p ⇒ q) ∧ p] ⇒ q 4. (∼ p ⇒ q) ⇔ (∼ p∧∼ q)2. F F ก F 1. F ก F {−1, 0, 1} F ∀x∃y[x 2 + x = y 2 + y] 2. F ก F F ∃x[3 x = log 3 x] 3. F ก F F ∀x∃y[(x > 0 ∧ y ≤ 0) ∧ (xy < 0)] ∃x∀y[(xy < 0) ⇒ (x ≤ 0 ∨ y > 0)] 4. F ก F F ∀x[x > 0 ⇒ x 3 ≥ x 2 ] ∃x[(x ≤ 0) ∧ (x 3 < x)]3. F A = {1, {1}} P(A) F A F F 1. ก P(A) − A F ก 3 2. ก P(P(A)) F ก 16 3. {{1}} ∈ P(A) − A 4. {∅, A} ∈ P(A) 1
• 2. F F F4. ก F A = {x ∈ R x 2 − 6x + 9 ≤ 4} R F F ก F 1. A = {x ∈ R 3 − x > 4} 2. A ⊂ (−1, ∞) 3. A = {x ∈ R x ≤ 7} 4. A ⊂ {x ∈ R 2x − 3 < 7}5. ก F y 1 = f(x) = x + 1 x−1 x F Fก 1 y 2 = f(y 1 ) , y 3 = f(y 2 ), ..... y n = f(y n − 1 ) n = 2, 3, 4, ..... y 2553 + y 2010 Fก F F 1. x−1 x+1 2. x2 + 1 x−1 3. x2 + 1 2x 4. 1 + 2x − x 2 x−16. Ff g กF ก f(x) = x2− 1 g(x) = f(x) − x − 1 x −4 F F ก. D g = (2, ∞) . F x>0 F g(x) = 0 1 F F F F ก F 1. ก. ก . ก 2. ก. ก F . 3. ก. F . ก 4. ก. . 2
• 3. F F F7. ก Fx F sin x + cos x = a sin x − cos x = b F F sin 4x F ก F F 1. 1 (a 3 b − ab 3 ) 2 2. 1 2 (ab 3 − a 3 b) 3. ab 3 − a 3 b 4. a 3 b − ab 38. ก F ก 25x 2 + 21y 2 + 100x − 42y − 404 = 0 F F F ก F (−3, 1 + 8 ) ก ก F F 1. 5y 2 − 4x 2 − 10 8 y − 32x − 25 = 0 2. 3y 2 − 2x 2 − 6 8 y − 8x + 15 = 0 3. y 2 − 4x 2 − 2y − 16x − 19 = 0 4. y 2 − 7x 2 − 2y − 28x − 28 = 09. A(−3, 1) B(1, 5) C(8, 3) D(2, −3) ABCD F F 1. F AB ก F DC 2. ก F AB ก DC F ก 10 2 F 3. ก ก A F F C D F Fก 9 2 2 F 4. ก ก B F F C D F Fก 9 2 F10. ก Fx y ก y≠1 F log y 2x = a 2y = b F x F Fก F F 1. 1 2 (log 2 b) a 2. 2(log 2 b) a 3. a 2 (log 2 b) 4. 2a(log 2 b) 3
• 4. F F F11. ก 72 x + 72 < 2 3x + 3 + 3 2x + 2 F F 1. (log 8 7 , log 9 8) 2. (log 9 8 , log 8 9) 3. (log 8 9 , log 7 8) 4. (log 9 10 , log 8 9)12. F ก 1 x + 1 x−1 + a = 0 ก 4 2 F F a F F F F F 1. (−∞, −3) 2. (−3, 0) 3. (0, 1) 4. (1, 3)13. ก F f x − 1   x  1 = x x≠0 x≠1 F 0<θ< π 2 F f(sec 2 θ) Fก F F 1. sin 2 θ 2. cos 2 θ 3. tan 2 θ 4. cot 2 θ14. F a b ก Fก a = i + 1 j − 3pk 2 b = − 2pi + 2j + pk p F a กก b b Fก 3 F F p F F F F 1. (−3, − 3 ) 2 2. (− 3 , 0) 2 3. (0, 3 ) 2 4. ( 3 , 3) 2 4
• 5. F F F15. ก F ABC A(0, 0) B(2, 2) C(x, y) (quadrant) 2 F F AC Fก F BC F ABC F Fก 4 F F C F F F F 1. x−y+4 = 0 2. 4x + 3y − 1 = 0 3. 2x − y − 3 = 0 4. x+y−5 = 016. F Z 1 , Z 2 , Z 3 , ..... F Z 1 = 0, Zn + 1 = Z2 + i n n = 1, 2, 3, ..... i= −1 F F Z 111 Fก F F 1. 1 2. 2 3. 3 4. 11017. ก ก 3 + 11 + 33 + ∧+ 3 +n2 1− 2 + ..... Fก F F n n 4 16 − 4 1. 20 3 2. 29 3 3. 31 3 4. 40 318. ก FR F f:R→R g:R→R กF 2 f(x) = 3x 3 , g(1) = 8 g (1) = 2 3 F (fog) (1) Fก F F 1. 1 3 2. 2 3 3. 1 4. 4 3 5
• 6. F F F19. ก F 13 4 F S, M, L XL F กก F 3 F ก F F ก 2 Fก F F 1. 72 425 2. 72 5525 3. 3 221 4. 3 2210020. ก FS A, B ก F S F F ก. P(A) = P(A ∩ B) + P(A ∩ B ) . F P(A) = 0.5 , P(B) = 0.6 P(A ∪ B ) = 0.7 F P(A − B) = 0.4 F F ก F 1. ก. ก . ก 2. ก. ก F . 3. ก. F . ก 4. ก. .21. ก F F F F ก 40 F ก F 35 ก F 50 F ก F ก ก F F 1. 3 : 2 2. 2 : 3 3. 2 : 1 4. 1 : 222. ก F A = 7 (7 7 ) , B = 7 77 , C = 77 7 D = (77 7 ) 7 F F ก F 1. B < A < C < D 2. B < C < A < D 3. C < B < D < A 4. C < A < D < B 6
• 7. F F F23. F กF " PAT" 16325, 34721, 12347, 52163, 90341, 50381 F F PAT 2564, 12345, 854, 12635, 34325, 45026 F F " PAT" 1. 75401 2. 13562 3. 72341 4. 8305124. F N ก F a ∗ b = ab a, b ∈ N F F a, b, c ∈ N ก. a∗b = b∗a . (a ∗ b) ∗ c = a ∗ (b ∗ c) . a ∗ (b + c) = (a ∗ b) + (a ∗ c) . (a + b) ∗ c = (a ∗ c) + (b ∗ c) F F ก F 1. ก 2 F . . 2. ก 2 F . . 3. ก 1 F . 4. ก. . . . กF 7
• 8. F F F25. F F F 5 A, B, C, D E A ก F "C D ก ก" B ก F "A C " C ก F "D ก ก" D ก F "E ก ก" E ก F "B ก ก" กF กF F F F F F F F 1. A, B, D C E 2. B D A C 3. A, B C D E 4. B E A C26. ก F A, B C F n(A ∪ B ∪ C) = 91 , n(A ∩ B ∩ C ) = 11, n((B − A) ∩ (B − C)) = 15 , n(A ∩ B ∩ C) = 20 n((A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C)) = 47 n(C) = 59 F n(A ∩ B ∩ C) Fก F27. F S = {x ∈ R 3x + 1 + x − 1 = 7x + 1 } R F ก ก S Fก F28. F A ก F F กF F ก 10 B ก F F กF F ก 10 C กF f:A→B กF F .. . a f(a) F Fก 1 กF a∈A ก C Fก F29. F α β ก tan α = a b       F cos  arcsin  a   + sin  arccos  a  = 1   a2 + b2     a2 + b2   F sin β F Fก F 8
• 9. F F F30. F cos 36 − cos 72 sin 36 tan 18 + cos 36 Fก F31. FA B ก F 2×2  −4 −4   −5 −8  2A − B =   A − 2B =    5 6   4 0  F det(A 4 B −1 ) Fก F32. F x, y, z w F ก ก  1 0   x −1   2y −1   1 0     =     −1 w   0 y   z 2   −1 w  F 4w − 3z + 2y − x Fก F33. F u, ν w ก Fก u = i + 2j + 3k , ν = 2i − dj + k , w = ai + bj + ck a, b, c d F u ⋅ w = 2 , u ⋅ (ν + w) = 3 , ν + w = i + qj + rk q,r w ก −2 i + 1 j + 1 k 3 2 3 F F a + 4b + 2c Fก F34. F Z1 Z2 F Z2 (conjugate) Z2 F 5Z 1 + 2Z 2 = 5 Z 2 = 1 + 2i i 2 = −1 F F 5Z −1 1 Fก F35. F {a n } a n = 2 + 4 + 6 2 K + 2n + n ก กn F lim a n n→∞ F Fก F  36. ก F n = 1, 2, 3, ..... n Sn = Σ  1  k = 1  k (k + 1) + k k + 1  F n→∞ lim S n Fก F 9
• 10. F F F37. ก Fa b f กF ก  x 3 − 3x − 2  , x<2  x−2 f(x) =  a−b , x=2   2  x + ax + 1 , x > 2 F f กF F F F a2 + b2 Fก F38. ก FR F f:R→R กF f (x) = 3 x + 5 ก x f(1) = 5 f(x 2 ) − 2 F F lim f(x) Fก F x→439. ก FR F f:R→R กF f (x) = 6x + 4 ก x F F F y = f(x) (2, 19) F ก 19 F F f(1) F ก F40. ก F A = {0, 1, 2, 3, 4} ก F F ก F 300 F ก A F ก F ก Fก F41. ก ก 7 ก F ก ก ก ก4 กF 7 Fก F ก F ก F ก Fก F42. F ก กF F ก 72 ( ก ) F ก 600 F ก ก1 F 60 F F 70 F F Fก F43. กก ก ก กF 4 2 ก Fก ก F กF ก2 F ก ก 4 45, 46 6ก ก F ก ก 4 Fก F44. ก ก F ก F F F F 700 F F4 FF F 400 F F −2 F ก Fก F F 10
• 11. F F F45. F F 4 กF 4 F F 20 ก ( F F 1 F 2 F 3 F 4 กF F 5 F F 6 F F 7)46. ก ก 221 ก ก ก 260 ก F ก F ก ก ก ก ก (1) F ก ก (2) ก F ก Fก F F ก F ก ก ก F ก F F กก F47. ก FR F f:R→R g:R→R กF ก ก ก ⊗ f g (f ⊗ g)(x) = f(g(x)) − g(f(x)) ก x F g(x) = 2x + 1 f(x) = x 2 − 1 ก x F (f ⊗ g)(1) F ก F48. F a, b, c, d ก F ก F 4 ก dcba F ก 9 F abcd F b F ก F 11
• 12. F F F49. F x F ก 1, 2, 3,....., 11 F F 1 F ก F ก 43 ก F ก 28 x F Fก F50. ก 2, 3, 4, 5, 6,..... F 1 9 17 ... 2 2 8 10 16 ... 3 3 7 11 15 ... 4 4 6 12 14 ... 5 5 13 ... 2400 F F ************************* 12
• 13. F F F F PAT 1 ( . .53)F 1 4 1 ∼ p ∨ q ∨ p ≡ (∼ p ∨ p) ∨ q ≡ T ∨ q ≡ T 2 F→q≡T 3 ≡ ∼ ((p → q) ∧ p) ∨ q ≡ ∼ (p → q)∨∼ p ∨ q ≡ ∼ (p → q) ∨ (p → q) ∼A∨A≡T ≡ T 4 ≡ ∼ (∼ p) ∨ q ↔ ∼ (p ∨ q) ≡ (p ∨ q) ↔ ∼ (p ∨ q) ≡ F A↔ ∼A≡FF 2 3 F 1 x = − 1, y = − 1 (−1) 2 + (−1) = (−1) 2 + (−1) T x = 0, y = 0 02 + 0 = 02 + 0 T x = 1, y = 1 12 + 1 = 12 + 1 T F 2 y y = 3x y = log 3 x F ก F ∴ F x F 3 x = log 3 x x F 3 ก ∼ ∀x∃y[p ∧ q ∧ r] ≡ ∃x∀y[∼ p∨∼ q)∨∼ r] ≡ ∃x∀y[r → (~p∨∼ q)] ≡ ∃x∀y[xy < 0 → (x ≤ 0 ∨ y > 0)] F 4 ∼ ∀x[p → q] ≡ ∼ ∀x[∼ p ∨ q] ≡ ∃x[p∧∼ q] ≡ ∃x[x > 0 ∧ x 3 < x 2 ] 1
• 14. F F FF 3 4 ก F A = {1, {1}} F P(A) = {∅, {1}, {{1}}, {1, {1}}} P(A) − A = {∅, {{1}}, {1, {1}}} ก 1 ก n(P(A) − A) = 3 ก 2 ก n(P(P(A)) = 2 2 n(A) 2 = 2 2 = 16 ก 3 ก {{1}} ∈ P(A) − A ก 4 {∅, A} = {∅, {1, {1}}} ∉ P(A)F 4 1 A (x − 3) 2 ≤ 4 x−3 ≤4 → −4 ≤x−3≤4 −1 ≤ x ≤ 7 A = [−1, 7] → A = (−∞, −1) ∪ (7, ∞) 1 A 3−x > 4 → x−3 > 4 x−3 > 4 x−3 < −4 x>7 x < −1 A = (−∞, −1) ∪ (7, ∞)F 5 2  x+1 +1 x+1+x−1  x−1 y2 = f x + 1  = x−1 = x−1 x + 1 − (x − 1) = 2x = x  x+1 −1 2  x−1 x−1 y 3 = f(y 2 ) = f(x) = x + 1 x−1 y 4 = f(y 3 ) = f  x − 1  = x + x 1 F y = x+1 , x−1 yF = x ∴ y 2553 + y 2010 = x + 1 + x x−1 = x + 1 + x2 − x x−1 = x2 + 1 x−1 2
• 15. F F FF 6 4 x−1 g(x) = − x−1 x2 − 4 x−1 Dg ≥0 x−1≥ 0 x2 − 4 (x − 1) (x − 2)(x + 2) ≥0 ∩ x≥1 -2 1 2 1 D g = {1} ∪ (2, ∞) ∴ ก. x−1 g(x) = 0 = x − 1 ⇒ x2− 1 = x − 1 x2 − 4 x −4 ∴x = 1 1 x2 − 4 = 1 → x2 − 4 = 1 → x2 − 5 = 0 → x = 5 ,− 5 x>0 2 F 1, 5 ∴ .F 7 3 sin x + cos x = a (1) sin x − cos x = b (2) (1) + (2), 2 sin x = a + b (1) − (2), 2 cos x = a − b (1) × (2), sin 2 x − cos 2 x = ab → cos 2 x − sin 2 x = − ab ∴ sin 4x = sin 2(2x) = 2 sin 2x cos 2x = 2(2 sin x cos x)(cos 2 x − sin 2 x) = (a + b)(a − b)(−ab) = (a 2 − b 2 )(−ab) = ab 3 − a 3 b 3
• 16. F F FF 8 3 ก 25x 2 + 21y 2 + 100x − 42y − 404 = 0 ก F 25x 2 + 100x + 21y 2 − 42y = 404 25(x 2 + 4x + 4) + 21(y 2 − 2y + 1) = 404 + 100 + 21 F 25(x + 2) 2 + 21(y − 1) 2 = 525 (x + 2) 2 (y − 1) 2 21 + 25 = 1 F ก c= ก- F = 25 − 21 = 2F HYPER ก F (−3, 1 + 8 ) HYPER ก F (y − 1) 2 (x + 2) 2 y ก HYPER 22 − b2 = 1 HYPER F (−3, 1 + 8 )(-3,1 + 8) (1 + 8 − 1) 2 (−3 + 2) 2 F1 F 4 − b2 = 1 c = 2 = a HYPER 2 − 12 = 1 → b 2 = 1 x b (-2,1) ก HYPER (y − 1) (x + 2) − 1 = 1 2 2 4 F2 4 F (y − 1) 2 − 4(x + 2) 2 = 4 y 2 − 2y + 1 − 4(x 2 + 4x + 4) = 4 y 2 − 2y + 1 − 4x 2 − 16x − 16 = 4 y 2 − 4x 2 − 2y − 16x − 19 = 0 4
• 17. F F FF 9 4 ก1 ก m AB = 5 − 1 = 1 1 − (−3) m DC = −3− 3 = 1 − 2 8 m AB = m DC F AB ก F DC ก2 ก AB = [1 − (−3)] 2 + (5 − 1) 2 = 32 = 4 2 DC = (8 − 2) 2 + [3 − (−3)] 2 = 72 = 6 2 ∴ AB + DC = 4 2 + 6 2 = 10 2 ก3 ก m CD = 1 CD ก x−y−5 = 0 ก ก A CD −3 − 1−5 9 9 2 9 2 = = ⋅ = 2 2 2 2 2 ก4 ก ก B CD 1−5−5 9 = 2 2 ก F ABCD D C A B ก FF ก A CD = ก B CD F 4F 10 1 ก 2y = b ก log y 2x = a log 2 2 y = log 2 b 2x = y a ∴ y = log 2 b x = 1ya 2 y = log 2 b F x = 1 (log 2 b) a 2 5
• 18. F F FF 11 2 72 x + 72 < 2 3(x + 1) + 3 2(x + 1) 9 x ⋅ 8 x + 72 − 8 x + 1 − 9 x + 1 < 0 9 x 8 x − 8 x ⋅ 8 − 9 x ⋅ 9 + 72 < 0 8 x (9 x − 8) − 9(9 x − 8) < 0 (9 x − 8)(8 x − 9) < 0 F 9x − 8 = 0 → 9 x = 8 → log 9 9 x = log 9 8 → x = log 9 8 F 8x − 9 = 0 → 8 x = 9 → log 8 8 x = log 8 9 → x = log 8 9 ก (9 x − 8)(8 x − 9) < 0 (log 9 8, log 8 9) log 9 8 log8 9F 12 2  1  x +  1  x − 1 + a = 0 →  1  2x + 2  1  x + a = 0 4 2 2 2 2  1  2x + 2  1  x + 1 = 1 − a →   1  x + 1  = 1 − a 2  2 2   y กก F x ∈ R+ x 0 < 1 < 1 2 1 < 1−a < 4 x (0,1) y = ( 1 )x 1 < 1 + 1 < 2 2 0 < −a < 3 2 x x 2 1 <   1  + 1 < 4 2  0 > a > −3   ∴ a ∈ (−3, 0) ก 1 x + 1 x−1 + a = 0 ∴ a<0 F F 3, 4 4 2 ก F x = 1, 1 + 1 + a = 0 → a = − 5 4 4 F 1 ∴ F 2 6
• 19. F F FF 13 1 x x−1 = sec 2 θ x = sec 2 θx − sec 2 θ sec 2 θ = sec 2 θ ⋅ x − x sec 2 θ = x(sec 2 θ − 1) sec 2 θ x= sec 2 θ − 1 ∴ f(sec 2 θ) = = sec θ − 1 1 2 sec 2 θ sec θ 2 sec 2 θ − 1 = 1− 1 sec 2 θ = 1 − cos 2 θ = sin 2 θF 14 2 ก b = 3 F a กก b F a⋅b = 0 (−2p) 2 + 2 2 + p 2 = 3 (1)(−2p) +  1  (2) + (−3p)(p) 2 = 0 5p 2 + 4 = 3 −2p + 1 − 3p 2 = 0 กก F 3p 2 + 2p − 1 = 0 F 5p 2 + 4 = 9 3(1) + 2p − 1 = 0 5p 2 = 5 p = −1 p2 = 1 ∴ F p F a กก b b = 3 −1 F F − 3 , 0  2 F 15 1 กF ก F F F ก m AB = 2 − 0 = 1 y 2−0 CD กก AB F m CD = − 1 C(x,y) y−1 ก m CD = x−1 B(2,2) y−1 D(1,1) F −1 = x−1 x A(0,0) −1(x − 1) = y − 1 −x + 1 = y − 1 F y = 2−x (1) 7
• 20. F F F ก ∆ABC = 4 F 1 2 (AB)(CD) = 4 1 (2 2 )CD = 4 → CD = 2 2 2 ก CD = (x − 1) 2 + (y − 1) 2 2 2 = (x − 1) 2 + (2 − x − 1) 2 2 2 = 2(x − 1) 2 3 F F F 2 2 = 2 x−1 C F Q2 2 = x − 1 → x = − 1, 3 F x = −1 (1) F y = 2 − (−1) = 3 ∴ C ก (−1, 3) C ก F Choice F ก Choice 1F 16 2 z1 = 0 n = 1, z2 = z2 + i = i 1 n = 2, z3 = z2 + i = i2 + i = − 1 + i 2 n = 3, z 4 = z 2 + i = (−1 + i) 2 + i = − 2i + i = − i 3 n = 4, z 5 = z 2 + i = (−i) 2 + i = − 1 + i 4 2 n = 5, z 6 = (−1 + i) 2 + i = − i ∴ z 111 = −1+i = 2F 17 4 ∞ ∞ ∞ Σ  3 +n2− 1− 2  = Σ   n n 3n 2n S∞ = Σ an =    + n−1 − 2 n=1 n=1 4 n=1 4n − 1 4 4n − 1  ∞ 3n ∞ 2n ∞ S∞ = Σ n−1 + Σ n−1 − Σ 2 n=1 4 n=1 4 n = 1 4n − 1 S ∞ =  3 + 9 + 27 + .....  +  2 + 1 + 1 + .....  −  2 + 1 + 1 + .....   4 16   2   2 8  3 S∞ = + 2 1 − 2 1 = 12 + 4 − 8 = 40 1− 3 1− 1− 3 3 4 2 4 8
• 21. F F FF 18 2 2 ก f(x) = 3x 3 −1 −1 f (x) = 2x 3 ∴ f (8) = 2(2 3 ) 3 = 2 1  = 1 2 F (fog) (1) = f (g(1)) ⋅ g (1) = f (8) ⋅  2  3 = (1)  2  = 2 3 3F 19 1 ก F 13 4 13 × 4 = 52 n(S) = F 3 ก 52 F  52  3 = 22100 n(E) = F 3 F ก 2 F  13   2   4   4  = 3744  2  1 2 1 ก2 2 2 OR  13   4   48  = 3744  1  2  1  2 2 n(E) 3744 72 P(E) = n(S) = 22100 = 425F 20 2 ก. ก F A B n(A) = n(A ∩ B) + n(A ∩ B ) F P(A) = P(A ∩ B) + P(A ∩ B ) ก. ก A B A B ⊂ ⊂ . ก F A B P(A − B) = 0.2 . 0.2 0.3 0.3 0.2 P(A B ) ⊂ 9
• 22. F F FF 21 3 N 1 µ1 + N 2 µ2 ก µ = N1 + N2 35N + 50N F 40 = N +N 40N + 40N = 35N + 50N 50N = 10N N N = 10 5 = 2 1F 22 3 A = 7 (7 ) 7 A>B F ∴ F 4 B = 7 77 = (7 11 ) 7 B>C F ∴ F 1, 2 C = 77 7 ∴ F 3F 23 3 กก ก PAT 1. ก 5 ก 17 2. F ก , F, , F, 3. F ก F ก 3 F F F F 3F 24 4 F ก. a ∗ b = a b , b ∗ a = b a ab ≠ ba F ก. F . (a ∗ b) ∗ c = (a b ) ∗ c = (a b ) c = a bc c = / F . a ∗ (b ∗ c) = a ∗ (b c ) = a b F . a ∗ (b + c) = a (b + c) = F / . (a ∗ b) + (a ∗ c) = ab + ac F . (a + b) ∗ c = (a + b) c = F / . (a ∗ c) + (b ∗ c) = a c + b c 10
• 23. F F FF 25 1 1. F A C ก ก D D ก ก F F A ก ก 2. A ก ก F B ก ก 3. B ก ก F E 4. E F D ก ก 5. D ก ก F CF 26 18 กF ก F F F- F F n(A ∩ B ∩ C ) = 11 A B n((B − A) ∩ (B − C)) = 15 A ∩B ∩C n((A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C)) = 47 C = F ∴ n(A ∩ B ∩ C) = n(A ∪ B ∪ C) − n(A ∪ B) = 91 − 11 − 15 − 47 = 18F 27 5 กก 2 (3x + 1) + 2 3x + 1 ⋅ x − 1 + x − 1 = 7x + 1 2 3x + 1 ⋅ x − 1 = 3x + 1 กก 2 4(3x + 1)(x − 1) = (3x + 1) 2 0 = (3x + 1) 2 − 4(3x + 1)(x − 1) 0 = (3x + 1) ⋅ [3x + 1 − 4(x − 1)] 0 = (3x + 1)(−x + 5) ∴ x = − 1, 5 3 F x = −1 3 F F F ( − 1 − 1 ∉ R) 3 x = 5 F F ( 16 + 4 = 36 ) 11
• 24. F F FF 28 25 A = {2, 3, 5, 7} B = {1, 2, 3, ....., 10} A B 2 1 (a, f(a)) ≠ 1 ก a∈A 3 2 A B 5 3 2 → 2, 4, 6, 8, 10 7 . . . 3 → 3, 6, 9 10 5 → 5, 10 7→7 1 7 F f(7) = 7 F (7, 7) = 7 ≠ 1 2 5 F f(5) = 5 10 F (5, 5) = 5 ≠ 1 (5, 10) = 5 ≠ 1 3 3 F f(3) = 3 6 9 F (3, 3) = 3 ≠ 1, (3, 6) = 3 ≠ 1 (3, 9) = 3 ≠ 1 4 2 F ก 1 F f(5) ≠ 10 f(3) ≠ 6 F f(2) = 2, 4, 6, 8, 10 ก 2 F f(5) ≠ 10 f(3) = 6 F f(2) = 2, 4, 8, 10 ก 3 F f(5) = 10 f(3) ≠ 6 F f(2) = 2, 4, 6, 8 ก 4 F f(5) = 10 f(3) = 6 F f(2) = 2, 4, 8 12
• 25. F F F 2 2 (1) 2 4 (2) 2 6 (3) 2 8 (4) 2 10 (5) 3 3 2 2 (6) 2 4 (7) 2 8 (8) 5 5 3 6 2 10 (9) 2 2 (10) 2 4 (11) 3 9 2 6 (12) 2 8 (13)7 7 2 10 (14) 3 3 2 2 (15) 2 4 (16) 2 6 (17) 5 10 3 6 2 8 (18) 2 2 (19) 2 4 (20) 3 9 2 8 (21) 2 2 (22) 2 4 (23) 2 6 (24) 2 8 (25)∴ ก F F 25 13
• 26. F F FF 29 1 2 β a a a 2 + b2 sin α = cos β = a a2 + b2 , a2 + b2 α b cos [arcsin (sin α)] + sin [arccos (cos β)] = 1 cos α + sin β = 1 cos (90 − β) + sin β = 1 sin β + sin β = 1 ∴ sin β = 1 2F 30 1 2 sin 54 − sin 18 = 2 sin 18 cos 18 sin 18 + 1 − 2 sin 2 18 cos 18 = 2 cos 36 sin 18 2 sin 18 cos 18 cos 36 = cos 18 2 sin 36 cos 36 = 2 cos 18 = sin 72 = 1 2 2 sin 72F 31 32  −4 −4  2A − B =   (1)  5 6   −5 −8  A − 2B =   (2)  4 0   −8 −8  (1) × 2 : 4A − 2B =   (3)  10 12   −3 0  1  −3 0   −1 0  (3) − (2) : 3A =   → A = 3  =    6 12   6 12   2 4  0 −1 0 det A = = −4 2 4 -4  −1 0   −4 −4  A (1) F 2 −B =    2 4   5 6   −2 0   −4 −4   2 4  B =  −   =    4 8   5 6   −1 2  4 2 4 det B = = 8 −1 2 4 det(A 4 B −1 ) = (det A 4 )(det B −1 ) = (det A) 4  1   det B  = (−4) 4  1  = 32   8 14
• 27. F F FF 32 6  1 0   x −1   2y −1   1 0     =     −1 w   0 y   z 2   −1 w   x −1   2y + 1 −w    =    −x 1 + yw   z − 2 2w  ก 1 ก 2 F −1 = − w → w = 1 ก 2 ก 2 F 1 + yw = 2w → 1 + y(1) = 2(1) → y = 1 ก 1 ก 1 F x = 2y + 1 = 2(1) + 1 = 3 → x = 3 ก 2 ก 1 F −x = z − 2 → − 3 = z − 2 → z = −1 ∴ F 4w − 3z + 2y − x = 4(1) − 3(−1) + 2(1) − 3 = 6F 33 3 ก F u⋅w = 2 F 1a + 2b + 3c = 2 (1) ก w = ai + bj + ck ก −2i + 1j + 1k 3 2 3  −2  a   3     1  F  b  = m 2    m F   c   1   3  a = m− 2   3 m = a2 = b = c 1 1 − 3 2 3 b = m 1  2 FF − 3a = 2b = 3c 2 (2) c = m 1  3 F 2b 3c ก (2) (1) F 1a +  − 3a  +  − 3a  =  2  2 2→ a = −1 F a (2) F 3 2 = 2b = 3c → b = 3 , c = 1 4 2 ∴ a + 4b + 2c = − 1 + 4 3  + 2 1  = 3 4 2 15
• 28. F F FF 34 5 z 2 = 1 + 2i , z 2 = 1 − 2i 5z 1 + 2z 2 = 5 5z 1 + 2(1 − 2i) = 5 5z 1 = 3 + 4i 5 z 1 = 3 + 4i → z 1 = 1 ∴ 5z −1 = 5 = 5 = 5 = 5 z z 1F 35 1 n (2 + 2n) = n +n 2 an = 2 n2 n2 ∴ nlim∞ a n lim n + n = 1 2 = → n→∞ 2 nF 36 1   k+1 − k 1 = 1  1  k (k + 1) + k k + 1 k k+1  k+1 + k ( k+1 − k ) = 1 k k+1 ( k+1 − k ) = 1 k − 1 k+1  1 n  Sn = Σ  − 1  = 1 − 1 + 1 − 1 + 1 − 1 + ..... + 1 − 1 k=1 k k+1  2 2 3 3 4 n n+1   ∴ nlim∞ S n → = lim  1 − 1  = 1 n→∞  n+1 F 37 53 Con x = 2 f(2) = a − b lim x x − 2− 2 = lim 3x 1− 3 = 9 3 − 3x 2 lim f(x) = x → 2− x → 2− x → 2− = lim f(x) = lim x 2 + ax + 1 = 2a + 5 x → 2+ x → 2+ 2a + 5 = 9 → a = 2 a−b = 9 → 2−b = 9 → b = −7 ∴ a 2 + b 2 = 4 + 49 = 53 16
• 29. F F FF 38 6 3 1 3x 2 f(x) = ∫ f (x)dx = ∫(3x 2 + 5)dx = 3 + 5x + c 2 3 f(x) = 2x 2 + 5x + c f(1) = 2 + 5 + c = 5 → c = − 2 3 ∴ f(x) = 2x 2 + 5x − 2 f(x 2 ) = 2x 3 + 5x 2 − 2 ∴ f(x 2 ) − 2 lim f(x) = lim (2x 3 + 5x 2 − 2) − 2 3 x→4 x→4 2x 2 + 5x − 2 = 128 + 80 − 4 16 + 20 − 2 = 204 = 6 34F 39 7 ก F F F y = f(x) (2, 19) F ก 19 F f (2) = 19 f(2) = 19 ก f (x) = 6x + 4 f (x) = ∫ (6x + 4)dx = 3x 2 + 4x + c f (2) = 12 + 8 + c = 19 → c = − 1 ∴ f (x) = 3x 2 + 4x − 1 f(x) = ∫(3x 2 + 4x − 1)dx = x 3 + 2x 2 − x + c f(2) = 8 + 8 − 2 + c = 19 → c = 5 ∴ f(x) = x 3 + 2x 2 − x + 5 f(1) = 1 + 2 − 1 + 5 = 7F 40 44 ก 3 ก 1,2 2 × 4 × 3 = 24 ก 2 ก 4 × 4 = 16 ก 1 ก 4 = 4 44 17
• 30. F F FF 41 192 ก F F 4!2!4 = 192F 42 520 ก µ = Σ x N Σx F 72 = N → Σx = 72 N Σx + 60 70 = N+1 70N + 70 = Σ x + 60 70N + 70 = 72N + 60 N = 5 ก σ 2 = ΣN − µ 2 x2 Σ x2 600 = 5 − 72 2 → Σ x 2 = 28920 = 28920 + 60 − 70 2 2 σ2 F 6 = 520F 43 6 กก F FF 45 45 47 51 Med = 46 µ = 45 + 45 + 47 + 51 4 = 47 Σ(x − µ) 2 σ2 = N = 4 + 4 + 0 + 16 4 = 6 18
• 31. F F F F 44 10 x−µ ก z = σ 700 − µ F 4 = σ (1) 400 − µ −2 = σ (2) (1) − (2) 6 = 300 σ σ = 50 µ = 500 F ก = σ = 50 × 100 = 10% µ 500 F 45 7 31 F 3 F ก ก 10 F กก 1: 1 ก F F F กF F 1 2 3 4 5 6 7 29 30 31 ก F F F5 กก ก 1 ก Fก F F F กก 2: 1 ก F F กF F 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Fก F ก ก F กF F 4∴ 20 ก F 19
• 32. F F FF 46 37 ก F ก = .. . 221 ก 260 221 = 13 × 17 . . . 221 ก 260 13 260 = 13 × 20 ∴ F ก ก 13 ก Fก 17 ก , 20 ก ∴ F F 37 กF 47 7 (f ⊗ g)(1) = f(g(1)) − g(f(1)) = f(3) − g(0) = 8−1 = 7F 48 b = 0 abcd ∴ a= 1 d = 9 9 ( F a>1 F abcd × 9 ก 4 ก) dcba ก 9cb 1 = 9 × (1bc9) 9001 + 100c + 10b = 9(1009 + 100b + 10c) 9001 + 100c + 10b = 9081 + 900b + 90c 10c − 890b = 80 ∴ c = 8 + 89b (1) c, b F F0→9 F b = 0 c = 8 F F ก (1) ( ก b>0 F c>9 F F b = 1 → c = 97)F 49 ก6 + ก5 = 1 + 2 + 3 + ..... + 11 43 + 28 − x = 11 (11 + 1) = 66 2 x = 5 20
• 33. F F FF 50 I. 2 F 2 10 F 2 3 F 3 11 F 3 4 F 4 12 F 4 5 F 5 13 F 5 6 F 4 14 F 4 7 F 3 15 F 3 8 F 2 16 F 2 9 F 1 17 F 1 2400 F 8 ∴ 2400 F 2 II. 2400 = 8 × 300 8 ก ∴ 2400 F 2 ************************* 21