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SUEC 高中 Adv Maths Revision (Earth as Sphere, Locus, Permutation, Combination, Binominial).pptx
- 1. 𝒑𝒈 𝟖𝟕 1000 𝑘𝑚 𝑟 60 ° 𝑁 ? = 1000 60 ° 𝑁 𝑟 = 6370 𝑐𝑜𝑠 60° 𝜃 360 ° × 2𝜋 × 6370 𝑐𝑜𝑠 60° = 1000 𝜃 = 18 ° 15° → 1 ℎ𝑟 18 ° → 18 15 ℎ𝑟 = 1 ℎ𝑟 12 𝑚𝑖𝑛
- 2. 𝒑𝒈 𝟖𝟕 1000 𝑘𝑚 𝑟 60 ° 𝑁 ? 60 ° 𝑁 1.853 𝑘𝑚 → 1 海里 1000 𝑘𝑚 → 1000 1.853 海里 = 539.67 海里 𝜃 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 60 𝑐𝑜𝑠 60° = 539.67 60 𝑐𝑜𝑠 60° = 18 ° 15° → 1 ℎ𝑟 18 ° → 18 15 ℎ𝑟 = 1 ℎ𝑟 12 𝑚𝑖𝑛
- 3. 𝒑𝒈 𝟖𝟕 500 𝑛𝑚 𝑟 60 ° 𝑁 ? 60 ° 𝑁 𝜃 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 60 𝑐𝑜𝑠 60° = 500 60 𝑐𝑜𝑠 60° = 16.67 ° 15° → 1 ℎ𝑟 16.67 ° → 16.67 15 ℎ𝑟 = 1 ℎ𝑟 7 𝑚𝑖𝑛 1 𝑝𝑚 𝑊, 𝑡𝑖𝑚𝑒 =? 𝑄 𝑙𝑜𝑐𝑎𝑙 𝑡𝑖𝑚𝑒 = 13 ℎ𝑟 − 1 ℎ𝑟 7 𝑚𝑖𝑛 𝑸 𝒊𝒔 𝑾 𝒕𝒐 𝑷, 𝒘𝒆 𝒔𝒖𝒃𝒕𝒓𝒂𝒄𝒕 𝒕𝒊𝒎𝒆 = 11 ℎ𝑟 53 𝑚𝑖𝑛
- 4. 𝒑𝒈 𝟖𝟕 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑙𝑎𝑡 = 𝑇𝑜𝑘𝑦𝑜 𝑙𝑎𝑡 + 𝐴𝑑𝑒𝑙𝑎𝑖𝑑𝑒 𝑙𝑎𝑡 = 35.67 ° + 34.92 ° = 70.59 ° 𝑇𝑜𝑘𝑦𝑜 (35 ° 40’ 𝑁, 139 ° 𝐸) 𝐴𝑑𝑒𝑙𝑎𝑖𝑑𝑒 (34 ° 55’ 𝑆, 139 ° 𝐸) = 70.59 × 60 = 4,235 海里 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 70.59 ° 360 ° × 2𝜋 × 6,370 𝑘𝑚 = 7,849 𝑘𝑚 (𝑎) (𝑏)
- 5. 𝒑𝒈 𝟖𝟖 𝐵 (43 ° 40’ 𝑁) 𝐴 (9 ° 𝑁) 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒𝑠 = 𝐵 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 − 𝐴 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 = 43 ° 40’ − 9 ° = 34 ° 40’ = 34.67 ° = 34.67 × 60 = 2,080 海里
- 6. 𝒑𝒈 𝟖𝟖 𝑃 (15 ° 𝑁, 30 ° 𝐸) 𝐵 (?, 𝟑𝟎 ° 𝑬) = 2000 海里 𝜃 = 33.33 ° 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒𝑠 = 𝑃 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 + 𝐵 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 33.33 ° = 15 ° + 𝐵 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝐵 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 = 33 ° 20’ − 15 ° = 18 ° 20’ ∴ 𝐵 (18 ° 20’ 𝑆, 30 ° 𝐸)
- 7. 𝒑𝒈 𝟖𝟖 3000 𝑛𝑚 𝑟 15 ° 𝑁 ? 15 ° 𝑁 𝜃 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 60 𝑐𝑜𝑠 15° = 3000 60 𝑐𝑜𝑠 15° = 51.76 ° 𝐶 𝐶 (15 ° 𝑁, ? ) 𝑃 (15 ° 𝑁, 30 ° 𝐸) 51.76 ° = 30 ° − 𝐶 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑒𝑠 = 𝑃 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑒 − 𝐶 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑒 𝐶 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑒 = 81 ° 46’ 𝐸
- 8. 𝒑𝒈 𝟖𝟖 = 1000 海里 𝜃 = 16 ° 40’ 𝑁 ∴ 𝐴 (16 ° 40’ 𝑁, 10 ° 𝐸) 𝑙𝑎𝑡𝑖𝑡𝑢𝑑𝑒 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑒 = 600 海里 𝜃 = 10 ° 𝐸
- 9. 𝒑𝒈 𝟖𝟖 = (22 + 30) × 60 = 3,120 海里
- 10. 𝒑𝒈 𝟏𝟏𝟒 𝑥2 + 𝑦2 + 4𝑥 − 6𝑦 − 3 = 0 𝑃𝐴 = 4 𝑥 + 2 2 + 𝑦 − 3 2 = 4 𝑥2 + 4𝑥 + 4 + 𝑦2 − 6𝑦 + 9 = 16 𝐴 = (−2, 3) 5𝑥 + 2𝑦 − 16 = 0 𝑙𝑒𝑡 𝑃 = (𝑥, 𝑦) 𝐴 = (−3, 1) 𝐵 = (7, 5) 𝑃𝐴 = 𝑃𝐵 𝑥 + 3 2 + 𝑦 − 1 2 = 𝑥 − 7 2 + 𝑦 − 5 2 𝑥2 + 6𝑥 + 9 + 𝑦2 − 2𝑦 + 1 = 𝑥2 − 14𝑥 + 49 + 𝑦2 − 10𝑦 + 25 𝑥 + 3 2 + 𝑦 − 1 2 = 𝑥 − 7 2 + 𝑦 − 5 2 𝑙𝑒𝑡 𝑃 = (𝑥, 𝑦)
- 11. 𝒑𝒈 𝟏𝟏𝟒 16𝑥2 + 7𝑦2 − 64𝑥 − 48 = 0 𝑃𝐴 + 𝑃𝐵 = 8 𝑥 − 2 2 + 𝑦 − 3 2 + 𝑥 − 2 2 + 𝑦 + 3 2 = 8 𝑙𝑒𝑡 𝑃 = (𝑥, 𝑦) 𝐴 = (2, 3) 𝐵 = (2, −3) 𝑥 − 2 2 + 𝑦 − 3 2 = 64 − 16 𝑥 − 2 2 + 𝑦 + 3 2 + 𝑥 − 2 2 + 𝑦 + 3 2 𝑥 − 2 2 + 𝑦 − 3 2 = 8 − 𝑥 − 2 2 + 𝑦 + 3 2 𝑦2 − 6𝑦 + 9 = 64 − 16 𝑥 − 2 2 + 𝑦 + 3 2 + 𝑦2 + 6𝑦 + 9 12𝑦 + 64 = 16 𝑥 − 2 2 + 𝑦 + 3 2 3 4 𝑦 + 4 2 = 𝑥 − 2 2 + 𝑦 + 3 2 9 16 𝑦2 + 6𝑦 + 16 = 𝑥2 − 4𝑥 + 4 + 𝑦2 + 6𝑦 + 9
- 12. 𝒑𝒈 𝟏𝟏𝟒 𝐴𝑀 = 6 𝑥2 + 𝑦2 = 6 𝑥2 + 𝑦2 = 36 𝐿𝑒𝑡 𝐴 & 𝐵 = 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑥 & 𝑦 𝑎𝑥𝑒𝑠. 𝐴 = (𝑎, 0) , 𝐵 = (0, 𝑏) 𝑀 = (𝑥, 𝑦) = 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐴𝐵 ∴ 𝐴 = (2𝑥, 0), 𝐵 = (0, 2𝑦)
- 13. 𝒑𝒈 𝟏𝟏𝟒 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 𝑙𝑖𝑛𝑒 𝑒𝑞 ∶ 𝑦 − 𝑦1 = 𝑚 (𝑥 − 𝑥1) 𝑦 = 𝑚 (𝑥 − 4) 𝐴 = (4,0) = (𝑥1 , 𝑦1) 𝑀 = 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐵𝐶 = (𝑥 , 𝑦) 𝑙𝑒𝑡 𝐵 = (𝑥𝐵 , 𝑦𝐵) 𝐶 = (𝑥𝐶 , 𝑦𝐶) 𝑥𝐵 2 + 𝑦𝐵 2 = 4 − −① 𝑥𝐶 2 + 𝑦𝐶 2 = 4 − −② ① − ② (𝑥𝐵 + 𝑥𝐶)(𝑥𝐵 − 𝑥𝐶) + (𝑦𝐵 + 𝑦𝐶)(𝑦𝐵 − 𝑦𝐶) = 0 2𝑥 (𝑥𝐵 − 𝑥𝐶) + 2𝑦 (𝑦𝐵 − 𝑦𝐶) = 0 (𝑥𝐵 2 − 𝑥𝐶 2 ) + (𝑦𝐵 2 − 𝑦𝐶 2 ) = 0 2𝑥 (𝑥𝐵 − 𝑥𝐶) = − 2𝑦 (𝑦𝐵 − 𝑦𝐶) 𝑥 𝑦 = − (𝑦𝐵 − 𝑦𝐶) (𝑥𝐵 − 𝑥𝐶) 𝑚 = 𝑦 (𝑥 − 4) = − 𝑚 𝑥2 + 𝑦2 − 4𝑥 = 0
- 14. 𝒑𝒈 𝟏𝟔𝟔 (𝑎) (𝑥 + 𝑎1)(𝑥 + 𝑎2) = 𝑥2 + 𝑎2𝑥 + 𝑎1𝑥 + 𝑎1𝑎2 = 𝑥2 + (𝑎1 + 𝑎2) 𝑥 + 𝑎1𝑎2 (𝑏) (𝑥 + 𝑎1)(𝑥 + 𝑎2)(𝑥 + 𝑎3) = 𝑥2 + (𝑎1 + 𝑎2) 𝑥 + 𝑎1𝑎2 (𝑥 + 𝑎3) = 𝑥3 + (𝑎1 + 𝑎2) 𝑥2 + 𝑎1𝑎2 𝑥 + 𝑎3 𝑥2 + (𝑎1 + 𝑎2) 𝑎3 𝑥 + 𝑎1𝑎2𝑎3 = 𝑥3 + (𝑎1 + 𝑎2 + 𝑎3) 𝑥2 + (𝑎1𝑎2 + 𝑎1𝑎3 + 𝑎2𝑎3) 𝑥 + 𝑎1𝑎2𝑎3
- 15. 𝒑𝒈 𝟏𝟔𝟔 (𝑐) (𝑥 + 𝑎1)(𝑥 + 𝑎2)(𝑥 + 𝑎3)(𝑥 + 𝑎4) = 𝑥3 + (𝑎1 + 𝑎2 + 𝑎3) 𝑥2 + (𝑎1𝑎2 + 𝑎1𝑎3 + 𝑎2𝑎3) 𝑥 + 𝑎1𝑎2𝑎3 (𝑥 + 𝑎4) = 𝑥4 +(𝑎1 + 𝑎2 + 𝑎3) 𝑥3 + (𝑎1𝑎2 + 𝑎1𝑎3 + 𝑎2𝑎3) 𝑥2 + 𝑎1𝑎2𝑎3𝑥 + 𝑎4𝑥3 + (𝑎1 + 𝑎2 + 𝑎3) 𝑎4 𝑥2 + (𝑎1𝑎2 + 𝑎1𝑎3 + 𝑎2𝑎3) 𝑎4𝑥 + 𝑎1𝑎2𝑎3𝑎4 = 𝑥4 +(𝑎1 + 𝑎2 + 𝑎3 + 𝑎4) 𝑥3 + (𝑎1𝑎2 + 𝑎1𝑎3 + 𝑎1𝑎4 + 𝑎2𝑎3 + 𝑎2𝑎4 + 𝑎3𝑎4) 𝑥2 + (𝑎1𝑎2𝑎3 + 𝑎1𝑎2𝑎4 + 𝑎1𝑎3𝑎4 + 𝑎2𝑎3𝑎4)𝑥 + 𝑎1𝑎2𝑎3𝑎4 (𝑑) (𝑥 − 1)(𝑥 + 3)(𝑥 − 5)(𝑥 + 7) = 𝑥4 +(−1 + 3 − 5 + 7) 𝑥3 + (−3 + 5 − 7 − 15 + 21 − 35) 𝑥2 + (15 − 21 − 35 − 105)𝑥 + (−1)(3)(−5)(7) = 𝑥4 +4𝑥3 − 34𝑥2 − 76𝑥 + 105
- 16. 𝒑𝒈 𝟏𝟔𝟕 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥3 =? 𝑐𝑜𝑒𝑓𝑓 𝑜𝑓 𝑥3 = −2 + 3 − 4 − 6 + 8 − 12 + 5(1 − 2 + 3 − 4) = 𝑥4 +(𝑎1 + 𝑎2 + 𝑎3 + 𝑎4) 𝑥3 + (𝑎1𝑎2 + 𝑎1𝑎3 + 𝑎1𝑎4 + 𝑎2𝑎3 + 𝑎2𝑎4 + 𝑎3𝑎4) 𝑥2 + (𝑎1𝑎2𝑎3 + 𝑎1𝑎2𝑎4 + 𝑎1𝑎3𝑎4 + 𝑎2𝑎3𝑎4)𝑥 + 𝑎1𝑎2𝑎3𝑎4 (𝑥 + 𝑎1)(𝑥 + 𝑎2)(𝑥 + 𝑎3)(𝑥 + 𝑎4)(𝑥 + 5) = 𝑥4 + (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4) 𝑥3 + (𝑎1𝑎2 + 𝑎1𝑎3 + 𝑎1𝑎4 + 𝑎2𝑎3 + 𝑎2𝑎4 + 𝑎3𝑎4) 𝑥2+ . . . (𝑥 + 5) =. . . +(𝑎1𝑎2 + 𝑎1𝑎3 + 𝑎1𝑎4 + 𝑎2𝑎3 + 𝑎2𝑎4 + 𝑎3𝑎4) 𝑥3 + (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4) 5𝑥3 +. . . = −23
- 17. 𝒑𝒈 𝟏𝟔𝟕 (𝑎) (1 − 2𝑥)5 = 1 + 5𝐶1 (− 2𝑥) + 5𝐶2 − 2𝑥 2 +5𝐶3 − 2𝑥 3 + 5𝐶4 − 2𝑥 4 + 5𝐶5 − 2𝑥 5 = 1 − 10𝑥 + 40 𝑥2 − 80 𝑥3 + 80 𝑥4 − 32 𝑥5 (𝑏) 𝑥2 − 1 𝑥 4 = 4𝐶0 (𝑥2 )4 + 4𝐶1 𝑥2 3 − 1 𝑥 +4𝐶2 𝑥2 2 − 1 𝑥 2 + 4𝐶3 𝑥2 − 1 𝑥 3 + 4𝐶4 − 1 𝑥 4 = 𝑥8 − 4 𝑥5 + 6 𝑥2 − 4 𝑥 + 1 𝑥4
- 18. 𝒑𝒈 𝟏𝟔𝟕 (𝑎) (1 + 2𝑥) 1 2 = 1 + 1 2 2𝑥 + 1 2 − 1 2 2 ! 2𝑥 2 + 1 2 − 1 2 − 3 2 3 ! 2𝑥 3 + ⋯ = 1 + 𝑥 − 1 2 𝑥2 + 1 2 𝑥3 + ⋯
- 19. 𝒑𝒈 𝟏𝟔𝟕 (𝑏) 1 + 𝑥 1 − 2𝑥 = (1 + 𝑥) 1 − 2𝑥 − 1 = (1 + 𝑥) 1 + −1 −2𝑥 + −1 −2 2 ! −2𝑥 2 + −1 −2 (−3) 3 ! −2𝑥 3+ . . . = (1 + 𝑥) 1 + 2𝑥 + 4𝑥2 + 8𝑥3 . . . = 1 + 2𝑥 + 4𝑥2 + 8𝑥3 + 𝑥 + 2𝑥2 + 4𝑥3+. . . = 1 + 3𝑥 + 6𝑥2 + 12𝑥3 . . .
- 20. 𝒑𝒈 𝟏𝟔𝟕 (𝑐) (1 + 𝑥) −1 + 1 + 2𝑥 − 1 2 = 1 + −1 (𝑥) + (−1)(−2) 2 ! 𝑥 2 + (−1)(−2)(−3) 3 ! 𝑥 3+. . . + 1 + − 1 2 (2𝑥) + − 1 2 − 3 2 2 ! 2𝑥 2 + − 1 2 − 3 2 − 5 2 3 ! 2𝑥 3 +. . . = (1 − 𝑥 + 𝑥2 − 𝑥3 +. . . ) + 1 − 𝑥 + 3 2 𝑥2 − 15 6 𝑥3 +. . . = 2 − 2𝑥 + 5 2 𝑥2 − 7 2 𝑥3 +. . .
Editor's Notes
- 2 places on latitide circle at 60 N are 1000 km apart. Try to find their time difference.
- 2 places on latitide circle at 60 N are 1000 km apart. Try to find their time difference.
- P & Q are at 60N latitude and 500 nm apart. The time at the place of P = 1 pm. If Q is due W of P, try to find the place of Q.
- How many nm is the distance from Panama (9N, 79 20 W) to Toronto (43 40 N, 79 20 W)?
- An aircraft flew 2000 nm southward from its base P (15 N, 30 E) to reach B. Find the latitude and longitude of B. Another plane flew away from the same base P, flew 3000 nm eastwards and arrived at C. Find the latitude and longitude of B.
- An aircraft flew 2000 nm southward from its base P (15 N, 30 E) to reach B. Find the latitude and longitude of B. Another plane flew away from the same base P, flew 3000 nm eastwards and arrived at C. Find the latitude and longitude of B.
- A is 1000 nm N of the Equator and 600 nm E of Greenwich. Find the latitude and longitude of A.
- An airplane flies along the equator from A (42 E) to B (20 E) and then N to C (30 N). Find the distance traveled by this aircraft.
- 1. The distance between a moving point P and a fixed point (-2,3) is equal to 4. Find the trajectory equation of point P. 2. The distance between a moving point and 2 known points (-3,1) and (7.5) is equal. Find the trajectory equation of this moving point.
- The sum of the distance between a moving point P and 2 fixed points (2,3) and (2,-3) is always 8 and find the trajectory equation of point P.
- The 2 ends of a line segment with a length of 12 units often move on 2 axes. Find the trajectory equation of its midpoint.
- The line passing through the point A (4,0) intersects the circle x^2 + y^2 = 4 at the point B, C. Find the trajectory equation of the midpoint M in the chord BC.