11X1 T14 11 some different types (2011)

Download 11X1 T14 11 some different types (2011)

Post on 19-Jun-2015

503 views

Category:

Education

0 download

TRANSCRIPT

  • 1. Some Different Looking InductionProblems2006 Extension 1 HSC Q4d) tan tan i Use the fact that tan to show that1 tan tan 1 tan n tann 1 cot tann 1 tan n

2. Some Different Looking InductionProblems2006 Extension 1 HSC Q4d) tan tan i Use the fact that tan to show that1 tan tan 1 tan n tann 1 cot tann 1 tan n tann 1 tan n tann 1 n 1 tann 1 tan n 3. Some Different Looking InductionProblems2006 Extension 1 HSC Q4d) tan tan i Use the fact that tan to show that1 tan tan 1 tan n tann 1 cot tann 1 tan n tann 1 tan n tann 1 n 1 tann 1 tan ntann 1 tan ntan 1 tann 1 tan n 4. Some Different Looking InductionProblems2006 Extension 1 HSC Q4d) tan tan i Use the fact that tan to show that1 tan tan 1 tan n tann 1 cot tann 1 tan n tann 1 tan n tann 1 n 1 tann 1 tan ntann 1 tan ntan 1 tann 1 tan n1 tann 1 tan n tan tann 1 tan n 5. Some Different Looking InductionProblems2006 Extension 1 HSC Q4d) tan tan i Use the fact that tan to show that1 tan tan 1 tan n tann 1 cot tann 1 tan n tann 1 tan n tann 1 n 1 tann 1 tan ntann 1 tan ntan 1 tann 1 tan n1 tann 1 tan n tan tann 1 tan n tann 1 tan n1 tann 1 tan n tan 6. Some Different Looking InductionProblems2006 Extension 1 HSC Q4d) tan tan i Use the fact that tan to show that1 tan tan 1 tan n tann 1 cot tann 1 tan n tann 1 tan n tann 1 n 1 tann 1 tan ntann 1 tan ntan 1 tann 1 tan n1 tann 1 tan n tan tann 1 tan n tann 1 tan n1 tann 1 tan n tan 1 tann 1 tan n cot tann 1 tan n 7. 2006 Extension 1 Examiners Comments; (d) (i) This part was not done well, with many candidates presenting circular arguments. 8. 2006 Extension 1 Examiners Comments; (d) (i) This part was not done well, with many candidates presenting circular arguments.However, a number of candidates successfully rearrangedthe given fact and easily saw the substitution required. 9. 2006 Extension 1 Examiners Comments; (d) (i) This part was not done well, with many candidates presenting circular arguments.However, a number of candidates successfully rearrangedthe given fact and easily saw the substitution required.Many candidates spent considerable time on this part,which was worth only one mark. 10. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1 11. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1 12. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1 RHS 2 cot tan 2 13. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 14. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 1 tan tan 2 1 15. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 1 tan tan 2 1 cot tan 2 tan 1 16. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 1 tan tan 2 1 cot tan 2 tan 1 cot tan 2 1 1 17. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 1 tan tan 2 1 cot tan 2 tan 1 cot tan 2 1 1 cot tan 2 2 18. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 1 tan tan 2 1 cot tan 2 tan 1 cot tan 2 1 1 cot tan 2 2 Hence the result is true for n = 1 19. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 1 tan tan 2 1 cot tan 2 tan 1 cot tan 2 1 1 cot tan 2 2 Hence the result is true for n = 1Step 2: Assume the result is true for n = k, where k is a positiveinteger 20. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 1 tan tan 2 1 cot tan 2 tan 1 cot tan 2 1 1 cot tan 2 2 Hence the result is true for n = 1Step 2: Assume the result is true for n = k, where k is a positiveintegeri.e. tan tan 2 tan 2 tan 3 tan k tank 1 k 1 cot tank 1 21. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 1 tan tan 2 1 cot tan 2 tan 1 cot tan 2 1 1 cot tan 2 2 Hence the result is true for n = 1Step 2: Assume the result is true for n = k, where k is a positiveintegeri.e. tan tan 2 tan 2 tan 3 tan k tank 1 k 1 cot tank 1Step 3: Prove the result is true for n = k + 1 22. ii Use mathematical induction to prove that for all integers n 1tan tan 2 tan 2 tan 3 tan n tann 1 n 1 cot tann 1Step 1: Prove the result is true for n = 1LHS tan tan 2RHS 2 cot tan 2 1 tan tan 2 1 cot tan 2 tan 1 cot tan 2 1 1 cot tan 2 2 Hence the result is true for n = 1Step 2: Assume the result is true for n = k, where k is a positiveintegeri.e. tan tan 2 tan 2 tan 3 tan k tank 1 k 1 cot tank 1Step 3: Prove the result is true for n = k + 1 i.e. tan tan 2 tan 2 tan 3 tank 1 tank 2 k 2 cot tank 2 23. Proof:tan tan 2 tan 2 tan 3 tank 1 tank 2 tan tan 2 tan 2 tan 3 tan k tank 1 tank 1 tank 2 24. Proof:tan tan 2 tan 2 tan 3 tank 1 tank 2 tan tan 2 tan 2 tan 3 tan k tank 1 tank 1 tank 2 k 1 cot tank 1 tank 1 tank 2 25. Proof:tan tan 2 tan 2 tan 3 tank 1 tank 2 tan tan 2 tan 2 tan 3 tan k tank 1 tank 1 tank 2 k 1 cot tank 1 tank 1 tank 2 k 1 cot tank 1 cot tank 2 tank 1 1 26. Proof:tan tan 2 tan 2 tan 3 tank 1 tank 2 tan tan 2 tan 2 tan 3 tan k tank 1 tank 1 tank 2 k 1 cot tank 1 tank 1 tank 2 k 1 cot tank 1 cot tank 2 tank 1 1 k 2 cot tank 1 cot tank 2 cot tank 1 27. Proof:tan tan 2 tan 2 tan 3 tank 1 tank 2 tan tan 2 tan 2 tan 3 tan k tank 1 tank 1 tank 2 k 1 cot tank 1 tank 1 tank 2 k 1 cot tank 1 cot tank 2 tank 1 1 k 2 cot tank 1 cot tank 2 cot tank 1 k 2 cot tank 2 28. Proof:tan tan 2 tan 2 tan 3 tank 1 tank 2 tan tan 2 tan 2 tan 3 tan k tank 1 tank 1 tank 2 k 1 cot tank 1 tank 1 tank 2 k 1 cot tank 1 cot tank 2 tank 1 1 k 2 cot tank 1 cot tank 2 cot tank 1 k 2 cot tank 2 Hence the result is true for n = k + 1 if it is also true for n = k 29. Proof:tan tan 2 tan 2 tan 3 tank 1 tank 2 tan tan 2 tan 2 tan 3 tan k tank 1 tank 1 tank 2 k 1 cot tank 1 tank 1 tank 2 k 1 cot tank 1 cot tank 2 tank 1 1 k 2 cot tank 1 cot tank 2 cot tank 1 k 2 cot tank 2 Hence the result is true for n = k + 1 if it is also true for n = k Step 4: Since the result is true for n= 1, then the result is true for all positive integral values of n, by induction 30. 2006 Extension 1 Examiners Comments; (ii) Most candidates attempted this part. However, manyonly earned the mark for stating the inductive step. 31. 2006 Extension 1 Examiners Comments; (ii) Most candidates attempted this part. However, manyonly earned the mark for stating the inductive step.They could not prove the expression true for n=1 as theywere not able to use part (i), and consequently not able toprove the inductive step. 32. 2004 Extension 1 HSC Q4a)Use mathematical induction to prove that for all integers n 3; 2 2 2 2 2 1 - 1 1 1 3 4 5 n nn 1 33. 2004 Extension 1 HSC Q4a)Use mathematical induction to prove that for all integers n 3; 2 2 2 2 2 1 - 1 1 1 3 4 5 n nn 1Step 1: Prove the result is true for n = 3 34. 2004 Extension 1 HSC Q4a)Use mathematical induction to prove that for all integers n 3; 2 2 2 2 2 1 - 1 1 1 3 4 5 n nn 1Step 1: Prove the result is true for n = 3 2 LHS 1 313 35. 2004 Extension 1 HSC Q4a)Use mathematical induction to prove that for all integers n 3; 2 2 2 2 2 1 - 1 1 1 3 4 5 n nn 1Step 1: Prove the result is true for n = 3 2 2 LHS 1 RHS 3 32 1133 36. 2004 Extension 1 HSC Q4a)Use mathematical induction to prove that for all integers n 3; 2 2 2 2 2 1 - 1 1 1 3 4 5 n nn 1Step 1: Prove the result is true for n = 3 2 2 LHS 1 RHS 3 32 1133 LHS RHS 37. 2004 Extension 1 HSC Q4a)Use mathematical induction to prove that for all integers n 3; 2 2 2 2 2 1 - 1 1 1 3 4 5 n nn 1Step 1: Prove the result is true for n = 3 2 2 LHS 1 RHS 3 32 1133 LHS RHSHence the result is true for n = 3 38. 2004 Extension 1 HSC Q4a)Use mathematical induction to prove that for all integers n 3; 2 2 2 2 2 1 - 1 1 1 3 4 5 n nn 1Step 1: Prove the result is true for n = 3 2 2 LHS 1 RHS 3 32 1133 LHS RHSHence the result is true for n = 3Step 2: Assume the result is true for n = k, where k is a positiveinteger 39. 2004 Extension 1 HSC Q4a)Use mathematical induction to prove that for all integers n 3; 2 2 2 2 2 1 - 1 1 1 3 4 5 n nn 1Step 1: Prove the result is true for n = 3 2 2 LHS 1 RHS 3 32 1133 LHS RHSHence the result is true for n = 3Step 2: Assume the result is true for n = k, where k is a positiveinteger 2 2 2 2 2 i.e. 1 - 1 1 1 3 4 5 k k k 1 40. Step 3: Prove the result is true for n = k + 1 41. Step 3: Prove the result is true for n = k + 1 2 2 2 2 2i.e. Prove 1 - 1 1 1 3 4 5 k 1 k 1k 42. Step 3: Prove the result is true for n = k + 1 2 2 2 2 2i.e. Prove 1 - 1 1 1 3 4 5 k 1 k 1kProof: 2 2 2 2 1 - 1 1 1 3 4 5 k 1 1 - 1 1 1 1 222 22 3 4 5 k k 1 43. Step 3: Prove the result is true for n = k + 1 2 2 2 2 2i.e. Prove 1 - 1 1 1 3 4 5 k 1 k 1kProof: 2 2 2 2 1 - 1 1 1 3 4 5 k 1 1 - 1 1 1 1 2 2 222 3 4 5 k k 1 2 2 1 k k 1 k 1 44. Step 3: Prove the result is true for n = k + 1 2 2 2 2 2i.e. Prove 1 - 1 1 1 3 4 5 k 1 k 1kProof: 2 2 2 2 1 - 1 1 1 3 4 5 k 1 1 - 1 1 1 1 2 2 222 3 4 5 k k 1 2 2 1 k k 1 k 1 2k 1 2 k k 1 k 1 45. Step 3: Prove the result is true for n = k + 1 2 2 2 2 2i.e. Prove 1 - 1 1 1 3 4 5 k 1 k 1kProof: 2 2 2 2 1 - 1 1 1 3 4 5 k 1 1 - 1 1 1 1 2 22 22 3 4 5 k k 1 2 2 1 k k 1 k 1 2k 1 2 k k 1 k 1 2k 1 k k 1 k 1 46. Step 3: Prove the result is true for n = k + 1 2 2 2 2 2i.e. Prove 1 - 1 1 1 3 4 5 k 1 k 1kProof: 2 2 2 2 1 - 1 1 1 3 4 5 k 1 1 - 1 1 1 1 22222 3 4 5 k k 1 2 2 1 k k 1 k 1 2 k 1 2 k k 1 k 1 2 k 1 k k 1 k 1 2 k k 1 47. Step 3: Prove the result is true for n = k + 1 2 2 2 2 2i.e. Prove 1 - 1 1 1 3 4 5 k 1 k 1kProof: 2 2 2 2 1 - 1 1 1 3 4 5 k 1 1 - 1 1 1 1 22222 3 4 5 k k 1 2 2 1 k k 1 k 1 2 k 1 2 k k 1 k 1 2 k 1 k k 1 k 1 2 k k 1Step 4: Since the result is true for n = 3, then the result is true for all positive integral values of n > 2 by induction. 48. Step 3: Prove the result is true for n = k + 1 2 2 2 2 2i.e. Prove 1 - 1 1 1 3 4 5 k 1 k 1kProof: 2 2 2 2 1 - 1 1 1 3 4 5 k 1 1 - 1 1 1 1 22222 3 4 5 k k 1 2 2 1 k k 1 k 1 2 k 1 2 k k 1 k 1 2 k 1Hence the result is true for k k 1 k 1n = k + 1 if it is also true 2for n = k k k 1Step 4: Since the result is true for n = 3, then the result is true for all positive integral values of n > 2 by induction. 49. 2004 Extension 1 Examiners Comments; (a) Many candidates submitted excellent responses including all the necessary components of an induction proof. 50. 2004 Extension 1 Examiners Comments; (a) Many candidates submitted excellent responses including all the necessary components of an induction proof.Weaker responses were those that did not verify for n = 3,had an incorrect statement of the assumption for n = k orhad an incorrect use of the assumption in the attempt toprove the statement. 51. 2004 Extension 1 Examiners Comments; (a) Many candidates submitted excellent responses including all the necessary components of an induction proof.Weaker responses were those that did not verify for n = 3,had an incorrect statement of the assumption for n = k orhad an incorrect use of the assumption in the attempt toprove the statement. Poor algebraic skills made it difficult or even impossible for some candidates to complete the proof. 52. 2004 Extension 1 Examiners Comments; (a) Many candidates submitted excellent responses including all the necessary components of an induction proof.Weaker responses were those that did not verify for n = 3,had an incorrect statement of the assumption for n = k orhad an incorrect use of the assumption in the attempt toprove the statement. Poor algebraic skills made it difficult or even impossible for some candidates to complete the proof. Rote learning of the formula S(k) +T(k+1) = S(k+1) led many candidates to treat the expression as a sum rather than a product, thus making completion of the proof impossible.