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Discussion of paper

Discussion of paper. F.6 Pure Mathematics 2009-06-29. JOHN NG. http://johnmayhk.wordpress.com. Q.1 (b) Satisfactory. . y = f( x ). Slope of tangent at x = 0 is f’ (0). a. O. . should be found by definition. Q.1(c) Satisfactory. Prove. is continuous at x = 0,. check:.

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Discussion of paper

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  1. Discussion of paper F.6 Pure Mathematics 2009-06-29 JOHN NG http://johnmayhk.wordpress.com

  2. Q.1 (b) Satisfactory

  3. y = f(x) Slope of tangent at x = 0 is f’ (0) a O 

  4. should be found by definition.

  5. Q.1(c) Satisfactory Prove is continuous at x = 0, check: It is not enough to check We need

  6. Q.2 (a) Good

  7. Q.2 (b) Satisfactory is valid when n is a POSITIVE INTEGER. Hence, putting x = 0, is valid when n is a POSITIVE INTEGER. Thus, it is WRONG to write The iterance should stop at

  8. Q.3 (a) Satisfactory  Also, students just used l’hôpital’s rule without using so-called very important limit to simplify the work.

  9. Alternative method

  10. Q.3 (b) Not satisfactory  

  11. Make sure to make indeterminate forms like 0/0, / etc. before using l’hôpital’s rule. It is a 0/0 form, because

  12. Q.4 (b) Satisfactory Prove f(x) is bounded on R, we need to find a FIXED number M such that |f(x)|  M for ALL x in R. Hence, it is wrong to say  f(x) is bounded Just one step further, we can get rid of the x,

  13. Q.5 (a) Not satisfactory Students may know what [x] is, but not familiarized with the operations.  Some wrote [2x] = 2[x] or [2x] = 2n - x  Some claimed Note:

  14. Q.5 (b) Not satisfactory y y = 3/x y = 2/x y = 1/x x The graph of y = [2x]/x is strange to students. Students should be taught that, for integer n, for Hence,

  15. Q.5 (b)(ii) Students tried to prove g(x) is injective. Just tried some values of x and see that g(x) is NOT. e.g. For 0 < x < 1/2, g(x) = 0/x – 0 = 0 showing that is NOT injective.

  16. Q.6 (a)  Mistook that Taking logarithm is the trick.

  17. Q.6 (a) Not satisfactory Not many students could show that f’(x) > 0. A negative sign is in the expression, and it is not clear enough that f’(x) > 0. Better

  18. Q.6 (b)(i) Satisfactory Alternative method (By (a))

  19. Q.6 (b)(ii) Not satisfactory Alternative method Hence

  20. Q.6 (b)(ii) Not satisfactory is increasing (as n increases) and is bounded from above by 1, does not imply that There is a misunderstanding.

  21. Q.7 (a)(i) Good f(x) = 0 has a triple root , some mistook that   Some weak in basic differentiation rules, e.g. 

  22. Q.7 (a)(ii) Satisfactory Also Hence, C = 0, Some used integration in this part. They claimed  Thus,  is a repeated root of f(x) = 0 Note: the integration is invalid.

  23. Q.7 (b)(i) Good Students tried to solve the repeated root to prove a2 b. It may be easier to consider that the quadratic equation g’(x) = 0 has a real root, hence   0.

  24. Q.7 (b)(ii) Not satisfactory Many students set up Viète‘s formulas (韋達定理) and obtained complicated relations. Not many students could solve the repeated root by elimination: Eliminate 3, Eliminate 2, yield

  25. Q.7 (c)(i)(ii) Good Many students could solve (c) without using the result in (b)(ii). Students may find it easy to cope with concrete numerical problems.

  26. Q.8 (a)(i) Satisfactory Some students divided cases wrongly like Instead of To sketch the graph of 

  27. Q.8 (a)(ii) Satisfactory and Some students obtained the following wrongly and cannot draw the conclusion where f is differentiable at x = -1 or not.

  28. Q.8 (b) Satisfactory If students could obtain the first and second derivatives correctly, it is likely that they could perform very well in this part

  29. Q.8 (c) Satisfactory Many students ignored that (-1,0) is also a minimum point. Some said there was no inflection point.

  30. Q.9 (a)(ii) Satisfactory Students should pay attention that g(t) means g is a function of t ONLY, other indeterminate like x and c, are constants with respect to t. Hence g’(t) is differentiating g with respect to t.

  31. Q.9 (a)(ii) Satisfactory Also, by applying the mean value theorem, state clearly the range of the value of the d, i.e. for some d lying between x and c.

  32. Q.9 (b) Not satisfactory Put x = ak (k = 1,2,…,n) into result in (a)(ii) for some dk lying between ak and c. Not many students could complete this part. Summing up and use the f’’(x) < 0 on (a,b), result follows.

  33. Q.9 (c) Not satisfactory Many students took instead of to obtain

  34. Q.9 (c) Note that does not satisfy the condition that

  35. Q.10 (a)(ii) Not satisfactory Many students tried to prove by M.I. that but not success in most of the case. Solution:

  36. Q.10 (a)(ii) The easiest way should use the increasing of {an}, that is and solve the quadratic inequality above.

  37. Q.10 (b)(i) Satisfactory Students may find question in this type is not-so-familiarized, many could not use the fact that for all n N, to derive

  38. Q.10 (b)(ii) Not satisfactory Few could complete this part. Many used the previous result wrongly, they wrote Instead of the fact that (b)(i) valid only for n N, i.e.

  39. Q.10 (c) Not satisfactory No one could complete this part. No one could show the following key step.

  40. Q.10 (c) Now check the condition:

  41. Q.10 (c) By (a), it is easy to have Hence, we have Thus, {cn} converges by (b)(ii), i.e. the following sum converges.

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