tutorinseoul

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tutorinseoul last won the day on March 22

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About tutorinseoul

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    May 2009
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    South Korea
  1. Link http://m.imgur.com/Rxpgevz,rhLNNT5,tO2yIVu
  2. Hey, I've written up my solution; see if you can understand and ask me if there unclear part!
  3. Very well done :)!! With this technique, we can pretty much calculate EVERY trig equation; when they come in product of two different trig, we can use null factor law to work out the x. Even if we get an equation of sum of trig, we can combine them into product of trig functions to apply null factor law again. Good job :)!
  4. My triick will be to compare with the closest angle I know! For example, we have to get cos 7pi/10. The closest angle I know is 5pi/10 with exact angle. cos(pi/2) being zero, and since it's decreasing till pi, we know for my y=cos(7pi/10) value it must be bounded by 0 and -1. And then I can compare the value. I will write it up nearly tonight! Been busy in Seoul too lol.
  5. Hi all! I like this question a lot as it requires understanding of 'Discrete Random Variables!' You need to remember how to get expected value from discrete random variables. Enjoy this question :)!
  6. http://imgur.com/2gApK7O Hi everyone! I like this question a lot as it shows that you need to understand what function is and how to work on the modulus domain and range! I love it and please enjoy the question as much as I did :)!
  7. I like this question a lot as it goes over an intense substitution. Overall, standard question for paper 3 option 9 :)! Enjoy!
  8. HI! I like this question a lot as it shows the phenomenon of sum of roots of unity with a nice guideline of geometric series! Overall, intermediate level of section B :)! Enjoy!
  9. Hey thanks for the reply :)!. It gets simple if you use factor formula from compound angle. Beauty of compound angle is that you can make whatever the sum of two trig equations become a product,which is nice to solve due to null factor theorem. Give it a try again :)!
  10. Heya, this is SL and paper 2 :)! share your answer and i will check them!
  11. Hi! I like this question as it tackles on polynomials and coefficients of polynomials and application of series! Enjoy!
  12. Hi! I like this question as it deals with compound angle AND factor formula. Please enjoy and discuss your answer :)!
  13. aha, very well done except for some mistake with algebra! Try again for the quadratic substitution and derive a correct answer! (The value was wrong.) Also, for the last part, there could be more than one solution in the second quadrant. Ask me if you get stuck for the second part :)!
  14. Why not only rely on past papers? In my opinion, the best preparation with past papers is not to look over mark scheme. Essentially, as the name suggests, mark schemes only aim to aid teachers to mark student’s work on his paper. Not only does it lack with the detailed explanation, but also students lose their opportunity to encounter real IB style questions if they rely too much on mark scheme. For section A questions it relatively gives a good explanation; for Section B questions, however, when things get lengthy, explanations tend to be very concise and often do not make sense for students. It solely aims to guide teachers to base their grading criteria; it is not built for a model of ‘worked solutions.’ I encourage students more to discuss their solutions with friends. Also, IB community is very strong among students via internet, such as IB survival and IB Facebook group page. It’s a good place to start a discussion. Let me give you an example why you should avoid the habit of relying your study with mark schemes: Example: This is one of my favorite questions from IB HL. It not only tackles on definition of functions, but also it requires students to understand the concepts on Topic 1 and apply their skills to ‘unseen’ questions. If we look at Part (d) (iii), it asks about giving area in terms of a certain function A. The first part of the questions asks us to find out the relation between Fn and its inverse; here students are required to know what inverse relation is. For the geometrical interpretation, functions that are inverse to each other are symmetric along y=x. Then the second part tests our inequality skills with algebraic and geometrical interpretation: For Part (d), if you realize in the square box of 1 by 1, Fn is less than x, which means it is below y=x from 0 to 1, and since we know its inverse must be symmetric by y=x, the inverse must be above y=x. From this aspect, we can draw a random 1:1 function and work with the area to derive an answer. I believe this is how a student should approach to this type of question, since this ‘interpretation’ of function’s inverse with its algebraic property of composition and geometrical property of symmetry is the key for the unseen question. Let us compare our approach with the mark scheme of the question: As you can see, it’s painfully concise, and students cannot obtain their logical processing of this question from the mark scheme. Yes, when you apply the numbers and methods given by the mark scheme you could get the answer. But you will lose the essence of ‘interpretation’ skills as well as the half-independent attempt on this nice question. You can always elaborate your answer if your logic is correct; that is one of the greatest tactics for IB maths exam. PARTIAL mark!