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kw0573 last won the day on March 5

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About kw0573

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    May 2015
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  1. I think the question is saying how equivalent expressions can change the difficulty of evaluating the integrand. For example if we substitute part a) into b), then integral becomes much more difficult, with an hairy denominator. Part a) shows how that hard integral can be simplified to b).
  2. @tutorinseoul I think the testing values is unnecessary. @SC2Player. You are right. The typo you made mistakenly allowed you to reject 5 - 5sqrt (5)) / 8 as a value of cos2θ because it is negative. It is always important to make sure the roots you reject actually can be rejected.
  3. @SC2Player What did @tutorinseoul meant by the value was wrong, anyways?
  4. Keep looking you can find it! On the exam you may have to troubleshoot like this too.
  5. Hint: What's cos (9 pi / 10)?
  6. I find part 2 ridiculously difficult. I gave up, but I think the logic is something like have a quadratic in sin x or cos x, then find the other root using Viete's theorem. I looked up the answer in a solver and it's not a familiar angle like pi/4 or pi/3. Thus it's also very difficult to guess and check. I graduated already but it is interesting to see if anyone in the program can figure this out!
  7. I like this question. It lists polynomial terms in ascending order as opposed to the common descending, and gives an: not too deceiving but may trick the student rushing for time. It's a perfect example of those Part A questions with which one needs to be careful but it's overall a straightforward question.
  8. You essentially said the expression is equal to both 5, and 10. In any case you should try your best at such "hence" questions because they appear quite often in exams. Just try something using both results in a) and b). Good luck you can do it
  9. The question is supposed to say (z+1)/(z-1) = cos x + i sin x. Using identities cot (x) = (cot2 (x/2) - 1) / (2 cot (x/2)), and cos θ = 1/2 * (eiθ + e-iθ) and i sin θ = 1/2 * (eiθ - e-iθ) I used z = reiθ. where z* = re-iθ. 1) Multiply (z+1)/(z-1) by (z*-1) / (z*- 1), and using above identities I get (r2 - 2r i sin θ - 1) / (r2 - 2 r cos θ + 1) = cos x + i sin x 2) Match imaginary part to imaginary part to get sin x = (-2r sin θ) / (r2 - 2 r cos θ + 1) 3) now divide both sides by sin x, we get (r2 - 2r i sin θ - 1) / (-2r sin θ) = cot x + i, which simplifies to (r2 - 1) / (-2r sin θ) = cot x 4) Match again with the cot (x) identity above, we see that -2r sin θ = cot (x / 2) ==> r sin θ = - cot (x / 2) and from numerator matching that r2 = cot2 (x / 2) which implies Re(z) = 0. Hence z = i r sin θ = -i cot (x/2). An alternative solution can be found here, which is a more or less equivalent using z = cos θ + i sin θ instead. Question 2 is a direct binomial expansion, without any substitution, just (z+1)^2 - (z-1)^2 = 0, odd power terms cancel. Divide by 10 you get the answer. Question 3 you made a typo somewhere.
  10. There are many discussions on this previously, for example here: I can't say Math Studies SL + Physics HL can guarantee success but there have been many precedents.
  11. Looks very good; very concise. Remember that in the top chart, forces should be "intermolecular forces". IB like to ask students to discern between intramolecular and intermolecular forces. A more comprehensive summary should also include tips to balance equations, find limiting reagents, concentration <--> mol(and mass) conversions. These occur much more frequently on exams than say homogeneous/hetereogeneous mixtures or distinguishing between Boyle/Gay-Lussac/Charles.
  12. I am not sure if it's my encoding, but many equal signs and theta's are showing up as other characters.
  13. I would say that since most schools of a region (this this case, UK) have similar admission standards, it may be better to apply to just 3 of these 5 schools of which you are most interested in going. Then you should also apply for few other less competitive schools, where you would consider to attend for proximity to home, lower costs, good environment etc. For example I applied to Carnegie Mellon University (CMU) and it was a bad decision because I would never go there because of how pricey it was (2-3x my current tuition). With that in mind, getting admitted to only one university is not a bad idea if you are interested in attending any of the universities you apply to.
  14. If you are able to use your own approaches, then you should score well in Personal Engagement. Any type of exploratory IA, such as yours of functions of complex numbers), is likely to score high in this criterion. You just have to convince that not everything you say will be likely found in a textbook on the topic, because you could be mentioning personal interpretations, etc.
  15. Complexity-wise it is definitely enough for HL. But for HL you also need to use multiple areas of math (see Criterion E Use of Mathematics), such as trig and complex numbers, permutations and calculus, etc.