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1. ## Curious about where you all applied

AHHAHAHA. I'm going to HKUST to pursue the same course and I also got the same score as you ! Look forward to meeting you in August
2. ## MAY13 Chemistry HL Paper 1 and 2 TZ2

I did the HF question. Well urhh, i could have been wrong but basically what i did was i converted the Ka to the pKa and then I deduced that it was a weak acid (which HF is). Then I stated the Kb for the other one, converted it to its pKb, deduced that it was a weak base. Said that the weak acid would form a strong conjugate base and the weak base would form a strong conjugate acid, therefore, they would both be neutralized and hence the resulting solution is neutral.
3. ## MAY13 Maths HL TZ2 Paper 2 Discussion

For the binomial one, in the question, it said one experiment was carried out, hence n=1 ? Or at least that's what I thought. 1 experiment = 1 trial ? And that's what the other person who's commenting thought as well apparently, so I'm not alone, the English was quite ambiguous
4. ## MAY13 Maths HL TZ2 Paper 2 Discussion

I can't remember the question but I think it was binomial and I just did P(X=1) + P(X=16). I wasn't really sure what it was asking for too, and generally I'm not too confident on the probability questions in this paper I thought it was quite weird. I think 1 experiment should be n=1. So p(x=3) is not possible. What did others put? I thought the same thing as you, for n=1, x=3 gave 0, and x=1 was 0.25, so i just put 0.25, but I had honestly no clue Oh damn, for that I thought the 16 trials constituted "1 experiment" so I got something like 0.261. You're probably correct. It didn't say "experi
5. ## MAY13 Maths HL TZ2 Paper 2 Discussion

Shouldn't the integration of cos2x give (1/2)sin(2x) + c ? And I think IB wanted us to just leave it in the square root form, because when you think about it, y = underroot x has both the positive and negative values in it's range. So when you write sqrt(1/2 sin2x + C), you're not limiting your range, unless you went and specifically wrote (y>=0) which I'm assuming you didn't. But that's simply my opinion, I could be wrong. Yes thats the integration cos2x, and that ultimateley gives y²=sin2x+c since the integration of y is also y²/2. What do u mean leave it just in the square root form? y=
6. ## MAY13 Maths HL TZ2 Paper 2 Discussion

Shouldn't the integration of cos2x give (1/2)sin(2x) + c ? And I think IB wanted us to just leave it in the square root form, because when you think about it, y = underroot x has both the positive and negative values in it's range. So when you write sqrt(1/2 sin2x + C), you're not limiting your range, unless you went and specifically wrote (y>=0) which I'm assuming you didn't. But that's simply my opinion, I could be wrong.
7. ## MAY13 Maths HL P1 TZ2 Discussion

You mean in P2 or P1? If p2, then we can only discuss in 20 minutes If p1, which one? No, i meant question 10 in paper 1, the arctan(1/8) one! I thought that question was quite straightforward, yeah p=3 and the value was pi/4. Just double angle formula. Can you go a few posts upwards and check my response to q.10 in paper 2? I had this weird confusion there, maybe you can explain it. And dont forget the asymptotes That graph had both a vertical and horizontal asymptote if I'm not wrong. You're wrong. y=3/2 was not a horizontal asymptote because it was included in the range when x=8; and x=
8. ## MAY13 Maths HL P1 TZ2 Discussion

You mean in P2 or P1? If p2, then we can only discuss in 20 minutes If p1, which one? No, i meant question 10 in paper 1, the arctan(1/8) one! pi/2 or something like that ? I remember that I got a perfect value for the arctan in the end :/
9. ## MAY13 Maths HL P1 TZ2 Discussion

And dont forget the asymptotes That graph had both a vertical and horizontal asymptote if I'm not wrong. You're wrong. y=3/2 was not a horizontal asymptote because it was included in the range when x=8; and x=-2 was not a vertical asymptote, because it was not included in the domain. Urhhh, (2x-1)/(X+2) right ? I'm not talking about the inverse function, just the original function. They had specified a domain ? Oh well. But yeah, im pretty sure that y=2 was a horizontal asymptote.
10. ## MAY13 Maths HL P1 TZ2 Discussion

And dont forget the asymptotes That graph had both a vertical and horizontal asymptote if I'm not wrong.
11. ## MAY13 Maths HL P1 TZ2 Discussion

Haha a rhombus? That's very creative I just used tanx=im/re and used some funky manipulations to get tanx=tanpi/7 You had to draw a graph, the function 2-5/(x+2) and its inverse, and then f(|x|). For f(|x|)=-1/4 i got x=+- 2/9 you? You said you missed out on the minus sign for the second derivative for that question, but uhhh, why would you need a derivative there, if i'm wrong, excuse my ignorance. All I did for that one was draw the original graph, draw a line y=x and draw the reflection of the original function ? Just out of curiosity, how did you manage to do that with a second deri
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