Jump to content # Rohan Madhwal

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1. 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. 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. 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. 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. 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. 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. 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. 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. 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. And dont forget the asymptotes That graph had both a vertical and horizontal asymptote if I'm not wrong.
11. 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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