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IB Math SL paper 1


Outis

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As for two equal roots, i think i got k= 3 & -1

too equal roots basically means the same as one answer! so you had to make Delta=0.. which makes k=3

that's why "equal" was in bold font

I took two equal roots to mean the discriminant = 0 (b^2 - 4ac) because that gives one repeated root, so there would, technically, be two equal roots. So that would mean that the x values have to be the same, not the k ones - right? Since k is the constant, not the variable.

I could be completely wrong here, but that's what I took it to mean and how I approached it. I found that both k = 3 and k = -1 led to one repeated root (they led to (x-1)^2 and (x+1)^2 I think), so I kept both of them in.

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As for two equal roots, i think i got k= 3 & -1

too equal roots basically means the same as one answer! so you had to make Delta=0.. which makes k=3

that's why "equal" was in bold font

I took two equal roots to mean the discriminant = 0 (b^2 - 4ac) because that gives one repeated root, so there would, technically, be two equal roots. So that would mean that the x values have to be the same, not the k ones - right? Since k is the constant, not the variable.

I could be completely wrong here, but that's what I took it to mean and how I approached it. I found that both k = 3 and k = -1 led to one repeated root (they led to (x-1)^2 and (x+1)^2 I think), so I kept both of them in.

Hey.. I did the same thing.. So is it correct?

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Paper 1 in TZ1 wasn't too bad. Question 10 was killer; my teacher hasn't even figured it out yet.

My schools in Canada, so i guess that's time zone 1. I thought that Paper 1 was relatively straight forward, other than question 10 of course. That's the one about velocity and distance i believe? although I forgot to simplify my final answer I think I got it reasonably correct.

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I had TZ1. :headchop:

Yeah.

But in all seriousness, it could go either way. :D I had trouble with 6 and 10. Can't even remember what 6 was about.

Reading this thread, I was panicking thinking I bombed the entire test...until I realized these were TZ2 questions.

Edited by 007imagine
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haha..this is TZ2...

and oh my god, i think this binomial expansion question has screwed everyone's head up...like me..:)

Hey guys, how was p2?

I found section B to be very tough...Did anyone get the last question number 10??????? :(

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The funny thing is I think I did the expansion question differently then everyone and still ended up with the right answer :D I started out the same way but my final calculation looked different but I think that's because I made it more complicated for myself.

And on top of that my paper is a mess because I went in a completely wrong direction at first before realizing I had to cross it all out, so I wonder if they will even be able to read what I have done apart from the giant n = 14 Anyhow even if I lose like 3 marks it's not going to hurt. Usually at least a 80/90 is a 7 right?

TZ2 was pretty easy... TZ1 has supposedly been known to be harder :P Sucks for North America.

haha..this is TZ2...

and oh my god, i think this binomial expansion question has screwed everyone's head up...like me.. :)

Hey guys, how was p2?

I found section B to be very tough...Did anyone get the last question number 10??????? :(

It's not been 24 hours yet... can't discuss paper 2.

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As for two equal roots, i think i got k= 3 & -1

too equal roots basically means the same as one answer! so you had to make Delta=0.. which makes k=3

that's why "equal" was in bold font

I took two equal roots to mean the discriminant = 0 (b^2 - 4ac) because that gives one repeated root, so there would, technically, be two equal roots. So that would mean that the x values have to be the same, not the k ones - right? Since k is the constant, not the variable.

I could be completely wrong here, but that's what I took it to mean and how I approached it. I found that both k = 3 and k = -1 led to one repeated root (they led to (x-1)^2 and (x+1)^2 I think), so I kept both of them in.

yeah yeah that's exactly what I meant! you did it right

to get a 7 in 2011 you had to be over 80, in 2010 it was 77, and in 2008 it was 67! You can never tell! :D

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yeah yeah that's exactly what I meant! you did it right

Okay, great! I really freaked out for a second thinking I had gone the complete wrong direction!

I'm really hoping for a 7 now... P1 and my IAs were good, so hopefully it'll all turn out well! (although I feel like after that P1 the boundaries will be on the high side for TZ2 [at least for that paper] but oh well)

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Great to hear, that k was 3 AND -1.. :) Hoping for a 7 as well, even though there were some tricky parts in Paper2.

That's correct :), paper 1 wasn't hard at all (i only messed up the arithmetic's on probability, area under graph and binomial, so i could get a -2 on each of those questions) Still hoping for a 7 though.

However, I believe it's not good to dwell in this ultimate ego. It could affect our performance on other subjects, so stay sharp!

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I got one as 5/11, I think it was c, as for b....I think it was 11/21....not sure though...

how did you get 11/21? this is the one with the two white balls from bag A right?? I just multiplied the probabilities of getting white balls along the branches of the tree diagram so I got 3/7 * 2/6 = 1/7

not sure if I read the question wrong now, do you remember how many white balls there were in total in bag A?

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I got one as 5/11, I think it was c, as for b....I think it was 11/21....not sure though...

how did you get 11/21? this is the one with the two white balls from bag A right?? I just multiplied the probabilities of getting white balls along the branches of the tree diagram so I got 3/7 * 2/6 = 1/7

not sure if I read the question wrong now, do you remember how many white balls there were in total in bag A?

Yeah I got 1/7 too

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I got one as 5/11, I think it was c, as for b....I think it was 11/21....not sure though...

how did you get 11/21? this is the one with the two white balls from bag A right?? I just multiplied the probabilities of getting white balls along the branches of the tree diagram so I got 3/7 * 2/6 = 1/7

not sure if I read the question wrong now, do you remember how many white balls there were in total in bag A?

Yeah I got 1/7 too

I got one as 5/11, I think it was c, as for b....I think it was 11/21....not sure though...

how did you get 11/21? this is the one with the two white balls from bag A right?? I just multiplied the probabilities of getting white balls along the branches of the tree diagram so I got 3/7 * 2/6 = 1/7

not sure if I read the question wrong now, do you remember how many white balls there were in total in bag A?

Yeah I got 1/7 too

That's right! even i got that.

That's because .. there were total 7 balls, from which 4 were red and 3 white in Bag A.

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