MAY13 Maths HL P1 TZ2 Discussion

[Zuler]

For the minimum I had a quadratic under the square root, and I remember my 'a' was 1 and 'b' was -2, so my -b/2a = -(-2)/2(1) = 1, not square root2?

[hrach]

I actually haven't made any kind of mistake (honestly), but I haven't answered about 15 marks. I think this was one of the most hardest exams. What do you think about that?

[Giveme45]

yep, same answer. but for the w+1 question, i drew a rhombus, and since the diagonal of a rhombus bisects the angle, arg(w+1)=pi/7. wait what was the question with the graph and second derivative?

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?

[Giveme45]

For the minimum I had a quadratic under the square root, and I remember my 'a' was 1 and 'b' was -2, so my -b/2a = -(-2)/2(1) = 1, not square root2?

Where in that question was a quadratic o.O?

[Giveme45]

Did you guys actually manage a lot in the complex question?! I thought it was harder than the usual (that particular question): while most of it was standard, I thought the z1^3n + z2^3n = 2z3^3n proof was hard (though I figured it out right in the end that I had to separate the 2pi's, but didn't get enough time to write it down). Also, the last last question, god that was insane at first (the z^2 - 2z cos2pi/7 + 1 is a factor). I didn't write anything for it and didn't have a clue, but funny enough today in the shower I figured it out (God, what has IB done to me?!)

For anyone who's curious, you had to realise the roots occur in conjugate pairs (a little unfair at first considering my 7 roots had positive arguments since the standard is to usually substitute for cis 2kpi/7 for positive k), so you would realise that two roots were cis 2pi/7 and cis -2pi/7. The sum of these roots are 2 cos 2pi/7 (-2c = -b/a) and the product of these roots were 1 (c^2 + d^2 = c/a). And then yeah, the other two factors were z^2 -2z cos4pi/7 + 1 and z^2 - 2z cos 6pi/7 + 1.

I realised this in the shower and not in the exam, I seriously need to shut off my brain soon.

I left these two (I did a bit of work in the proof one but didn't finish it as I said), and the sum to infinity one I reached the quadratic but was getting a negative discrimant by mistake so left it for time. I also didn't fully sketch my f(|x|) because of time. How many marks do you think I left out? (I was thinking 2 for the sum to infinity, 2 for the graph, 2 for the proof and 4 for the z^2 - 2z cos 2pi/7 + 1 thing)

I think the boundaries for paper 1 will be 88+ or so, would you agree?

If that's the only mistakes you did then i'd suppose 2 for the sum to infinity, 3 or 4 for the graph, 7 for the proof and 9 for the z² thing. That was pretty much the grading for this, sorry to disappoint ya buddy.

[Zuler]

Did you guys actually manage a lot in the complex question?! I thought it was harder than the usual (that particular question): while most of it was standard, I thought the z1^3n + z2^3n = 2z3^3n proof was hard (though I figured it out right in the end that I had to separate the 2pi's, but didn't get enough time to write it down). Also, the last last question, god that was insane at first (the z^2 - 2z cos2pi/7 + 1 is a factor). I didn't write anything for it and didn't have a clue, but funny enough today in the shower I figured it out (God, what has IB done to me?!)

For anyone who's curious, you had to realise the roots occur in conjugate pairs (a little unfair at first considering my 7 roots had positive arguments since the standard is to usually substitute for cis 2kpi/7 for positive k), so you would realise that two roots were cis 2pi/7 and cis -2pi/7. The sum of these roots are 2 cos 2pi/7 (-2c = -b/a) and the product of these roots were 1 (c^2 + d^2 = c/a). And then yeah, the other two factors were z^2 -2z cos4pi/7 + 1 and z^2 - 2z cos 6pi/7 + 1.

I realised this in the shower and not in the exam, I seriously need to shut off my brain soon.

I left these two (I did a bit of work in the proof one but didn't finish it as I said), and the sum to infinity one I reached the quadratic but was getting a negative discrimant by mistake so left it for time. I also didn't fully sketch my f(|x|) because of time. How many marks do you think I left out? (I was thinking 2 for the sum to infinity, 2 for the graph, 2 for the proof and 4 for the z^2 - 2z cos 2pi/7 + 1 thing)

I think the boundaries for paper 1 will be 88+ or so, would you agree?

If that's the only mistakes you did then i'd suppose 2 for the sum to infinity, 3 or 4 for the graph, 7 for the proof and 9 for the z² thing. That was pretty much the grading for this, sorry to disappoint ya buddy.

Umm, I don't remember the exact mark, but I know for SURE the proof and the z^2 thing were not 7 and 9 marks on their own. The proof thing I nearly got, I had about 2-3 steps left which is why I'm thinking that's -2 (out of the 4 or maybe 5 marks it was out of) and I KNOW the whole seventh roots of unity thing was 9 marks. I did the 7th roots of 1, I did the argument of z+1, which I think was 5/9, so that's roughly 4 marks I left there?

Also, for your f(|x|) = -0.25, I got the same answers

And for the | z1 + az2 | one, I had sqrt ( (a+something)^2 + (a+something)^), the real and imaginary parts, and it gave me a quadratic in terms of a under the root, and and so I figured the minimum value of this was -b/2a hence 1? If you can remember the exact question I'll show you my exact working Sounds like I was wrong though anyway

Edited May 10, 2013 by Zuler

[Giveme45]

Did you guys actually manage a lot in the complex question?! I thought it was harder than the usual (that particular question): while most of it was standard, I thought the z1^3n + z2^3n = 2z3^3n proof was hard (though I figured it out right in the end that I had to separate the 2pi's, but didn't get enough time to write it down). Also, the last last question, god that was insane at first (the z^2 - 2z cos2pi/7 + 1 is a factor). I didn't write anything for it and didn't have a clue, but funny enough today in the shower I figured it out (God, what has IB done to me?!)

For anyone who's curious, you had to realise the roots occur in conjugate pairs (a little unfair at first considering my 7 roots had positive arguments since the standard is to usually substitute for cis 2kpi/7 for positive k), so you would realise that two roots were cis 2pi/7 and cis -2pi/7. The sum of these roots are 2 cos 2pi/7 (-2c = -b/a) and the product of these roots were 1 (c^2 + d^2 = c/a). And then yeah, the other two factors were z^2 -2z cos4pi/7 + 1 and z^2 - 2z cos 6pi/7 + 1.

I realised this in the shower and not in the exam, I seriously need to shut off my brain soon.

I left these two (I did a bit of work in the proof one but didn't finish it as I said), and the sum to infinity one I reached the quadratic but was getting a negative discrimant by mistake so left it for time. I also didn't fully sketch my f(|x|) because of time. How many marks do you think I left out? (I was thinking 2 for the sum to infinity, 2 for the graph, 2 for the proof and 4 for the z^2 - 2z cos 2pi/7 + 1 thing)

I think the boundaries for paper 1 will be 88+ or so, would you agree?

If that's the only mistakes you did then i'd suppose 2 for the sum to infinity, 3 or 4 for the graph, 7 for the proof and 9 for the z² thing. That was pretty much the grading for this, sorry to disappoint ya buddy.

Umm, I don't remember the exact mark, but I know for SURE the proof and the z^2 thing were not 7 and 9 marks on their own. The proof thing I nearly got, I had about 2-3 steps left which is why I'm thinking that's -2 (out of the 4 or maybe 5 marks it was out of) and I KNOW the whole seventh roots of unity thing was 9 marks. I did the 7th roots of 1, I did the argument of z+1, which I think was 5/9, so that's roughly 4 marks I left there?

Also, for your f(|x|) = -0.25, I got the same answers

And for the | z1 + az2 | one, I had sqrt ( (a+something)^2 + (a+something)^), the real and imaginary parts, and it gave me a quadratic in terms of a under the root, and and so I figured the minimum value of this was -b/2a hence 1? If you can remember the exact question I'll show you my exact working Sounds like I was wrong though anyway

Hm not quite sure about the marks tbh, so i dont wanna give you wrong information, but i think that i saw that the z² thing was at least 7 marks and the proof about 5 or something. Awesome that we got the same answers ^^

I cant even remember how i aprochaed that question, but im quite sure sqrt 2 was right haha ^^ I dont even remember the question at all

[Rohan Madhwal]

yep, same answer. but for the w+1 question, i drew a rhombus, and since the diagonal of a rhombus bisects the angle, arg(w+1)=pi/7. wait what was the question with the graph and second derivative?

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 derivative ? haha

[Giveme45]

yep, same answer. but for the w+1 question, i drew a rhombus, and since the diagonal of a rhombus bisects the angle, arg(w+1)=pi/7. wait what was the question with the graph and second derivative?

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 derivative ? haha

I calculated the second derivative for drawing the graph of f(x), in order to see the shape of the function, whether its concave up or down

[IB-Adam]

I calculated the second derivative for drawing the graph of f(x), in order to see the shape of the function, whether its concave up or down

Waste of time, you were supposed to sketch the function. Sketch = general shape, including y, x-intercepts, max, and min.

I remember that that question alone was worth 15 marks

[ApotheoCarp]

I think Section A was mostly ok, and Section B wasn't that bad either. I did, however, only figure out the complex number question (the last part of it) literally 2 minutes after walking out of the classroom. Can someone tell me if what I think is right?

The roots of z^7=1 should also be the roots of z^7-1=0, so to find the two remaining quadratic functions just use the sum and product of roots?

[Rohan Madhwal]

Waste of time, you were supposed to sketch the function. Sketch = general shape, including y, x-intercepts, max, and min.

I remember that that question alone was worth 15 marks

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

[Giveme45]

I calculated the second derivative for drawing the graph of f(x), in order to see the shape of the function, whether its concave up or down

Waste of time, you were supposed to sketch the function. Sketch = general shape, including y, x-intercepts, max, and min.

I remember that that question alone was worth 15 marks

Yeah and I calculated the derivative to see the general shape, so hows that a waste of time o.O ?

And no, it was not 15 marks. The marking for the graphs will be approx. like this: graph of f(x). A1 Shape A1 x-intersect A1 y-intersect

graph of inverse A1 Shape (with ft since reflection) A1 Intercepts. Graph of f(|x|) A1 Shape A1 intersects.

Why this marking? Well because first part of the question was 8 marks, which included finding the inverse (3), drawing both graphs (5). Second part was also 7 marks which would be something like graph (2-3), finding x=+-2/9 (4-5).

So yeah, i might lose 3-4 marks here.

Waste of time, you were supposed to sketch the function. Sketch = general shape, including y, x-intercepts, max, and min.

I remember that that question alone was worth 15 marks

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.

Edited May 11, 2013 by Giveme45

[Giveme45]

I think Section A was mostly ok, and Section B wasn't that bad either. I did, however, only figure out the complex number question (the last part of it) literally 2 minutes after walking out of the classroom. Can someone tell me if what I think is right?

The roots of z^7=1 should also be the roots of z^7-1=0, so to find the two remaining quadratic functions just use the sum and product of roots?

The remaining two where the same expression as given, but with 4pi/7 and 6pi/7.

[Garmr]

Nice to see everyone thought it was a good paper, although I still missed out at least 15-20 marks worth of questions so I won't expect to break the 100s. But Section A was reasonable and Section B was pretty good, with the exception of question 13. I only managed to do the easy parts of that question due to time constraints so I probably lost 12~ marks there already.

For the |z1 + az2| question I got an answer but then I realized my a was not a rational number so I did something wrong somewhere but I stopped thinking about it since I would run out of even more time if I did >_>

Overall I'm hoping for 80-90/120 in this paper. I did think it was easier than past papers I've done (except maybe M12) so I hope the boundaries aren't significantly higher

[Zuler]

Nice to see everyone thought it was a good paper, although I still missed out at least 15-20 marks worth of questions so I won't expect to break the 100s. But Section A was reasonable and Section B was pretty good, with the exception of question 13. I only managed to do the easy parts of that question due to time constraints so I probably lost 12~ marks there already.

For the |z1 + az2| question I got an answer but then I realized my a was not a rational number so I did something wrong somewhere but I stopped thinking about it since I would run out of even more time if I did >_>

Overall I'm hoping for 80-90/120 in this paper. I did think it was easier than past papers I've done (except maybe M12) so I hope the boundaries aren't significantly higher

Haha exactly, agreed with you there. Even I thought only M12 was an easier paper than that, so what would you think are the boundaries? 87+ for 7 maybe? Or possibly as low as 85?

[Giveme45]

Nice to see everyone thought it was a good paper, although I still missed out at least 15-20 marks worth of questions so I won't expect to break the 100s. But Section A was reasonable and Section B was pretty good, with the exception of question 13. I only managed to do the easy parts of that question due to time constraints so I probably lost 12~ marks there already.

For the |z1 + az2| question I got an answer but then I realized my a was not a rational number so I did something wrong somewhere but I stopped thinking about it since I would run out of even more time if I did >_>

Overall I'm hoping for 80-90/120 in this paper. I did think it was easier than past papers I've done (except maybe M12) so I hope the boundaries aren't significantly higher

Haha exactly, agreed with you there. Even I thought only M12 was an easier paper than that, so what would you think are the boundaries? 87+ for 7 maybe? Or possibly as low as 85?

That would be epic but I doubt that tbh, Ill expect them to be about 92-90. Can't wait to discuss P2!!!!

[Zuler]

Nice to see everyone thought it was a good paper, although I still missed out at least 15-20 marks worth of questions so I won't expect to break the 100s. But Section A was reasonable and Section B was pretty good, with the exception of question 13. I only managed to do the easy parts of that question due to time constraints so I probably lost 12~ marks there already.

For the |z1 + az2| question I got an answer but then I realized my a was not a rational number so I did something wrong somewhere but I stopped thinking about it since I would run out of even more time if I did >_>

Overall I'm hoping for 80-90/120 in this paper. I did think it was easier than past papers I've done (except maybe M12) so I hope the boundaries aren't significantly higher

Haha exactly, agreed with you there. Even I thought only M12 was an easier paper than that, so what would you think are the boundaries? 87+ for 7 maybe? Or possibly as low as 85?

That would be epic but I doubt that tbh, Ill expect them to be about 92-90. Can't wait to discuss P2!!!!

But considering we all agree that Paper 1 2012 TZ2 was easier, and that had a boundary of 91 (http://www.globaljaya.net/secondary/IB/Subjects%20Report/May%202012%20subject%20report/Maths%20HL%20subject%20report%202012%20TZ2.pdf), isn't it safe to assume our P1 boundaries will be atleast a few marks lower? So like 88+ or what not?

[Giveme45]

Nice to see everyone thought it was a good paper, although I still missed out at least 15-20 marks worth of questions so I won't expect to break the 100s. But Section A was reasonable and Section B was pretty good, with the exception of question 13. I only managed to do the easy parts of that question due to time constraints so I probably lost 12~ marks there already.

For the |z1 + az2| question I got an answer but then I realized my a was not a rational number so I did something wrong somewhere but I stopped thinking about it since I would run out of even more time if I did >_>

Overall I'm hoping for 80-90/120 in this paper. I did think it was easier than past papers I've done (except maybe M12) so I hope the boundaries aren't significantly higher

Haha exactly, agreed with you there. Even I thought only M12 was an easier paper than that, so what would you think are the boundaries? 87+ for 7 maybe? Or possibly as low as 85?

That would be epic but I doubt that tbh, Ill expect them to be about 92-90. Can't wait to discuss P2!!!!

But considering we all agree that Paper 1 2012 TZ2 was easier, and that had a boundary of 91 (http://www.globaljaya.net/secondary/IB/Subjects%20Report/May%202012%20subject%20report/Maths%20HL%20subject%20report%202012%20TZ2.pdf), isn't it safe to assume our P1 boundaries will be atleast a few marks lower? So like 88+ or what not?

I found both ismilar, tbh, but lets hope so

[bomaha]

q10 was a good question btw. what did u guys get?