MAY13 Maths HL P1 TZ2 Discussion

[Giveme45]

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

You mean in P2 or P1? If p2, then we can only discuss in 20 minutes If p1, which one?

[kinguncaged1]

Can we discuss p2 now?!

[Giveme45]

Can we discuss p2 now?!

Yes evne though i opened a seperate thread for that, so that P1 and P2 questions wont get confused here. But the moderator still needs to accept it so i guess we can start here then move over ^^

[kinguncaged1]

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

[Giveme45]

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40

(60-v)dv/dt=1/40

Integrating both sides would give

-ln|60-v|=t/40+c

so at t=0, v=0 gives c=-ln60.

We have ln|60-v|=ln60-t/40

so at t=30

ln|60-v|=ln60-3/4

|60-v|=e^(ln60-3/4)

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that

there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because

I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

Edited May 11, 2013 by Giveme45

[kinguncaged1]

I definitely remember n=1416...

I am honestly unsure about the absolute value thing, but I didn't even use the absolute values.

Regarding the one with, when x=0 .99 rad, and when x=20 1.16 rad... I thought there wouldn't be a solution for both... As I remember it, the equation in question was

Theta= 180 - arctan(13/20-x) - arctan(8/x)

[Giveme45]

I definitely remember n=1416...

I am honestly unsure about the absolute value thing, but I didn't even use the absolute values.

Regarding the one with, when x=0 .99 rad, and when x=20 1.16 rad... I thought there wouldn't be a solution for both... As I remember it, the equation in question was

Theta= 180 - arctan(13/20-x) - arctan(8/x)

Yeah you had to think outside the box because for these particular values you could not use this equation.

When x=0 , it was pi/2 - arctan(13/20) for example

[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?

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

Yeah, that would probably have been it. Darn. Ah well, I still think I did pretty well in the rest of the exam. Hopefully my feelings translate into reality.

[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

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?

Yeah, I hope it will be in the mid 80s, cause then I'd have a reasonably good shot at a 7 (or at least a 6 thanks to my really bad IAs )

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

I believe that was the arctan (1/8) + arctan (1/5) = arctan (1/p), find p and the hence find arctan (1/8) + arctan (1/5) + arctan (1/2) question. I forgot the method so I skipped that question altogether and made a 6 mark sacrifice but I did it later at home and found p=3 haha

[Garmr]

kk awesome, what questions do you remember?

I remember getting 37.1 meters per second for a differential equation question with velocity and acceleration...

That is the question which i was mega ambigious about. I hope you can answer my amibguity, because then i might get 120/120 marks, and I would feel ****ing good.

Q.10: Acceleration was given to be (60-v)/40, and initially the particle was at rest so at t=0, v=0. We were supposed to find the value of v at t=30.

dv/dt=(60-v)/40

(60-v)dv/dt=1/40

Integrating both sides would give

-ln|60-v|=t/40+c

so at t=0, v=0 gives c=-ln60.

We have ln|60-v|=ln60-t/40

so at t=30

ln|60-v|=ln60-3/4

|60-v|=e^(ln60-3/4)

This gives v=31.7 or v=88.3 because of the absolute value sign. Now this really confused me during the exams, since its impossible that

there are two velocities at one particular time (if we ignore heisenberg, which probably doesnt apply here). Ultimately, I chose v=31.7 because

I assumed that you can ignore the absolute value signs. Could you please explain?

Apart from that values that I remember are:

Income 137 dollars, age of cats 28, 2013, n=1416, k=1, k=3 (not sure bout this one), volume: 10.05, area 2.99, theta when x=0 0.99 rad,theta when x=20 1.16 something rad,

I can't remember my answer but 31.7 and your method seems very familiar so I hope we all did the right thing

[bomaha]

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

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!

[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.

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.

Edited May 11, 2013 by Rohan Madhwal

[Rohan Madhwal]

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

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 :/

[bomaha]

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

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 :/

arctan1=pi/4

[Giveme45]

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

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.

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.

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.

Yes they had a specified domain, and how is y=2 a horizontal asymptote in (2x-1)/(X+2)?

[Garmr]

Giveme45 - on your velocity question, I got 31.7 as well, because I didn't think the absolute sign mattered much. Hope it's the right thing.

[Giveme45]

Giveme45 - on your velocity question, I got 31.7 as well, because I didn't think the absolute sign mattered much. Hope it's the right thing.

Yeah 31.7 is defnitely one of the values which would solve the equation. I'm just trying to understand how to justify that the absolute value signs didnt matter / justify why 88.3 was impossible. I hope someone here can elaborate.

Also for q12:

Q.12

We had the equation: ydy/dx=cos2x.

We were supposed to find the general expression in the form of y=f(x) (with the constant of integration).

So if we integrate it by dx, then we get

y²/2=sin2x/2+c

so

y²=sin2x+k

y=+- sqrt(sin2x+k)

And there we go again, I was baffled how we were supposed to know whether to chose the positive or negative expression. Ultimately I chose the positive, was that right or were

we free to choose.

[Garmr]

Giveme45 - on your velocity question, I got 31.7 as well, because I didn't think the absolute sign mattered much. Hope it's the right thing.

Yeah 31.7 is defnitely one of the values which would solve the equation. I'm just trying to understand how to justify that the absolute value signs didnt matter / justify why 88.3 was impossible. I hope someone here can elaborate.

Also for q12:

Q.12

We had the equation: ydy/dx=cos2x.

We were supposed to find the general expression in the form of y=f(x) (with the constant of integration).

So if we integrate it by dx, then we get

y²/2=sin2x/2+c

so

y²=sin2x+k

y=+- sqrt(sin2x+k)

And there we go again, I was baffled how we were supposed to know whether to chose the positive or negative expression. Ultimately I chose the positive, was that right or were

we free to choose.

I have to admit I was careless and didn't consider the plus/minus ambiguity, but I chose positive as well. I don't think it could be both so hopefully that was also the right thing.

[Rohan Madhwal]

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

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.

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.

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.

Yes they had a specified domain, and how is y=2 a horizontal asymptote in (2x-1)/(X+2)?

Hahaha uhh cause when you put y=2 into the equation, the x will cancel out That's the rule for horizontal asymptote's isnt it ? try graphing it with your ti-84.

[Giveme45]

I really don't see an explanation for these. And if i'm correct, then these questions were just simply unprofesionally written...

Does no one else here know?