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Math HL P2 TZ2 - 24 hours passed

Adalia

Of course stats is on the syllabus, but Q12 doesn't fit in nicely anywhere.

Oh I was SO pissed off when leaving the exam hall. The only combination questions I understand is when its a circle, because you do (n-1)!
I forgot the -1. I realised it just as I handed my paper to the IBDPC. FML.

@ running from demons: hahaha. yes, that is the answer to Q12 :biggrin:

@ mum10mum: i can help you with sets if you want :smile:

I'm not sure, I think it does fit to 3.1 in the syllabus. If you turn to pg 248 in the H&H book, you will find a very similar problem with Q12. Well it askes you to find an angle, find a side using pythagoras theorem, finding perimeter. I assume some part of the question is presumed knowledge.

Reply 61

Adalia

2

Doing 6! and dividing by six gives you the same answer as doing (n-1)! where n=6 (because there were 6 people).
Because in (6x5x4x3x2x1)/6 the 6s cancel so you're left with 5x4x3x2x1 = 5!, exactly like using the circle method thing.

Reply 62

Roundabout

11

jc_bach

I'm not sure, I think it does fit to 3.1 in the syllabus. If you turn to pg 248 in the H&H book, you will find a very similar problem with Q12. Well it askes you to find an angle, find a side using pythagoras theorem, finding perimeter. I assume some part of the question is presumed knowledge.

The H&H book is perhaps the only one with a question like that. Both the Oxford and Cambridge HL Maths books don't cover it.

Reply 63

Adalia

2

Another two that I was really annoyed about seeing were the

- using the substitution u=some sine term, integrate.
The day before a friend of mine who's retaking this year came up to me (its in the H & H under Further Integration) w/ a question about something similar and we went and asked every single HL student we could find and not a single one thought we needed them or anything.

- the T' reflection one. I heard from one exam author for Nov - my teacher - and another HL examiner/teacher that that was no longer on the syllabus. If it came, we would be guided through. So, the night before, a question came up and I passed it of course. There was no guide, and I couldn't remember how to properly do it. gah.

Reply 64

Tae-Hwan Lee

answers for Q 12 (well this is what i wrote)

1. because PQ = 50, x+10<50. therefore x<40

2. well... draw extra line from Q to a radius on the other circle so that the line is parallel to MN.

then cos ((x-10)/50)

3. (a) square root of (2500 - (x-10)^2) or (50^2) - (x-10)^2

(b) when x = 10

4. (a) (alpha) = 2pi - 2(theta)

theta = arccos((x-10)/50)
therefore, (alpha) = 2pi - 2(theta)

(b) (beta) = 2pi - 2(pi/2) - 2(pi/2 - theta) = 2 (theta)
= 2 arccos((x-10)/50)

5. since arc = r(theta) where theta is angle in radius,
perimeter = x(2pi - 2(theta)) + 10(2 arccos((x-10)/50)) + 2 (square root of (2500 - (x-10)^2) or (50^2) - (x-10)^2)

ii and iii... well. put it in your calculator, and do it!

Reply 65

shinejuly

1

Adalia

Another two that I was really annoyed about seeing were the

- using the substitution u=some sine term, integrate.
The day before a friend of mine who's retaking this year came up to me (its in the H & H under Further Integration) w/ a question about something similar and we went and asked every single HL student we could find and not a single one thought we needed them or anything.

- the T' reflection one. I heard from one exam author for Nov - my teacher - and another HL examiner/teacher that that was no longer on the syllabus. If it came, we would be guided through. So, the night before, a question came up and I passed it of course. There was no guide, and I couldn't remember how to properly do it. gah.

UGH the substitution one... I did part of it: I got like something like sin2@ then i dont know what to do.. but before our teacher handed out a booklet of caculus questions, and there is one question very very similar to that.. the day before I was revisin,lookin at the question was like" ok, i know this,, next.."

Oh yeah, could sombody explain to me about the acceleration one? :s-smilie:

Reply 66

Tae-Hwan Lee

shinejuly

Oh yeah, could sombody explain to me about the acceleration one? :s-smilie:

a = dv/dt

therefore dv/v = (-1/2)dt
therefore ln v = (-1/2)t + C
which means v = e^((-1/2)t + C) = Ae^((-1/2)t)
By substituting v=40 and t=0, A = 40

therefore, v = 40e^((-1/2)t)

Reply 67

mum10mum

Tae-Hwan Lee

answers for Q 12 (well this is what i wrote)

1. because PQ = 50, x+10<50. therefore x<40

2. well... draw extra line from Q to a radius on the other circle so that the line is parallel to MN.

then cos ((x-10)/50)

3. (a) square root of (2500 - (x-10)^2) or (50^2) - (x-10)^2

(b) when x = 10

4. (a) (alpha) = 2pi - 2(theta)

theta = arccos((x-10)/50)
therefore, (alpha) = 2pi - 2(theta)

(b) (beta) = 2pi - 2(pi/2) - 2(pi/2 - theta) = 2 (theta)
= 2 arccos((x-10)/50)

5. since arc = r(theta) where theta is angle in radius,
perimeter = x(2pi - 2(theta)) + 10(2 arccos((x-10)/50)) + 2 (square root of (2500 - (x-10)^2) or (50^2) - (x-10)^2)

ii and iii... well. put it in your calculator, and do it!

Thanks
If i had time i would have definitely solved this one..anyway paper 3 is more important now
But i cant understand this set relation thing properly
Especially when they ask me to find the equivalence classes..
Any tips?

Reply 68

shinejuly

1

Tae-Hwan Lee

a = dv/dt

therefore dv/v = (-1/2)dt
therefore ln v = (-1/2)t + C
which means v = e^((-1/2)t + C) = Ae^((-1/2)t)
By substituting v=40 and t=0, A = 40

therefore, v = 40e^((-1/2)t)

dont really get it but thanks !! :]

Reply 69

proudofmyself

Tae-Hwan Lee

a = dv/dt

therefore dv/v = (-1/2)dt
therefore ln v = (-1/2)t + C
which means v = e^((-1/2)t + C) = Ae^((-1/2)t)
By substituting v=40 and t=0, A = 40

therefore, v = 40e^((-1/2)t)

Is there anybody who does not agree with this?

Becuase I don't agree..mine doesn't contain e

Reply 70

running from demons

9

Tae-Hwan Lee

answers for Q 12 (well this is what i wrote)

1. because PQ = 50, x+10<50. therefore x<40

2. well... draw extra line from Q to a radius on the other circle so that the line is parallel to MN.

then cos ((x-10)/50)

3. (a) square root of (2500 - (x-10)^2) or (50^2) - (x-10)^2

(b) when x = 10

4. (a) (alpha) = 2pi - 2(theta)

theta = arccos((x-10)/50)
therefore, (alpha) = 2pi - 2(theta)

(b) (beta) = 2pi - 2(pi/2) - 2(pi/2 - theta) = 2 (theta)
= 2 arccos((x-10)/50)

5. since arc = r(theta) where theta is angle in radius,
perimeter = x(2pi - 2(theta)) + 10(2 arccos((x-10)/50)) + 2 (square root of (2500 - (x-10)^2) or (50^2) - (x-10)^2)

ii and iii... well. put it in your calculator, and do it!

Wow. I actually did all of that!I just didn't put it in the calculator right. Method marks at least I suppose.

proudofmyself

Is there anybody who does not agree with this?

Becuase I don't agree..mine doesn't contain e

That method is correct. There was definitely an exponential term.

Reply 71

mum10mum

proudofmyself

Is there anybody who does not agree with this?

Becuase I don't agree..mine doesn't contain e

therefore dv/v = (-1/2)dt

if this state is right [as i dont remember this question]
then its correct method

Reply 72

Roundabout

11

proudofmyself

Is there anybody who does not agree with this?

Becuase I don't agree..mine doesn't contain e

I didn't get the same result.

I think it started off with a=-0.5v and the job was to find an expression for v in terms of t given that v=40 when t=0.
This is what I did:

dv/dt=-0.5v
invert it to get:
dt/dv=-2/v
integrate to get:
t = -2ln(v) +C

with boundary conditions, C=2ln(40).

Hence t = -2ln(v) + 2ln(40) and by using the rules of logarithms:
t=ln(1600/(v^2))
e^t=1600/(v^2)
v^2 =1600/(e^t)
v = square root of (1600/e^t).

edit:
and this of course can be further simplified to v=40e^(-1/2). Which I didn't do on the exam. I hope they won't penalise. :biggrin:

Reply 73

_dp_

proudofmyself

Is there anybody who does not agree with this?

Becuase I don't agree..mine doesn't contain e

AFAIK this is the only correct solution: it is a differential equation, and the rate of change of the velocity (a) is proportional to the velocity itself. Hence it can only be exponential.

Reply 74

shinejuly

1

mum10mum

Thanks
If i had time i would have definitely solved this one..anyway paper 3 is more important now
But i cant understand this set relation thing properly
Especially when they ask me to find the equivalence classes..
Any tips?

in the oxford revision guide it says " related elements are said to be quivalent and the various groupings are called equivalence classes." there is an example if ad=bc is the equivalece relation then a/b=c/d will be the equivalence class as sets of equivalent fractions. i know its quite confusin but i think its like a condition or something like that... things you can find in other areas in math that you can totally relate to the equivalence relation given..
am i right?

edit : am i right anyone?

Reply 75

proudofmyself

oh s***, were we asked to solve for v? I stopped after getting t = -2ln(v) + C
Damn it

I was like, "oh this question is quite simple and straightforward..thanks"
I was too wrong

Reply 76

Aghobrial

I didn't do well! I did past papers easily but this was by far the most difficult! Do they grade it based on a curve? Does anyone think they could really get a 7? I don't I though I could but Now i don't!

Reply 77

mum10mum

I dont get this
after looking at the answer then only i 'feel' like i get it
simple relations like y - x is divisible by 5, i can find the quivalence class
but the question in the paper would be much harder than this....
I just chose sets and relation because it sounds easy
but i think i was wrong :s

Reply 78

mum10mum

Any guess for new grade boundary for this years paper?
We have to see the difficulty of paper 3 but i really hope it is between 70~73%

Reply 79

Roundabout

11

mum10mum

Any guess for new grade boundary for this years paper?
We have to see the difficulty of paper 3 but i really hope it is between 70~73%

See my earlier post:

Roundabout

Well for easy exams the boundary for 7 has been around 80, but I don't consider this year's exam to be amongst the easiest ones in the recent past, and others seem to agree. Maybe overall the same standard as M08 TZ1, perhaps a bit harder. In M08 TZ1 the boundaries for 7 overall varied between 74 and 75 out of 100, so I'd expect the boundary for 7 this year in TZ2 to be around 73-75, depending on P3.

Math HL P2 TZ2 - 24 hours passed

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