maths PaperII Answers

Mutant_Fruit

These are not neceseraliy right, this is just what i got.

Q1
a) x^2 + y^2 +2x - 10y +13 = 0
b) (i) p=-6 (ii) (1,0) (4,3)
c) x^2 + y^2 + 6x - 6y +9 = 0 and x^2 + y^2 + 30x - 30y + 225 = 0 *corrected, wasn;t paying attention and just wrote +c *

Q2
A) 0.324i + .946j (i decimilised, fraction answer= 12/37i -35/37j)
b) d= (a + 3c)/4 (ii) db = (15c - 3a)/4 (iii) prove it
c) Just prove it...

Q3 a) (3,2) I think
b) 3
ii -32
c) 3+- 2root2

Q4
a) CosA= 15/17 (ii) sin2a = 240/289
b) Prove it (ii) 209degrees
c) r = 4 - 2(squareroot of 2) (ii) area = 8 - 2pi[2 - 2(squareroot of 2)]

Q5
a) Prove it
b) 2 (ii) Sin2A
c) 5 (ii) 13 (iii) 45

Q6
a) 126 (ii) 60
b) -3[(-1/3)^n] (ii) 6
c) 24/105 (ii) 18/135 (iii) 432/3075 or could be 432/3675. I can;t read it.

Q7
a) 8! (ii) (8C55 )(5!)
b) 3/8 (ii) 2/13 (iii) 27/110 (iv) 13/25
c) prove it


forgot to write the answers for these, so i'm doing these from memory
Q8
a) -Cosx + Sinx +c (or something similar)
b) 1 + x/1! + (X^2)/2! + (x^3)/3! + (x^4)/4! + (x^5)/5!

b) (ii) too lazy to write this out...

c) Volume = (pi)(r^2)[3 - 2(Pi)(r)] (ii) 1/pi

EDIT: Once again, don't know if all these are right, but they seemed to mostly match up to my friends answers...

popinfresh

hmm, what did evry1 think of the paper. The trig questions were handy. Particularily the identitys. As was question 1. Was stuck for time though left out a few C parts and had to do a probability Q instead of Q2 due to time shortage:(

PrecariousNuts

Yeah I had these done out earlier before boards.ie died. They are all right with a few corrections

Q1 c) should be +225 instead of +c
Q2 A) is that 12/37i -35/37j?

Q3 a) (3,2) I think
b) 3
ii -32
c) 3+- 2root2

Q4 bii 209 degrees


Q8 a sinx - xcosx + c
b iii Un = (x^2)^n-1 over (n+1)! and the answer to the second bit is 0 as n-> infinity

all the rest are right.
I'm so relieved to see that you got the same answers for Q4.

Q5 was waaay too easy for honors. The only bit I didn't get out is Q2 c ii not sure why but it was almost right, 7/10. Otherwise everything else is right. A1 overall for sure!


Finished at eleven, stayed till 12 "doubly redoubling strokes upon my (foe) paper"

subway_ie

Did anybody else use the angle between two lines formula for Q5.(c) iii? I started to use it, ran out of time and just wrote down "Theta = pi/4" at the end...

My fúck ups:
Q1:
(b) - must have fecked up finding p, got (-2,1) and (-2,0) for the points of intersection.
(c) - (ii) - only got one equation, with mad figures. Stuck the two points into the general circle equation, used the fact that f^2 = c to get the simultaneous. Don’t know what happened after that...

Q2:
(b) - Messed it up. Totally. Might get an attempt mark. Maybe. Hopefully. (Wishful thinking).
(c) - I said p = ai + bj and q = ci + dj. Tried to sub them in, but didn't work out.

Q3:
(a) Don't know if I used the right formula... point looked ok though.
(b) (ii) - Wrote down K=2. Feckin Retard. Think I had K=27 or 37 or something like that.
(c) (ii) I used the fact that K is perpendicular to L in the first part, with an equation y = (-1/x), since the y-intercept is at the origin. Then for part 2, I said that slope of f(L).f(L) = -1. Don't think it came out right.

Q4:
b) (ii) Got it out at the last minute, but got a value of 270 degrees, when everybody else got 60 degrees.
c) (ii) Got a weird answer of 8 - 2pi - 3pi/2 + pi/root2. Assumed that ¦bc¦ = root =32 and that the two angles of the big triangle are 45, and then used bcSin45 to get the length ¦ac¦, added it to radius. Then, for part (ii), got the area of each sector individually and took it away from the area of the triangle (8 I think?).

Q5:
a) - Feck. Used pg 9 identities to prove it, when I should have used the unit circle. I didn't prove, I "showed" which is just feckin retarded since I knew how to do it. :mad:
c) - (iii) - tried to use the angle between two lines formula for some unknown reason (probably because I was dreaming about it all night) so I messed about with that forawhile and then ran out of time, so I just stuck down theta = pi/4 = 45 degrees. Total guess.

Q8:
a) Forgot the feckin " + c ". Arghhhhhhh. 1/3 of the marks for one feckin letter.
b) (iii) Don't know if it's right, but I guessed Un = X^(2n-2) / (n + 1)!.... most people in my centre got different answers though.
c) (ii) - r = volume = 1/pi. Seemed a bit dodgy in the exam, but apparently nearly everybody got it.

Overall, a paper riddled with stupid mistakes. Bye bye "A".

PrecariousNuts

Yeah you shouldn't have used ai + bj for part c on question 2, you are told the moduluses are equal and it would have been such a mess trying to work it out although I didn't get it all out myself so I can't really say much.

If you got nice whole number answers for 3a you should be grand.

Q4 b answer is 209 sub it into the equation, it works out as zero.(almost due to rounding). Part c: I bisected the right angle and just used basic trig to work out the answer. Your area is correct.

Mutant_Fruit

(c) - I said p = ai + bj and q = ci + dj. Tried to sub them in, but didn't work out.

Thats right. You then had to get pq (c-a)i + (d-b)j and dot product that with (p+q)= (a+c)i + (b+d)j

You then fill into your dot product formula, and get [(p+q).(pq)] / (mod p+q)(mod PQ) = CosO (just moving it around)

For it to be perpendicular, Cos0 = zero, therefore the top part of the fraction = 0

Work out just the top part and you end up with a^2 - c^2 + b^2 - c^2 = 0

Therefore a^2 + b^2 = c^2 + d^2.

The question told you that (mod p) = (mod q), and a^2 + b^2 = (Mod p) and (c^2 + d^2)= (mod q).

Therfore they are perpendicular. q.e.d.

subway_ie

Originally posted by Mutant_Fruit
Thats right. You then had to get pq (c-a)i + (d-b)j and dot product that with (p+q)= (a+c)i + (b+d)j

You then fill into your dot product formula, and get [(p+q).(pq)] / (mod p+q)(mod PQ) = CosO (just moving it around)

I dunno why, since what you said seems like the obvious thing to do, but I tried messing around with slopes, saying that the j value divided by the i value, when multiplied together should equal -1.... which they didnt. Well, not for me anyway.

PrecariousNuts

Oh, I managed the first part without giving them values using only my wit and cunning

Mutant_Fruit

well, you're mother was a hamster and your father smelled of elderberry! that what i tihnk of your cunning!

I used hard work to get that answer, about 10 minutes of it. (cos i didn;t spot the (mod P+Q) = (Mod PQ) for ages, i never read the question properly)

DS

Originally posted by Mutant_Fruit
These are not neceseraliy right, this is just what i got.

Q1
a) x^2 + y^2 +2x - 10y +13 = 0
b) (i) p=-6 (ii) (1,0) (4,3)
c) x^2 + y^2 + 6x - 6y +9 = 0 and x^2 + y^2 + 30x - 30y + 225 = 0 *corrected, wasn;t paying attention and just wrote +c *

Same here.

Q2
A) 0.324i + .946j (i decimilised, fraction answer= 12/37i -35/37j)
b) d= (a + 3c)/4 (ii) db = (15c - 3a)/4 (iii) prove it
c) Just prove it...

b ii is (3a + 9c)/4, has to be all positive due to direction of db

Q3 a) (3,2) I think
b) 3
ii -32
c) 3+- 2root2

Pretty sure c part ii is (3 +- root(13))/2

Q5
a) Prove it
b) 2 (ii) Sin2A
c) 5 (ii) 13 (iii) 45

All same.

Q6
a) 126 (ii) 60
b) -3[(-1/3)^n] (ii) 6
c) 24/105 (ii) 18/135 (iii) 432/3075 or could be 432/3675. I can;t read it.

Ditto apart from c part iii, for which I got 2/35. Can you explain your logic for that, I haven't seen 2 people with the same answer yet. I said that to get one of each colour is 8/35, from (i), and when you take one of each colour, the chances that 2 will be odd and two will be even are 2^2/2^4=1/4. 8/35 by 1/4 = 2/35. Seemed nice to me.

Q7
a) 8! (ii) (8C55 )(5!)
b) 3/8 (ii) 2/13 (iii) 27/110 (iv) 12/25
c) prove it

Same apart from b part iv I got 13/25, so did most people: (13C1 x 12C1)/25C2

Grand paper, few toughies but I pulled through. Pretty sure of an A1.

Mutant_Fruit

Ditto apart from c part iii, for which I got 2/35.

Well you got the same answer as me for part 1 and 2, so you're 9/10ths of the way there.

Part one made you get the odds of all different colours, part 2 made you get the odds that you get 2 odd and 2 even.

Part III asks for all different AND 2odd 2 even, therefore all you had to do was multiply the answer from part 1 and part 2!

Even if you got the wrong asnwers for 1 and 2, if you multiplied em, you still get the marks for 3.

EDIT: You're right, it is 13/25. I must have either gotten it wrong, or just wrote down the wrong answer on the question paper.

claire h

I am suddenly overcome with a great sense of despair concerning my mark for Paper 2. Maybe I just won't look at my maths grade when the results come out. Sounds like a plan...

DEmeant0r

Here are my paper II pass maths answers, so don't take them to be 100% right:

Q1:
(a) 24cm
(b): 1682.8 m
(c): (i) 2.1 m (ii) 0.81pie (iii) 0488666667pie (not sure if this is way off)

Q2:
(a): (8,1)
(b): (i) 5 (ii) 10.60660172 (again not sure) (iii) 3x-4y+16=0 (iv): shown (v): Not a parallelogram

Q3:
(a): (i): 6 (ii): xsquared + ysquared = 144
(b): (i) Verified (ii) Verified
(c): (i): (x+2)squared + (y-1)squared = 10 (ii): t=-1 and t=1

Q6:
(a): (i): 40320 (ii): 5040
(b): (i): 2925 (ii): 117 (iii): 205 (iv): 25
(c): (i): 1/24 (ii): 1/6 (iii): 1/32

Q7:
(a): 3.2 to the nearest one decimal place
(b): 400 (ii): graph (iii): 25 minutes (iv): 220 (v): 60

Q10:
(a): 1+4x+6xsquared+4xcubed+xtothepowerof4
(b): (i): 4 (ii): -16 (iii): shown
(c): (i): 2/3 (ii): shown

PrecariousNuts

I think we should stop discussing right now, with every post I read I feel my A1 slowly slipping out of my clutches.

DS

Originally posted by Mutant_Fruit
Well you got the same answer as me for part 1 and 2, so you're 9/10ths of the way there.

Part one made you get the odds of all different colours, part 2 made you get the odds that you get 2 odd and 2 even.

Part III asks for all different AND 2odd 2 even, therefore all you had to do was multiply the answer from part 1 and part 2!

Even if you got the wrong asnwers for 1 and 2, if you multiplied em, you still get the marks for 3.

Hmm. I actually put that down on my paper as a first attempt, and then wrote down that I was making a second attempt. They have to mark them both separately and give me the marks for the best one. Which is nice of them (it pays to read the marking schemes, my teacher never told me this and it's REALLY useful). I'm convinced it's not that simple though (as we hereby enter post mortem heaven), because the 2 events depend on each other don't they? If you take one of each colour, you can't, for example, take 2 reds and 2 yellows to make up your 2 odds and 2 evens, which you assume by using 18/35. The actual answer should be less probable, and 2/35 is roughly half as likely as 432/3675. Tricky question. Any other post mortemists want to add to this? C'mon, it's max 10 marks you won't lose any sleep over it

penguincakes

Originally posted by Discharger Snake
Ditto apart from c part iii, for which I got 2/35. Can you explain your logic for that, I haven't seen 2 people with the same answer yet. I said that to get one of each colour is 8/35, from (i), and when you take one of each colour, the chances that 2 will be odd and two will be even are 2^2/2^4=1/4. 8/35 by 1/4 = 2/35. Seemed nice to me.

I got 3 / 35 like so:

there are 6 ways of arranging odds & evens:

oeoe
oeeo
ooee

eoeo
eooe
eeoo

let's take the first case:

oeoe

Now there are 4 odds available, then 3 more evens of different colours, then 2 more odds, then 1 more even. ie, 4/8*3/7*2/6*1/5, which = 4!/1680.

There are 6 ways of arranging this, therefore

P = 6*24/1680 = 24/280 = 12/140 = 6/70 = 3/35.

Seifer

I was oozing with confidence before reading this; like everyone else so I must assume that a few of your answers are wrong.:p

Q1 I got the same

Q2 bii i got (3/4a + 9/4c) but I'm not worried cause I'm not counting it as I f00ked up the c part.

Q3 All the same except for c part. f(L): y(2-m)=x+c
f(K): y(2+m)=x were the equations I got and
m =+- root5

Q4 bii 152 degrees

Q5 I got the same

Q6 ci 8/35 iii 144/1225 After looking more closely I've discovered I got the same as you, you just didn't simplify.:)

Q7 I got the same

Q8 I got the same; my general term was x^2n/(2n)! anyone else get that?:)

claire h

Originally posted by Seifer
Q8 I got the same; my general term was x^2n/(2n)! anyone else get that?

I got that! Yay! I like your answers, they *must* be right. Also got the same for Q2 bii and Q4 bii.

DS

I got 3 / 35 like so:

That seems more solid than mine. Bloody probability. There's always 5 perfectly plausible solutions, and they all give different answers.

Q8 I got the same; my general term was x^2n/(2n)! anyone else get that?

That's the (n+1)th term. If you put 1 in there you'll get the second term. It's x^(2n-2)/(2n-2)! Not a big deal by the way, your ratio test was still valid.

penguincakes

Ah that depends if you start with u1 or u0 doesn't it? I *ahem* assumed that we were starting with u1 so I have 2^2n/(2n)! as well. I don't think they can mark you down for that...?

Seifer

I was working on the basis that n=0 was the first term...don't know why.

DS

They probably won't mark you down, but the first term is n=1. There is no u0, it's theoretical, it doesn't exist in the series. The value of the first term is 1.
So, test:
x^(2n-2)/(2n-2)! = x^(2-2)/(2-2)! = x^0/0! = 1 = first term
x^(2n)/(2n)! = x^2/2! = second term, when you sub 1 in for n, which isn't right. But it has no effect on the result because you don't care about the first/second terms, only the terms where n tends to infinity. So everyone's happy I think.

Mutant_Fruit

(2n-2)! = 1 when n=1... if only i knew that earlier, woul dhave made the question easier. I still got it out perfectly though

subway_ie

Originally posted by Mutant_Fruit
(2n-2)! = 1 when n=1... if only i knew that earlier, woul dhave made the question easier. I still got it out perfectly though

I can't believe I got that... maybe I should have guessed answers for more questions?

Mutant_Fruit

Originally posted by Seifer
I was working on the basis that n=0 was the first term...don't know why.

it doesn;t really matter whether you use n=0 as first term or n=1. they'll accept both answers.

Cathy

Originally posted by subway_ie
Did anybody else use the angle between two lines formula for Q5.(c) iii? I started to use it, ran out of time and just wrote down "Theta = pi/4" at the end...

I got 45 too, but I didn't use the formula - I said that the point of intersection of ag and fd was p. |ap|=|ag|=|pd|=13/2. |af|=|dg|=5.
By the cosine rule, |dg|^2 = |pd|^2 + |pg|^2 - 2.|pd|.|pg|.cos<dpg
Subbing in: 5^2 = (13/2)^2 + (13/2)^2 - (2)(13/2)(13/2)(cos<dpg)
25 = 169/2 - 169/2(cos<dpg)
169/2(cos<dpg) = 169/2 - 25
169/2(cos<dpg) = 119/2
169(cos<dpg) = 119
cos <dpg = 119/169
<dpg = 45 degrees (to the nearest degree)

Originally posted by Mutant_Fruit
well, you're mother was a hamster and your father smelled of elderberry!

Monty Python Always good.

Swifty

Thanks for completely destroying my confidence in my performance on this paper today. Thanks.. a.. BUNCH