Maths paper 1

PrecariousNuts

The induction question: 2^n >= n^2

test for 4

16=16... grand

assume true for k

2^k >= k^2

we want to get

2^(k+1) >= (k+1)^2

multiply the original by 2

2.2^k >= 2k^2

that becomes

2^(k+1) >= 2k^2

thus

2^(k+1) >= (k+1)^2

provided

2k^2>= (k+1)^2

this works out to have k = 1+- root two >= 0
and since you're told n>=4 this holds as true

Mutant_Fruit

yeah, i showed it was true for n=4, assumed it was true for k, then let n=k+1 and got a bit stuck there, and i did a bit of waffle, and then said this is true. Then a little more of the usual waffle of "Since its true for k+1, then it must be true for 5, therefore its true for 6... therefore true for all n E reals. And wrote Q.E.D. and left it. I'll get at least the attempt marks for that

PrecariousNuts

Originally posted by claire h
I could actually do the questions! Really wasn't expecting that.

Q2(c)(i) was tricky, I got k=3 but I wasn't sure about it.

Q3(c), got weird numbers in my matrix but I think my methods were okay.

Q5(a), couldn't do it, I had the general formula but it wasn't working out for me. Did this as an extra question though so I'm not too bothered.

Q6(c)(ii), didn't have a clue. The 'hence' threw me; couldn't see how you'd connect the two.

Q8(c)(ii), not so great.

Q2(c) is correct
Q3(c) same
Q5(a) very easy, I don't see how you could have gone wrong
Q6(c) is must admit that this was quite hard however I proved that 2 of the roots were imaginary and I hope it was sufficient.

3[(x+2)^2 + 1] = 0
(x+2)^2 = -1
+-(x+2) = i
x = -2 + i
x = -2 - i

hopefully thats in someway right

PrecariousNuts

and lets all pray that paper 2 is as easy

Mutant_Fruit

Q6(c)(ii),

That got me too, but not for long. For one real root, it can only cross the axis at one point, ok? Remember the way it asked you to get the derivitive of f(x). The derivitive = 0 at a max or min.

So all you had to do was set the derivitive equal to 0, and show that it couldn't equal 0 for any real value for X. That meant there was no maximum or minimum, and since there was no max or min, it was just like a striaght line to infinity, meaning it could only cross the axis once. Q.E.D.

PrecariousNuts

haha! brilliant I wrote those exact words down at the last minute for the sake of it!

DS

Loved that paper, 6 Q's done in an hour, did an extra one and then just went over them a few times, didn't bother with 3 for some reason. I wrote all my answers in my paper and so did my friend, and we got all the same answers, so I think we're both heading pretty close to 200 marks on paper 1, which is a bloody fantastic way to be relaxed going in for paper 2, which after the controversy last year can't be too bad.

Someone asked about the induction proof, the last bit was a bit tricky. Here's what I did starting from P(k+1):

2.2^k > (k+1)^2
2^k > 1/2(k+1)^2
Which is true if:
k^2 > 1/2(k+1)^2 -> from P(k)
2k^2 > (k+1)^2
2k^2 > k^2 + 2k + 1
k^2 - 2k - 1 > 0 -> cue me staring blankly at the page for a couple of minutes before thinking of completing the square
(k-1)^2 - 2 > 0
(k-1)^2 > 2
k-1 > root(2)
k > 1 + root(2)
which is true, as k > 4

Obviously I used > there instead of >= for convenience.

The only question mark for me was 6c(ii). I made a pretty good attempt I think, but I'm not sure if I proved it properly. Anyway:

You have an expression for f'(x) from part (i) which you have to use.
What I said was that f(x) having only one root means it only crosses the x-axis once.
So, the first thing you do for that type of question is find the turning points.
You should have gotten 3[(x+2)^2+1] for f'(x) yeah?
Well put that = 0
And you get (x+2)^2 = -1
x+2 = root(-1) -> unreal roots
Therefore there are no turning points (I couldn't get my head around this, I was so fixed in the notion that a cubic equation always has 2 turning points, never seen anything like this before)
It can only cross the x-axis once, because there are no turning points.
That's the best I could do.

As for Q1c(ii)...
It's clear from the "Hence", and from examining it a bit that p=a-x, q=b-x, and p+q=a+b-2x
So you have the same equation here, with one difference:
In your equation -3pq(p+q)=0
So sub in your values:
-3(a-x)(b-x)(a+b-2x)=0
Which is a terrifyingly simple cubic equation simplified for you (some people in my class went multiplying that out without realising the answer was right there)
So from that equation, it's clear that:
a-x=0 x=a
b-x=0 x=b
a+b-2x=0 x=(a+b)/2

Stench Blossoms

Originally posted by c0rk3r
Ordinary level Paper one answers,my answers. dispute away.

Q1 (a)6 (b)i 7000 ii 16660 iii44740 (c)i2703 ii2252.5 iii2297.5



5 (a)40,35,30,25,20

sorry but i think ya have them wrong... in q1 part (c)iii it asked how much MORE wud ya have to pay...

and in question 5 i think that bits wrong aswell

Raskolnikov

A friend showed my the this years paper, wow, it was even easier than the joke of a paper we had in 2002.

c0rk3r

aye 1 (c)iii is wrong! misread the question, probably lose 5 marks.no biggy. can afford to lose 60marks overall and still get an A1.

If 5 (a)i is wrong i'll take my own life! its not wrong

Crucifix

I was happy with the paper. I'm not saying I did well, but the paper could've been far worse.

Stench Blossoms

Originally posted by c0rk3r
aye 1 (c)iii is wrong! misread the question, probably lose 5 marks.no biggy. can afford to lose 60marks overall and still get an A1.

If 5 (a)i is wrong i'll take my own life! its not wrong

i dunno but if d=5 which is a positive how cum ut number decrease? doesnt that only happen wen "d" is negative??

i could be mistaken but i dont think i am...

conZ

Ordinary level paper was decent enough. I'd say I hit a B.

C0rk3r,
For question 1 c part 3, you got Q1 (c) €2297.50. That's wrong. The question went : "If the pump had registered the correct volume delivered, how much more would have been paid?"

You were meant to take the actual payment, €2252.50 from €2297.50, and give the difference.

c0rk3r

Originally posted by orla
i dunno but if d=5 which is a positive how cum ut number decrease? doesnt that only happen wen "d" is negative??

i could be mistaken but i dont think i am...

d=-5

aye conz as i said in a previous post i was mistaken although it was no huge mathematical blunder.May only cost me 5marks.

Mutant_Fruit

blunder=3 mark loss, and slip=1 mark.

popinfresh

I got the whole of Q8 C to work out, I was well chuffed
what u had to do for part b was sub in the f(-h) f(o) anf f(h). When u subbed in f(0) u gt=ot c=y2, then you make a simultanious equation out of the other 2. then 2a will work out as having a h^2 at the bottom which canceled out. Also I couldn't get 5 (c) to work ****in induction

Healio

ordinary level q1 easy marks= 50/600 x 100/1 = 8.33333%

Barbie_666

Originally posted by conZ
Ordinary level paper was decent enough. I'd say I hit a B.

C0rk3r,
For question 1 c part 3, you got Q1 (c) €2297.50. That's wrong. The question went : "If the pump had registered the correct volume delivered, how much more would have been paid?"

You were meant to take the actual payment, €2252.50 from €2297.50, and give the difference.

oh man.........

Seifer

I did 7 questions, got all the a's and b's. I know I got two of the c's completely right and I got one part at least of the others. I'd say I got an A on it but not sure if I made it to A1. Hope to secure that on paper 2 which is a hard thing to do.
I'm only doing maths this weekend as my next proper test isn't till wednesday; biology. Hate to have to study for geography too:p