Higher Level Maths 2009(Paper 1): How was it for you?

challengemaster

was nice enough alright - all I can say is, fcuking simultanious logs!

I managed to get 2c out fully, if its right or not is another question

From what I can recall, my answer to c (i) was r = (q²-4q²n²p³) / 2n²p² or along those lines. Tbh, my brain is half-fried lol.

Elli-May

whadabouchasir wrote: »

All you had to do was to show that the root is between 2 and 2.5 . Exactly the same as part(i) you just use 2.5 instead of 3.

I did that!
I was all confused coz I thought you had to use the newton raphson method

Fair paper all in all I think

How did you guys show that the tangents couldn't be perpendicular

whadabouchasir

Delta Kilo wrote: »

I used the minus b formula and let one root equal n times the other root and tried to solve.

I tried that too,but it just ended up in a big mess,sum and product simultaneous equations didn't give me much joy either.

Fringe

Did really good. It was tricky but I made it. Definitely got an A.

2(c) was a stupid question. Ended up with something ridiculous. I figured out how to do (ii) but I wasn't bothered since it was the seventh question I was doing. I think you get r, put n and p in. Then you say sqrt(q^2 - 4r) > 0. Sub in r and you can then get the range of values for q. I didn't get a nice quadratic though so I moved on.

4(c) was also really tricky. It was a nice question though.

5(c) was very awkward. Took me so long to figure out how to change the base9. I ended up messing with logs for a good while until I got it.

7(c) was badly phrased. They added way too much information to make it seem cool and trendy like all the cool kids use Newton-Raphson today. Turns out it was much easier than it appeared to be.

SligoBrewer

Why should 7C be easier marked?

The question was basically spelled out for you.

Fringe

Elli-May wrote: »

I did that!
I was all confused coz I thought you had to use the newton raphson method

Fair paper all in all I think

How did you guys show that the tangents couldn't be perpendicular

Differentiate the equation. You get something like -1/(x-1)^2.
This means that the slope will always be negative so if you multiply them, they'll never equal -1.

Acid_Violet

Fringe wrote: »

5(c) was very awkward. Took me so long to figure out how to change the base9. I ended up messing with logs for a good while until I got it.

How'd ya manage that? What was your answer?

amacachi

For 2cii I got q<0, q>0. Have a feeling that'll be wrong.

Also I decided during the paper that 25+1=24 and only realised when I was in the car home. :mad:
Also I didn't read a question properly and didn't draw a graph, but got the rest right.
Looked over volume of a cone last night but couldn't remember it.

All in all I got 75-80%, no chance of the A1 I pretty much needed now.

Acid_Violet

SligoBrewer wrote: »

Why should 7C be easier marked?

The question was basically spelled out for you.

They asked you that Anne's root is less than 2.5 but it was 2.66, which in turn was greater than Barry's. I know I'd my method right.

ayapatrick

Elli-May wrote: »

I did that!
I was all confused coz I thought you had to use the newton raphson method

Fair paper all in all I think

How did you guys show that the tangents couldn't be perpendicular

i think ya had not sure?
i thin we had to get the second approx for barry which i didnt fekin do:( otherwise think it went well!:D hope fully between 80 and 90 % all goin well! didnt get 2 c though! as someone mentioned i used the -b formula on it!

timmywex

Elli-May wrote: »

I did that!


How did you guys show that the tangents couldn't be perpendicular

I diff'd the formula and got somthing over (x-3)^2


Then i said since a square will be positive, its not possible to get a negative slope, thus its not possible to have m1.m2=-1

amacachi

timmywex wrote: »

I diff'd the formula and got somthing over (x-3)^2


Then i said since a square will be positive, its not possible to get a negative slope, thus its not possible to have m1.m2=-1

I just said dy/dx times dy/dx = whatever it was and it ended up that it couldn't be a negative number.

Fringe

Acid_Violet wrote: »

How'd ya manage that? What was your answer?

I did something weird by reasoning that log9 is 1/2log3. I did it by playing with log10 and log100 and after experimenting, found out.

I only realised now though that an easier way is just to use change of base to get log3 x/log3 9 which is 1/2 log3 x.

Nihilist21

I've approximated that I got in the region of 90%, I can't see how I'd get less than 85% - since I had time to go over all my answers. Anyhoo here's a breakdown of how it was per question.

1 = Got it all except C, which I did via the right method just must have made a slip (so should get most of the marks).

2= All except C (ii), clearly others had problems with this too.

3= All of this right I think, I wasn't sure about C (ii) but my friend and I both did it the same way, using the equation for sin3a in the prior question.

6= All correct, except for perhaps a small slip - unsure.

7= Everything up to C (i) was fine, then (ii) and (iii) I attempted.

8= Everything right, was a lovely integration question. So happy the volume of a cone came up instead of an area question.

Delta Kilo

whadabouchasir wrote: »

I tried that too,but it just ended up in a big mess,sum and product simultaneous equations didn't give me much joy either.

Ya its really annoying when you know you have the correct methods but you end up with a big jungle of letters and your like: Wha?

SligoBrewer

Acid_Violet wrote: »

They asked you that Anne's root is less than 2.5 but it was 2.66, which in turn was greater than Barry's. I know I'd my method right.

There you had to sub in 2.5 into f(x) and prove that the roots crossed before 2.5

What you did there was for part iii

sheehy898

Wasn't too bad, I didn't get an A or anything but maybe a B, at worst a C.I didn't like 4c or 2c:)

kevogy

1a) cross multiplyd got 6x = 9y whatever that ment
1c) divided in in disaster

2b) complete guess
2c) complete guess

3c) the bastards and their de moivre

6) ok
7c) got anne = 2.6667 anthough it wax supposed to b less than 2.5

8) attempted disaster


26 % out the entire maths according to me

Acid_Violet

Hang on..... Maybe I misread 7c(ii)...

Were they looking for you to show that f(2)< 2.5 whilst f(3)>2.5?

whadabouchasir

Fringe wrote: »

Did really good. cool and trendy like all the cool kids use Newton-Raphson today.

Do they not?

Acid_Violet

Oh lerd..... Well at least I got 15/20 for that question so....

I'd say most students had the same prob as me so they might give attempt marks just for the craic. Let's hope I made few enough slips to ensure >B1

MrPain

i found it fine, but i didnt get 7c, but everything else worked out

rcaz

20 marks for a Cone!! Gift!!

I'm really pleased with it! I know I got one of those little diagrams in Q2 B wrong (I said 'irrational roots' instead of 'complex roots' :mad:), but other than that, simple stuff I think!

The other one that tripped me up most was 3B, A^17, but my answers were the same as the answers some other guys got, so maybe we're all right? Or all wrong...

My teacher was looking through the paper almost laughing at how easy some of the questions were

If paper two is anything like that, I think I'll be in A territory

OxfordComma

I was fairly happy. I would've preferred the 2008 or 2007 paper 1 but it was grand. They don't normally give algebra questions that are that hard but what can ya do...

Question 2(c) was dire! I got nq^2/p(n^2+2n+1) ...probably wrong. That's the worst part (c) i've seen in algebra in a long time!

Loved all 3 calculus questions...integration was shockingly easy! 7(c) was poorly phrased but not particularly challenging really.

I got really lucky in complex numbers. I did part (c) (i) literally just before I went to bed last night! The same question came up a few years ago... Didn't know they could put an integral in a complex number question tho!!

Overall, fairly happy... Kinda need an A in maths for points so fingers crossed!

Synods

2c said set of values as opposed to range.....does that mean if it was just two values it's ok? got q=+/-6 root of r over5....

Mongey

With Regards to 7 c ii)
It asked to show that Anne's starting approximation is less than 2.5.

Given that her starting approximation was 2, 2 is less than 2.5

Then iii) asks you to show the X2 of Barrys is closer to the root than Annes X2

So it was a trick question, was it not?

HQvhs

For 7c) ii) I subbed Anne's approximation and Barry's approximation into f(x) and as Anne's was closer to 0 than Barry's I said it was more accurate. And then same method for part iii). Not sure if this is right, everyone else seems to have done different things!

This is because the f(root) = 0.

All in all, very fair paper. A bit more practical than usual but that was good. Not sure how I did.

dermo1990

Question 1: grand...b ii) wasn't completely sure on

Question 2: was alright except for the infamous part c ii). As far as I know no one in my class got it.

Questions 5: Full marks defo. Part b was a pain having to muliply out (k+1)^5

Question 6: was alright, forgot to get around to cii). That part was probably worth 5 or so marks anyway

Question 7: Went well, part c was horrendously phrased.

Question 8: ok, wasn't sure on part c, still attempted it though

GA361

Qs 2(c) and 3 were quite a bit harder than I had expected . . . . Qs 6 and 8 were nice enough . . . but I triped up on 7(c) . . . either because it was incredibly easy or more likely it was badly phrased . . .

SligoBrewer

Mongey wrote: »

With Regards to 7 c ii)
It asked to show that Anne's starting approximation is less than 2.5.

Given that her starting approximation was 2, 2 is less than 2.5

Then iii) asks you to show the X2 of Barrys is closer to the root than Annes X2

So it was a trick question, was it not?

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:rolleyes: