Project Maths Sample Papers

lostatsea

The sample papers for 2013 are now online. My teacher said the big change was on Paper 1, Section C. There is no longer a choice. Last years pilot schools could do 2 out of 3 which was a choice of 2 differentiation and 1 integration for the Higher Level. Now it appears to be 1 diff and 1 integration. So anybody who thought they could leave out integration now have to do it.

Moody_mona

Have you a link, I can't find it.

lostatsea

http://www.examinations.ie/index.php?l=en&mc=sc&sc=ma

little sis...

lostatsea wrote: »

http://www.examinations.ie/index.php?l=en&mc=sc&sc=ma

oh my god they're horrible !

kevin12345

My teacher told us these wont be out until after the mocks? Oh well, I'll take a look at them now. Thanks OP.

Edit: Parts of Paper 1 makes me sad.

Slow Show

little sis... wrote: »

oh my god they're horrible !

Aren't you in 5th year? of course they're going to look horrible so, chill!

I think they look OK...definitely some tricky parts and I really hate how in the long questions one question leads onto the next so if you can't do the first you might not be able to keep going. But overall it's not too terrifying, the first part looks kinda lovely, the contexts and applications a fair bit less lovely but it could be worse, and the functions and calculus is all good. Talking about paper 1 here seeing as I had a fair idea of what P2 would be like already.

little sis...

Slow Show wrote: »

Aren't you in 5th year? of course they're going to look horrible so, chill!

yeh but I should be able to do SOME of it like hopefully my view will change by next year!

BrittneyC_xo

I honestly really really like the paper for Paper 1, Ordinary level. I'm a repeat and so I did the old paper this year, and this sample paper looks so much more easier.

JackTheGrinder

does anyone know if there is a marking scheme? or alternatively anyone got the answer to question 3 part b and c paper 1? youd be doing me a big favour!

MathsManiac

JackTheGrinder wrote: »

does anyone know if there is a marking scheme? or alternatively anyone got the answer to question 3 part b and c paper 1? youd be doing me a big favour!

Higher, Ordinary or Foundation?

MathsManiac

If it's higher, then to do part (b), you divide the known factor (x+k) into the f(x) to get a quadtratic. (You should get x^2 -kx +1). Then you can use your favourite quadratic-solving method to find the remaining two roots in terms of k.

Then, for part (c), think about what it would take to make these two roots not be real (since you know -k is going to be a root from part (a)).

JackTheGrinder

MathsManiac wrote: »

If it's higher, then to do part (b), you divide the known factor (x+k) into the f(x) to get a quadtratic. (You should get x^2 -kx +1). Then you can use your favourite quadratic-solving method to find the remaining two roots in terms of k.

Then, for part (c), think about what it would take to make these two roots not be real (since you know -k is going to be a root from part (a)).

Hey thanks for replying man, I have been looking at part b and trying long division for like half an hour now with no luck though.. shouldnt the remainder be zero? I keep getting k^3+k^2 x as the remainder.. frustrating. I dont think im doing the divison wrong but maybe I am... is there something im missing? you just divide x+k into x^3+(1-k^2)x +k, right? and the answer should be a quadratic?

little sis...

JackTheGrinder wrote: »

Hey thanks for replying man, I have been looking at part b and trying long division for like half an hour now with no luck though.. shouldnt the remainder be zero? I keep getting k^3+k^2 x as the remainder.. frustrating. I dont think im doing the divison wrong but maybe I am... is there something im missing? you just divide x+k into x^3+(1-k^2)x +k, right? and the answer should be a quadratic?

you dont actually get zero you make whatever your remainder is EQUAL zero.

MathsManiac

little sis... wrote: »

you dont actually get zero you make whatever your remainder is EQUAL zero.

That's not correct. You should get 0 when you do the division, because we already know that x+k is a factor.

JackTheGrinder wrote: »

Hey thanks for replying man, I have been looking at part b and trying long division for like half an hour now with no luck though.. shouldnt the remainder be zero? I keep getting k^3+k^2 x as the remainder.. frustrating. I dont think im doing the divison wrong but maybe I am... is there something im missing? you just divide x+k into x^3+(1-k^2)x +k, right? and the answer should be a quadratic?

The fact that there' no x^2 term in the cubic mught be causing you to make a mistake in your long division. Try writing it as x^3 + 0x^2 + (1-k^2)x + k, and see how it goes.

JackTheGrinder

MathsManiac wrote: »

That's not correct. You should get 0 when you do the division, because we already know that x+k is a factor.




The fact that there' no x^2 term in the cubic mught be causing you to make a mistake in your long division. Try writing it as x^3 + 0x^2 + (1-k^2)x + k, and see how it goes.

yep that did it! thanks a million man! now part c hmm... btw you a student?

JackTheGrinder

JackTheGrinder wrote: »

yep that did it! thanks a million man! now part c hmm... btw you a student?

sorry... just to confirm... what were the answers for part b in the end.. i got x^2 -kx +1 after dividing. Im using the formula to be safe and it is giving me indefinite answer... sorry im a little rusty on algebra how do I get the answers??

MathsManiac

JackTheGrinder wrote: »

sorry... just to confirm... what were the answers for part b in the end.. i got x^2 -kx +1 after dividing. Im using the formula to be safe and it is giving me indefinite answer... sorry im a little rusty on algebra how do I get the answers??

The two remaining roots are (k +/- sqrt(k^2 - 4))/2.

These roots will be complex if the thing inside the square root is negative, real if it's not.

In answer to your other question, I'm not a student; I used to be a teacher.

JackTheGrinder

MathsManiac wrote: »

The two remaining roots are (k +/- sqrt(k^2 - 4))/2.

These roots will be complex if the thing inside the square root is negative, real if it's not.

In answer to your other question, I'm not a student; I used to be a teacher.

Ah ok thank you, thats a big relief to get that q sorted! I am looking at c at the minute... so in order for the root to not be complex k would need to be equal or grater than 2? is that correct? how do I get the answer from this?

JackTheGrinder

ah k would need to be +/-2 for there to be exactly one real root is it? because the squared term is zero then and +/- 0 is the same thing?

MathsManiac

You have the right idea now, but remember that you started with a cubic that definitely has at least one real root, namely -k. So you want the quadratic part to produce no more real roots. For this to happen, you need (k^2 - 4) to be less than 0. So k^2 < 4. So |k|<2. That is, -2<k<2.

JackTheGrinder

MathsManiac wrote: »

You have the right idea now, but remember that you started with a cubic that definitely has at least one real root, namely -k. So you want the quadratic part to produce no more real roots. For this to happen, you need (k^2 - 4) to be less than 0. So k^2 < 4. So |k|<2. That is, -2<k<2.

ah ok so if the square piece is negative, it isnt a real root... and k cant be 2 or -2 because if the square gave an answer of 0, this would still make it a real root? so it has to be between -2 and 2... think I am getting this now, thanks a million man I am definitely clearer on it anyway! got a shock when I looked at the paper first, not as bad now I m trawling through it slowly

JackTheGrinder

apologies if this next question comes across overly basic but inequalities have always been kind of grey for me... for question 4 part (b)(ii), what exactly is an inequality? I guess from the graph the answer is 1<x<5 but how is this mathematically worked out? the "greather than" sign can confuse me...

Moody_mona

Those lines mean absolute value, so you're always referring to the positive aspect of what's inside it. With the inequality you're given you'll find a value that's less than 2, but you have to also find the value for when it's greater than -2.

Alternatively you can square both sides, both sides are positive so that won't affect the inequality sign, and then solve the resulting quadratic. As a rule, if there's a < sign in your quadratic or absolute value inequality, your answers will be an "inbetween" answer, whereas if there is a > sign, your two answers will be individual and "outside".

Inequalities are very similar to equations but instead of me saying x=4, if I say x<4 it means x can be any number less than four. An INEQUALity is just that, when they're not equal.

JackTheGrinder

Moody_mona wrote: »

Those lines mean absolute value, so you're always referring to the positive aspect of what's inside it. With the inequality you're given you'll find a value that's less than 2, but you have to also find the value for when it's greater than -2.

Alternatively you can square both sides, both sides are positive so that won't affect the inequality sign, and then solve the resulting quadratic. As a rule, if there's a < sign in your quadratic or absolute value inequality, your answers will be an "inbetween" answer, whereas if there is a > sign, your two answers will be individual and "outside".

Inequalities are very similar to equations but instead of me saying x=4, if I say x<4 it means x can be any number less than four. An INEQUALity is just that, when they're not equal.

cheers, clears that up a lot

leinster99

What question number is algebra and statistics?

JackTheGrinder

leinster99 wrote: »

What question number is algebra and statistics?

stats is most of paper 2, algebra is everywhere?