Post your Maths Answers (higher)

Nanaki

Cokehead Mother wrote: »

What do you mean? You let u = x^3, so du/dx = 3x^2 and then you multiply that by 1/2rootu.

The quick way is to you use your junior cert knowledge of indices and then differentiate. It's just x to the power of 3 halves.

I wrote it ax x^(3/2)

orangetictac

What did y'all get for 4c)???

orangetictac

tabouli wrote: »

4)a) 3

c)ii) 2n^3 +4n^2 +7n +2^n -1

I think i got 2n^3 +4n^2 +7n +2(2^n -1)
???

King Ludvig

Q1(a) 2/(2-x)

Q2(a) (x+5)^2 +7
(b)(i) 47

Q3(b)(i) (1-i)

Q5(a) -2<x<5
(b)(i) x=-3 and x=9
(ii) x= -2 and x=11

Q6(a) (1/2)(x)^-3/2
(c)(i) -36

Q7(a) 2+2cos2x
(b)(i) (-5x-3y)/(3x+5y)

Q8(a) 2+(sin3x)/3
(b)(i) e-1
(c) 22.92 units squared

tabouli

orangetictac wrote: »

I think i got 2n^3 +4n^2 +7n +2(2^n -1)
???

Not sure, I tried to use Sn of a GP for that part, but got confused.

Calorimeterman

question 8 c, is around about 10 units, I drew it out, and I can draw graphs well, I got 6 in the calculation, but by counting the squares I managed to get it to about 10.5 units (give or take)

orangetictac

well i got 32/3 so ur not 2 far off i guess

Cunning

a/b+b/a = (a^2+b^2)/(ab)

now let b/a=C so that b=C.a , substitute C.a for b

now we get (a^2+C^2.a^2)/(a.C.a)

cancel the a^2 above and below the lines to get
(1+C^2) / C

let this = X, which is the anser to the sum a/b+b/a

now X.C = 1+C^2

rearrange for a quadratic
C^2 - C.X + 1 = 0


using the standard formula C= -X +/- sqrt(X^2 -4) all over 2

Remeber that b=C.A and that a and/or b must NOT be complex
now look at the parts under the square root
in order for C not to be a complex/imaginary number X^2 > 4
thus X must be < -2
or X must be > 2

nice
B2 in LC pass maths 2001

padkav

you could also get the answer in the form 3/2*square Root x, same as yours!

Peleus

Cunning wrote: »

rearrange for a quadratic
C^2 - C.X + 1 = 0


using the standard formula C= -X +/- sqrt(X^2 -4) all over 2

you got it right but [C^2 - C.X + 1 = 0] isn't a quadratic. you're missing an X squared.

Peleus

Cokehead Mother wrote: »

What do you mean? You let u = x^3, so du/dx = 3x^2 and then you multiply that by 1/2rootu.

The quick way is to you use your junior cert knowledge of indices and then differentiate. It's just x to the power of 3 halves.

ye, sorry you can use substitution. i just thought you couldn't, in a moment of silliness.

t-mobile1892

Woah im in JC nd im just lost readin dis......

Cokehead Mother

Peleus wrote: »

you got it right but [C^2 - C.X + 1 = 0] isn't a quadratic. you're missing an X squared.

Equations can have variables other than x!

fivetwenty

BTW - Did everyone here reach a stage of depression in this exam?

Mine came at the way too early 1(B) - I tried f(0),f(1),f(-1),f(2),(f-2), and said "**** this - It's never went this high up and made sense" . . . Lo and behold, 3 suddenly became my favourite number

Cokehead Mother wrote: »

Yep. Did it the exact same way and got n = -1 like you. Messiest question ever.

Yes! I did the same, except forgot to finish it off, I just realised no values they wanted would work, so stupidly assumed the examiner would!

ZorbaTehZ

Cunning, how can you use the quadratic formula when you have two variables in there, if you write it as an explicit function you can see that it isn't a quadratic. Plus I don't think you're taking into account the situation where a and b have opposite signs.

Peslo

Peleus wrote: »

[C^2 - C.X + 1 = 0] isn't a quadratic. you're missing an X squared.

Is this guy doing HL LC Maths??

For pete's sake loike.

Peslo

fivetwenty wrote: »

BTW - Did everyone here reach a stage of depression in this exam?

Mine came at the way too early 1(B) - I tried f(0),f(1),f(-1),f(2),(f-2), and said "**** this - It's never went this high up and made sense" . . . Lo and behold, 3 suddenly became my favourite number



Yes! I did the same, except forgot to finish it off, I just realised no values they wanted would work, so stupidly assumed the examiner would!

You are supposed to use one of the factors of the term indepdant of x!!!!
it was 9, so 3 should have been the first one you try! if not after 1 and -1 like!

Cokehead Mother

ZorbaTehZ wrote: »

Cunning, how can you use the quadratic formula when you have two variables in there, if you write it as an explicit function you can see that it isn't a quadratic. Plus I don't think you're taking into account the situation where a and b have opposite signs.

I don't understand what you mean. He's just solving a quadratic in terms of a/b. Why can't you do that?

His solution works on the basis of a and b being any real numbers so I think he's got the possibility of them having opposite signs covered.

fivetwenty

Peslo wrote: »

You are supposed to use one of the factors of the term indepdant of x!!!!
it was 9, so 3 should have been the first one you try! if not after 1 and -1 like!

Ooooh, Ah well, s/he can't take anything off me now!

ZorbaTehZ

Well if you have a function like this x^2 + xy - 3 = 0, could you use the quadratic equation on that? You need to write it explicitly, ie y = (something) - ie it would need to look like y = ax^2 + bx + c. If you think about an equation like cunning described, it wouldn't even look like a typical quadratic, explicitly it would be something like x = 1/C + C, which I think would be asymptotic.

Regarding the as and bs, if they are of opposite signs, then you have negative on the left -a/b-b/a, there's a solution given either on this thread or the other one which takes into account.

Cokehead Mother

I'm certain his solution is fine... 1:30am is a bad time for this. I'll show it to an alert american or something... I want a second person to tell me it's wrong.

Peleus

Cokehead Mother wrote: »

Equations can have variables other than x!

lol, i know, the way he wrote it I thought it was a quadratic in X not C.

SimpleLogic

could someone post the answer for for 5C

stev2604

Does anyone know when the marking schemes come out?

Peleus

stev2604 wrote: »

Does anyone know when the marking schemes come out?

I'm pretty sure they come out when the results are out. The marking schemes are definately not made until after the exams. theyre made over the summer and are probably available to the public after the results. maybe before, but not anytime soon.

Cokehead Mother

Burner- wrote: »

could someone post the answer for for 5C

Well if a, b and c are consecutive terms in a geometric sequence, then b/a = c/b.

So you get this



And then go on a cancelling spree and end up with n = -1, which obviously isn't a natural number.

Challenged

The Algebra solutions (Questions 1 & 2) have just been posted up:

http://www.studentxpress.ie/papers/algsoln2008.pdf

Challenged

The solutions to Questions 4 & 5 have just been posted up:

http://www.studentxpress.ie/papers/seqsoln2008.pdf

Cunning

Peleus wrote: »

you got it right but [C^2 - C.X + 1 = 0] isn't a quadratic. you're missing an X squared.

ermmm i think you'll find its a quadratic.
not in x but in C?
ppl often have difficulty with this for some reason

MathsManiac

Cokehead Mother wrote: »

I'm certain his solution is fine... 1:30am is a bad time for this. I'll show it to an alert american or something... I want a second person to tell me it's wrong.

Agreed, and a very nice solution it is too.

Well done, Cunning! Thou art veritably eponymous!