Higher Mathematics : Paper Two Opinion

Thomas_S_Hunterson

cocoa wrote:

First of all, thanks a lot for typing it out, I know it can be a bit of a pain.
I'm with you up to here but I would have said u=1+x^2 (as you did) and integration of 1/u=lnu not 1/lnu
Can you tell me what the twist is?

It should read (1/2)(lnu), that should make it clearer

charlie1234

ah yeah, i see what you mean, its just the way i wrote it, thats (.5)(ln u) not 1/2lnu, sorry should've made that clearer

cocoa

well in that case, I got it right! Thanks again.

carlowboy

ere it is, we had it as part (a) too lol

u = tan-1 x
du = (1/1+x^2).1.dx

dv = 1.dx
v = x + c

Sudv = uv - Svdu

Stan-1xdx = (tan-1x)(x) - S(x)(1/(1+x^2))dx

F*** I got it down to there but i said it integrated to tan^-1 x

T-b0n3

Didnt see any one post the answers to 7b is 5/18 and 7/18 right??

Adam

carlowboy

can anyone explain to me how I was wrong as I'm willing to concede I was.

JC 2K3

What question?

cocoa

T-b0n3 wrote:

Didnt see any one post the answers to 7b is 5/18 and 7/18 right??

Adam

5/18 and 11/36.
part (ii) there may be another way, but i just drew a sample space and found that 11 out of the 36 options fitted the conditions. they were:
(3,6)(4,4)(4,5)(4,6)(5,6)(6,6)(5,5)(6,3)(5,4)(6,4)(6,5)

carlowboy, 1/(1+x^2) integrates to tan^-1(x) but x/(1+x^2) does not.

carlowboy

Uh oh...I just realised.....

I got tan^-1x instead ln(1 +x^2) but I see th problem as it was du not dx. A blunder perhaps?

analyse this

Help me out here people!!

8(c)

Integration of tan^-1(x) dx

u=tan^-1(x)
tanu=x
d/dx (tanu=x)=sec^2(u) du/dx=1
du/dx=1/sec^2 u
du/dx=cos^2 u
= (1/x^2 + 1)
du=dx/x^2+1

dv=dx
v=integration of 1dx
v=x

=>Integration of tan^-1(x) dx= xtan^-1(x) - integration of x/x^2+1 dx

integration of x/x^2+1 dx= integration of x dx by integration of 1/x^2+1 dx
= (x^2/2)(tan^-1x)

=>integration of tan^-1 x dx= xtan^-1(x)- (x^2/2)(tan^-1(x))

[xtan^-1(x)-(x^2/2)(tan^-1(x))] with limits 1 and 0

= tan^-1(1) -(1/2)tan^-1(1)-0
=pi/4-pi/8
=pi/8


What did i do wrong??!!:(

cocoa

analyse this wrote:

integration of x/x^2+1 dx= integration of x dx by integration of 1/x^2+1 dx
= (x^2/2)(tan^-1x)

right there. you can't integrate the two separately, it's not allowed. (don't ask me why, it's one of the basic facts of integration)

analyse this

darn!!

Shox

Does anyone know how much i'd lose in question 8 part (b) for thinkin the general term is Ur not Ur+1. . .

Think there was 10 marks for that bit.

MadSambuka

who cares what the solution is! you're never going to have to do them ever again..

JC 2K3

None, I don't think it really mattered in that question tbh.

Like, you get the same answer by evaluating lim n->inf ¦Un+2/Un+1¦

seandoiler

Shox wrote:

Does anyone know how much i'd lose in question 8 part (b) for thinkin the general term is Ur not Ur+1. . .

maybe i'm missing something and some one can fill me in but from what i know the term given in the paper is the rth term and hence the general term of the series......i'm not sure why people think that it is the (r+1)th term...my only guess is that people do not start their series from zero and have a problem referring to the zeroeth term...am i right in thinking this??

however as JC said it really doesn't matter, which terms you take as long as you take the limit as r -> infinity for the ratio of two consecutive terms

cocoa

seandoiler wrote:

maybe i'm missing something and some one can fill me in but from what i know the term given in the paper is the rth term and hence the general term of the series......i'm not sure why people think that it is the (r+1)th term...my only guess is that people do not start their series from zero and have a problem referring to the zeroeth term...am i right in thinking this??

however as JC said it really doesn't matter, which terms you take as long as you take the limit as r -> infinity for the ratio of two consecutive terms

you've summed it up pretty well there, and I suppose it is correct to say that the general term is either Ur+1 when r starts at 1 or Ur when r starts at zero. Just to let you know, the reason why people use Ur+1 and r starting at 1 is because this avoids the confusion of having r start at different places in different series.

seandoiler

cocoa wrote:

you've summed it up pretty well there, and I suppose it is correct to say that the general term is either Ur+1 when r starts at 1 or Ur when r starts at zero. Just to let you know, the reason why people use Ur+1 and r starting at 1 is because this avoids the confusion of having r start at different places in different series.

cheers cocoa....i never really had any problems with series starting in different places, just always start at zero and alter my general term to suit...thanks for clearing that up....to the other poster, you will not loose any marks at all

cocoa

yay, I finished making a full list of fleshed out solutions check it out: solutions.
whoa, looking back, that was quite the OCD of me... :S

seandoiler

cocoa wrote:

yay, I finished making a full list of fleshed out solutions check it out: solutions.
whoa, looking back, that was quite the OCD of me... :S

fair play...mistake in question 2 (c) (i) though !!

SamHamilton

Anyone know if the marks for Irish are the same every year?

carlowboy

SamHamilton wrote:

Anyone know if the marks for Irish are the same every year?

Ba chóir go mbeadh.

cocoa

seandoiler wrote:

fair play...mistake in question 2 (c) (i) though !!

not disagreeing with you here, but could you tell me where? (I'm also assuming you mean 2(c)(i) paper 2...)

or you could just edit it yourself and fix it

JC 2K3

You're overcomplicaing things a bit in P1 Q8 (b)(i) IMO, why not take u = x^2 + 9 ?

adam_ccfc

JC 2K3 wrote:

You're overcomplicaing things a bit in P1 Q8 (b)(i) IMO, why not take u = x^2 + 9 ?

Yeah, that's what I did. Easiest way of going about it.

SamHamilton

Hey! For Q.8 on paper 2 you take The general term as Ur+1 and use Ur when proving that the series in convergent. I took the general term as Ur and used Ur+1 with that.

My Ur+1 was

m(m-1)(m-2)...(m-r+1)(m-r+2)Xr+1/r+1!

Does that matter? I got -x at the end of it all and pu it in between -1 and 1 and multiplied it all by -1 and got -1<X<1.

Is that acceptable?

Nehpets

JC 2K3 wrote:

You're overcomplicaing things a bit in P1 Q8 (b)(i) IMO, why not take u = x^2 + 9 ?

I did it that way too

cocoa

JC, you're completely right, I just never really did that much integration, I'll change it now...

Samhamilton. If you took what they gave you as Ur then your Ur+1 should have looked like this:
(m(m-1)(m-2)...(m-r))/(r+1)! all by x^(r+1)

you probably made another mistake later on and the two cancled out. (in fact, there is such a "double mistake" in texts and tests 5) Depending on how vigilant your examiner is, you may or may not lose marks for it.

SamHamilton

(m(m-1)(m-2)...(m-r))/(r+1)!

That's what I thought but then I saw somewhere, probably in Texts and Tests 5 that the first "r part" was left in it with another "r part" + 1 put after it. I didn't look at series at all but I assumed tha, because it was a series, the first "r part" should stay there and the "r+1 part" put after it. Isn't there something like that in induction of a series?

Also, it didn't help that the first time I saw the maclaurin series of the binomial expansion was the night before the exam!:D (Thanks boards.ie predictions!)

Bluefox21

lol I took n=m so and then Um and Um+1 thought it was phrased badly