Higher Mathematics : Paper Two Opinion

Gangsta

ah lads that was soft I'm pretty sure I got everything right except Q4c(ii) and I think I ballsed up the probability proof!! Hopefully Still an A1.

JC 2k3 would you mind posting the solution to Q4c(ii) or anyone else who got it? Much appreciated. I spent an hour on that and I couldn't do it!:(

Bluefox21

JC 2K3 wrote:

Answers:
1(a):
x^2 + y^2 - 10x - 24 = 0


3(a):
3/2 (I think I got this wrong, fúck)
oh and 7b part two thought that was ten over 36....

3/2 looks right and would I get the marks for writing (X-5)"2+(Y^2)=49 for part 1

Tomlowe

yeah it was

k(ax + by + c) + l(px + qy + r)=0

px + qy + r =0 is only possible when k=0.

for the b part i said let x1, y1, be the coordinates that satisfy both equations, ie make both equns = 0, then sub in x1, y1 into the K and L equation and got K(0) + L(0) = 0

Theres nothing wrong with that but i have a feeling im not going to get the marks... i did it anyway cos i owned the other 5 section A questions i did... ****ers

SamHamilton

My Answers: (thanks JC2K3)

Answers:
1(a):
x^2+y^2 - 10X + 25

(b):
(i) r1 = 2root2, r2 = root2 and d = root2
Since d = r1 - r2 they touch internally
(ii) (4,5)

(c):
X^2+y^2-6X-10y+9

2(a):
-8i -3j

(b):
(i) ab.b is not equal to zero
(ii) 5root3/12

(c):
(i) t(4i + 7j)
(ii) p.r = 40t ; q.r = 104t
(iii) <rop = <roq therefore r is on the line that bisects poq

3(a):
3/2

(b):
(i)yeah
(ii)1:2

(c):
(i)Get point of intersection, plug it in and it = 0
(ii)9k + 7l = 0
(iii) A big white space on my page

4(a):
Yeah

(b):
x = 30, 150, 199, 341

(c):
(i)2r^2 + 2r^2Cos^2(90-alpha)! Don't even ask...
(ii)With my answer from (i) it was a nightmare!

5(a):
2/3

(b):
First part fine. Forget the proof for Sin(A+B)!

(c):
120 yay! In the last 5 minutes went over it and saw a - where there should have been a +! That was the best moment of my life.

8(a):
1/2

(b):
Isn't the Ur they give wrong? Anyway, I think I got it. It all cancelled to -X and then I put it between brackets, changed the signs and it worked out.

(c):
pi/4 - ln2/2

All in all it was a beautiful paper.

Adam We

did anyone else do all 8 Questions??

i did on both papers. but i'm worried i've fallen to a B1. Few iffy mistakes here and there

And that fecking binomial on Question 8. Only found out it was on the course this morning. Dunno if i got it out right or not, think i was a bit off

chas_88

Shox wrote:

DAMN!! for 4 part (c) (i)

i thought it was the length of the arc ab ya had to measure... dats defo a 0

What did ye get for the real number k in the vector question??

i got k=5Root3/12 ????? Dont think this is rite

i think it is, that's what i got anyway!

WildCardDoW

For anyone else doing enginerring that is worried, here's an last year Maynooth exam Q1:
1.
(a) Find the three roots of the equation x3-4x2+9x-10=0 which has one integer root.

(b) Solve the following simultaneous equations without using a calculator:

(c) Show that if the roots of x2+bx+c = 0 differ by 1, then b2-4c=1.

(d) Using the result from part (c), find the value of k in the following equation, and the roots of the equation, given that the roots are consecutive integers.



You do four questions and they give you a list of formulas...:eek:

I also got annoyed at my question 3, I wrote out the dot product wrong and wasted five minutes fixing everything.

cocoa

in answer to Sam, they don't give Ur, they give the general term, which is Ur+1

Thomas_S_Hunterson

Shox wrote:

did anyone know how to do the very last part of the line question... when k=1 what line is impossible or somethin.... i hadnt a clue??

Ididn't do the question, but it should be whatever line had the coefficient of L

EamonnKeane

8c: pi/4 - ln2/2

I got an infinite power series: 1/2 - 1/12 + 1/30 - 1/56 + 1/90 ...

but had no idea where it converged. Probably wrong.

JC 2K3

Gangsta wrote:

JC 2k3 would you mind posting the solution to Q4c(ii) or anyone else who got it? Much appreciated. I spent an hour on that and I couldn't do it!:(

This is going to be depressing typing it out now. I only realised what a massive mistake I had made with 2 mins to go and could only draw a rough diagram of what you were supposed to do.

Basically get the area of the triangle with a and 2 sides of r with (1/2)(r^2)sin2A and get the area of the sector that makes up the rest of that section with (1/2)(r^2)(2A) and equate them with (pi)(r^2)/4.

(1/2)(r^2)sin2A + (1/2)(r^2)(2A) = (pi)(r^2)/4

sin2A + 2A = pi/2

So simple....

So fúcking simple....

WildCardDoW

I aws an idiot and got al confused with the ratio test, simply because they didn't give n's. I wasn't sure what to replace and eventually gave up in frustration, luckily I did well in the rest of the question, bugger, bugger, bugger.

Adam We

WildCardDoW wrote:

I aws an idiot and got al confused with the ratio test, simply because they didn't give n's. I wasn't sure what to replace and eventually gave up in frustration, luckily I did well in the rest of the question, bugger, bugger, bugger.

same. i took it as Um and Um+1 rather than Ur and Ur+1

Couldn't remember ever seeing the feckin Binomial as a Calculus Question in my life, until someone pointed it out to me that morning

got all 8 Qs done, both Trigs 50/50, the Cs were tricky but do-able, and Q7 was 50/50 too. then i've about 3 40ish to 40+ish/50 and a 30/50 for Calculus

altogether it seems 260ish

along with a 245ish paper 1, i could be so pissed off with just missing out on 510. Fingers crossed

WildCardDoW

I mean, they can't do that can they, there were two possiblities for n to be, it should have stated which they were. I wonder if we kick up a fuss will we get default marks!

rjt

Really happy with that - maybe even 100%. I found 1,2,5,6 and 7 grand (and the answers I remember match with those given by Cocoa, JC2k3 and others). 4(c)(ii) stumped me though - I put down area of sector boc = (1/2)(r^2)sin2A, forgetting it was in radians. I skipped 3, would've liked to have done it but ran out of time. It looked harder than usual, and it's my weakest question.

Also, answers for 10 in case anyone wants them. I think/hope I got this question right:
10
(a)
(i) Subtraction of natural numbers isn't closed (e. 1-5=-4, not a natural number)
(ii) 0 has no inverse.
(b)(i)Draw Cayley table, see that all elements are members of the set {I_{pi}, R180, Sx and Sy}. Therefore it is closed. I_{pi}*a=a*i_{pi} for all a in G, so I_{pi} is the identity. I_{pi} is present in each row and column, so we have inverses, and we may assume associativity.
(ii)Group is abelian/commutitive (Cayley table is symmetric), so Z(G)=G.
(c)
(i)Standard proof.
(ii)Standard proof.

I actually forgot what Lagrange's Theorem was at the start, and begun to think that I'd lose 20 marks immediately for 10(c), but thankfully I remembered at the end.

cocoa

WildCardDoW wrote:

I mean, they can't do that can they, there were two possiblities for n to be, it should have stated which they were. I wonder if we kick up a fuss will we get default marks!

no, they said general term which means Ur+1.

Tomlowe

anyway ratio test should work for any consecutive terms

WildCardDoW

I thought it was Un = equation, and then Un+1? I don't care, I know I made the mistake anyway.

Nehpets

I did this for (b) (i)

Found Centre of large circle.
Proved it was a point on the smaller circle.
Found radius of larger circle and stated it should be equal to the diameter of the smaller circle.
Then I found the diameter of the smaller circle and it was the same as the radius of the big circle (therefor touching internally)

This is not the regular method because I was making it up as I went along. Will I get the marks for this? It seems right to me.

EDIT: delighted to see I got q.1 fully right. I thought I had it fully wrong :O

Gangsta

JC 2K3 wrote:

This is going to be depressing typing it out now. I only realised what a massive mistake I had made with 2 mins to go and could only draw a rough diagram of what you were supposed to do.

Basically get the area of the triangle with a and 2 sides of r with (1/2)(r^2)sin2A and get the area of the sector that makes up the rest of that section with (1/2)(r^2)(2A) and equate them with (pi)(r^2)/4.

(1/2)(r^2)sin2A + (1/2)(r^2)(2A) = (pi)(r^2)/4

sin2A + 2A = pi/2

So simple....

So fúcking simple....

cheers mate. Ya it's fcuking soft!damn that could be 10marks down the drain.

WildCardDoW

Meh, I found the radii of the circles, took them away and then proved it was equal to the distance of the centres, I think that's right.

In the end both were root2, so I figured I would move on!

DtotheK

****in messed up permutations and combinations and i'm pissed off cause that was only part A. that was a tricky ish paper but i got most of the answers others are gettin.

I really wanted an A but prob got a B... ****
guess time will tell

JC 2K3

Gangsta wrote:

cheers mate. Ya it's fcuking soft!damn that could be 10marks down the drain.

I'm wishfully hoping for 7/10 - 5/5 for drawing a correct diagram and 2/5 for attempt marks??

T-b0n3

Can some one put my mind to ease with 7(a) I thought it was 7c4 like some one previously posted but am I right in saying it is 7x6x5x4 = 840?? or??

Adam

Gangsta

JC 2K3 wrote:

I'm wishfully hoping for 7/10 - 5/5 for drawing a correct diagram and 2/5 for attempt marks??

well last year for the trig q with the circle and the sector I wrote loads of bull**** and got 15marks!hopin for the same this time!

cocoa

T-b0n3 wrote:

Can some one put my mind to ease with 7(a) I thought it was 7c4 like some one previously posted but am I right in saying it is 7x6x5x4 = 840?? or??

Adam

divide that by 4! and you have 35.

DtotheK

i got 840 and 816(i think) but apparently that's wrong...

The answer is 35 and 34 i heard...

although i amnt definate and i hope we're right T

EDIT damn... atleast we'll get attempt marks...

carlowboy

Christ I thought you were talking about 6a there and I was bricking it.

DtotheK

well atleast maths is all over!!!!!!!!!!!


No scrap that... physics and applied maths......
I wish this shít would end fast.
I wish we got our results straight away and not have them lurking in the backround of the summer..

T-b0n3

I had7c4 down but then scribbled it out.. Sigh I see it now its cuase it says 4 letter selections so "flor" would be the same as "rolf" and we worked out every different combination...

Adam