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EDEXCEL P5 JUN 05 - HOW DID IT GO?

Gaz031

I found that to be a 'friendly' paper but the other candidate at my centre found it very hard.

Q1(a) 0.5arcsin2x-(1/4)rt[1-4x^2]+C
(b)0.372

Q2:Proof?

Q3 (6pia^2)/5

Q4(a)Proof
(b)(1/4)(e^2-1)

Q5:Proof?

Q6: Proof?

Q7:Proof? 2rt3 for c.

Q8: Proof but the answer to the last part was +-ln[(1/2)(3+rt5)]

Looks good!

Wait.. Q8, you sure it was ±?

Reply 81

yazan_l

5

dvs

I'm dreading thursday. I've got M4 and M6. :/

i've got Chemistry synoptic, p4 p6 on 28th, and m4 m6 on 30th.. so ..... :confused:

Reply 82

drw25

I can remember the basic premise of every question apart from 2. Any ideas?

In case anyone wants to know (from an MSN convo):
"warren says:
q1
warren says:
a) integrate x+1 over root 1-x^2
warren says:
b) find integral between 0.3 and 0 of above
warren says:
q2 was finding the area of the arcosh graph between 2 and 1, i think
warren says:
q3 was surface area of x=asin^3t y=acos^3t between 0 and pi by 2
warren says:
q4 a was reduction formula: integral of x^n.e^2x dx and b was finding intergral of x^2.e^2x between 1 and 0
warren says:
5 was co-ordinate geometry - a was find one tangent, b was find the other, c was show that their intersection point makes a parbola, d was focus and directrix of new parabole and e was sketch both parabolas
warren says:
hmm...i think 6 might be what i said was question 2
warren says:
in fact i'm pretty sure it is
warren says:
7 was intrinsic: something like x=t-cost y=-sin t
warren says:
had to find s in terms of t
warren says:
for a
warren says:
then s in terms of psi for b
warren says:
then radius of curvature for c
warren says:
8 was hyperbolic: first part was show the log thing, 2nd was finding sech x as a log, 3rd was solving the hyperbolic equation
warren says:
which leaves 2..."

Reply 83

Font

man i'm depressed - went miserabley :frown:

- hope the boundaries are low - anyone know around what they will be?

phil?

think i got 70ish, but may be higher if i get method marks, or lower if i messed more stuff up than i think i did! :frown:

Reply 84

dvs

10

2 was cosh(2x) = f(k) for part a, and then you had to find the value of p for part b.

Reply 85

Gaz031

15

dvs

Looks good!
Wait.. Q8, you sure it was ±?

I'm fairly sure. sech is an even function after all.
Nice to hear the others look okay :smile:

Reply 86

Gaz031

15

master-chafe

I thought it went suprisingly well. I managed to prove q6 (integral of x arcosh x dx between x=1 and x=2 iirc) (I used substitution x=cosh u after realising parts would take all night).

x=coshu sounds good. I did it slightly different to you though in that i got Area=.......some values....∫(x^2-1)^(1/2) dx using IBP and some rearranging, then proceeded to evaluate the final integral using x=coshu and collecting terms to prove the statement.

Reply 87

dvs

10

Gaz031

I'm fairly sure. sech is an even function after all.
Nice to hear the others look okay :smile:

Hmm! You don't happen to remember what the equation was, do you?

Reply 88

mackin boi

OP

1

o another winna thread 4 me :smile:

muwahahah

Reply 89

Gaz031

15

dvs

Hmm! You don't happen to remember what the equation was, do you?

It was something that involved (tanhx)^2, sechx and an integral which transformed to a quadratic in sechx but i can't remember the coefficients. Sorry.

Reply 90

Syncman

well that was an experience.
First 5 q's done in bout 30 mins. Last 3 Q's never done lol. I got at least half, but i doubt i picked much up after Q 5.

Reply 91

dvs

10

Gaz031

It was something that involved (tanhx)^2, sechx and an integral which transformed to a quadratic in sechx but i can't remember the coefficients. Sorry.

Yeah. I remember getting x=arsech(2/3) and arsech(-2), or at least something along similar! I forgot to use symmetry to get the other root, even though I should've since the question said that answers should be left in logarithmic form. It's pretty cool that you managed to spot that. :biggrin:

I'm gonna go bed now.

Reply 92

Daft Wader

yazan_l

only thing tht i'm afraid of is varying mass!

Damn right! I've managed to do a grand total of 3 varying mass questions... as a friend put it: "it's a horrible game of chase the variable". I'm prepared for it though: cross produsts aren't hard, and as long as the integrations for moments of inertia are nothing like the ones in p5 i'll be fine. Mind you moments of momentum... now those are tricky.

Reply 93

inequality

7

riks2004

i reckon itll be bout 56-58/75 for an A

But that's about 75%. Can someone explain exactly how UMS marks work please.

Oh ye and q.6 was a integral using a double integration by parts, first for ∫xarcoshx dx and then for the ∫√(x²-1) bit.

Question 8.a) was definately unusual for an edexcel P5 paper. It involved conditions and proof :eek:

Reply 94

inequality

7

dvs

Yeah. I remember getting x=arsech(2/3) and arsech(-2), or at least something along similar! I forgot to use symmetry to get the other root, even though I should've since the question said that answers should be left in logarithmic form. It's pretty cool that you managed to spot that. :biggrin:

I'm gonna go bed now.

Hmm answers, I'm sure one of them was to be ignored as sechx > 0 for all x.

Reply 95

lesser weevil

1

that was the most HATEFUL paper I've had this year! Argh, I only answered at best HALF of the questions, and those were the low mark ones...

Forget the estimated grade, I got a Z...

Reply 96

herman368

I somehow find this paper quite straightforward. The only answer I fear I might have got wrong was Q1b) I got 0.09 something. Also I couldnt do 8.a) The rest of the answers were right so *maybe* 70/75 if im lucky.

Reply 97

sarz777

hi....i thought t he paper was quite good overall....there was nothing too difficult about it, especially if you solved the solomon papers. ....... well thank god for that......best of luck to everyone doing p6!

Reply 98

JohnSPals

drw25

I think it was a pretty nasty paper: there were lots of tricks that were pretty much essential, such as the +1-1 thing, hyperbolic substitutions and half angles. I just had time to answer everything apart from 8a (the logs thing) - and now I've been told how to do that bit is seems so obvious...

Some answers I remember:
.25(e^2-1)
(6pi/5)a^2
128/259
focus was at [(25/16)a, 0]

For the very last part of the very last question, it mentioned answer*s*, but I found that one of the solutions to my quadratic was out of the range of arsech - anyone know how many answers were expected?

Those answers you got = YAY! For me. I remember getting identcal answers.

I noticed that one, too. I couldn't do 8(a) (ahh well - only three marks), so I justified it using all the "arcosh x > 1" rubbish. I wondered if you used the result in part (a) but I don't think you did because the √(1-x²) within both logs didn't alter.

My overall opinion was that it was pretty easy. 1-5 were fairly bog-standard and could be quickly done. Question six, I saw what to do with about five minutes left (I was such a rebel - I was writing way after they told us to stop and still didn't quite finish it :-D). Question 7 was quite nice, I thought, and all the double-angle formulae cancelling out was quite satisfying when equating dy/dx = tan(psi) = sinx / (1 + cosx). Question 8 was a bit of a bugger but the proving arsech bit were easy marks to be got at least, as was getting that equation into a quadratic in sechx.

If anyone wondered, I did parts on question 6 (u = arcoshx and dv/dx = x - very unusual that it's done in that order), and then used a substitution in the form of x = sinhθ I think because the integrand had an x² as numerator.

Reply 99

Womble548

For 8a you had to think of 'rationalising the denominator'. Mulitply the top and bottom by (1- √(1-x²)) or whatever it was and simplify :d

I think this paper was very doable if you had more time. I usually have about 20 mins at the end of every maths paper. But this one and D1 I used every bit of time available to just find the answers. No time to check!

EDEXCEL P5 JUN 05 - HOW DID IT GO?

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