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1. Re: STEP maths I, II, III 1990 solutions
I will rep anyone who does STEP II, Question 16. My eyes are bleeding just reading it
2. Re: STEP maths I, II, III 1990 solutions
(Original post by Dystopia)
Attempt at STEP III, Q8.

Spoiler:
Show

The integrating factor is .

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\Rightarrow (x^{3}-1)y = -\int{x^{2}e^{x} + \frac{c}{x} \; \mathrm{d}x = -x^{2}e^{x} + 2xe^{x} - 2e^{x} - c\ln{x} + k

Initial conditions: x=2, y=2e2, y'=0

From this I get

Which is utterly horrific, so presumably I've made an error.
I tried this one and ended up with y(1 - x^3) = e^x(x^2 - 2x + 2) + 56e(x + 2) - 16e^2/7 which is different from your answer but still very complicated. Would anyone mind having a crack at this one? Would be greatly appreciated!
3. Re: STEP maths I, II, III 1990 solutions
1990 Paper 1 number 3
Attached Files
4. 1990PAPER1.no.3.pdf (35.3 KB, 80 views)
5. Re: STEP maths I, II, III 1990 solutions
1990 Paper 1 numbers 11 & 13
Attached Files
6. 1990PAPER1.no.11.pdf (30.8 KB, 47 views)
7. 1990PAPER1.no.13.pdf (35.8 KB, 34 views)
8. Re: STEP maths I, II, III 1990 solutions
(Original post by brianeverit)
paper I/4 & 7
Can anyone help me with I Question 7. I don't understand the validity conditions. When I solve INT 1/u du = INT 2/(x + 1) dx i get...

ln|u| = 2ln|x + 1| + C.

Im not sure why this isn't valid for x < -1 but I confess to not really knowing much about why we really use mod signs on logs so thats clearly the key to me not understanding. Any help is much appreciated
9. Re: STEP maths I, II, III 1990 solutions
(Original post by maltodextrin)
Can anyone help me with I Question 7. I don't understand the validity conditions. When I solve INT 1/u du = INT 2/(x + 1) dx i get...

ln|u| = 2ln|x + 1| + C.

Im not sure why this isn't valid for x < -1 but I confess to not really knowing much about why we really use mod signs on logs so thats clearly the key to me not understanding. Any help is much appreciated
Well ln(0) or ln(-a) , a > 0 not defined. When integrating, in all fairness we can be kind of loose with our mod signs

ln| x | = ln(x) , x > 0 or ln(-x) , x < 0

but ln(-x) = ln((-1)(x)) = ln(x) + ln(-1) which gets taken up in our arbitary constant.
10. Re: STEP maths I, II, III 1990 solutions
1990 Paper II number 3
Attached Files
11. 1990PAPERII.3.pdf (54.8 KB, 81 views)
12. Re: STEP maths I, II, III 1990 solutions
1990 Paper II numbers 12,13,14,16
Attached Files
13. 1990PAPERII.14.pdf (37.5 KB, 10 views)
14. 1990PAPERII.16.pdf (21.4 KB, 14 views)
15. Re: STEP maths I, II, III 1990 solutions
1990 PAPER ii NUMBERS 12,13,14 & 16
Attached Files
16. 1990PAPERII.12.pdf (22.7 KB, 31 views)
17. 1990PAPERII.13.pdf (43.4 KB, 32 views)
18. 1990PAPERII.14.pdf (37.5 KB, 24 views)
19. 1990PAPERII.16.pdf (21.4 KB, 43 views)
20. Re: STEP maths I, II, III 1990 solutions
(Original post by brianeverit)
1990 Paper II numbers 12,13,14,16
sORRY, TWO FILES DID NOT UPLOAD CORRECTLY SO HAVE SENT AGAIN.
Last edited by brianeverit; 03-05-2009 at 15:29.
21. Re: STEP maths I, II, III 1990 solutions
1990 Paper III number 10
Attached Files
22. 1990PAPERIII.10.pdf (41.6 KB, 58 views)
23. Re: STEP maths I, II, III 1990 solutions
1990 Paper III number 11
Attached Files
24. 1990PAPERIII.11.pdf (36.7 KB, 31 views)
25. Re: STEP maths I, II, III 1990 solutions
1990 Paper III number 12
Attached Files
26. 1990PAPERIII.12.pdf (46.6 KB, 27 views)
27. Re: STEP maths I, II, III 1990 solutions
1990 Paper III number 14
Attached Files
28. 1990PAPERIII.14.pdf (42.7 KB, 26 views)
29. Re: STEP maths I, II, III 1990 solutions
(Original post by Dystopia)
Attempt at STEP III, Q8.

Spoiler:
Show

The integrating factor is .

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\Rightarrow (x^{3}-1)y = -\int{x^{2}e^{x} + \frac{c}{x} \; \mathrm{d}x = -x^{2}e^{x} + 2xe^{x} - 2e^{x} - c\ln{x} + k

Initial conditions: x=2, y=2e2, y'=0

From this I get

Which is utterly horrific, so presumably I've made an error.
I got to the end of this, went to put the initial conditions in and thought "haha I've gone totally wrong here" and didn't even bother to finish it, but you got the same so that's hopeful. Also, do the double implication signs at the top need some justification, ie for example explain why with no constants? I wasn't sure why it's ok not to have constants.
30. Re: STEP maths I, II, III 1990 solutions
How do I get access to the Further mathematics A papers? I found MAthematics and Further MAthematics B papers on paperbank... but not the second one..
31. Re: STEP maths I, II, III 1990 solutions
I do like this alternative to the third part of III/4:

. And now we have a geometric series, common ratio .

The sum is . The denominator turns out to be i sec theta sin theta, and multiplying top and bottom by i gives
.
32. Re: STEP maths I, II, III 1990 solutions
(Original post by SimonM)
STEP I, Question 7

Let

So

Therefore, we can express in terms of if

That is .

Solving

Let , to find a suitable u, solve , that is

Some separating variables gives

We must now solve , that is , so

Therefore

Therefore

Valid for all (I think)
But we are given y(1)=0 NOT y(0)=1
33. Re: STEP maths I, II, III 1990 solutions
(Original post by *bobo*)
STEP I
10) i)total resistive forces= 50+2(v-3)= 2(22 +v) where v is pos. in direction of turning point from starting point.
using P=Fv, where F=resistive forces due to no acceleration
300=2v(22+v)
2v^2 + 44v- 300=0
v=0.25 (in forward journey), v=1.25(backward journey) to 2dp
using v=s/t:
total time for race= 5000/0.25 + 5000/1.25
= 24000 seconds (supposed to answer to nearest second but don't have a calc on me)

ii) total energy expended using P=W/t
W=300x24000=7200000

x(1)t(1)+x(2)t(2)= 7200000

x(1)= v(1)(50+2(v(1)-3))= v(1)(44 + 2v)
t(1)= 5000/x(1)

x(2)= v(2)(50+2(v(2) +3))=v(2)(56+2v(2))
t(2)=5000/v(2)

=> 5000(44+2v(1))+5000(56+2v(2))=72 00000

=> 100+ 2v(1) + 2v(2)= 1440
=> v(1) +v(2)=670 so is independant of x(1) and x(2) values.

I'm afraid there is something seriously wrong here. The solution fails the "sanity" test on a number of counts
Firstly, surely he runs faster on the outward journey rather than the return.
Secondly, is 3/4 metre per second a sensible answer for his speed.
Thirdly, 24000 secs, i.e almost 7 hours to run 10 km ??
Last edited by brianeverit; 14-07-2009 at 10:27.
34. Re: STEP maths I, II, III 1990 solutions
(Original post by nota bene)
I just had a quick go at I/14, and have not checked my working and reasoning, so likely loads of mistakes. (Also I'm not 100% sure what the last part wants)

With Juggins strategy to draw trice and then stop there are eight possible ways of not ending up with a score of 0, WWW, RRR, WRR (in any order) and WWR (in any order).
P(WWW)=1/8
P(RRR)=(3/10)^3=27/1000
P(WRR in any order)=
P(WWR in any order)=

So 1-(64/125)=61/125, which is the probability he ends up with zero points in total.

The average score is

Muggins with his obtained N points will have a probability of 1/2 to obtain N+1 points, probability 3/10 to obtain N+2 points and probability 1/5 to end up at zero points. Therefore the average score is

The greatest possible average final score can be obtained by applying the above formula after each try and see if the average increases, and in that case continue - until the formula suggests the average will not increase by taking part in one more draw.

Edit: As David points out in post #86 the following should probably be added at the end for a complete solution :
There is an error in the calculation of the expected score. It should be 2.031 NOT 264/125