STEP maths I, II, III 1990 solutions
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
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Re: STEP maths I, II, III 1990 solutionsI 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!(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.
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Re: STEP maths I, II, III 1990 solutionsCan 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...(Original post by brianeverit)
paper I/4 & 7
Comments and /or corrections welcomed.
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 -
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Re: STEP maths I, II, III 1990 solutionsWell ln(0) or ln(-a) , a > 0 not defined. When integrating, in all fairness we can be kind of loose with our mod signs(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
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. -
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Re: STEP maths I, II, III 1990 solutionsI 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(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.
with no constants? I wasn't sure why it's ok not to have constants.
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Re: STEP maths I, II, III 1990 solutionsI do like this alternative to the third part of III/4:
![1 + \cos \theta \sec \theta + \cos 2\theta \sec^2 \theta + ... + \cos n\theta \sec^n \theta = \text{Re} \left[1 + \sec \theta e^{i \theta} + ... + e^{ni \theta} \sec^n \theta \right]](http://www.thestudentroom.co.uk/latexrender/pictures/0d/0de3d0c949dabe065de7a61138bf605f.png "1 + \cos \theta \sec \theta + \cos 2\theta \sec^2 \theta + ... + \cos n\theta \sec^n \theta = \text{Re} \left[1 + \sec \theta e^{i \theta} + ... + e^{ni \theta} \sec^n \theta \right]")
. And now we have a geometric series, common ratio 
.
The sum is![\text{Re} \left[ \dfrac{\sec^{n+1} \theta e^{i(n+1) \theta} - 1}{\sec \theta e^{i \theta} - 1} \right]](http://www.thestudentroom.co.uk/latexrender/pictures/99/99311dd2f300fd010d890a91e6374d61.png "\text{Re} \left[ \dfrac{\sec^{n+1} \theta e^{i(n+1) \theta} - 1}{\sec \theta e^{i \theta} - 1} \right]")
. The denominator turns out to be i sec theta sin theta, and multiplying top and bottom by i gives
![\text{Re} \left[ \dfrac{-i \sec^{n+1} \theta e^{i(n+1)\theta} + i}{\sec \theta \sin \theta} \right] = \dfrac{\sec^{n+1} \theta \sin(n+1) \theta}{\sec \theta \sin \theta} = \dfrac{\sec^n \theta \sin(n+1)\theta}{\sin \theta}](http://www.thestudentroom.co.uk/latexrender/pictures/66/66b46b8b9dad6e59d1f050af3f951b5a.png "\text{Re} \left[ \dfrac{-i \sec^{n+1} \theta e^{i(n+1)\theta} + i}{\sec \theta \sin \theta} \right] = \dfrac{\sec^{n+1} \theta \sin(n+1) \theta}{\sec \theta \sin \theta} = \dfrac{\sec^n \theta \sin(n+1)\theta}{\sin \theta}")
. 
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Re: STEP maths I, II, III 1990 solutionsBut we are given y(1)=0 NOT y(0)=1(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)
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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. -
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Re: STEP maths I, II, III 1990 solutionsThere is an error in the calculation of the expected score. It should be 2.031 NOT 264/125(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 :
