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# STEP,I,II,III 1998 Mark Schemes

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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1. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by Sauron)
I was looking at this question, and I couldn't understand where the complications were. What I mean is, why can't the probability of picking a red pastille only on the last draw not simply be expressed as (n-1) draws of non-red pastilles and one draw of a red pastille on the final draw. I.e. (b/r+b)^(n-1) multiplied by r/(r+b) ?
Not sure what you're trying to say. The question asks for the probability of the picking a red pastel on the last draw given that only one red pastel has been eaten.
2. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by bogstandardname)
I believe may be a little mistaken, at the beginning you say "OK, the formula is clearly wrong, since we must have P=0 when p=1 and vice versa" but with your given formula at the end, substituting in p=0, gives P=0 and p=1 gives P=1. Having worked through the question I get , Please do tell me if I am talking rubbish.
I may well have reversed the sense of P somewhere. IIRC, there was a negation issue between the given answer and the correct one and I was trying to make that clear. It was a long time ago...
3. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by Sauron)
This solution is great from a more advanced point of view, but someone taking STEP I really isn't going to do it that way, so it would be better if you could post a more in-touch solution.
Thanks
Someone doing STEP I today really isn't going to answer a 1998 STEP I question that is intrinsically off-syllabus.

I mean, the question is about complex numbers and argand diagrams, and you're seriously objecting to my using basic e^it manipulations?
4. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by DFranklin)
Not sure what you're trying to say. The question asks for the probability of the picking a red pastel on the last draw given that only one red pastel has been eaten.
But isn't the probability of eating that one red pastille on the last draw the same as saying that you draw (n-1) non-red pastilles, which all get put back in the bag, then only on the last draw do you take out the red pastille, with probability r/(r+b). So, multiplying these probabilities, you get [b/r+b]^(n-1)[r/r+b]. I know I'm wrong but I don't see why this isn't what the question is asking
5. Re: STEP,I,II,III 1998 Mark Schemes
Take a simple case. Two R, one G, two draws. So 4 possibilities: RR (p=2/3 x 1/2 = 1/3), RG (p = 1/3), GR (p=1/3 x 2/3 = 2/9), GG (p = 1/3 x 1/3 = 1/9).

The only case where you only eat one red pastille and it's on the last draw is GR. Given no other information, this occurs with probability 2/9. (This is what you are arguing).

But you are also given the information that one red pastel has been eaten. This means that cases RR and GG are impossible. So the probability we seek is actually

P(GR) / (P(RG) + P(GR)) = (2/9) / (1/3 + 2/9) = 2/5.

If this still confuses you, you need to revise conditional probabilities.
6. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by DFranklin)
Step I, 1998, Q10.

When A is at the top, the total mass of the system is . The net downwards force is g times the difference in the masses of A and B = and so the downwards acceleration on A is . Using , the time for the downward trip is given by
.
On the trip back up, A is massless and so B is in free fall. So the time for the upward trip is just . So the total time is

Now we have ore transfered each trip, and we are taking the long term average, so the rate of ore transfer will just be "mass per trip" / "time per trip". So the rate of ore transfer is
which is just times a constant. So we maximise the rate by maximising f.

Now f is maximised either at an end point of possible lambda or where . But the end points are and f(0)=f(1) =0, while f(0.5) > 0. So the maximum doesn't occur at an end point and so .
This is question 11, not question 10
7. Re: STEP,I,II,III 1998 Mark Schemes
STEP II, Q1
(Original post by Trangulor)

i) Similar thing to part a). If n is odd, all three terms are odd, and since the LHS' odd+odd gives even, and RHS is odd, there is a contradiction. If n is even, all the terms are even, so it works.

Also, by expanding, you get

If n is an integer, then RHS is even, so LHS is even. If is even, then n is also even

ii)
, 432>0
so n-36>0 or n>36
Thus
If n is a positive integer greater than 36, so cannot equal 432, so there are no integer values of n which satisfy the above equation

Thanks to khaixang for spotting my dodgy arithmetic

Shorter way for the last part:
Since n is even, it can be expressed as . Sub back into the equation gives

Cancel gives the equation in the first part. No solution for first part equation => no for solution for this
Last edited by hypercube; 22-02-2013 at 22:06.
8. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by dvs)
STEP III Q7.
The sketch is manageable. Since f(3) = 3 > 0, we need only try m=1,2 - and both fail.

If b'(x) = 0, then
3x^2 (e^(x/T) - 1) = x^3 (1/T) e^(x/T)

Since x>0, then
e^(x/T) (x/T - 3) + 3 = 0
=> f(x/T) = 0

So it has one turning point, namely the root of f(x/T) = 0. So the graph of y = b(x) for x>=0 looks like it increasing to the turning point, then decreases and has an asymptote x=0.

Now we do the integral. (Prepare for some dirty maths.)

(!!!)

So n=4.

For the last bit, we consider a triangle of base 3 and height b(m) = b(3). Then:
K T^4 = 2 * 1/2 * 3 * 27/(e^(3/T) - 1)
= 81/(e^(3/T) - 1)

Therefore,
K = 81/(T^4 (e^(3/T) - 1))
----------------------------

Of course, all of this could be wrong!

I think we have all the pure maths STEP III questions now.
Is there an easier way of evaluating the final integral other than repeated integration by parts? I think that i'm missing something.
Also i don't think that there should have been a negative in the final summation, or the b(x) has an asymptote at x=0
9. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by Nick_)
Is there an easier way of evaluating the final integral other than repeated integration by parts? I think that i'm missing something.
Also i don't think that there should have been a negative in the final summation, or the b(x) has an asymptote at x=0
Well, you don't actually need to evaluate the integral.

Use the substitution y = x / T to rewrite the integral in terms of

and the way the original integral depends on T should be obvious.
10. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by DFranklin)
Well, you don't actually need to evaluate the integral.

Use the substitution y = x / T to rewrite the integral in terms of

and the way the original integral depends on T should be obvious.
Thanks, thats a much nicer method
Last edited by Nick_; 01-03-2013 at 19:49. Reason: typo
11. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by Nick_)
Thanks, thanks a much nicer method
I have a suspicion the method dvs posted was based something he'd done at university where you actually want to evaluate the integral as well.
12. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by Nick_)
Thanks, thats a much nicer method
Paper III question 7
Just to finish this off
so and
area of triangle with base and height

hence,
13. Re: STEP,I,II,III 1998 Mark Schemes
Back in those days the syllabus was different and the students were expected to have read further into the subject. They may well have met the method shown on the document....I certainly did in my day (way back!!!!)...
Attached Files
14. Step1998Paper3Question7BIT.pdf (46.5 KB, 15 views)
15. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by mikelbird)
Back in those days the syllabus was different and the students were expected to have read further into the subject. They may well have met the method shown on the document....I certainly did in my day (way back!!!!)...
That's a useful method for more intractable problems, but as DFranklin points out a simple substitution like x = Tu transforms the integral immediately giving the answer as , where k is the value of the integral that is now totally independent of T.
16. Re: STEP,I,II,III 1998 Mark Schemes
I really liked Q3 of STEP II here, but I'm just a sucker for series questions. However I split the expression into partial fractions and used a kind of muddled method of differences. Would that still be valid, or was the question mainly leaning for you to use induction (like in the solution on this thread)? Which would be considered less time consuming?
17. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by dvs)
STEP III Q7.
The sketch is manageable. Since f(3) = 3 > 0, we need only try m=1,2 - and both fail.

If b'(x) = 0, then
3x^2 (e^(x/T) - 1) = x^3 (1/T) e^(x/T)

Since x>0, then
e^(x/T) (x/T - 3) + 3 = 0
=> f(x/T) = 0

So it has one turning point, namely the root of f(x/T) = 0. So the graph of y = b(x) for x>=0 looks like it increasing to the turning point, then decreases and has an asymptote x=0.

Now we do the integral. (Prepare for some dirty maths.)

(!!!)

So n=4.

For the last bit, we consider a triangle of base 3 and height b(m) = b(3). Then:
K T^4 = 2 * 1/2 * 3 * 27/(e^(3/T) - 1)
= 81/(e^(3/T) - 1)

Therefore,
K = 81/(T^4 (e^(3/T) - 1))
----------------------------

Of course, all of this could be wrong!

I think we have all the pure maths STEP III questions now.

the area is positive and the sum is clearly negative, i think u did parts wrong..
but, very nice solution, i like it.
Last edited by FunctionOfX; 25-06-2013 at 17:09.
18. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by Speleo)
STEP I Question 8

(1/r).d/dr(rdv/dr) = -k
d/dr(rdv/dr) = -kr
rdv/dr = -k(r^2/2) + C
dv/dr = -(1/2)kr + C/r
v = -(1/4)kr^2 + Clnr + D

r->0
lnr -> - infinity
r^2 -> 0

Therefore |v| -> infinity as long as C =/= 0.

v = -(1/4)kr^2 + D
When r=a, v=0
0 = -(k/4)a^2 + D
D = (k/4)a^2

v = (k/4)(a^2 - r^2)

F = 2pi INT rv dr
= 2pi INT (k/4)a^2r - (k/4)r^3 dr
= 2pi [(k/8)a^2r^2 - (k/16)r^4]
limits 0 and a.
-> 2kpi[a^4/8 - a^4/16]
= (a^4).kpi/8

NB: I left a minus sign out early on and had to go through changing it right at the end, so there is a small possibility that there is a mistake somewhere, I would appreciate if someone could check it.
I reached the same answer on all parts of the question, so it's probably good. Nice question. Thanks for the solution.
19. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by ukgea)
STEP I Question 3

(iii) False.

A polynomial must have a finite degree. Let the degree of P(theta) be n. Consider the nth-degree term of P(theta). Since the polynomial has degree n, the coefficient of this term cannot be zero. So, as theta becomes very large, this term grows faster than all the other terms, and thus the polynomial will tend to either positive or negative infinity, and thus further than 10^-6 away from the interval [-1, 1] which is the range of the cos(theta) function.

The only case when the highest-degree term of a polynomial is indeed zero, is for the trivial polynomial P(theta) = 0 for all theta. But this does not satisfy the requirement either, for say theta = 0,

which is obviously larger than 10^-6.

Therefore, there does not exist any polynomial satisfying the requirement.
What about a Maclaurin expansion of cos(theta)? If you use sufficient terms it will be within 10^-6 of cos(theta)? Of course the bigger theta is the more terms you need to allow the factorials to have an impact. But they only want less than 10^-6 so a finite number of terms must be possible.
20. Re: STEP,I,II,III 1998 Mark Schemes
(Original post by TimPowell)
What about a Maclaurin expansion of cos(theta)? If you use sufficient terms it will be within 10^-6 of cos(theta)? Of course the bigger theta is the more terms you need to allow the factorials to have an impact. But they only want less than 10^-6 so a finite number of terms must be possible.
Tim, the point is just one polynomial is required....not one that you have to adjust depending on theta.

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