STEP 2018 Solutions Watch

username3720230
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#61
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#61
(Original post by DFranklin)
(iii) Candidate has probability (1-p) of answering a question. So p(answers 3) = \binom{5}{2} p^2(1-p)^3 = 10 p^2(1-p)^3. So p(Pass AND answers 3) = 10p^2(1-p)^3 P3 = 10p^2(1-p)^3 p^3 = 10p^5(1-p)^3.
Similarly, p(Pass AND answers 4) = 5 p(1-p)^4 P4 = 20p^4(1-p)^5 + 5p^5(1-p)^4
And p(Pass AND answers 5) = (1-p)^5 P5 = 5p^4(1-p)^6 + p^5(1-p)^5

So p(Pass) is the sum of these, that is:

p^4(1-p)^3(10p + 20(1-p)^2+5p(1-p)+5(1-p)^3+p(1-p)^2)

=\frac{1}{n^4}\frac{(n-1)^3}{n^3}\left(\frac{10}{n} + 20\frac{(n-1)^2}{n^2} + 5\frac{n(n-1)}{n^2} + 5 \frac{(n-1)^3}{n^3} + \frac{(n-1)^2}{n^3}\right)

=\frac{(n-1)^3}{n^{10}}(10n^2 +20n(n-1)^2+5n^2(n-1)+5(n-1)^3+(n-1)^2)

=\frac{(n-1)^3}{n^{10}}(10n^2 + 20n^3-40n^2 + 20n +5n^3-5n^2+5n^3-15n^2+15n-5+n^2-2n+1)

=\frac{(n-1)^3}{n^{10}}(30n^3-49n^2+33n-4)

(Wouldn't be surprised at all if there are algebra errors in this - direct to LaTeX and to be honest I really found it hard to care given the messiness of it all).
For the last result I got \frac{25n-9}{(n+1)^5} could someone check which is right im not that confident because of all the algebra and sorry if someone posted it further down in the thread
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DFranklin
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#62
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(Original post by sameactuallylol)
For the last result I got \frac{25n-9}{(n+1)^5} could someone check which is right im not that confident because of all the algebra and sorry if someone posted it further down in the thread
In case you missed it, I made a big slip (taking p(answers a question) to be (n-1)/n instead of n/(n+1)), which (of course) fundamentally messes up my answer to the last part. Someone else did say they'd post their solution (and that they'd checked it against a computer simulation).

I'm not sure I can face redoing the last part myself, but I might do if no-one else does post a correct version.
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Lordkpaj
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#63
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Here is my attempt on Q13. I got the same results as DFranklin on (i) and (ii) and 25n-9/(n+1)^5 on (iii) as sameactuallylol suggested.



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Gregorius
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#64
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(Original post by sameactuallylol)
For the last result I got \frac{25n-9}{(n+1)^5} could someone check which is right im not that confident because of all the algebra and sorry if someone posted it further down in the thread
Yes, I've just ground through this horror myself and agree with your answer!
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Notnek
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#65
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DFranklin do you mind checking the last part of my Q9 solution above when you have time? The last part seemed too simple and I feel uneasy not having used x\sin\alpha \geq d\sin \beta anywhere. Thanks.

EDIT: Answered below.
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JeremyC
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#66
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#66
(Original post by Notnek)
DFranklin do you mind checking the last part of my Q9 solution above when you have time? The last part seemed too simple and I feel uneasy not having used x\sin\alpha \geq d\sin \beta anywhere. Thanks.
I think it looks good. In the first line the second term of the derivative needs to be multiplied by \frac{1}{2}, but you clearly meant to do that as the next line is correct.

The condition x\sin{\alpha} \geq d \sin{\beta} is required to ensure that k^2x - kd is non-negative, and hence that we can take the square root of it in the expression for the total time.
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Notnek
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#67
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(Original post by JeremyC)
I think it looks good. In the first line the second term of the derivative needs to be multiplied by \frac{1}{2}, but you clearly meant to do that as the next line is correct.

The condition x\sin{\alpha} \geq d \sin{\beta} is required to ensure that k^2x - kd is non-negative, and hence that we can take the square root of it in the expression for the total time.
Thanks I made the correction. Do you think this explanation of it being non-negative is required anywhere throughout the working to get full marks? I'm still not used to STEP questions, hence why I'm starting to practice them
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JeremyC
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#68
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(Original post by Notnek)
Thanks I made the correction. Do you think this explanation of it being non-negative is required anywhere throughout the working to get full marks? I'm still not used to STEP questions, hence why I'm starting to practice them
I don't know exactly how the questions get marked, but if I were marking that question I'd probably give one mark for showing the expression inside the square root is non-negative.

For full marks, I'd look out for particular cases which are not allowed - e.g. dividing by zero, or if working with the reals taking the square root of a negative number or the logarithm of a non-positive number. Try to find some reason why that problem case cannot occur based on the conditions given in the question. If this is non-obvious, the question will often ask you to show a preliminary step (as with question 9 and showing that x\sin{\alpha} \geq d\sin{\beta}) which will then need to be applied later on. The STEP examiners have been writing the 2018 papers since at least 2016, and have thought very carefully about how to lead very talented mathematicians to the answers.
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Cai Jianlin
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#69
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I have got about 80-85, what grade should I have?
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mikelbird
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#70
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Question 10 of paper 1
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KingCrepe
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#71
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(Original post by Cai Jianlin)
I have got about 80-85, what grade should I have?
1.
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Cai Jianlin
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#72
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(Original post by KingCrepe)
1.
thx, I dun think I can have an S in this paper
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I hate maths
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#73
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Step 2 question 5 (if there are mistakes let me know)

Edit: corrected a grammatical thing and a missing minus sign.
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I hate maths
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#74
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Step 2 question 2 (if there are mistakes let me know)

I might do the sketch for the stem of the question later if I'm feeling it. Edit: sketch is here, and thank you A02 for making a suggestion which reduces workload in part i.

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Commentary on the sketch:
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Hopefully you can see that f(x) is concave when the "y-coordinate" of Y  \geqslant "y-coordinate" of X. A small amount of algebra leads to the desired inequality.

Also, f''(x)<0 for a<x<b means the gradient of f(x) is decreasing for a<x<b, meaning it is concave over a<x<b.
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zeldor711
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#75
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I'll probably put on Q4 when I get home, maybe Q8 as well and Q1 if by some chance no one has done it.
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Oskarrrrrrr
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#76
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(Original post by zeldor711)
I'll probably put on Q4 when I get home, maybe Q8 as well and Q1 if by some chance no one has done it.
I was gonna put up q1 but if you are I’ll drop q8 the differential eq
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Zacken
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#77
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(Original post by Oskarrrrrrr)
I was gonna put up q1 but if you are I’ll drop q8 the differential eq
I'm just about to post Q8
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zeldor711
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#78
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(Original post by Oskarrrrrrr)
I was gonna put up q1 but if you are I’ll drop q8 the differential eq
Nah, its fine, put up whatever you want to. Maybe I'll do question 9 if I can figure out what I did wrong.
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Zacken
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#79
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STEP II, Question 8

i) Let v = \sqrt{y} so that \frac{dv}{dt} = \frac{dv}{dy} \frac{dy}{dt} \implies \frac{dy}{dt} = 2\sqrt{y}\frac{dv}{dt} and the DE becomes

\displaystyle

\begin{equation*}2v \frac{dv}{dt} = \alpha v - \beta v^2  \iff \frac{dv}{dt} = \frac{1}{2}(\alpha - \beta v) \iff \log(\alpha-\beta v) = c-\frac{\beta t}{2}\end{equation*}

Hence \displaystyle \alpha-\beta v = \alpha\exp\left(-\frac{\beta t}{2}\right) after imposing boundary conditions and so we get \displaystyle y = \left(\frac{\alpha}{\beta}\right  )^2\left(1-\exp\left(-\frac{\beta t}{2}\right)\right)^2

ii) Sub in v = y^{1/3} so that you get

\displaystyle 

\begin{equation*}3v^2 \frac{dv}{dt} = \alpha v^2 - \beta v^3 \iff 3 \frac{dv}{dt} = \alpha - \beta v\end{equation*}

and so we get

\displaystyle 

\begin{equation*}y = \left(\frac{\alpha}{\beta}\right  )^3 \left(1-\exp\left(-\frac{\beta t}{3}\right)\right)^3\end{equati  on*}

iii) Check Desmos. They should both intersect at 0, and some easy size analysis of the exponential for various values of t will get you the graph. (y1 > y2 for t > 0)
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Oskarrrrrrr
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#80
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(Original post by Zacken)
I'm just about to post Q8
Okay wicked you post yours then I’ll put mine up too I had quite a funky method for the graphing ! Best of luck with results
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