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# STEP 2016 Solutions

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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
1. (Original post by Number Nine)
bagsy q1

Using the substitution we have

So that:

So:

i) Find

as required
Cleaning up the LaTeX a little for you.
..
2. (Original post by zacken)
cleaning up the latex a little for you.
..
yeah can
3. Anyone got the pdf? Willing to have a bash at any unpopular questions
4. wow everyone here is so clever. have fun at oxbridge you guys
5. (Original post by fefssdf)
wow everyone here is so clever. have fun at oxbridge you guys
We don't talk about Oxford round these parts

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6. (Original post by drandy76)
We don't talk about Oxford round these parts

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ok im backing off
7. (Original post by Llewellyn)
Anyone got the pdf? Willing to have a bash at any unpopular questions
http://www.thestudentroom.co.uk/show...9#post66077019
9. (Original post by Zacken)
Cleaning up the LaTeX a little for you.
..
http://www.thestudentroom.co.uk/show...9#post66074739
10. (Original post by student0042)
..
11. (Original post by Number Nine)
bagsy q1

apparently my latex is ugly
How many marks do you reckon the first part will be? 5-6?
12. (Original post by Shrek1234)
How many marks do you reckon the first part will be? 5-6?
4-5 id say Zacken opinion
13. (Original post by Shrek1234)
How many marks do you reckon the first part will be? 5-6?
(Original post by Number Nine)
4-5 id say Zacken opinion
14. For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1-x)+(1/1-x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2-x would be the solution as the -x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x
15. (Original post by fefssdf)
wow everyone here is so clever. have fun at oxbridge you guys
thanks fam ill try ot enjoy it
16. (Original post by Ewanclementson)
For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1-x)+(1/1-x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2-x would be the solution as the -x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x
Nice! Yes I think it would be.
17. (Original post by Zacken)
STEP III: Question 4

Sum to N
Note that

So, it follows, by re-writing and telescoping, we have:

Sum to infinity

Now, take , so that for we have meaning that our sum is:

Hyperbolic sum

Write the hyperbolic sum in exponential form:

So take and use the previous result, noting that we have:

Now, for the final result, it's simply double the sum and adding on to account for the term. This cleans up to
OMG OMG OMG OMG I Totally forgot that I had done most of this question too!! I think I'm gonna get 13 marks out of it because I used x=e^2y and didn't do the lastest bit. You know what i might actually get a 1 out of that messy exam.
18. (Original post by Hext)
You would not be deducted more than 4 marks.
Try quote people

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19. Quite liked the simple harmonic question (q9) so I thought I'd write up and post a solution:
Attached Images

20. (Original post by L-Tyrosine)
Quite liked the simple harmonic question (q9) so I thought I'd write up and post a solution:
How many marks do you think I'd get for getting all the way to the F=ma part and then stoping because I was too scared of the algebra hahaha ?

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