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STEP 2016 Solutions

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Original post by Mathemagicien
Best of luck then. Still, wouldn't they be likely to accept you even if you got 1,1?


Duno, will found it in august.
My college is pretty lenient given they accepted every person with a 1 in either paper. And only around hakf /10 got 1,1>=


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For the a,b,c,d question i.e. q2, was the substitution you were supposed to make for the last part x,1,2,3? (I did this) How many marks do you think would be received for completing that question entirely but messing up the final substitution of factors and so getting the wrong final solutions for x (the actual numbers which I think are actually x = 0,2,4). I hope that made sense.
Original post by savetheworms
oh then yeah i agree with what you put
did you remember to put +c i remembered literally seconds after walking out the exam hall urgh how much do you reckon they''ll penalise that?


It was all definite integrals.... Sorry


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Original post by physicsmaths
I have a nice answer aswell. It was integral 0 to pi/4 of x/(cosx)(cosx+sinx)
I am saying what other ways were there other then mine cause First i tried t sub then after divide top bottom by cos^2(x) giving a nice solution.


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I divided by cos(x) on a whim and then realised I could divide by it again to get an integral in ln(...) form. Pretty nice if you ask me.
Original post by physicsmaths
It was all definite integrals.... Sorry


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i;m a moron hahahaa how did i not realise that ytf am i doing maths at uni lol thanks
~70 for a grade 1 or higher?
I had a look at the paper and thought 5-8, 12-13 looked alright. 1-4 looked less inviting / longer than usual (but not by much). Overall I thought the paper was on the easier side - don't shoot me.

My estimates last night were 99 / 70 / 60 (I think - check with Insight) for the S / 1 / 2 boundaries i.e. fairly similar to 2009. I hadn't tried many of the questions though, so I might have to revise that estimate.
@physicsmaths @Zacken How many marks do you think I will lose for the first question for misreading the question and treating the question as if the tangents were perpendicular. oops! Got the rest of the question right but obviously I got a different condition on p and q therefore couldn't show that the intersection coordinates satisfied the curve c2
(edited 7 years ago)
Original post by Mathemagicien
I'm having difficulty reading your solution, so perhaps this is the method you did, however this is roughly what I did (IIRC):

Note the last integral J is xcosxcos(xpi/4)sqrt2\int \dfrac{x}{cosxcos(x-pi/4)*sqrt{2}}

so use formula to say J = pi/4xcosxcos(xpi/4)sqrt2\int \dfrac{pi/4 - x}{cosxcos(x-pi/4)*sqrt{2}}

Then 2J = pi/4 * 1cosxcos(xpi/4)sqrt2\int \dfrac{1}{cosx cos (x-pi/4)*sqrt{2}} from 0 to pi/8
then J = pi/8 * 1cos2x+sinxcosx\int \dfrac{1}{cos^2 x + sinxcosx}
then J = pi/4 * 1cos2x+sin2x+1\int \dfrac{1}{cos 2x + sin 2x + 1} from 0 to pi/8then J = pi/8 * 1cosx+sinx+1\int \dfrac{1}{cos x + sin x + 1} from 0 to pi/4then use t sub


I got to your last line, but couldn't get any further.
Reply 89
Original post by shamika
I had a look at the paper and thought 5-8, 12-13 looked alright. 1-4 looked less inviting / longer than usual (but not by much). Overall I thought the paper was on the easier side - don't shoot me.


Q1 and 3 were trivial. Q6 was near impossible. I've heard of only 1 person managing the last part of Q6 out of 30 or so of the more able people. Any guesses for the mark distribution of Q6?

Edit to add: you've said it was on the easier side and then given lower boundaries for a 1 than 2010-2014?
(edited 7 years ago)
General thoughts on the difficulty? I thought it was fairly average, no obvious gift questions but quite a few questions that it was easyish to get started with.
I think I've got two ~fulls (Q3 and Q12), significant chunks of Q1, Q7 (I think that was the integration one?), and the differential equation question, and 2-3 marks on Q5. On the integration, I did everything up to getting an integral of 1/cos x + sin x + 1 in the last part, so hopefully that will be 15ish, and on d.e. I did everything up to the last part. Hopefully that'll be a 1.
Original post by sweeneyrod
I got to your last line, but couldn't get any further.


before you use the double angle formulae you can just divide through by (cos)^2 and then use sub u = tanx i think
Original post by sweeneyrod
General thoughts on the difficulty? I thought it was fairly average, no obvious gift questions but quite a few questions that it was easyish to get started with.


Slightly easier then last year. I guess
95/70 for S/1.
I was horrified tbh, no vectors and geometry set me back massively. ****inf ********.


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Original post by Zacken
Q1 and 3 were trivial. Q6 was near impossible. I've heard of only 1 person managing the last part of Q6 out of 30 or so of the more able people. Any guesses for the mark distribution of Q6?

Edit to add: you've said it was on the easier side and then given lower boundaries for a 1 than 2010-2014?


Sorry, what was q3 again? Was that the f_n(x) one and proving that there are one or no solutions.

Also, which one was q6? I thought q6 was the DE one but I may be completely wrong.

On another note, how did people find the question where for the last part you had to find tan(theta) and x given y =(3+sqrt(7))/2, or something like that.
Reply 95
Q2 STEP II Solution
(edited 7 years ago)
Reply 96
Original post by tridianprime
Sorry, what was q3 again? Was that the f_n(x) one and proving that there are one or no solutions.

Also, which one was q6? I thought q6 was the DE one but I may be completely wrong.

On another note, how did people find the question where for the last part you had to find tan(theta) and x given y =(3+sqrt(7))/2, or something like that.


Yep, that was Q3. And Q6 was the DE, the last part of Q6 was the bit about y_(mn).
Original post by physicsmaths
Slightly easier then last year. I guess
95/70 for S/1.
I was horrified tbh, no vectors and geometry set me back massively. ****inf ********.


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I was actually sad about no vectors, because I did some vectors questions the day before and felt super prepared.
For the last integration I found a neat trick, split 1/cosx(cosx+sinx) into (sinx-cosx)/cosx + 2cosx/(sinx+cosx)
Original post by tridianprime
Sorry, what was q3 again? Was that the f_n(x) one and proving that there are one or no solutions.

Also, which one was q6? I thought q6 was the DE one but I may be completely wrong.

On another note, how did people find the question where for the last part you had to find tan(theta) and x given y =(3+sqrt(7))/2, or something like that.

i liked it, what was your answer?

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