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    Using an energy method I got the same value of b. My c was different, but I think I was wrong on this part (Question 9)
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    (Original post by savetheworms)
    I thought the last part was something like ln(1+tanx) ? sorry I have a rubbish memory lol
    That was second to last.


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    (Original post by physicsmaths)
    There may be other ways to do the very last part but I couldn't find any that worked. I tried a t sub but it required use of tan(pi/8) which is.. Ugly.


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    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 \int \dfrac{x}{cosxcos(x-pi/4)*sqrt{2}}

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

    Then 2J = pi/4 * \int \dfrac{1}{cosx cos (x-pi/4)*sqrt{2}} from 0 to pi/8
    then J = pi/8 * \int \dfrac{1}{cos^2 x + sinxcosx}
    then J = pi/4 * \int \dfrac{1}{cos 2x + sin 2x + 1} from 0 to pi/8
    then J = pi/8 * \int \dfrac{1}{cos x + sin x + 1} from 0 to pi/4
    then use t sub
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    How did everyone find the paper compared to last year?

    For me it was absolutely dire. I freaked out under the pressure and did really badly. I got a full solution to one question, and all of the first mechanics question (but stupidly left my answer in terms of M, m, u, R instead of a). I got all of the integration question except I got stuck at integrating 1/[cosx(cosx + sinx)] - I used a t substitution and was left floundering around. Then the rest of my "solutions" were basically me desperately trying to scrape marks together.

    Unless grade boundaries are *really* low, I think that's me out
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    (Original post by physicsmaths)
    Need S,1. Will be close in II.Posted from TSR Mobile
    Best of luck then. Still, wouldn't they be likely to accept you even if you got 1,1?
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    (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:

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

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

    Then 2J = pi/4 * \int \dfrac{1}{cosx cos (x-pi/4)*sqrt{2}} from 0 to pi/8
    then J = pi/8 * \int \dfrac{1}{cos^2 x + sinxcosx}
    then J = pi/4 * \int \dfrac{1}{cos 2x + sin 2x + 1} from 0 to pi/8then J = pi/8 * \int \dfrac{1}{cos x + sin x + 1} from 0 to pi/4then use t sub
    I got to your line
    J=pi/8 int (1/(cos^2(x)+sinxcosx)) then divide top and bottom by cos^2(x) giving sec^2(x)/(1+tanx) which is ln(1+tanx) yielding same answer as the line before that.


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    (Original post by jjsnyder)
    For sure, I only got 1 full


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    Me too. Q7 integration and the rest were either mostly full or just halfs. Q3,8 were nearly full and Q4,1,2 were halfs.
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    (Original post by physicsmaths)
    That was second to last.


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    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?
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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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    (Original post by physicsmaths)
    I got to your line
    J=pi/8 int (1/(cos^2(x)+sinxcosx)) then divide top and bottom by cos^2(x) giving sec^2(x)/(1+tanx) which is ln(1+tanx) yielding same answer as the line before that. Posted from TSR Mobile
    That's probably the slicker answer they were looking for
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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.
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    (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.
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    (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
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    ~70 for a grade 1 or higher?
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    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.
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    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
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    (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 \int \dfrac{x}{cosxcos(x-pi/4)*sqrt{2}}

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

    Then 2J = pi/4 * \int \dfrac{1}{cosx cos (x-pi/4)*sqrt{2}} from 0 to pi/8
    then J = pi/8 * \int \dfrac{1}{cos^2 x + sinxcosx}
    then J = pi/4 * \int \dfrac{1}{cos 2x + sin 2x + 1} from 0 to pi/8then J = pi/8 * \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.
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    (Original post by sweeneyrod)
    I got to your last line, but couldn't get any further.
    Lol, the LaTex compiler only working for your quote

    It falls out with t-sub (Weierstrauss) - you get (pi/8) * int 2/(2+2t) dt from t=0 to t=1, and result follows
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    (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?
 
 
 
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