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    (Original post by Firestartc)
    Ikr trickier, i could answer all the questions but i think i lost 4 marks in total so 71/75.
    One for not writing 3sec^3 (theta) or wtv, as i wrote the sec^2(theta) term over cos(theta). Another for putting 1/y^3 not with the -1. Another for not simplifying 1/(3+(x/3)) to remove the fraction in the denominator and another for not writing +c in the integrals (tbh i can't remember if i wrote them so i'll assume i didn't). Im so done with my life tbh, i can't stop making silly errors in maths..
    I created a poll to see what the grade boundaries will be like compared to last year's, plz help me by filling it out, ty:
    http://www.strawpoll.me/10431970
    Done what were the grade boundaries last year anyhow
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    (Original post by IrrationalRoot)
    Some answers:

    1. 0.74379 to 5dp
    Multiply this by e: 2.02183 to 5dp.

    2. 30, 150, 196.6, 343.4 degrees

    146.3,326.3 degrees

    3. \dfrac{2}{3}

    4. -\frac{1}{2}\tan 3t
    -\frac{1}{8}\sec^3 3t
    -\dfrac{1}{y^3}

    5. Perimeter = Arc Length + Chord Length, arc length is easy, chord length can be found with a right-angled triangle with angle \dfrac{\theta}{2}.
    Usual stuff with iterations etc; for function use f(\theta)=\theta +2\sin\frac{\theta}{2}-4.5

    6. An easy counterexample is a=b=1, c=2, d=3

    7. x=-\dfrac{1}{3} (x=2 doesn't work)

    8. a=-3,5
    b=-\frac{2}{3}

    9. f^{-1}(x)=12-3\ln(x-8)
    Domain is [9,+\infty)

    10. To show hh(x)=x simply expand out the left.
    So it is clear that h is its own inverse and so h^{-1}(-1)=\frac{1}{9}

    That's all I can remember, please correct anything that's wrong .

    EDIT: Just realised I left out diff/int questions.

    Differentiation: \ln(\cos x) gives -\tan x,

    \tan^{-1}(\frac{x}{3}) gives \dfrac{3}{9+x^2}

    e^{6x}(3x-2)^4 gives 18xe^{6x}(3x-2)^3

    Integration: a=2.3
    Im sorry but could you please walk me through how you got all of your 4 roots in 2a?
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    (Original post by caitanna98)
    Done what were the grade boundaries last year anyhow
    55/75 for 80/100 UMS, 62/75 for 90/100 UMS and 69/75 for 100/100 UMS
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    (Original post by JessyMarie)
    Idea on grade boundaries?
    Hard to tell without a large enough number of people answering the survey i posted I guess, haha
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    (Original post by caitanna98)
    Unofficial mark scheme?
    I made one in post 89!
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    (Original post by Ceridwen101)
    your answer to the differential involving e^x is wrong, you need to use the product rule
    If you're talking about the post above, I did use the product rule.
    Unfortunately if you think it's wrong that must mean you got the wrong answer.
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    (Original post by Firestartc)
    55/75 for 80/100 UMS, 62/75 for 90/100 UMS and 69/75 for 100/100 UMS
    I think I probably scraped my A* then, as I think I got about 63/75, and with slightly lower grade boundaries I bet will happen, I should be okay thank god (even if I was a couple of marks lower than that). C4'll be the deciding factor
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    (Original post by IrrationalRoot)
    If you're talking about the post above, I did use the product rule.
    Unfortunately if you think it's wrong that must mean you got the wrong answer.
    Or you could have got the answer wrong ? :/
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    (Original post by JessyMarie)
    Or you could have got the answer wrong ? :/
    nah, he got it correct
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    (Original post by Maths is Life)
    Im sorry but could you please walk me through how you got all of your 4 roots in 2a?
    I did say
    "First part of Q2 is rearranging to get an equation in using , which is a standard question."
    in a previous post but I'll give you the full working:

    3\csc\theta(\csc\theta -1)=5\cot^2\theta-9



\Rightarrow 3\csc^2\theta-3\csc\theta=5(\csc^2\theta-1)-9=5\csc^2\theta-5-9=5\csc^2\theta-14



\Rightarrow 2\csc^2\theta+3\csc\theta-14=(2\csc\theta+7)(\csc\theta-2)=0



\Rightarrow \csc\theta=-\frac{7}{2} \mathrm{or} \csc\theta=2



\Rightarrow \sin\theta=-\frac{2}{7} \mathrm{or} \sin\theta=\frac{1}{2}

    Hence solutions.
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    (Original post by JessyMarie)
    Or you could have got the answer wrong ? :/
    No my answer is correct.
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    has any one got a copy of the paper?
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    (Original post by Hi:))
    has any one got a copy of the paper?
    See post 179 on the previous page.
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    (Original post by JessyMarie)
    Any news on a unofficial mark scheme?
    theres a mark scheme up [email protected] twitter
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    (Original post by Hi:))
    theres a mark scheme up [email protected] twitter
    Yh I know I got it lol
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    (Original post by IrrationalRoot)
    If you're talking about the post above, I did use the product rule.
    Unfortunately if you think it's wrong that must mean you got the wrong answer.
    product rule is Vdu + Udv so the answer would be 12e^6x(3x-2)^3 + 6e^6x(3x-2)^4
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    (Original post by Ceridwen101)
    product rule is Vdu + Udv so the answer would be 12e^6x(3x-2)^3 + 6e^6x(3x-2)^4
    Which drops a mark or two since it's not simplified.
    Btw don't straight up call someone's answer 'wrong' without making sure it actually is wrong first.
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    (Original post by IrrationalRoot)
    Which drops a mark or two since it's not simplified.
    Btw don't straight up call someone's answer 'wrong' without making sure it actually is wrong first.
    how can you simplify that, the brackets aren't to the same power
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    (Original post by Ceridwen101)
    how can you simplify that, the brackets aren't to the same power
    Factorise.
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    (Original post by IrrationalRoot)
    Factorise.
    Would it count if I took 6e^6x out and put the rest in brackets? Or would that not be considered fully simplified?
 
 
 
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