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Arsey's C3 Edexcel 25th Jan 2013 Model Answers Thread (attached to first post)

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    Here are my model answers

    The Core 3 paper wasn't anywhere near as bad as I expect some of you think it was. There were a couple of nightmare parts to questions but most of it was fairly standard and a well prepared candidate should have been able to achieve a decent mark out of 75.

    Q1 Differentiation a) easy b) pretty easy - 7 marks (all okay)


    Q2 Numerical methods a) easy (you don't need the bit I did in square brackets) b) easy - 8 marks (all okay)


    Q3 Transforming graphs. This was a pretty easy question - 9 marks (all okay)


    Q4 Trigonometry a) very easy, it is on virtually every paper b) This is pretty hard, understanding what to choose to lead to the maximum of p wasn't easy - 8 marks (4 easy, 4 hard)

    Q5
    Differentiation. Pretty easy, the proof was fairly easy - 11 marks (all okay)


    Q6 i) Trigonometry i) I think this will be poorly answered but it is pretty easy ii) and b) easy - 11 marks (all okay)


    Q7 Algebraic fractions and differentiation. Fairly easy - 12 marks (all okay)


    Q8 Differentiation a) stupidly easy b) very hard - if you got this right, well done c) reasonable - 9 marks (5 okay - 4 hard)


    So I think there are 8 hard marks that anyone taking the exam will have done well to get right.

    In terms of boundaries I would be guessing at

    52 = 70ums
    59 = 80ums
    66 = 90ums
    73 = 100ums
    Attached Files
  2. File Type: pdf C3 Jan 13 Model red.pdf (698.1 KB, 28641 views)
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    Thanks ARSEY
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    Omg ive been waiting all morning i love arsey
  5. Offline

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    (Original post by Arsey)
    Here are my model answers
    hi arsey i got 1+sin45 ?

    what do u predict the grade boundaries cause everyone got 2/11 not 2 for that question
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    (Original post by Arsey)
    Here are my model answers
    Arsey:

    Would I get all the marks? Question 6i)

     (sin(22.5) + cos(22.5))^2

    Now I know this is straight forward but I went about it in a completely different way:

    using the half angle formula:
     cos^2(22.5) = \displaystyle\frac{cos(45)+1}{2}
     cos(22.5) = \displaystyle\sqrt{\frac{cos(45)  +1}{2}}
     sin^2(22.5) = \displaystyle\frac{1-cos(45)}{2}
     sin(22.5) = \displaystyle\sqrt{\frac{1-cos(45)}{2}}

    expanding:

     sin^2(22.5) + 2cos(22.5)(sin(22.5)) + cos^2(22.5)
    then I sub in all the values found and get the answer to be  1 + \frac{\sqrt{2}}{2} which is correct, but would I get all the marks?

    I've shown every step in my working, and didn't use a calculator at all to get to the final step.
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    Argh for cos2theta +sintheta equals to one, i put as 2sintheta minus sintheta equals to one. NOOOOOO I lost all the marks for that.
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    (Original post by Arsey)
    Here are my model answers
    thank you so much arsey! wanna ask, if i got 594 pounds instead of 593 for the last question, will i still get full marks?
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    Thanks alot arsey! DDDDDD
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    (Original post by bubblegummer)
    thank you so much arsey! wanna ask, if i got 594 pounds instead of 593 for the last question, will i still get full marks?
    I think you will lose 1 accuracy
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    I loved that paper, found 8b and the sin22.5 one but solved them both in the final half hour so pretty sure I got at least 92%


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    I completely missed the simple method for evaluating Q6a), and ended up butchering the method with a method that I am 90% sure isn’t the expected one, but also 90% sure that it’s completely valid. Would the following be awarded all of the marks?

    Spoiler:
    Show


     cos2\theta \equiv 1-2sin^2\theta \equiv 2cos^2\theta – 1

     \rightarrow sin\theta = \sqrt{\dfrac{1}{2}-\dfrac{1}{2}cos2\theta}

     \rightarrow cos\theta =  \sqrt{\dfrac{1}{2}cos2\theta +\dfrac{1}{2}}

     \rightarrow (sin(22.5^{\circ}) +cos(22.5^{\circ}))^2 = \left(\sqrt{\dfrac{1}{2}-\dfrac{1}{2}cos45^{\circ}}+ \sqrt{\dfrac{1}{2}cos45^{\circ}+  \dfrac{1}{2}}\ \right)^2

     = \left(\sqrt{\dfrac{2-\sqrt{2}}{4}}+ \sqrt{\dfrac{2+\sqrt{2}}{4}} \ \right)^2

     = \dfrac{2-\sqrt{2}}{4} +\dfrac{2+\sqrt{2}}{4} +2\sqrt{\dfrac{2-\sqrt{2}}{4}} \sqrt{\dfrac{2+\sqrt{2}}{4}}

     = 1 + 2\sqrt{\dfrac{1}{8}}

     = 1+\dfrac{1}{\sqrt{2}}

     = \dfrac{2+\sqrt{2}}{2}



    (Note: May this be a lesson to us all to think about the question before launching into algebra.)
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    (Original post by otrivine)
    what do u think the grade boundaries will be? I am sure majority will have got 2/11 wrong well everyone in my school got 2/11.
    what was difficult about Q2?
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    I think this paper was hard purely because of exam cond. under pressure its sooo much more difficult to think around a problem like 8b. if it wasnt exam cond. i probably would have got it right...
    If its 59 out of 75 for an A then Ill be over the moon
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    (Original post by thorn0123)
    ....
    Glad I'm not the only one who completely missed the easy method for Q6.
  16. Offline

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    (Original post by DJMayes)
    I completely missed the simple method for evaluating Q6a), and ended up butchering the method with a method that I am 90% sure isn’t the expected one, but also 90% sure that it’s completely valid. Would the following be awarded all of the marks?

    Spoiler:
    Show


     cos2\theta \equiv 1-2sin^2\theta \equiv 2cos^2\theta – 1

     \rightarrow sin\theta = \sqrt{\dfrac{1}{2}-\dfrac{1}{2}cos2\theta}

     \rightarrow cos\theta =  \sqrt{\dfrac{1}{2}cos2\theta +\dfrac{1}{2}}

     \rightarrow (sin(22.5^{\circ}) +cos(22.5^{\circ}))^2 = \left(\sqrt{\dfrac{1}{2}-\dfrac{1}{2}cos45^{\circ}}+ \sqrt{\dfrac{1}{2}cos45^{\circ}+  \dfrac{1}{2}}\ \right)^2

     = \left(\sqrt{\dfrac{2-\sqrt{2}}{4}}+ \sqrt{\dfrac{2+\sqrt{2}}{4}} \ \right)^2

     = \dfrac{2-\sqrt{2}}{4} +\dfrac{2+\sqrt{2}}{4} +2\sqrt{\dfrac{2-\sqrt{2}}{4}} \sqrt{\dfrac{2+\sqrt{2}}{4}}

     = 1 + 2\sqrt{\dfrac{1}{8}}

     = 1+\dfrac{1}{\sqrt{2}}

     = \dfrac{2+\sqrt{2}}{2}



    (Note: May this be a lesson to us all to think about the question before launching into algebra.)
    haha

    I done the exact same thing - I walked out the exam thinking "I must be the only person to have gotten that question, it was so hard" Then to realise it was so easy :|
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    (Original post by DJMayes)
    I completely missed the simple method for evaluating Q6a), and ended up butchering the method with a method that I am 90% sure isn’t the expected one, but also 90% sure that it’s completely valid. Would the following be awarded all of the marks?

    Spoiler:
    Show


     cos2\theta \equiv 1-2sin^2\theta \equiv 2cos^2\theta – 1

     \rightarrow sin\theta = \sqrt{\dfrac{1}{2}-\dfrac{1}{2}cos2\theta}

     \rightarrow cos\theta =  \sqrt{\dfrac{1}{2}cos2\theta +\dfrac{1}{2}}

     \rightarrow (sin(22.5^{\circ}) +cos(22.5^{\circ}))^2 = \left(\sqrt{\dfrac{1}{2}-\dfrac{1}{2}cos45^{\circ}}+ \sqrt{\dfrac{1}{2}cos45^{\circ}+  \dfrac{1}{2}}\ \right)^2

     = \left(\sqrt{\dfrac{2-\sqrt{2}}{4}}+ \sqrt{\dfrac{2+\sqrt{2}}{4}} \ \right)^2

     = \dfrac{2-\sqrt{2}}{4} +\dfrac{2+\sqrt{2}}{4} +2\sqrt{\dfrac{2-\sqrt{2}}{4}} \sqrt{\dfrac{2+\sqrt{2}}{4}}

     = 1 + 2\sqrt{\dfrac{1}{8}}

     = 1+\dfrac{1}{\sqrt{2}}

     = \dfrac{2+\sqrt{2}}{2}



    (Note: May this be a lesson to us all to think about the question before launching into algebra.)
    Very impressive and mental in equal measures. Well done.
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    (Original post by Arsey)
    what was difficult about Q2?
    No the fraction one cause I assumed R=10 we just put that into the fraction to give 2/11
  19. Offline

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    (Original post by Arsey)
    what was difficult about Q2?
    A lot of people thought maximum therefore cos(theta - 0.92...) = 1 and subbed that in, as it's usually like that.
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    (Original post by Arsey)
    Very impressive and mental in equal measures. Well done.
    Would we get all the marks?
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    (Original post by bubblegummer)
    thank you so much arsey! wanna ask, if i got 594 pounds instead of 593 for the last question, will i still get full marks?
    if you had the correct full answers and then just rounded incorrectly you may get full marks

    it must be 2 marks for differentiating, method for substituting then an answer mark
Updated: June 13, 2013
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