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    What did everyone get for the vectors questions?


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    I sketched the graphs in the answer booklet, not on the graph paper. On the graph paper, I wrote: "See answer booklet". Will my graph sketches still be marked?
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    Easy exam. Except last part of the geometry question which i skipped lol
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    (Original post by K-Man_PhysCheM)
    I sketched the graphs in the answer booklet, not on the graph paper. On the graph paper, I wrote: "See answer booklet". Will my graph sketches still be marked?
    So we were meant to do it on graph paper?? I did the opposite and wrote on the answer booklet that my sketch was on the graph paper!

    I'm sure they'll mark it - it's only a sketch anyway so graph paper isn't really necessary for it


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    (Original post by LaurenLovesMaths)
    What did everyone get for the vectors questions?


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    For part (c), I got area = 7 \sqrt{101}; can't remember the vectors, but I think I got p = -8
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    (Original post by K-Man_PhysCheM)
    For part (c), I got area = 7 \sqrt{101}; can't remember the vectors, but I think I got p = -8
    Good because I got the same! Can't remember my vectors for C and B though but we needed that for the last part so I'm guessing it was right. I was really confused about the root 101 though so I'm glad it seems right 😂


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    (Original post by LaurenLovesMaths)
    So we were meant to do it on graph paper?? I did the opposite and wrote on the answer booklet that my sketch was on the graph paper!

    I'm sure they'll mark it - it's only a sketch anyway so graph paper isn't really necessary for it


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    I've not done an Edexcel Maths exam before, but everyone else I know used the graph paper. I didn't see any need to use it tbh, and I made sure the graphs looked clear. The second one was symmetric with mirror line on the y-axis right?
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    (Original post by K-Man_PhysCheM)
    I've not done an Edexcel Maths exam before, but everyone else I know used the graph paper. I didn't see any need to use it tbh, and I made sure the graphs looked clear. The second one was symmetric with mirror line on the y-axis right?
    Yeah it was. My graph wasn't great so I added a label saying it should be symmetrical about the y-axis to avoid any ambiguity 😂😂


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    (Original post by LaurenLovesMaths)
    Good because I got the same! Can't remember my vectors for C and B though but we needed that for the last part so I'm guessing it was right. I was really confused about the root 101 though so I'm glad it seems right 😂


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    For OC, t was 3 I think.
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    (Original post by LaurenLovesMaths)
    Yeah it was. My graph wasn't great so I added a label saying it should be symmetrical about the y-axis to avoid any ambiguity 😂😂


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    Haha 😂 that's good, how did you find the rest of the paper?? I thought it was pretty nice tbh, and I answered every question! (probably not all right tho lol)
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    (Original post by K-Man_PhysCheM)
    Haha 😂 that's good, how did you find the rest of the paper?? I thought it was pretty nice tbh, and I answered every question! (probably not all right tho lol)
    Yeah I liked the paper! I answered all of them too. The only question that I didn't like was the last part of the triangle one, where I got like the square root of 3piroot3 which is definitely not right haha

    Hope grade boundaries aren't too high though eek


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    (Original post by LaurenLovesMaths)
    Yeah I liked the paper! I answered all of them too. The only question that I didn't like was the last part of the triangle one, where I got like the square root of 3piroot3 which is definitely not right haha

    Hope grade boundaries aren't too high though eek


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    For the triangle question, I drew the 6 equilateral triangles into the shape of a hexagon, with each triangle having a vertex in the middle of the hexagon. Let P be the point on LM and Q be the point on LN through which the curve crosses. I used the information given in the question to conclude that the smallest length PQ would be part of the arc of a circle drawn through all six triangles with centre at the centre-point of the hexagon.

    The big circle would have half the area of the whole shape.

    See my working below:
    Spoiler:
    Show





    Area of each triangle, A_t = \frac{1}{2} \times 2 \times 2 \times \sin60 = 2 \times \frac{1}{\sqrt2} = \sqrt2

    \Rightarrow A_{total} = 6\sqrt{2}
    \Rightarrow A_{circle} = 3\sqrt{2} = \pi r^2

    \Rightarrow r = \left( \dfrac{3 \sqrt2}{\pi} \right)^{\frac{1}{2}}

    PQ = \frac{1}{6} \times 2 \pi r = \dfrac{\pi r}{3} = \dfrac{\pi}{3} \times \left( \dfrac{3 \sqrt2}{\pi} \right)^{\frac{1}{2}} \equiv \left( \dfrac{3 \sqrt2 \pi^2}{9 \pi} \right)^{\frac{1}{2}}

    \Rightarrow PQ = \left( \dfrac{\pi \sqrt2}{3} \right)^{\frac{1}{2}}


    EDIT: \sin60 = \frac{\sqrt3}{2}, not \frac{1}{\sqrt2}, so all \sqrt2 above should be replaced with \sqrt{3}. And there I was thinking I could do A-level maths... probably got the rest of the paper wrong too then lol



    Yeah, this paper felt more straightforward than some of the others, so hopefully boundaries aren't too high!!
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    (Original post by K-Man_PhysCheM)
    For the triangle question, I drew the 6 equilateral triangles into the shape of a hexagon, with each triangle having a vertex in the middle of the hexagon. Let P be the point on LM and Q be the point on LN through which the curve crosses. I used the information given in the question to conclude that the smallest length PQ would be part of the arc of a circle drawn through all six triangles with centre at the centre-point of the hexagon.

    The big circle would have half the area of the whole shape.

    See my working below:
    Spoiler:
    Show




    Area of each triangle, A_t = \frac{1}{2} \times 2 \times 2 \times \sin60 = 2 \times \frac{1}{\sqrt2} = \sqrt2

    \Rightarrow A_{total} = 6\sqrt{2}
    \Rightarrow A_{circle} = 3\sqrt{2} = \pi r^2

    \Rightarrow r = \left( \dfrac{3 \sqrt2}{\pi} \right)^{\frac{1}{2}}

    PQ = \frac{1}{6} \times 2 \pi r = \dfrac{\pi r}{3} = \dfrac{\pi}{3} \times \left( \dfrac{3 \sqrt2}{\pi} \right)^{\frac{1}{2}} \equiv \left( \dfrac{3 \sqrt2 \pi^2}{9 \pi} \right)^{\frac{1}{2}}

    \Rightarrow PQ = \left( \dfrac{\pi \sqrt2}{3} \right)^{\frac{1}{2}}





    Yeah, this paper felt more straightforward than some of the others, so hopefully boundaries aren't too high!!
    I tried the hexagon thing & drew the right diagram, do you think I'll get any marks for that? (Well done btw!)

    & yeah I hope so too! I think last year's was okay too, maybe slightly harder, but I didn't like 2015's (although the 2015&2016 boundaries were very similar) so who knows? The boundaries always seem to be about 70 for a distinction - they never seem to change much so let's hope they don't this year😂


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    (Original post by LaurenLovesMaths)
    I tried the hexagon thing & drew the right diagram, do you think I'll get any marks for that? (Well done btw!)

    & yeah I hope so too! I think last year's was okay too, maybe slightly harder, but I didn't like 2015's (although the 2015&2016 boundaries were very similar) so who knows? The boundaries always seem to be about 70 for a distinction - they never seem to change much so let's hope they don't this year😂


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    For the diagram alone, you might get a mark though I suspect some sort of written plan may be required. Some of the working you did might get a couple method marks. It was certainly a tough question, probably the hardest on the paper (maybe tied with the integration question), and it sounds like you got the rest done so I wouldn't worry. And thanks!

    Yeah, now you mention it, the 2015 paper was pretty hard, I remember getting the integration question completely wrong when I tried it. However, in 2010 they had that freak 81/100 boundary for Distinction, which was crazy, so let's hope it's not crazy this year!!
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    (Original post by K-Man_PhysCheM)
    For the diagram alone, you might get a mark though I suspect some sort of written plan may be required. Some of the working you did might get a couple method marks. It was certainly a tough question, probably the hardest on the paper (maybe tied with the integration question), and it sounds like you got the rest done so I wouldn't worry. And thanks!

    Yeah, now you mention it, the 2015 paper was pretty hard, I remember getting the integration question completely wrong when I tried it. However, in 2010 they had that freak 81/100 boundary for Distinction, which was crazy, so let's hope it's not crazy this year!!
    Yeah I agree I think the triangle one was the hardest. & oh wow 81 is very high... really hope that's not the case! Not sure if it was easier or harder than 2010


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    [QUOTE=K-Man_PhysCheM;72512528]For the triangle question, I drew the 6 equilateral triangles into the shape of a hexagon, with each triangle having a vertex in the middle of the hexagon. Let P be the point on LM and Q be the point on LN through which the curve crosses. I used the information given in the question to conclude that the smallest length PQ would be part of the arc of a circle drawn through all six triangles with centre at the centre-point of the hexagon.

    The big circle would have half the area of the whole shape.

    See my working below:
    Spoiler:
    Show





    Area of each triangle, A_t = \frac{1}{2} \times 2 \times 2 \times \sin60 = 2 \times \frac{1}{\sqrt2} = \sqrt2

    \Rightarrow A_{total} = 6\sqrt{2}
    \Rightarrow A_{circle} = 3\sqrt{2} = \pi r^2

    \Rightarrow r = \left( \dfrac{3 \sqrt2}{\pi} \right)^{\frac{1}{2}}

    PQ = \frac{1}{6} \times 2 \pi r = \dfrac{\pi r}{3} = \dfrac{\pi}{3} \times \left( \dfrac{3 \sqrt2}{\pi} \right)^{\frac{1}{2}} \equiv \left( \dfrac{3 \sqrt2 \pi^2}{9 \pi} \right)^{\frac{1}{2}}

    \Rightarrow PQ = \left( \dfrac{\pi \sqrt2}{3} \right)^{\frac{1}{2}}






    Those should all be route 3s no?
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    [QUOTE=Planet Thunk;72512900]
    (Original post by K-Man_PhysCheM)
    For the triangle question, I drew the 6 equilateral triangles into the shape of a hexagon, with each triangle having a vertex in the middle of the hexagon. Let P be the point on LM and Q be the point on LN through which the curve crosses. I used the information given in the question to conclude that the smallest length PQ would be part of the arc of a circle drawn through all six triangles with centre at the centre-point of the hexagon.

    The big circle would have half the area of the whole shape.

    See my working below:
    Spoiler:
    Show







    Area of each triangle, A_t = \frac{1}{2} \times 2 \times 2 \times \sin60 = 2 \times \frac{1}{\sqrt2} = \sqrt2

    \Rightarrow A_{total} = 6\sqrt{2}
    \Rightarrow A_{circle} = 3\sqrt{2} = \pi r^2

    \Rightarrow r = \left( \dfrac{3 \sqrt2}{\pi} \right)^{\frac{1}{2}}

    PQ = \frac{1}{6} \times 2 \pi r = \dfrac{\pi r}{3} = \dfrac{\pi}{3} \times \left( \dfrac{3 \sqrt2}{\pi} \right)^{\frac{1}{2}} \equiv \left( \dfrac{3 \sqrt2 \pi^2}{9 \pi} \right)^{\frac{1}{2}}

    \Rightarrow PQ = \left( \dfrac{\pi \sqrt2}{3} \right)^{\frac{1}{2}}








    Those should all be route 3s no?
    I knew I'd have got something wrong somewhere!! Damn, you're right, \sin60 = \frac{\sqrt3}{2}, then as you said.
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    [QUOTE=K-Man_PhysCheM;72512946]
    (Original post by Planet Thunk)

    I knew I'd have got something wrong somewhere!! Damn, you're right, \sin60 = \frac{\sqrt3}{2}, then as you said.
    Omg in the exam I got a third x square root(pix3root3) which according to my calculator is the same answer as what you got, when you swap the root 2s with root 3s!! Mine's in a much more gross form though - I just hope they can tell it's the same answer😂


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    [QUOTE=LaurenLovesMaths;72513114]
    (Original post by K-Man_PhysCheM)

    Omg in the exam I got a third x square root(pix3root3) which according to my calculator is the same answer as what you got, when you swap the root 2s with root 3s!! Mine's in a much more gross form though - I just hope they can tell it's the same answer😂


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    Haha well done, yeah they should be able to tell so you should get all the marks for that question. Now it's me who's hoping they're generous with method marks!
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    [QUOTE=K-Man_PhysCheM;72513162]
    (Original post by LaurenLovesMaths)

    Haha well done, yeah they should be able to tell so you should get all the marks for that question. Now it's me who's hoping they're generous with method marks!
    I really hope they do😂 I can't believe I got it right - I did it in a different way to you too, which kind of worries me in case they do just think my answer is wrong... I'm sure you'll be fine because your method is good


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