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    [QUOTE=LaurenLovesMaths;72517016]
    (Original post by Redcoats)

    Again, really don't want to jinx it😂 but, providing that the examiner knows my answer to Q4c is the same but in a grosser form, I think/hope I've done all the questions pretty well

    What about you?


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    My absolute minimum is 69 but it could be anything above that (bear in mind I'm in year 12)
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    [QUOTE=Redcoats;72517074]
    (Original post by LaurenLovesMaths)

    My absolute minimum is 69 but it could be anything above that (bear in mind I'm in year 12)
    You should hopefully get a distinction then! & yeah that's pretty damn good for year 12. You could always resit it next year if you need it for uni too


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    [QUOTE=LaurenLovesMaths;72517112]
    (Original post by Redcoats)

    You should hopefully get a distinction then! & yeah that's pretty damn good for year 12. You could always resit it next year if you need it for uni too


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    I don't think I'll need it for university (I took it for fun). Thank you nonetheless and I'm sure you've done well
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    [QUOTE=Redcoats;72517284]
    (Original post by LaurenLovesMaths)

    I don't think I'll need it for university (I took it for fun). Thank you nonetheless and I'm sure you've done well
    Ahh okay, fair enough haha! & thank you


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    Is there any sort of unofficial mark scheme yet?


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    (Original post by LaurenLovesMaths)
    Is there any sort of unofficial mark scheme yet?


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    I'm going to make one as soon as I can get the paper. The board haven't uploaded it yet, so if anyone has got a copy of it they could send me I'd make an unofficial mark scheme. I'll definitely make one but it depends on when I can see the paper.
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    (Original post by A Slice of Pi)
    I'm going to make one as soon as I can get the paper. The board haven't uploaded it yet, so if anyone has got a copy of it they could send me I'd make an unofficial mark scheme. I'll definitely make one but it depends on when I can see the paper.
    Okay thank you
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    I've got hold of the paper and will post an unofficial MS tomorrow. First thoughts are that this was an unusually easy paper apart from the last part of Q4, which was very interesting. Q1-3 could have been on a regular core 3/4 paper
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    (Original post by A Slice of Pi)
    I've got hold of the paper and will post an unofficial MS tomorrow. First thoughts are that this was an unusually easy paper apart from the last part of Q4, which was very interesting. Q1-3 could have been on a regular core 3/4 paper
    Thank you! Yes I agree, it was surprisingly easy compared to some previous papers


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    Q1

    (a) Inverse: f^{-1}(x) = x^2 - 2

    Domain: x \in \mathbb{R}, x \geqslant \sqrt{2} \qquad [2]

    (b) Range: g \geqslant  1 \qquad [2]

    (c) x=\frac{7}{4} \qquad [3] \qquad \textbf{\large{/7}}

    Q2

    (a) Proof.  \qquad [4]

    (b) x = 20^\circ, 120^\circ, 140^\circ \qquad [5] \qquad \textbf{\large{/9}}

    Q3

    (a) (i)  p = -8

    (ii)  \textbf{b} = (-1,3,5)^{T} \qquad [5]

    (b)  \textbf{c} = (5,1,8)^{T} \qquad [5]

    (c)  7\sqrt{101} \qquad [3] \qquad \textbf{\large{/13}}

    Q4

    (a) Proof.  \qquad [2]

    (b)  \sqrt{2} \qquad [5]

    (c)  \big(\frac{\pi}{3^{1/2}}\big)^{1/2}\qquad [6] \qquad \textbf{\large{/13}}

    Q5

    (a)  a = 1, b = 3 \qquad [2]

    (b) Labelled sketches showing the translated curves  \qquad [12] \qquad \textbf{\large{/14}}

    Q6

    (a) Proof. \qquad [2]

    (b) \ln(x+3) - \ln(x+4+\sqrt{2x+7}) + c \qquad [6]

    (c) \frac{-2}{2x+7}+\frac{1}{x+3} \qquad [2]

    (d) \ln(\frac{4}{3})-\frac{4}{15} \qquad [6] \qquad \textbf{\large{/16}}

    Q7

    (a) Obviously x=p and x=q are roots of the equation LHS = 0, because these are the points where the line meets the curve. Thus, by the factor theorem, the LHS has (x-p) and (x-q) as factors. We see that each is a double root, because the curve and line share the same gradient at the two points of intersection. Once we take into account the double roots, there are no factors left to find (as the coefficient of x^4 is 1). Thus LHS = (x-p)^2(x-q)^2 \qquad [2]

    (b) Proof. \qquad [6]

    (c) S = p^2 + 4pq + q^2, T = 2pq(p+q), U = p^2q^2    \qquad [5]

    (d) p = 1, q = 4, y = 6x - 16    \qquad [8] \qquad \textbf{\large{/21}}

    General thoughts on the paper: Questions 1, 2 and 3 were much easier than usual, and wouldn't look out of place towards the end of a regular core 3 or 4 paper. Q4 is also a good source of marks, so long as the answers to parts (b) and (c) are justified. A sketch alone may not suffice in (c) if there is not sufficient explanation to go with it. Questions 5 and 6 were straightforward, as was Q7 to some extent. In Q7 (b), they may insist on some explanation as to why the integral gives the area of that particular region, so you could have talked about vertical translations for some S+ marks. In all, I'd say this was perhaps on the same level of difficulty as the June 2010 paper (maybe even slightly easier), so it could well go as high as 80 for a distinction.
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    (Original post by A Slice of Pi)
    Q1

    (a) Inverse: f^{-1}(x) = x^2 - 2

    Domain: x \in \mathbb{R}, x \geqslant \sqrt{2} \qquad [2]

    (b) Range: g \geqslant  1 \qquad [2]

    (c) x=\frac{7}{4} \qquad [3] \qquad \textbf{\large{/7}}

    Q2

    (a) Proof.  \qquad [4]

    (b) x = 20^\circ, 120^\circ, 140^\circ \qquad [5] \qquad \textbf{\large{/9}}

    Q3

    (a) (i)  p = -8

    (ii)  \textbf{b} = (-1,3,5)^{T} \qquad [5]

    (b)  \textbf{c} = (5,1,8)^{T} \qquad [5]

    (c)  7\sqrt{101} \qquad [3] \qquad \textbf{\large{/13}}

    Q4

    (a) Proof.  \qquad [2]

    (b)  \sqrt{2} \qquad [5]

    (c)  \big(\frac{\pi}{3^{1/2}}\big)^{1/2}\qquad [6] \qquad \textbf{\large{/13}}

    Q5

    (a)  a = 1, b = 3 \qquad [2]

    (b) Labelled sketches showing the translated curves  \qquad [12] \qquad \textbf{\large{/14}}

    Q6

    (a) Proof. \qquad [2]

    (b) \ln(x+3) - \ln(x+4+\sqrt{2x+7}) + c \qquad [6]

    (c) \frac{-2}{2x+7}+\frac{1}{x+3} \qquad [2]

    (d) \ln(\frac{4}{3})-\frac{4}{15} \qquad [6] \qquad \textbf{\large{/16}}

    Q7

    (a) Obviously x=p and x=q are roots of the equation LHS = 0, because these are the points where the line meets the curve. Thus, by the factor theorem, the LHS has (x-p) and (x-q) as factors. We see that each is a double root, because the curve and line share the same gradient at the two points of intersection. Once we take into account the double roots, there are no factors left to find (as the coefficient of x^4 is 1). Thus LHS = (x-p)^2(x-q)^2 \qquad [2]

    (b) Proof. \qquad [6]

    (c) S = p^2 + 4pq + q^2, T = 2pq(p+q), U = p^2q^2    \qquad [5]

    (d) p = 1, q = 4, y = 6x - 16    \qquad [8] \qquad \textbf{\large{/21}}

    General thoughts on the paper: Questions 1, 2 and 3 were much easier than usual, and wouldn't look out of place towards the end of a regular core 3 or 4 paper. Q4 is also a good source of marks, so long as the answers to parts (b) and (c) are justified. A sketch alone may not suffice in (c) if there is not sufficient explanation to go with it. Questions 5 and 6 were straightforward, as was Q7 to some extent. In Q7 (b), they may insist on some explanation as to why the integral gives the area of that particular region, so you could have talked about vertical translations for some S+ marks. In all, I'd say this was perhaps on the same level of difficulty as the June 2010 paper (maybe even slightly easier), so it could well go as high as 80 for a distinction.
    thanks for this, can you post the actual paper? just trying to remember question 5 :/
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    The 2017 AEA paper is available here
    AdvancedExtension Award 2017
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    Don't remind me of that paper lmao, Q4 was horrendous
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    (Original post by theeconomistkid)
    ...
    You'll kick yourself about Q4, terribly worded so not your fault. I couldn't decipher it until I saw KMan's posts. c) was also ambiguous and misleading.

    a & b) was ÷ triangle into 2 equal areas via a line. They didn't state PQ is a line, but more importantly calls it 'shortest distance'. Confusing as c) asks for that too.

    c) was poor of them, it's just this: PQ needn't be a line now, infact the shortest PQ is when it's the arc of a circle. So draw the arc.

    But why draw '6 copies to make a big shape', if you can calc arc PQ directly! Very strange & misleading. So you draw a circle via 6 PQ-arcs in a hex ... only to ÷ by 6

    I'm surprised complaints weren't made, poor of them to print that drivel of a Q.

    (Original post by K-Man_PhysChem)
    ...
    (Original post by LaurenLovesMaths)
    ...
    Sounds like you've both done well RE boundaries, last year it was 55/73 and a hard (albeit interesting) paper, based on that and 2010 being 63/81, could be 82-84. But I felt 2015 was nice yet oddly only 70, you get hard/easier years where they still stick it at ~70.

    In 2012 they gave twice the % of distinctions and 20-25% more merits, crazy! Competition for uni places and cohort quality, year to year, may help explain 'discrepancies'. So yes it depends
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    (Original post by Physics Enemy)
    You'll kick yourself about Q4, terribly worded so not your fault. I couldn't decipher it until I saw KMan's posts. c) was also ambiguous and misleading.

    a & b) was ÷ triangle into 2 equal areas via a line. They didn't state PQ is a line, but more importantly calls it 'shortest distance'. Confusing as c) asks for that too.

    c) was poor of them, it's just this: PQ needn't be a line now, infact the shortest PQ is when it's the arc of a circle. So draw the arc.

    But why draw '6 copies to make a big shape', if you can calc arc PQ directly! Very strange & misleading. So you draw a circle via 6 PQ-arcs in a hex ... only to ÷ by 6

    I'm surprised complaints weren't made, poor of them to print that drivel of a Q.
    I agree with most of your analysis here. The wording of Q4 was fine up to a point. The main body of Q4 stated that LPQ was a triangle, so it is implicit that PQ is a straight line up until part (c). I had to re-read part (c) a few times also to understand what they were looking for. I am still not sure how much detail was required in the sketch. I'm guessing a sketch showing the arrangement of the triangles inside a hexagon and an explanation based on the symmetry will suffice. It could've been worded a lot better.

    With regards to grade boundaries, I thought it was easier than last years (and 2015). I wouldn't be surprised if its the same as 2010
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    (Original post by A Slice of Pi)
    I agree with most of your analysis here. The wording of Q4 was fine up to a point. The main body of Q4 stated that LPQ was a triangle, so it is implicit that PQ is a straight line up until part (c). I had to re-read part (c) a few times also to understand what they were looking for. I am still not sure how much detail was required in the sketch. I'm guessing a sketch showing the arrangement of the triangles inside a hexagon and an explanation based on the symmetry will suffice. It could've been worded a lot better.
    It was implicit PQ was a line in a & b either from the diagram or 'triangle LPQ' - but they called it shortest distance, contradicting c).

    Not sure if you get my point RE c). It was very ambiguous/confusing as you can just calc arc PQ directly. Angle 60°, sides 2 cm, area known - easy. Drawing a circle inside a hex was pointless & weird.

    They goofed up Q4 big time. It was very simple, GCSE level, but a strange comprehension exercise in the end.

    I'm shocked nobody in the thread picked up on the fact arc PQ was sitting there waiting to be found directly.
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    (Original post by Physics Enemy)
    It was implicit PQ was a line in a & b either from the diagram or 'triangle LPQ' - but they called it shortest distance, contradicting c).

    Not sure if you get my point RE c). It was very ambiguous/confusing as you can just calc arc PQ directly. Angle 60, sides 2cm, area known - easy. Drawing a circle inside a hex was pointless & weird.

    They goofed up Q4 big time. It was very simple, GCSE level, but a strange comprehension exercise in the end.

    I'm shocked nobody in the thread picked up on the fact arc PQ was sitting there waited to be found directly.
    I used the sector argument in my calculation for (c), although I didn't post it on here. The question on the whole required little knowledge from A-Level. Cosine/sine rule are taught at GCSE and differentiation wasn't necessary in (b) as there are other arguments you can use. I guess they could have said that the distance is shortest when PQ is a circular arc, but then it would just be a regular C2/GCSE exercise. There was a lot of unnecessary wording in the question.
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    (Original post by A Slice of Pi)
    I agree with most of your analysis here. The wording of Q4 was fine up to a point. The main body of Q4 stated that LPQ was a triangle, so it is implicit that PQ is a straight line up until part (c). I had to re-read part (c) a few times also to understand what they were looking for. I am still not sure how much detail was required in the sketch. I'm guessing a sketch showing the arrangement of the triangles inside a hexagon and an explanation based on the symmetry will suffice. It could've been worded a lot better.

    With regards to grade boundaries, I thought it was easier than last years (and 2015). I wouldn't be surprised if its the same as 2010
    I hope it is not 2010 boundaries, I estimate i got somewhere between 70-80 and have a sinking suspicion I'm getting a merit (this feeling has been growing over the past week, perhaps paranoia as its closing in on D-Day!). I'm year 12 going into 13 and will be applying to Cambridge, so in your opinion would a merit make me less likely to get an offer, seeing as I didn't do spectacularly, which to Cambridge is more or less a failure. Im just hoping the boundaries are kind, I would hate to be disadvantaged for taking a risk this year, but feel that has happened...
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    (Original post by tomahawker314)
    I hope it is not 2010 boundaries, I estimate i got somewhere between 70-80 and have a sinking suspicion I'm getting a merit (this feeling has been growing over the past week, perhaps paranoia as its closing in on D-Day!). I'm year 12 going into 13 and will be applying to Cambridge, so in your opinion would a merit make me less likely to get an offer, seeing as I didn't do spectacularly, which to Cambridge is more or less a failure. Im just hoping the boundaries are kind, I would hate to be disadvantaged for taking a risk this year, but feel that has happened...
    I am a Year 12 as well! What course are you applying for?
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    (Original post by tomahawker314)
    I hope it is not 2010 boundaries, I estimate i got somewhere between 70-80 and have a sinking suspicion I'm getting a merit (this feeling has been growing over the past week, perhaps paranoia as its closing in on D-Day!). I'm year 12 going into 13 and will be applying to Cambridge, so in your opinion would a merit make me less likely to get an offer, seeing as I didn't do spectacularly, which to Cambridge is more or less a failure. Im just hoping the boundaries are kind, I would hate to be disadvantaged for taking a risk this year, but feel that has happened...
    Hey, I'm in a similar position to you in that I sat AEA this year in Y12 and will be applying to Cambridge (for physical Natsci) next year. As far as I'm aware, there is no obligation to declare AEA, so if you don't get the grade you're aiming for, you don't have to state it.

    If you're applying for maths at Cambridge, getting a merit might not be the best, however it's not the end of your application. Cambridge say that they understand that even the best students can have a bad day, and so long as your AS maths/FM has high UMS and you perform well at interview, I would suspect that you would still stand a good chance (then you just have to do well at STEP, which is a completely different beast). Of course, others may be better suited to answer the question, but that's my interpretation from the research I've done.

    Good luck!
 
 
 
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