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    (Original post by Whyishow)
    p is the translation to the left hence 3 as the asymptote move by 3 to the left. q is the shift up, but as the curve,after taking the modulus, is symmetrical in y it means C and D have the same y-coordinate, therefore, before taking the modulus the y coordinate of C is negative that of D, a value of 6 does that.
    For me I thought the paper was hard, and the grade boundaries would probably be around 70 as usualll.. I totally messed up on question 5 on the last bit, cus I thought that freaking thing we had to show was an identity, I feel so stupid nowww for not just subbing an a value innn.. I als messed up my calculations a bit in the vectors question in part d). Otherwise I think i did okay, good thing is i don't need it for an offer or anything haha..
    I wasn't asking. I just put the question mark because I wasn't 100% sure it was right

    I thought the paper was hard aswell. Completely messed up on 5.c and all of no. 6 and 1. Most of 7 aswell :sigh:
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    (Original post by ElMoro)
    Hoping for something like that aswell but DFranklin reckons there'll be normal boundaries :dontknow:
    That's still my gut feeling; I don't honestly think it was more difficult than usual, in fact I'd say it was easier that most years.

    Counteracting that, the vector question in particular had a first part that was both tricky and left you in poor shape for the rest of the question if you couldn't get it out. So that may push the marks down a bit.

    But if you press me, I'm guessing the distinction mark will be slightly over 70.
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    (Original post by DFranklin)
    That's still my gut feeling; I don't honestly think it was more difficult than usual, in fact I'd say it was easier that most years.

    Counteracting that, the vector question in particular had a first part that was both tricky and left you in poor shape for the rest of the question if you couldn't get it out. So that may push the marks down a bit.

    But if you press me, I'm guessing the distinction mark will be slightly over 70.
    But you're a much better mathematician than all of us and you could do it not under exam pressure. You may be comparing the extra difficulty of this than normal to this years extra difficulty of STEP than normal and deducing that it's nothing in comparison. I think the fact that most people here found it hard is saying something, and a lot of us said that we were easily getting distinctions in past papers.

    I'm not saying there were things that were conceptually hard, but some of the algebra was ugly as ****. I don't know anyone in my college that managed 5 c) i. (Although that's not to say no-one in my college did, I haven't spoken to all of them) I go to a very large college and the number of people taking this exam in my college was quite small, and I'd safely say they're the best of the year.
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    (Original post by DFranklin)
    That's still my gut feeling; I don't honestly think it was more difficult than usual, in fact I'd say it was easier that most years.

    Counteracting that, the vector question in particular had a first part that was both tricky and left you in poor shape for the rest of the question if you couldn't get it out. So that may push the marks down a bit.

    But if you press me, I'm guessing the distinction mark will be slightly over 70.
    I understand. I guess there's always next year... :sigh:
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    (Original post by DFranklin)
    That's still my gut feeling; I don't honestly think it was more difficult than usual, in fact I'd say it was easier that most years.

    Counteracting that, the vector question in particular had a first part that was both tricky and left you in poor shape for the rest of the question if you couldn't get it out. So that may push the marks down a bit.

    But if you press me, I'm guessing the distinction mark will be slightly over 70.
    I've to agree. I think the only tough qn in the paper is Q1; in fact some people manage to solve Q1 well. Other than Q1 ppl will lose marks due to algebraic mistakes for the rest of the qn, esp in Q6. Personally, I think this yr paper is slightly easier than 08, but more demanding in terms of algebra compared to last yr. So I felt the worst case is the grade boundaries will only be 3-4 m lower than last yr; the 'best' case is the boundaries are similar to 08.

    It's hard to predict the scores; personally I thought 04 is harder than 06 as 04 has the algebraically demanding 'prove that's wrong and write out the correct solution' qn and geometry qn, but 04 boundaries are 2 m higher than 06.

    That's my personal take. I've no local friend to talk with for AEA, as all of them took STEP to meet their Imperial offer; I guess there are only 2-4 from my country taking this AEA. Anyway everything is done so just hope for the best!
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    (Original post by Webbykun)
    But you're a much better mathematician than all of us and you could do it not under exam pressure.
    True. But that's the same scenario every year. So I can be relatively consistent. Unlike you, I'm not comparing the 2011 exam under exam pressure, to every other exam without.

    You may be comparing the extra difficulty of this than normal to this years extra difficulty of STEP than normal and deducing that it's nothing in comparison.
    That really doesn't affect my judgement - my memory's not that good.

    I think the fact that most people here found it hard is saying something, and a lot of us said that we were easily getting distinctions in past papers.
    Every single year, the AEA aftermath is filled with people saying it was the most difficult paper for many years. It even happened last year, which objectively was by some way the easiest AEA paper ever set.

    I do take on board what you say about "easily getting distinctions in past papers", but I have to say that this also seems to be the case year-in and year-out. I'm not really sure why this is the case - I wouldn't have expected exam nerves to be such a big factor, but possibly everyone is less used to "big scary exams" than in my day.

    I'm not saying there were things that were conceptually hard, but some of the algebra was ugly as ****. I don't know anyone in my college that managed 5 c) i. (Although that's not to say no-one in my college did, I haven't spoken to all of them) I go to a very large college and the number of people taking this exam in my college was quite small, and I'd safely say they're the best of the year.
    I don't think this was algebraically tough - what was mean (and a poor show on the examiner's part IMHO), is that it's not obvious that you're supposed to be switching between "P is an arbitrary point" to "P is the point touching the circle". (Or in other words, it's not obvious (i) isn't supposed to be a simple algebraic identity).

    Once you know that, if we let C be the center of the circle (and we've deduced that the normal goes through C, which is pretty heavily hinted), the the LHS in (i) is the distance CU (edit: squared) (which only requires a tiny bit of algebra to show \frac{a^2-2}{a^2-4} - \frac{1}{2} = \frac{a^2}{2(a^2-4)}). The RHS is the distance CP (edit:squared) (which requires almost no algebra at all).

    But yes, this was one of the tougher parts of the paper. Even so, it's only 10 marks, and you should be able to get some of those marks by follow through.

    As I said earlier I could see the geometry question pushing marks down a bit - although again, it's not that the question was particularly hard for an AEA question - the issue is that the hard bit was at the start.

    There are a lot of complexities in judging the "general difficulty" of the AEA; obviously it's an unusual exam in its own right, but the abilities of the people taking it are even more unusual. You only need C1-C4, but many will have covered the FM syllabus as well. Some people will have prepared a lot, some not much at all. A few will have been preparing for STEP II /III as well. Some are expecting to get 90%, and some are hoping to get 50%. And each category will tend to have a different view about whether a paper was hard or not. So, sure, I might be wrong here. It's just a guess, after all.
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    (Original post by DFranklin)
    True. But that's the same scenario every year. So I can be relatively consistent. Unlike you, I'm not comparing the 2011 exam under exam pressure, to every other exam without.

    That really doesn't affect my judgement - my memory's not that good.

    Every single year, the AEA aftermath is filled with people saying it was the most difficult paper for many years. It even happened last year, which objectively was by some way the easiest AEA paper ever set.

    I do take on board what you say about "easily getting distinctions in past papers", but I have to say that this also seems to be the case year-in and year-out. I'm not really sure why this is the case - I wouldn't have expected exam nerves to be such a big factor, but possibly everyone is less used to "big scary exams" than in my day.

    I don't think this was algebraically tough - what was mean (and a poor show on the examiner's part IMHO), is that it's not obvious that you're supposed to be switching between "P is an arbitrary point" to "P is the point touching the circle". (Or in other words, it's not obvious (i) isn't supposed to be a simple algebraic identity).

    Once you know that, if we let C be the center of the circle (and we've deduced that the normal goes through C, which is pretty heavily hinted), the the LHS in (i) is the distance CU (edit: squared) (which only requires a tiny bit of algebra to show \frac{a^2-2}{a^2-4} - \frac{1}{2} = \frac{a^2}{2(a^2-4)}). The RHS is the distance CP (edit:squared) (which requires almost no algebra at all).

    But yes, this was one of the tougher parts of the paper. Even so, it's only 10 marks, and you should be able to get some of those marks by follow through.

    As I said earlier I could see the geometry question pushing marks down a bit - although again, it's not that the question was particularly hard for an AEA question - the issue is that the hard bit was at the start.

    There are a lot of complexities in judging the "general difficulty" of the AEA; obviously it's an unusual exam in its own right, but the abilities of the people taking it are even more unusual. You only need C1-C4, but many will have covered the FM syllabus as well. Some people will have prepared a lot, some not much at all. A few will have been preparing for STEP II /III as well. Some are expecting to get 90%, and some are hoping to get 50%. And each category will tend to have a different view about whether a paper was hard or not. So, sure, I might be wrong here. It's just a guess, after all.
    Hmm to be honest I mostly agree that it wasn't a bad paper, I just made so many slips up that I expect to lose a lot of marks so was hoping to convince myself you were wrong :P. It just sucks knowing you could have done so much better and getting into your firm depended on this exam.
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    I don't want to sound like an ass, but I agree with DFranklin that it wasn't too bad as an AEA paper (it was pretty similar to a lot of past papers I've seen). Granted, I couldn't do the a^2 bit, but besides that, a lot of it seemed quite straight forward (or maybe just compared to step this year which was rather annoying). It seems everybody always says their exam was hardest ever (in most maths things these days) but maybe that's just the exam pressure making it seem harder. I know the other people at college who sat it found some bits hard, but things like the rectangle question were kind of given to you.

    Any body remember their answer for the position vector of B? I remember getting something like (17, -7, -1) or something with similar numbers?
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    (Original post by Webbykun)
    Hmm to be honest I mostly agree that it wasn't a bad paper, I just made so many slips up that I expect to lose a lot of marks so was hoping to convince myself you were wrong :P. It just sucks knowing you could have done so much better and getting into your firm depended on this exam.
    You may be worrying too much about the slips - because there's more work involved in an AEA question than an A-level one, a few slips aren't such a big deal, I suspect.

    But yes, having everything rest on one exam and not doing as you'd hoped does suck.

    How did you prepare for the exam, may I ask?
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    (Original post by DFranklin)
    You may be worrying too much about the slips - because there's more work involved in an AEA question than an A-level one, a few slips aren't such a big deal, I suspect.

    But yes, having everything rest on one exam and not doing as you'd hoped does suck.

    How did you prepare for the exam, may I ask?
    I did the 2002 - 2008 past papers.
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    (Original post by Webbykun)
    I did the 2002 - 2008 past papers.
    Did you do any under "exam conditions"? (3 hours, no calculator, no internet etc.)
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    (Original post by DFranklin)
    Did you do any under "exam conditions"? (3 hours, no calculator, no internet etc.)
    All of them bar two which I did passively. The last one I did was the June 2008 one which I finished in two and a half hours and got over 90% on =/.
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    (Original post by Webbykun)
    All of them bar two which I did passively. The last one I did was the June 2008 one which I finished in two and a half hours and got over 90% on =/.
    I can understand you're feeling hard done by then - I can't really suggest anything you could have done differently.
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    (Original post by DFranklin)
    I can understand you're feeling hard done by then - I can't really suggest anything you could have done differently.
    Do Warwick get to see your score? I estimate I've dropped around 30 marks so I may be quite close to the boundary, which could work in my favour when trying to beg for a position. They might have to let a couple of people in who didn't meet their grade because of how poorly STEP went.
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    (Original post by Webbykun)
    Hmm to be honest I mostly agree that it wasn't a bad paper, I just made so many slips up that I expect to lose a lot of marks so was hoping to convince myself you were wrong :P. It just sucks knowing you could have done so much better and getting into your firm depended on this exam.
    (Original post by DFranklin)
    You may be worrying too much about the slips - because there's more work involved in an AEA question than an A-level one, a few slips aren't such a big deal, I suspect.
    If the slips you are referring are careless math errors, even at the earlier part of the qn, the examiners shouldn't penalise heavily, as what DFranklin said. They are more interested on how you present yr logic than accuracy; if yr mathematical logic is correct despite yr negligence, they will give you the credit, like the error carried forward concept, so as to help good students who are careless because of exam conditions. (My internal school math exams does it differently by penalising careless mistakes heavily, so as to make the students felt they've done badly and the students in turn will work harder!)

    Don't worry too much that the examiners don't give credit for yr errors though yr logic is correct.
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    I've had a couple of hours free this morning so I decided to come up with some answers. Please note, I haven't painstakingly checked these so one or two may be wrong. I haven't done too much AEA but I didn't think this one was too hard, personally. It had 5c which was a pain to figure out and 6 which took a little work but other than that, the remaining questions fell out relatively straightforwardly. I haven't done Q1 because there are already some good solutions in this thread.

    Q2
    I=\displaystyle\int ^{\frac{\pi}{2}}_0 (1+\tan (x/2))^2 dx
    Using a substitution of t=\tan (x/2):

    I=2\displaystyle\int ^1_0 \dfrac{(1+t)^2}{1+t^2}dt

    =2[t+\ln (1+t^2)]^1_0

    =2+\ln 4

    Therefore \boxed{(a,b)=(2,4)}.

    Q3
    (a)
    u_1= k
    u_2= pk
    u_3= pqk
    u_4=p^2qk
    u_5=p^2q^2k
    u_6= p^3q^2k

    (b)
    Note that:
    S=\displaystyle\sum _{r=1}^{2n} u_r = \displaystyle\sum _{r=1}^{n} u_{2r} + \displaystyle\sum _{r=1}^{n} u_{2r-1}
    =pk\displaystyle\sum _{r=1}^{n} (pq)^{r-1} + k\displaystyle\sum _{r=1}^{n} (pq)^{r-1}

    Which are GPs with a=1, r=pq, therefore:

    S=pk\dfrac{(1-(pq)^n)}{1-pq} + k\dfrac{(1-(pq)^n)}{1-pq}
    \implies \boxed{S=\dfrac{k(1+p)(1-(pq)^n)}{1-pq}}, as required.

    (c)
    T=\displaystyle\sum _{r=1}^{\infty} 6\times \left(\dfrac{4}{3}\right)^{[\frac{r}{2}]}\times \left(\dfrac{3}{5}\right)^{[\frac{r-1}{2}]}

    Note that T is S with (k,p, q) = \left(6,\dfrac{4}{3}, \dfrac{3}{5}\right).

    Therefore T=\dfrac{6\times (1+\frac{4}{3})}{1-\frac{4}{5}}
    \implies \boxed{T=70}


    Q4
    (a)
    Note that x-y^2=x^2

    Therefore (x-\frac{1}{2})^2 - \frac{1}{4} + y^2 = 0

    \implies (x-\frac{1}{2})^2 + y^2 = \frac{1}{2}^2

    Therefore C has centre \boxed{\left(\dfrac{1}{2},0 \right) } and radius \boxed{\dfrac{1}{2}}

    (b)
    Multiplying the x and y-coordinates of the upper right vertex of the rectangle yields:

    \boxed{R=\sin \alpha \cos ^3\alpha}

    (c)
    Note that R =\sin\alpha \cos\alpha \times \cos ^2\alpha

    =\dfrac{1}{2}\sin 2\alpha \times \left(\dfrac{1}{2}(\cos 2\alpha +1)\right)

    =\dfrac{1}{8}(\sin 4\alpha +2\sin 2\alpha)

    Therefore \dfrac{dR}{d\alpha} = 0 \implies \cos 4\alpha + \cos 2\alpha =0

    Note that \cos 4\alpha + \cos 2\alpha = 2\cos \left(\dfrac{(4+2)\alpha}{2} \right) \cos \left(\dfrac{(4-2)\alpha}{2} \right ).

    Therefore \dfrac{dR}{d\alpha} =0 \implies \cos 3\alpha =0 or \cos \alpha =0.

    Note that \cos \alpha \not= 0 over 0 < \alpha < \dfrac{\pi}{2}.

    Therefore the only solution for \alpha over that interval is \alpha = \dfrac{\pi}{6}. The corresponding value of R is \dfrac{3\sqrt 3}{16}.

    Note that, at \alpha = 0 and \alpha =\dfrac{\pi}{2}, R=0 at both endpoints. This is clearly less than the value of R we found above and since the stationary point is unique, it follows that this area must be a maximum. (Cheers to DFranklin for reminding me of this simple trick from the STEP solutions).

    Therefore \boxed{R_{max}=\dfrac{3\sqrt 3}{16}}

    Q5
    (a)
    \boxed{U= \left( 0,\dfrac{1}{2} \right) }

    (b)
    Note that y=1+\dfrac{2}{x^2-4}. Therefore \dfrac{dy}{dx}=-\dfrac{4x}{(x^2-4)^2}. Therefore the normal at P is of the form:

    y=\dfrac{(a^2-4)^2}{4a}x+C

    Where C is the point at which the normal to P cuts the y-axis.

    Note that this line passes through \left(a,\dfrac{a^2-2}{a^2-4}\right).

    It follows that \boxed{C=\dfrac{a^2-2}{a^2-4} - \dfrac{(a^2-4)^2}{4}}, as required.

    (ci)
    It's clear that C is the centre of this circle, where C is defined as above. Note that CU = \dfrac{a^2-2}{a^2-4} - \dfrac{(a^2-4)^2}{4}} - \dfrac{1}{2} = \dfrac{a^2}{2(a^2-4)}-\dfrac{(a^2-4)^2}{4}.

    This is a radius of the circle. Note also that CP is a radius of the circle and it is of length:

    CP = \sqrt{a^2 + \left(\dfrac{a^2-2}{a^2-4} - \dfrac{a^2-2}{a^2-4} + \dfrac{(a^2-4)^2}{4}\right)^2} = \sqrt{a^2+\dfrac{(a^2-4)^4}{16}}.

    Since CP=CU, it follows that \boxed{\left[\dfrac{a^2}{2(a^2-4)}-\dfrac{(a^2-4)^2}{4}\right]^2 = a^2+\dfrac{(a^2-4)^4}{16}}, as required.

    (cii)
    Expanding out the LHS of the above:

    \dfrac{a^4}{4(a^2-4)^2} - \dfrac{a^2(a^2-4)}{4} + \dfrac{(a^2-4)^4}{16} = a^2 +\dfrac{(a^2-4)^4}{16}

    \implies \dfrac{a^2}{(a^2-4)^2} - (a^2-4)=4

    \implies \dfrac{1}{(a^2-4)^2} = 1

    \implies \boxed{(a^2-4)^2=1}, as required.

    (ciii)
    Note that this latter condition implies that a^2 is 3 or 5, but note that a^2=3 would mean that C is below U, which is not possible if it is to touch (key word here) all three branches. Therefore a^2=5.

    It follows that circle centre C is \boxed{\left(0, \dfrac{11}{4}\right)} and E has radius \boxed{\dfrac{9}{4}}.

    Q6
    (a)
    Let the point on L that is exactly halfway between P and P' be given by X where \mathbf{x}= \begin{pmatrix} 13-5t \\ -3+3t \\ 8+4t \end{pmatrix}.

    Note that the line through P, X and consequently P' has direction:

    \mathbf{x} - \mathbf{p} =  \begin{pmatrix} 20-5t \\ -5+3t \\ -15+4t\end{pmatrix}.

    Note that this is perpendicular to the direction of L, therefore:

    \begin{pmatrix} 20-5t \\ -5+3t \\ -15+4t\end{pmatrix} \cdot \begin{pmatrix} -5 \\ 3 \\ 4\end{pmatrix} = 0

    \implies t=\dfrac{7}{2}.

    Therefore, the line through P, X and P' is given by:

    \mathbf{r} = \begin{pmatrix} -7 \\ 2 \\ 7 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 11 \\ -2 \end{pmatrix}

    Also note that \mathbf{x} = \begin{pmatrix} -9/2 \\ 15/2 \\ 6\end{pmatrix}. By setting this vector equal to the line through P and X, it follows that the corresponding value of \lambda that yields \mathbf{x} is \lambda = \dfrac{1}{2}.

    Therefore P' is given by the point on the line with parameter \lambda = 1.

    It follows that \boxed{\mathbf {p'} = \begin{pmatrix} -2 \\ 13 \\ 5 \end{pmatrix}}

    (b)
    It's a simple case of plugging the point in the line and showing there's a parameter t that works, so I won't bother typing it out.

    (c)
    Note that, if \angle PAP' is 2\theta, then \angle PAX =\theta.

    Note that \vector{AP} = \begin{pmatrix} 0 \\ -7 \\ -1\end{pmatrix} and \vector{AX} = \begin{pmatrix} 5 \\ -3 \\ -4 \end{pmatrix}.

    Therefore \begin{pmatrix} 0 \\ -7 \\ -1\end{pmatrix} \cdot \begin{pmatrix} 5 \\ -3 \\ -4\end{pmatrix} = \sqrt{50}\sqrt{50}\cos \theta

    \implies \cos \theta = \dfrac{1}{2}

    \implies \theta = 60^{\circ}

    \implies \boxed{\angle PAP' = 120^{\circ}}, as required.

    (d)
    Let B have position vector \mathbf{b}=\begin{pmatrix} 13-5t \\ -3+3t\\ -8+4t \end{pmatrix}.

    Note that AB = \begin{vmatrix} 20-5t \\ -12+3t\\ -16+4t \end{vmatrix} =  \sqrt{50}|t-4|

    Also, note that XP = \frac{1}{2}PP' = \frac{1}{2} \begin{vmatrix} 5 \\ 11 \\ -2 \end{vmatrix} = \dfrac{sqrt{150}}{2}.

    By the symmetry of the kite in the line L, notice that triangle APB has area 25\sqrt 3.

    It follows that we seek t such that \dfrac{1}{2} \times \sqrt{50}|t-4| \times \dfrac{\sqrt{150}}{2} = 25\sqrt 3

    \implies |t-4|=2

    \implies t=2 or t=6.

    Note that the values of t which produce A and X are 4 and 3.5 respectively. From the diagram, it's clear that t is decreasing as we move in the direction from A to X and thus to B. Therefore t\not= 6 and thus t=2.

    It follows that \boxed{ \mathbf{b} = \begin{pmatrix} 3 \\ 3 \\ 0 \end{pmatrix} }.

    (e)
    Note that \vec{PA} = \begin{pmatrix} 0 \\ 7 \\ 1 \end{pmatrix} and \vec{PB} = \begin{pmatrix} 10 \\ 1 \\ -7 \end{pmatrix}.

    Note that \begin{pmatrix} 0 \\ 7 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ 1 \\ -7 \end{pmatrix} = 7-7 = 0, which implies that \boxed{\angle BPA = 90^{\circ}}.

    (f)
    Since angle BPA is a right angle, it follows that AB is a diameter. Therefore C is the midpoint of AB. Let the position that represents C be denoted by \mathbf{c}.

    Therefore \mathbf{c} = \dfrac{1}{2}(\mathbf{a} + \mathbf {b}) = \boxed{\begin{pmatrix} -2\\ 6\\ 4 \end{pmatrix}}



    Q7
    (a)
    Note that f'(x)=0 \implies x^2-6x+5=0
    \implies (x-1)(x-5)=0.

    It follows that \boxed{A = (1, -2)} and \boxed{B= (5,-10)}

    (I can't believe that this is almost worth the same amount of marks at Q6a...)

    (bi)

    Note that  h(fg(x)+q) = \left| \dfrac{(x+p)^2 -5}{3-(x+p)} +q \right|

    The vertical asymptote at x=0 implies that \boxed{p=3}.

    This means that h(fg(x)+q) = \left| \dfrac{x^2+(6-q)x+4}{x} \right|.

    Note that there are no real roots for the curve and the symmetry in the y-axis suggests that the function must be even. It follows that q=6 works fine for both of these situations. Note also that q represents the translation of the curve given by f(x+3) in the y-direction. In order for the minimums to share the same y-value once we take the modulus of f(x+3)+q, they must be equidistant from the y-axis to start with. This only occurs when the curve is shifted upwards by 6 units, which proves q's uniqueness.

    (bii)
    \boxed{D(2,4)}

    (ci)
    Note that m(x)=\dfrac{x^2-5}{3-x} \implies x=\dfrac{(m^{-1}(x))^2- 5}{3-m^{-1}(x)}

    \implies m^{-1}(x) = \dfrac{-x\pm \sqrt{(x+2)(x+10)}}{2}

    Note that  \forall x\geq -2; m^{-1}(x) \leq 1.

    Consider m^{-1}(0) = \pm \sqrt 5. It's clear that, since m(x) is one-one over it's domain, that only one of these version fit. Since m^{-1}(0) = \sqrt 5 > 1, it breaks our condition and thus it follows that:

    \boxed{m^{-1}(x) = -\dfrac{x+\sqrt{(x+2)(x+10)}}{2}}

    (cii)
    Domain: \boxed{x\geq -2}

    (ciii)
    Note that, if a function intersects the line y=x, then inverse must also pass through that point. It follows that m(t) = t \implies m(t)= m^{-1}(t), and thus it is sufficient to consider intersections with y=t to find the value of t.

    Considering m(t)=t:
    \implies 2t^2-3t-5=0
    \implies t=\dfrac{5}{2} or t=-1.

    Note that t\leq 1, therefore \boxed{m(t)=m^{-1}(t) \implies t=-1}



    I'll write out the rest of Q6 and Q7 later, if I get time. I've got complete solutions up to all the questions now.

    I hope these help ease the nerves a little. Good luck for results day!
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    (Original post by Farhan.Hanif93)
    ..
    If f(t) = -t, then f is its own inverse. It which case it is not true to say that f(t) = f^{-1}(t) \implies f(t) = t.

    Edit: the implication does work the other way, and that's enough here. On the one hand hoping you can find a m(t) = t solution feels like a bit of a cheat, on the other hand it's obvious you *can* find a solution to m(t) = t here, so once you realise that's enough you can move on.

    2nd Edit: for 7(b), the stationary points from (a) must end up as the minimum points in (b). I think this uniquely determines q.
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    (Original post by DFranklin)
    If f(t) = -t, then f is its own inverse. It which case it is not true to say that f(t) = f^{-1}(t) \implies f(t) = t.

    Edit: the implication does work the other way, and that's enough here. On the one hand hoping you can find a m(t) = t solution feels like a bit of a cheat, on the other hand it's obvious you *can* find a solution to m(t) = t here, so once you realise that's enough you can move on.

    2nd Edit: for 7(b), the stationary points from (a) must end up as the minimum points in (b). I think this uniquely determines q.
    Thanks, I had a feeling there was something wrong with that solution. I've edited them both.
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    Just quick Q, sorry if its stupid: do the AEA results automatically get forwarded to the unis through UCAS? :P I'm not sure what they class under is all ?
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    (Original post by sb2star)
    I had no problems with Q2,3,4 and 5... so hopefully I didn't make any mistakes in those...

    how did you slove the Q2 ? i counld not sloved it at all
 
 
 
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