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    (Original post by Piecewise)
    Far simpler: as \cot(x) = \tan(90^{\circ}-x), we have \cot(\theta-53^{\circ}) = \tan(143^{\circ}-\theta); hence what we were asked to solve is  \tan(\theta+35^{\circ}) = \tan~(143^{\circ}-\theta) from which \theta = 54^{\circ}+90^{\circ}n. Obviously n = 0, and n = 1 give values of \theta in the desired range!
    even better, nice :P
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    (Original post by superkinetic)
    What's yr firm? Are you majoring in math in future?
    My firm is Warwick and my insurance is Nottingham and yeah I'm 'majoring' in maths, how about you?

    I've got to be honest, I think I could have done much better in this exam, I guess nerves got to me .
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    (Original post by Webbykun)
    My firm is Warwick and my insurance is Nottingham and yeah I'm 'majoring' in maths, how about you?

    I've got to be honest, I think I could have done much better in this exam, I guess nerves got to me .
    I am not going into math. I've an offer to study engineering at Imperial. Though they didn't make AEA a requirement, the admission person encourages me to take that.

    I understand that sometimes under exam conditions, you can't seem to perform to yr max. It happens to me during all my school internal exams, and in fact for my A levels too, when I am worrying how much marks I lost due to my negligence.
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    (Original post by implus)
    This is how I did question 1:

    You are a genius my friend :hat2:

    I didn't think of that at all :sad: Realised I could've done tan x = cot (90 - x) today :facepalm:
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    (Original post by superkinetic)
    But that's not true that opposite angles in a kite add up to 180 deg; it's coincidental that both opposite angles add up to 180 deg, as one of the angle is 90 deg.

    Assume there are 2 similar angles of 100 deg in a kite. Do the remaining opposite angles add up to 180 deg?

    Anw, do u require AEA for yr offer?
    I thought it was true for 4 sided shapes.
    Oh well!

    Yeah I require it for my insurance. :rolleyes:
    FML
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    (Original post by AnonyMatt)
    I thought it was true for 4 sided shapes.
    Oh well!

    Yeah I require it for my insurance. :rolleyes:
    FML
    How'd you find it, man?
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    (Original post by Ultimate1)
    How'd you find it, man?
    Absolutely crap. :p:

    I too thought Q1 was just a typo.

    I found the vectors question especially difficult too.

    I thought it was a very difficult paper - don't remember any of the others being as difficult at all.
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    (Original post by AnonyMatt)
    Absolutely crap. :p:

    I too thought Q1 was just a typo.

    I found the vectors question especially difficult too.

    I thought it was a very difficult paper - don't remember any of the others being as difficult at all.
    Same.

    I really would have liked a Merit here. ffs why was this one of the hardest years.
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    (Original post by Ultimate1)
    Same.

    I really would have liked a Merit here. ffs why was this one of the hardest years.
    It was like the worst year ever for the AEA and STEP II. Just *******s really. :p:
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    (Original post by AnonyMatt)
    It was like the worst year ever for the AEA and STEP II. Just *******s really. :p:
    Yep. I was expecting a 2 for STEP I but I just froze in the exam. When I tried it again a day later I got quite a lot of the Qs. I hate that.
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    What do you guys reckon the grade boundaries will be for a distinction? I'm seriously hoping for sub-70.
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    (Original post by Webbykun)
    What do you guys reckon the grade boundaries will be for a distinction? I'm seriously hoping for sub-70.
    Hoping for something like that aswell but DFranklin reckons there'll be normal boundaries :dontknow:
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    On that last question, how the hell were you meant to find p and q?
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    (Original post by Pheylan)
    On that last question, how the hell were you meant to find p and q?
    Well, the vertical asymptote had been shifted to the y-axis, i.e. 3 to the left. So p = 3.

    If you look at the original graph, the y-values of the stationary points were -2 and -10 if I remember correctly. They're symmetrical about the line y=-6. If you look to the translated graph, they both lie the same distance above the x-axis, so the line of symettry y=-6 has been shifted up 6 to the x-axis. Hence q = 6.
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    (Original post by Webbykun)
    Well, the vertical asymptote had been shifted to the y-axis, i.e. 3 to the left. So p = 3.

    If you look at the original graph, the y-values of the stationary points were -2 and -10 if I remember correctly. They're symmetrical about the line y=-6. If you look to the translated graph, they both lie the same distance above the x-axis, so the line of symettry y=-6 has been shifted up 6 to the x-axis. Hence q = 6.
    Ugh. ****ing exams
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    (Original post by Pheylan)
    On that last question, how the hell were you meant to find p and q?
    I got p=3, q=6 ?
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    So is anyone doing the solutions?
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    (Original post by soczek322)
    So is anyone doing the solutions?
    Did you send the paper to Mr. M and ask him?
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    (Original post by ElMoro)
    I got p=3, q=6 ?
    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..
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    Sorry just realised someone made the same point about p and q earlier.
 
 
 
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