The AEA/Oxford/S Paper superthread

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  1. Unbounded's Avatar
    • TSR Demigod
    Re: The AEA/Oxford/S Paper superthread
    (Original post by darkness9999)
    And also why is ff^{-1}(x)=x\Longleftrightarrow f(x)=f^{-1}(x)
    edit: I just realised what you posted. The statement  f(x) = f^{-1}(x) \iff ff^{-1}(x)=x is not true. And the only function I think of, such that  f(x) = f^{-1}(x) (*), is  f(x) = x, although there may be more [edit: there are indeed more eg. f = 1/x and f = a-x, for some constant a]. And it should be obvious that ff^{-1}(x)=x\Longleftarrow f(x)=f^{-1}(x), if that function is the only function that obeys (*).

    Consider  g(x) = \log x and  g^{-1}(x) = e^x, noting that  g \not= g^{-1}. Then still gg^{-1}(x) = x

    The statement  ff^{-1}(x) = x is true though. I think it's more a way of defining the inverse function, somewhat:

    Let f(A) map to B. Therefore  f^{-1}(B) maps to A.

     \therefore ff^{-1}(A) = f(B) = A

    And of course the same is true for  f^{-1}f(A) = A

    Hmm. I have a feeling my logic is circular.

    And those theorems were a good plan :yep:

    One I'm knicking out of a STEP thread:

     \displaystyle\int_a^b f(x) \ \mathrm{d}x = \displaystyle\int_a^b f(b+a-x) \ \mathrm{d}x

    Which can be proved with a quick substitution of  x = a+b-y for some variable y.

    An example you might like to try:

     I = \displaystyle\int_2^7 \dfrac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} \ \mathrm{d}x
    Spoiler:
    Show
    By the theorem in the post, the integral is also equivalent to:

     I = \displaystyle\int_2^7 \dfrac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} \ \mathrm{d}x

     \therefore I + I = \displaystyle\int_2^7 \left( \dfrac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} + \dfrac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}\right) \ \mathrm{d}x

     \therefore 2I = \displaystyle\int_2^7 \dfrac{\sqrt{9-x}+\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} \ \mathrm{d}x

     \iff 2I = \displaystyle\int_2^7 1 \ \mathrm{d}x = \Left [ x \Right ] _2^7 = 5 \iff \boxed{I = \frac{5}{2}}
    Last edited by Unbounded; 12-02-2010 at 19:58.
  2. darkness9999's Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by GHOSH-5)
    edit: I just realised what you posted. The statement  f(x) = f^{-1}(x) \iff ff^{-1}(x)=x is not true. And the only function I think of, such that  f(x) = f^{-1}(x) (*), is  f(x) = x, although there may be more. And it should be obvious that ff^{-1}(x)=x\Longleftarrow f(x)=f^{-1}(x), if that function is the only function that obeys (*).

    Consider  g(x) = \log x and  g^{-1}(x) = e^x, noting that  g \not= g^{-1}. Then still gg^{-1}(x) = x

    The statement  ff^{-1}(x) = x is true though. I think it's more a way of defining the inverse function, somewhat:

    Let f(A) map to B. Therefore  f^{-1}(B) maps to A.

     \therefore ff^{-1}(A) = f(B) = A

    And of course the same is true for  f^{-1}f(A) = A

    Hmm. I have a feeling my logic is circular.

    And those theorems were a good plan :yep:

    One I'm knicking out of a STEP thread:

     \displaystyle\int_a^b f(x) \ \mathrm{d}x = \displaystyle\int_a^b f(b+a-x) \ \mathrm{d}x

    Which can be proved with a quick substitution of  x = a+b-y for some variable y.

    An example you might like to try:

     I = \displaystyle\int_2^7 \dfrac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} \ \mathrm{d}x
    Spoiler:
    Show
    By the theorem in the post, the integral is also equivalent to:

     I = \displaystyle\int_2^7 \dfrac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} \ \mathrm{d}x

     \therefore I + I = \displaystyle\int_2^7 \left( \dfrac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}} + \dfrac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}\right) \ \mathrm{d}x

     \therefore 2I = \displaystyle\int_2^7 \dfrac{\sqrt{9-x}+\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} \ \mathrm{d}x

     \iff 2I = \displaystyle\int_2^7 1 \ \mathrm{d}x = \Left [ x \Right ] _2^7 = 5 \iff \boxed{I = \frac{5}{2}}
    Thanks for this and settling my confusion :top:
  3. Unbounded's Avatar
    • TSR Demigod
    Re: The AEA/Oxford/S Paper superthread
    (Original post by darkness9999)
    Thanks for this and settling my confusion :top:
    No worries. So glad I've finished all my non-maths subjects. I'll be pounding down some STEP and AEA for the next couple weeks I think :tongue:
  4. darkness9999's Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by GHOSH-5)
    No worries. So glad I've finished all my non-maths subjects. I'll be pounding down some STEP and AEA for the next couple weeks I think :tongue:
    I still have to revise for physics :sad: ... but its all for AEA after next week :p:
  5. Mark13's Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by GHOSH-5)
    And the only function I think of, such that  f(x) = f^{-1}(x) (*), is  f(x) = x, although there may be more.
    Any function f(x) whose graph y=f(x) has reflective symmetry in the line y=x will have that property I think. For example, f(x)=-x+c, f(x)=\frac{1}{x}, etc.
    Last edited by Mark13; 06-06-2009 at 18:22.
  6. Unbounded's Avatar
    • TSR Demigod
    Re: The AEA/Oxford/S Paper superthread
    (Original post by Mark13)
    Any function f(x) whose graph y=f(x) has reflective symmetry in the line y=x will have that property I think. For example, f(x)=-x+c, f(x)=\frac{1}{x}, etc.
    Indeed. I think it's generally true for any function  f with an inverse function. As the inverse function is, crudely, the 'reverse' process. One could consider a function to be like a machine: put a number in one side of it, and out pops another number; the inverse function is just doing those processes in reverse. I'm not sure how one would prove explicitly that  ff^{-1}(x) = x as it seems like almost a way of defining the inverse function. edit: turns out it is a way of defining the inverse according to wiki.
    Last edited by Unbounded; 06-06-2009 at 20:56.
  7. Unbounded's Avatar
    • TSR Demigod
    Re: The AEA/Oxford/S Paper superthread
    (Original post by darkness9999)
    I still have to revise for physics :sad: ... but its all for AEA after next week :p:
    :tongue: Good luck!
  8. Swayum's Avatar
    • TSR Legend
    • Location: My head
    Re: The AEA/Oxford/S Paper superthread
    (Original post by GHOSH-5)
    edit: turns out it is a way of defining the inverse according to wiki.
    What other definition would you (an A-level student) use?
  9. Unbounded's Avatar
    • TSR Demigod
    Re: The AEA/Oxford/S Paper superthread
    (Original post by Swayum)
    What other definition would you (an A-level student) use?
    Fair point.
  10. Horizontal 8's Avatar
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    Re: The AEA/Oxford/S Paper superthread
    Was the 2001 specimen (available from the warwick website) an actual oficial specimen? I ask because I thought the questions were very straight-forward compared to other papers I've done so far. On this one I managed to finish all the questions in about 45 minutes, whereas normally I wouldn't a) Finish all the questions entirely b) Let alone so quickly c) Make no mistakes d)not have to stop to think before diving straight into anwering any of the questions. Note: also, the 2002 specimen was one of the trickiest papers I've seen so far.. It just doesn't fit the pattern...
    Last edited by Horizontal 8; 08-06-2009 at 20:27.
  11. NesQuiK.'s Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by Horizontal 8)
    Was the 2001 specimen (available from the warwick website) an actual oficial specimen? I ask because I thought the questions were very straight-forward compared to other papers I've done so far. On this one I managed to finish all the questions in about 45 minutes, whereas normally I wouldn't a) Finish all the questions entirely b) Let alone so quickly c) Make no mistakes d)not have to stop to think before diving straight into anwering any of the questions. Note: also, the 2002 specimen was one of the trickiest papers I've seen so far.. It just doesn't fit the pattern...
    Just did a quick check and 2001 specimen is in my past paper pack so yes it must have been an official specimen. This is normal however for specimen papers, many are much easier than the papers that follow in subsequent years
  12. darkness9999's Avatar
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    Re: The AEA/Oxford/S Paper superthread
    hmm it is really annoying to see how they increase the difficulty of the paper each year ... i mean the 2005 paper was sooo easy compared to the 07/08 paper... hmm
  13. Horizontal 8's Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by darkness9999)
    hmm it is really annoying to see how they increase the difficulty of the paper each year ... i mean the 2005 paper was sooo easy compared to the 07/08 paper... hmm
    Hmm.. I found the 2004 paper the worst so far and the 2002 the easiest (excluding specimens)... 2005/2003 were ok aswell though...
  14. Horizontal 8's Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by NesQuiK.)
    Just did a quick check and 2001 specimen is in my past paper pack so yes it must have been an official specimen. This is normal however for specimen papers, many are much easier than the papers that follow in subsequent years
    Well I only found the 2002 Specimen easier than the 2004 paper so far...

    Still got 06/07/08 to go though
  15. NesQuiK.'s Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by Horizontal 8)
    Well I only found the 2002 Specimen easier than the 2004 paper so far...

    Still got 06/07/08 to go though
    Sill got all of them to do :p:
    How long do you take to do each paper on avg?
  16. darkness9999's Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by Horizontal 8)
    Hmm.. I found the 2004 paper the worst so far and the 2002 the easiest (excluding specimens)... 2005/2003 were ok aswell though...
    oh yeah that was a taff one... the questions I couldn't do were 4(b) and 5(d) ... killer
  17. Horizontal 8's Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by NesQuiK.)
    Sill got all of them to do :p:
    How long do you take to do each paper on avg?
    Varies quite abit but usually between 2 hours and 2h30...


    (Original post by darkness9999)
    oh yeah that was a taff one... the questions I couldn't do were 4(b) and 5(d) ... killer
    Ye it was abit of a pain... for question 4 (i) they listed like 6 different ways of doing it yet I did it a different way.. I definately showed 'em :rolleyes:
  18. Unbounded's Avatar
    • TSR Demigod
    Re: The AEA/Oxford/S Paper superthread
    (Original post by NesQuiK.)
    Sill got all of them to do :p:
    How long do you take to do each paper on avg?
    :p: I'm saving 05-08 for the next two weeks to bash out.

    I find it really varies. Sometimes I spend ages on one of the 'easier' questions at the start, but blast through the other 3-4 earlier questions in no time at all. :s:
  19. NesQuiK.'s Avatar
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    Re: The AEA/Oxford/S Paper superthread
    (Original post by GHOSH-5)
    :p: I'm saving 05-08 for the next two weeks to bash out.

    I find it really varies. Sometimes I spend ages on one of the 'easier' questions at the start, but blast through the other 3-4 earlier questions in no time at all. :s:
    Wish i could do that. Still got another 9 'real' exams to bash out in the next two weeks :work:
  20. Unbounded's Avatar
    • TSR Demigod
    Re: The AEA/Oxford/S Paper superthread
    (Original post by NesQuiK.)
    Wish i could do that. Still got another 9 'real' exams to bash out in the next two weeks :work:
    :eek: Good luck. God, how many exams are you taking in total this May/June sitting? :s:
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