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Advanced Higher Maths 2011-2012 : Discussion and Help

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    (Original post by laughylolly)
    Had my 1st year class today. They seemed pretty nice. They were doing BODMAS... oh the good old days haha. I spent most the lesson helping this one boy who just kept on forgetting to use it! And I was trying to get him to write down each step as he went so he wouldn't miss anything out. Eventually got there and I was well proud. haha

    The 4th year class I was also supposed to help out in today had a test and the teacher didn't bother to tell me...

    In AH Maths we were doing Binomial theorem. Like learning how to use it to find the say x^5 term or whatever without writing it all out. It was a bit tedious cause I had already done it and my teacher was going super slow and he's pretty slow as it is. He's an alright teacher but just not my style of learning...
    That doesn't sound bad at all.

    Glad you sound like you enjoyed it (for the most part!).

    Binomial theorem isn't the most exciting of topics. But like was said, it helps for de Moivre's theorem (which I just started learning).
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    Is the sine of the inverse sine of x, x?
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    (Original post by soup)
    Is the sine of the inverse sine of x, x?
    Yes, although the other way round, inverse sine of sine of x, does not necessarily equal x. Can you think of why?

    Spoiler:
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    It is only true if we limit the domain of sine to be -pi/2 to pi/2, the codomain of inverse sine. Otherwise for example we can have sin^-1(sin(pi)) = 0
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    (Original post by ukdragon37)
    Yes, although the other way round, inverse sine of sine of x, does not necessarily equal x. Can you think of why?

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    It is only true if we limit the domain of sine to be -pi/2 to pi/2, the codomain of inverse sine. Otherwise for example we can have sin^-1(sin(pi)) = 0
    Are there also restrictions for the sine of the inverse sine of x? Those being -1 and 1?
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    (Original post by soup)
    Are there also restrictions for the sine of the inverse sine of x? Those being -1 and 1?
    No that explicit restriction is unnecessary, since it is impossible to obtain the inverse sine of anything outside of that range anyway (and still get a real number result). On the other hand, the restriction on inverse sine of sine is necessary since it's perfectly fine to obtain sines of values outside of -pi/2 to pi/2.
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    Could someone do me a favour and integrate this for me quickly, just so I can make sure I'm doing it correctly?

    (4x) / (2x-1)^1/2 dx

    I'm getting 4(2x-1)^1/2 + C but the answers are saying different and I'm confused!

    Thanks
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    (Original post by Quintro)
    Could someone do me a favour and integrate this for me quickly, just so I can make sure I'm doing it correctly?

    (4x) / (2x-1)^1/2 dx

    I'm getting 4(2x-1)^1/2 + C but the answers are saying different and I'm confused!

    Thanks
    Do the answers say \frac{4}{3}(2x-1)^{1/2}(x+1)+C by any chance, or something like that?
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    (Original post by Quintro)
    Could someone do me a favour and integrate this for me quickly, just so I can make sure I'm doing it correctly?

    (4x) / (2x-1)^1/2 dx

    I'm getting 4(2x-1)^1/2 + C but the answers are saying different and I'm confused!

    Thanks
    First off, take the 4 outside. Always make the coefficient of the numerator 1.

    So this becomes:

    4*\int \frac{x}{(2x-1)^{1/2}}

    Then let u = 2x-1 and du = 2dx

    Let me know if that helps. (I realise now that it was silly to factor out the 4 to begin with; try factoring out only 2 to begin with so your numerator is 2x.)


    (Original post by Beth1234)
    Do the answers say \frac{4}{3}(2x-1)^{1/2}(x+1)+C by any chance, or something like that?
    That's the answer I worked to. So I hope they do...
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    (Original post by JordanR)
    That's the answer I worked to. So I hope they do...
    That's reassuring

    I substituted u=(2x-1)^{1/2} though. Still trying to work it out the way you did it

    EDIT: Ah. Got it. Switching from scribbling on Paint to working it out with a pencil really does make a difference...
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    For that question would you have to simplify it down by taking out (2x-1)^1/2 as a common factor or could you just leave it as:

    2/3(2x-1)^3/2 + 2(2x-1)^1/2
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    (Original post by soup)
    For that question would you have to simplify it down by taking out (2x-1)^1/2 as a common factor or could you just leave it as:

    2/3(2x-1)^3/2 + 2(2x-1)^1/2
    I think you should take out the common factor. It looks a lot neater afterwards!

    ...and don't forget the constant!
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    (Original post by JordanR)
    First off, take the 4 outside. Always make the coefficient of the numerator 1.

    So this becomes:

    4*\int \frac{x}{(2x-1)^{1/2}}

    Then let u = 2x-1 and du = 2dx

    Let me know if that helps. (I realise now that it was silly to factor out the 4 to begin with; try factoring out only 2 to begin with so your numerator is 2x.)



    That's the answer I worked to. So I hope they do...
    Ahh! It was a completely daft mistake on my part! Sorry for wasting your time, but the help was great anyway if that's any consolation! Apparently the derivative of 2x-1 is 2x... *facepalm*. I looked over my working like four times and missed it every time.

    I have a question about factoring out though. Can any number be factored out before integrating, as long as I don't take out a variable such as x?

    God I'm an idiot..

    EDIT:
    (Original post by Beth1234)
    ..
    And yes, that's the answer, thanks.
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    (Original post by Quintro)
    Ahh! It was a completely daft mistake on my part! Sorry for wasting your time, but the help was great anyway if that's any consolation! Apparently the derivative of 2x-1 is 2x... *facepalm*. I looked over my working like four times and missed it every time.

    I have a question about factoring out though. Can any number be factored out before integrating, as long as I don't take out a variable such as x?

    God I'm an idiot..

    EDIT:

    And yes, that's the answer, thanks.
    It happens to everyone! Don't worry about it. I remember being stuck with a question and then realised that I was raising the power by one and then multiplying by it, in some sort of pseudo blend of integration and differentiation, and wondered why I was getting a crazy answer.

    Yes. Any constant can be factored out. That can make trig integrals a lot, lot simpler.


    (Original post by Beth1234)
    That's reassuring

    I substituted u=(2x-1)^{1/2} though. Still trying to work it out the way you did it

    EDIT: Ah. Got it. Switching from scribbling on Paint to working it out with a pencil really does make a difference...
    Ah, good. Saves me Latexing my answer...
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    I did make a stupid mistake earlier but I'm still not getting this.. I'd appreciate if someone could look through my working to see where I'm screwing up...



    EDIT: I know I made an error at the bottom, should be a minus sign not a plus.
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    (Original post by Quintro)
    I did make a stupid mistake earlier but I'm still not getting this.. I'd appreciate if someone could look through my working to see where I'm screwing up...



    EDIT: I know I made an error at the bottom, should be a minus sign not a plus.
    Mistake is going from third-to-bottom line to second-to-bottom. Taking out 4/3 from (4x - 2)/3 should take out a 4 from the 2 as well, which leaves 1/2.
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    (Original post by Quintro)
    I did make a stupid mistake earlier but I'm still not getting this.. I'd appreciate if someone could look through my working to see where I'm screwing up...



    EDIT: I know I made an error at the bottom, should be a minus sign not a plus.
    Third to second last line is incorrect.

    (4x-2)/3 + 2 can be rewritten as \frac{4(x+1)}{3}

    That should help.

    Edit: damn it. Beaten.
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    I´m only in fourth year just now studying Intermediate 2 Maths, currently I can just repeat questions out a textbook to understand a concept and move onto the next. Is this the same with Higher and Adv Higher? Is it all about practice?

    And what is everyone aiming for in their exam?
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    (Original post by ukdragon37)
    ...
    (Original post by JordanR)
    ...
    Thanks very much guys, I've got it now. I really need to work on manipulating equations.

    And I have no idea why I'm doing Maths at this time
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    (Original post by Quintro)
    Thanks very much guys, I've got it now. I really need to work on manipulating equations.

    And I have no idea why I'm doing Maths at this time
    It's what learning calculus did for me most. Remember that in an exam, they wouldn't penalise you for not putting it into that form (unless they asked you to show that it equalled that specifically).

    So don't worry about it. c:

    Maths at this time? Well, it's because maths is great. Or because you need something to bore you to sleep.
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    (Original post by nerd434)
    I´m only in fourth year just now studying Intermediate 2 Maths, currently I can just repeat questions out a textbook to understand a concept and move onto the next. Is this the same with Higher and Adv Higher? Is it all about practice?

    And what is everyone aiming for in their exam?
    It's similar for Higher. However for AH you are more likely to get some question whose style you have never seen before and you need to work out what method to use on it. Of course practising as much as possible helps. You also need to have a good memory as there are many formulae and identities to remember.
Updated: August 12, 2012
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