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    Does anyone know if there a shortcut to solving those matrices simultaneous eqn q's aha? Would appreciate any tips on how to tackle this type of question!
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    (Original post by Leechayy)
    what does ag mean on the mark scheme?
    answer given.

    In a 'show that' question where the result is in the question, you can't get a mark just for writing it! So any credit is for the working .. which means you should be generous with the working to show the examiner you are not blagging.
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    (Original post by Djangoo)
    Does anyone know if there a shortcut to solving those matrices simultaneous eqn q's aha? Would appreciate any tips on how to tackle this type of question!
    Which part do you find tricky? If look at say Q3 on Jan 2010:

    (i) Inverting a 3x3 matrix is just a fiddly process. You must show working, but can sometimes check the result on your calculator. This one has an 'a' in one cell. Expect an (a-4) in the determinant because of the a¬=4 in the question.

    (ii) Using the inverse to solve a set of 3 equations is also routine.

    (iii) is the part most would find hard. Where the determinant is zero the equations are degenerate so there is not a single point solution. Is this the part where you get stuck?
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    can anyone help me with Eigenvectors? I can work them out perfectly fine but iv just done a 12 mark question in January 2013 working out all the eigenvalues and corresponding vectors, I got them correct expect all the signs (negative and positive) wrong and I don't know why!

    Any advice on how to avoid this would be great!!
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    (Original post by drslimey)
    can anyone help me with Eigenvectors? I can work them out perfectly fine but iv just done a 12 mark question in January 2013 working out all the eigenvalues and corresponding vectors, I got them correct expect all the signs (negative and positive) wrong and I don't know why!

    Any advice on how to avoid this would be great!!
    do you let x=k?

    if you have 2x=-3y or something for one equation then you will get y=-2/3x and if x=k then y=-2/3k

    so for a whole number say k is 3 then you get x=3 and y=-2
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    Anyone else think that the mark scheme for Q4iii) Jan 12 shows unnecessary amounts of working? The last three marks are just basically repeating what was done in the first section, I didn't even realise that the question was asking us to include it.

    if tanhy = x, \frac{dy}{dx} = \frac{1}{1-x^2} using implicit differentiation.

    But surely this implies that \frac{d}{dx} (artanhx) = \frac{1}{1-x^2}, because:

    tanhy = x \Rightarrow y = artanhx

    Am I just being thick and missing something? Or have we just proved the same thing twice.

    http://www.mei.org.uk/files/papers/2012_Jan_fp2.pdf
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    what paper do you lot think is the hardest?
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    (Original post by lizard54142)
    Anyone else think that the mark scheme for Q4iii) Jan 12 shows unnecessary amounts of working? The last three marks are just basically repeating what was done in the first section, I didn't even realise that the question was asking us to include it.

    if tanhy = x, \frac{dy}{dx} = \frac{1}{1-x^2} using implicit differentiation.

    But surely this implies that \frac{d}{dx} (artanhx) = \frac{1}{1-x^2}, because:

    tanhy = x \Rightarrow y = artanhx

    Am I just being thick and missing something? Or have we just proved the same thing twice.

    http://www.mei.org.uk/files/papers/2012_Jan_fp2.pdf
    it wants you differentiate the logarithm version of artanh x from the earlier part
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    (Original post by Epistaxis)
    it wants you differentiate the logarithm version of artanh x from the earlier part
    I know that; but why? We are essentially doing the same thing twice. The implicit differentiation already told us the derivative of artanh x
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    (Original post by lizard54142)
    I know that; but why? We are essentially doing the same thing twice. The implicit differentiation already told us the derivative of artanh x
    No idea why, but I guess it's just showing that you could use one of two ways maybe?

    Although you're right, it is a bit pointless
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    (Original post by Epistaxis)
    No idea why, but I guess it's just showing that you could use one of two ways maybe?

    Although you're right, it is a bit pointless
    You see what I mean? I am very confused, glad it's not just me.

    It's like giving you a picture of a triangle and asking you to work out its area, but to get full marks you have to use two formulas:

    Area = \frac{1}{2} \times b \times h and say, Heron's formula. You only get half marks if you calculate the area using only one formula.

    It makes no sense
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    (Original post by drslimey)
    can anyone help me with Eigenvectors? I can work them out perfectly fine but iv just done a 12 mark question in January 2013 working out all the eigenvalues and corresponding vectors, I got them correct expect all the signs (negative and positive) wrong and I don't know why!

    Any advice on how to avoid this would be great!!
    An eigenvector just indicates a direction. Its length doesn't matter. By convention we take the smallest vector with integer coefficients. So how do you choose between (-1, 1) and (1,-1) say? Answer is that either is equally valid. Anyone marking the paper will know this. So if the mark scheme gives the other one - no problem.
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    Hey guys, just wondering is there some result we have to know regarding arcsin(cosx) and arccos(sinx) cause I've noticed it cropping up every now and again in a step in an integration which isn't explained in the mark scheme?

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    (Original post by Dan205)
    Hey guys, just wondering is there some result we have to know regarding arcsin(cosx) and arccos(sinx) cause I've noticed it cropping up every now and again in a step in an integration which isn't explained in the mark scheme?

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    Can you link the example?
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    (Original post by lizard54142)
    Can you link the example?
    I can't remember what paper it was in :/ but the jist was you differentiated/integrated and got arcsin(y) after originally having y=cosx (i'm completely naming this example up here,sorry ) then to put it in terms of x you obviously have to do arcsin(cosx) but the mark scheme just jumps to sqrt(1-x^2) I think?

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    (Original post by Dan205)
    I can't remember what paper it was in :/ but the jist was you differentiated/integrated and got arcsin(y) after originally having y=cosx (i'm completely naming this example up here,sorry ) then to put it in terms of x you obviously have to do arcsin(cosx) but the mark scheme just jumps to sqrt(1-x^2) I think?

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    to get that sqrt, wouldn't that be cos(arcsin(x)) ? in which case it's just Pythagoras.
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    (Original post by ian.slater)
    to get that sqrt, wouldn't that be cos(arcsin(x)) ? in which case it's just Pythagoras.
    Yes quite possibly! Could you explain how it's Pythag? Really sorry I feel like such an idiot not getting it haha

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    (Original post by Dan205)
    I can't remember what paper it was in :/ but the jist was you differentiated/integrated and got arcsin(y) after originally having y=cosx (i'm completely naming this example up here,sorry ) then to put it in terms of x you obviously have to do arcsin(cosx) but the mark scheme just jumps to sqrt(1-x^2) I think?

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    Do you mean something like this:

    

\sin y = x

\cos y \frac{dy}{dx} = 1

\dfrac{dy}{dx} = \dfrac{1}{\cos y}

= \dfrac{1}{\sqrt {1 - \sin ^2y}}

= \dfrac{1}{\sqrt {1 - x^2}}

    They use the identity \sin ^2 x + \cos ^2 x = 1
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    (Original post by lizard54142)
    Do you mean something like this:

    

\sin y = x

\cos y \frac{dy}{dx} = 1

\dfrac{dy}{dx} = \dfrac{1}{\cos y}

= \dfrac{1}{\sqrt {1 - \sin ^2y}}

= \dfrac{1}{\sqrt {1 - x^2}}

    They use the identity \sin ^2 x + \cos ^2 x = 1
    Yes! Thank you so much, I was being an idiot haha

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    (Original post by Dan205)
    Yes! Thank you so much, I was being an idiot haha

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    Awesome, glad we fixed it in time for tomorrow!

    I forgot to add, but you have to justify taking the positive root: to do this you can draw a sketch graph of y=arcsinx and say "as the graph shows, the gradient is positive at all points, so we take the positive root".

    There is usually one mark for saying this!
 
 
 

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