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    (Original post by molly221)
    Attachment 589272

    How do you do this question? 2013 paper.
    You would draw the graph of y=sinx, then draw the graph of y = sinx reflected in x= pi. Then reflect the resulting graph in y=2. It helps to draw out each stage instead of doing it all in one. The questions are always in radians as well so it is easy to reflect in pi.
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    (Original post by 11234)
    Can u use c3 c4 techniques or will u get penalised?
    I've seen a couple of questions from past papers where they say "some students may have done this using the quotient rule" or similar things - like in A levels, every mathematically rigorous technique is equally valid. You can use FP3 techniques if you think it's easier that way.
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    2011 paper Question 5(v).

    How many marks do you think I would get if I put this as my solution:

    Let Sn be the number of solutions in a semi-grid size n.

    When n is odd, the number of solutions is the same as that for n - 1, as the semi-grid simply has a blank diagonal added, i.e. Sn = Sn-1

    When n is even, Sn = Sn-2 + 2^(n-1), as a grid of even n has all the same solutions as that for n-2 apart from the main diagonal, of which there are 2^(n-1) solutions, as already found in (iii).

    We can now redefine it for odd n, Sn = Sn-3 + 2^(n-2), by substituting in n-1 for n to the above formula, as we know n-1 is even

    Therefore for n >= 3, Sn =
    {
    Sn-2 + 2^(n-1); n is even
    Sn-3 + 2^(n-2); n is odd
    }

    S0 = 0, S1 = 0, S2 = 2

    I should have noticed the geometric series, but I did this recursive way instead. Would I get any marks at all for this? It does work but isn't on the solutions at all. Thanks
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    2007 Q5 part v?
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    (Original post by Mystery.)
    2007 Q5 part v?
    Okay, so you should have found that the recursion depth of g(2^k) is k + 1.

    So for g(2^l + 2^k), as l is greater than k, the power of 2^k will "run out" before the power of 2^l. So you'll get an expression similar to (iv), where you can apply your answer to (iv) to get an expression. Some more detailed information:
    Spoiler:
    Show
    So you should have got in part (iv) that g(2^k) = k + g(1). So g(2^l + 2^k) = k + g(2^(l-k) +1). Now, 2^(l-k) will be even, so 2^(l-k) - 1 must be odd. So we apply the third rule to get that it equals k + 1 + g(2^(l-k)).

    Now we take the expression from (iv) and substitute l-k in for k, to get tha g(2^(l-k)) = (l-k) + 1. So overall, g(2^l + 2^k) = k + 1 + l - k + 1. = l + 2.
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    (Original post by lewman99)
    Okay, so you should have found that the recursion depth of g(2^k) is k + 1.

    So for g(2^l + 2^k), as l is greater than k, the power of 2^k will "run out" before the power of 2^l. So you'll get an expression similar to (iv), where you can apply your answer to (iv) to get an expression. Some more detailed information:
    Spoiler:
    Show
    So you should have got in part (iv) that g(2^k) = k + g(1). So g(2^l + 2^k) = k + g(2^(l-k) +1). Now, 2^(l-k) will be even, so 2^(l-k) - 1 must be odd. So we apply the third rule to get that it equals k + 1 + g(2^(l-k)).

    Now we take the expression from (iv) and substitute l-k in for k, to get tha g(2^(l-k)) = (l-k) + 1. So overall, g(2^l + 2^k) = k + 1 + l - k + 1. = l + 2.
    I get it till it becomes g(2^(l-k)) in the second line
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    (Original post by Mystery.)
    I get it till it becomes g(2^(l-k)) in the second line
    Okay, so we know from the previous part that g(2^k) = k + 1, where k is an integer. Now, both l and k are integers, and it'll be positive because l > k, so l-k will be an integer, so we can substitute l-k into our formula.

    There's also another way to think about it. Essentially, you've got 2 "types" of steps: dividing by 2 when it's even, and subtracting 1 when it's odd. Now, because l > k, if you keep dividing by 2, l will become 1 in more steps than k will. So for each of these steps, we add 1 to the value of the function. This step then repeats l times, because it divides by 2 each time. But we also have to consider the step of subtracting 1 when it's odd, this occurs twice - when 2^k reaches 1 after k steps, and when 2^l reaches 1 after l steps. So overall, the value of the function is l + 2.
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    (Original post by ShatnersBassoon)
    Thank you for this information. Do you have any idea how much emphasis is usually placed on explanations in MAT? I'm applying for Maths+CompSci so I'll be doing question 6, which seems like a particularly explanation-heavy question. If I've got the correct answer (e.g. "Alice and Bob are liars; Charlie told the truth", is this usually sufficient to gain all the marks for that part of the question, or could I be penalised if my explanation is a little ineloquent?
    We do want to see explanation, to check that you've got to the right answer by a valid method, rather than just guessing well. The explanation should be clear; a good rule of thumb is that it should be enough to convince one of your colleagues. However, it doesn't need to be very eloquent.

    Gavin
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    can someone explain to me part iv of Q4 of 2015. i dont understand how the lopsidedness is minimised when it is split into 3 equal regions since they didn't provide a diagram so i'm having problems visualising it. Thanks!
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    (Original post by danielhx)
    can someone explain to me part iv of Q4 of 2015. i dont understand how the lopsidedness is minimised when it is split into 3 equal regions since they didn't provide a diagram so i'm having problems visualising it. Thanks!
    Lopsidedness is the biggest area of the circle. Every time we make one area smaller, another area will get bigger and hence the lopsidedness will get bigger. To minimise the lopsidedness, we must split it into 3 equal regions, as if we have any bigger or smaller regions, the lopsidedness will be bigger.

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    Do you need to know about harmonic series?
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    (Original post by Mystery.)
    Do you need to know about harmonic series?
    Only C1 and C2 knowledge will be required, so I don't think it would be necessary. But, I suggest you to know more about different things, coz MAT will be more difficult in the later years, as a trend from the past few papers.
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    Is anybody aware of good resources where you can get a grip for how to tackle longer answer questions as a lot of the time i am struggling to see how to best answer these questions
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    Do you get marks for workings or just answers?
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    Could someone help me with 1J from the 2014 paper please? https://www.maths.ox.ac.uk/system/fi...nts/test14.pdf


    I am struggling to understand how I would solve this and particularly the integral is confusing me a little.
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    (Original post by alfmeister)
    Could someone help me with 1J from the 2014 paper please? https://www.maths.ox.ac.uk/system/fi...nts/test14.pdf


    I am struggling to understand how I would solve this and particularly the integral is confusing me a little.
    Integrate both sides of the given identity from -1 to 1, remember that the existing integral in that identity is just a number, you might find it helpful to call it u = \int_{-1}^1 f(t) \, \mathrm{d}t and then integrate both sides like so:

    \int_{-1}^1 6 \, \mathrm{d}x + \int_{-1}^{1} f(x) \, \mathrm{d}x = 2\int_{-1}^1 f(-x) \, \mathrm{d}x + u\int_{-1}^1 3x^2 \, \mathrm{d}x

    which then simplifies down to 12 + u = 2u + 2u where u is the answer you want.
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    (Original post by Zacken)
    Integrate both sides of the given identity from -1 to 1, remember that the existing integral in that identity is just a number, you might find it helpful to call it u = \int_{-1}^1 f(t) \, \mathrm{d}t and then integrate both sides like so:

    \int_{-1}^1 6 \, \mathrm{d}x + \int_{-1}^{1} f(x) \, \mathrm{d}x = 2\int_{-1}^1 f(-x) \, \mathrm{d}x + u\int_{-1}^1 3x^2 \, \mathrm{d}x

    which then simplifies down to 12 + u = 2u + 2u where u is the answer you want.
    Thanks, that makes more sense now. Making the integral = u makes it a lot simpler. I'll have to try to find some more of them but I am not too sure where to look. Do you know questions similar to this?
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    (Original post by alfmeister)
    Thanks, that makes more sense now. Making the integral = u makes it a lot simpler. I'll have to try to find some more of them but I am not too sure where to look. Do you know questions similar to this?
    I know a few STEP questions that take the idea to the extreme, if you're interested.
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    (Original post by Zacken)
    I know a few STEP questions that take the idea to the extreme, if you're interested.
    Yes please, I'll give them a go and see what I can do.
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    (Original post by alfmeister)
    Yes please, I'll give them a go and see what I can do.
    STEP II, 2004, Q5 and STEP II, 2000, Q5.
 
 
 
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