Year 13 Maths Help Thread

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    (Original post by ValerieKR)
    it means the modulus is sqrt39
    yes that does make sense, although I feel like it should have brackets or one of the vector arrows!!!
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    (Original post by k.russell)
    yes that does make sense, although I feel like it should have brackets or one of the vector arrows!!!
    it annoys me too
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    (Original post by ValerieKR)
    it annoys me too
    thanks for your help Zacken as well. next step is to remember the cosine rule, which I feel I genuinely haven't used in 2 years lol
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    (Original post by k.russell)
    thanks for your help Zacken as well. next step is to remember the cosine rule, which I feel I genuinely haven't used in 2 years lol
    I think that should be on your exam board formula booklet?
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    (Original post by ValerieKR)
    it annoys me too
    what a dork
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    (Original post by Zacken)
    I think that should be on your exam board formula booklet?
    well I'll be damned, it is there under C2. Maybe that's why I got 77 ums in C2...
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    Show that  \displaystyle \sum_{r=1}^n \sin r =\frac{1}{2} (\sin n +\cot \frac{1}{2}-\cot \frac{1}{2} \cos n) .

    Show also that  \displaystyle \sum_{r=1}^n \sin(r+1)\cos (r+3) =\frac{1}{4}( \cos (5) \csc (1) -\cos (2n+5)\csc (1) -2n\sin (2)) .

    :bebored::bored::hmmmm:
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    (Original post by Ano123)
    Show that  \displaystyle \sum_{r=1}^n \sin r =\frac{1}{2} (\sin n +\cot \frac{1}{2}-\cot \frac{1}{2} \cos n) .

    Show also that  \displaystyle \sum_{r=1}^n \sin(r+1)\cos (r+3) =\frac{1}{4}( \cos (5) \csc (1) -\cos (2n+5)\csc (1) -2n\sin (2)) .

    :bebored::bored::hmmmm:
    What have you tried? The first one is very standard, try telescoping.
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    Is there really any way of preparing for STEP other than doing it 'step by step': This is my strategy:

    1. Do all the easier STEP I questions first.

    2. Check the solutions to see if you could have done it more efficiently.

    3. When you have done enough of the easier questions, you will have picked up enough tricks to take a second look at the questions you originally struggled with.

    4. Climb your way up the difficulty level...
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    (Original post by Ano123)
    Show that  \displaystyle \sum_{r=1}^n \sin r =\frac{1}{2} (\sin n +\cot \frac{1}{2}-\cot \frac{1}{2} \cos n) .

    Show also that  \displaystyle \sum_{r=1}^n \sin(r+1)\cos (r+3) =\frac{1}{4}( \cos (5) \csc (1) -\cos (2n+5)\csc (1) -2n\sin (2)) .

    :bebored::bored::hmmmm:
    Id go, complex numbers for first one.
    Consider product pormulae for second. Cleans both up very quickly. Or induct


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    (Original post by Palette)
    Is there really any way of preparing for STEP other than doing it 'step by step': This is my strategy:

    1. Do all the easier STEP I questions first.

    2. Check the solutions to see if you could have done it more efficiently.

    3. When you have done enough of the easier questions, you will have picked up enough tricks to take a second look at the questions you originally struggled with.

    4. Climb your way up the difficulty level...
    Yeah I suppose that's the natural progression.
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    I'll be resitting Core 1 & Core 2 this year as I ****ed it up last year. Anyone who has experience of resitting maths modules, when is a good time to start practicing/revising for them?
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    (Original post by DarkEnergy)
    I'll be resitting Core 1 & Core 2 this year as I ****ed it up last year. Anyone who has experience of resitting maths modules, when is a good time to start practicing/revising for them?
    Shouldn't be too much of a hassle, if any at all, after doing C3 and C4. Just keep fresh on topics outside these modules.
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    (Original post by RDKGames)
    Shouldn't be too much of a hassle, if any at all, after doing C3 and C4. Just keep fresh on topics outside these modules.
    Cheers
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    I need help on a C3 domain and range questions .... again.
    Ive been over it twice and thought i understood but then got confused when i got this question wrong.

    In simple terms is the domain the x value and the range the y value you get out of the function?

    f(x) = (2x +5) / (x - 3)
    x E R, x is not equal to 3

    for the first part you are told to find the range. I got it as

    0< f(x)<= 13 but it should be f(x) E R, f(x) is not equal to 2.

    Where have i gone wrong?? (this isnt related to the question but how is R different to x E R)
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    (Original post by kiiten)
    I need help on a C3 domain and range questions .... again.
    Ive been over it twice and thought i understood but then got confused when i got this question wrong.

    In simple terms is the domain the x value and the range the y value you get out of the function?

    f(x) = (2x +5) / (x - 3)
    x E R, x is not equal to 3

    for the first part you are told to find the range. I got it as

    0< f(x)<= 13 but it should be f(x) E R, f(x) is not equal to 2.

    Where have i gone wrong?? (this isnt related to the question but how is R different to x E R)
    Well, how did you work out the range?

    If you say \frac{2x+5}{x-3} =
    \frac{2x - 6 + 11}{x-3}...

    I also don't understand what you mean by R being different to 'x is an element of R', they mean different things but chances are in your context R on its own doesn't make sense.



    (Original post by DarkEnergy)
    I'll be resitting Core 1 & Core 2 this year as I ****ed it up last year. Anyone who has experience of resitting maths modules, when is a good time to start practicing/revising for them?
    C2 requires very little practice.

    Just be aware of the topics that aren't in C3 (and maybe C4) and the usual rules of not cancelling solutions, checking range of solutions etc and then work through the textbook close to exam time, you should be fine.
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    (Original post by SeanFM)
    Well, how did you work out the range?

    If you say \frac{2x+5}{x-3} =
    \frac{2x - 6 + 11}{x-3}...





    C2 requires very little practice.

    Just be aware of the topics that aren't in C3 (and maybe C4) and the usual rules of not cancelling solutions, checking range of solutions etc and then work through the textbook close to exam time, you should be fine.
    im not sure why you put -6 +11

    To find the range i subbed in x=3 to get 0 so f(x) must be bigger than 0 because x is not equal to 3. The i subbed in x=4 (that would give 1 as the denominator - largest value of f(x)) = 13

    ?
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    (Original post by kiiten)
    im not sure why you put -6 +11

    To find the range i subbed in x=3 to get 0 so f(x) must be bigger than 0 because x is not equal to 3. The i subbed in x=4 (that would give 1 as the denominator - largest value of f(x)) = 13

    ?
    Sometimes that method works (for an increasing/decreasing function) but this isn't one of them. As you can see from sketching a graph on it.

    It's a way to manipulate the fraction so that you end up with (A - B/(x+c)) which is much easier to find the range of.

    Notice that if you add 11 then take away 6 next to the x, then you can divide the three terms by the denominator to give a constant (A) as well as B and C.

    If you can't see how adding the 11 and taking away 6 works, I can explain it in more detail when I get a chance.
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    (Original post by SeanFM)
    Sometimes that method works (for an increasing/decreasing function) but this isn't one of them. As you can see from sketching a graph on it.

    It's a way to manipulate the fraction so that you end up with (A - B/(x+c)) which is much easier to find the range of.

    Notice that if you add 11 then take away 6 next to the x, then you can divide the three terms by the denominator to give a constant (A) as well as B and C.
    Have i done it wrong - i did 2x+11-6 / (x-3)
    2(x-3)+11 / (x-3)
    x-3 cancels out and you get 13??

    I understand it would be a lot easier to draw out the graph but i dont have a graphing calculator and always seem to draw it wrong

    What method would you use for a function that isnt a fraction?
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    Another (different) question

    If y = 1 + sinx is stretched in the x-direction, scale factor 1/2
    why is it y = 1 + 2sinx
    instead of
    y = 1 + sin2x

    ??
 
 
 
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