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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016 watch

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    On question 7 https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf why does w equal x+iy not u+iv please? Thanks MS https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf
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    (Original post by economicss)
    On question 7 https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf why does w equal x+iy not u+iv please? Thanks MS https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf
    They're just two variables, in theory you could use any letter you want ☺️
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    (Original post by UnknownDude)
    So why does the [1-w] change to [w-1]?
    (Original post by Zacken)
    It's a modulus. |w-1| and |1-w| are the same thing.Just like |1-0| and |0-1| are the same.edit: this is 5 freaking years old mate. What the heck.

    (Original post by UnknownDude)
    Can someone help me with this:

    http://www.thestudentroom.co.uk/show....php?t=1612404

    I've commented on it so can someone plssssss help
    Did you really bump a 5 year old thread...?

    Anyway, |1-w| = |(-1)(w-1)| = |(-1)|*|w-1|
    Its a modulus so
    |(-1)| = sqrt( (-1)^2 ) = 1
    then clearly
    |(-1)|*|w-1| = |w-1|



    (Original post by economicss)
    On question 7 https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf why does w equal x+iy not u+iv please? Thanks MS https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf
    They're variables so you can call them anything you want. Generally its popular to use z=x+iy and w =u +iv but you can choose any letters you want
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    (Original post by TheMarshmallows)
    Did you really bump a 5 year old thread...?

    Anyway, |1-w| = |(-1)(w-1)| = |(-1)|*|w-1|
    Its a modulus so
    |(-1)| = sqrt( (-1)^2 ) = 1
    then clearly
    |(-1)|*|w-1| = |w-1|
    Not sure why you quoted me?
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    (Original post by TheMarshmallows)
    Did you really bump a 5 year old thread...?

    Anyway, |1-w| = |(-1)(w-1)| = |(-1)|*|w-1|
    Its a modulus so
    |(-1)| = sqrt( (-1)^2 ) = 1
    then clearly
    |(-1)|*|w-1| = |w-1|
    Thanks man, I get it, THANKS SOO MUCH <3333 Found this thread as I was searching for answers to a question I didn't know. Can you also show meh part b? (I've posted it in the thread).
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    Sorry for hijacking threads, need help on part b.


    https://gyazo.com/84a6009f72c4f60befff510159b3e63d

    Mark scheme:

    https://gyazo.com/30a25cecc618ea71adb7733754242e64
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    Zacken
    Exam solutions did this, and he gave an explanation at the end by saying that you have to place some restrictions on the function so you can get the half line (and not the full line).

    Question: surely [tex] y<-2 [\tex] shoudl be restriction too, no?
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    (Original post by UnknownDude)
    Thanks man, I get it, THANKS SOO MUCH <3333 Found this thread as I was searching for answers to a question I didn't know. Can you also show meh part b? (I've posted it in the thread).

    See here for an explanation about where you should shade:

    (Original post by TheMarshmallows)
    Pick a point (z) that satisfies whatever you have, then use the transformation to turn it into W and look at where that point is on the graph. For example in June 2015 5B:

    mod(z) <= 2

    pick a point that satisfies this. Easiest point is z = 0 + 0i

    w = z/(z+3i) = 0/(0+3i) = 0

    w = 0 + 0i is inside of the circle from part a, so you shade inside of the circle.
    In your case, testing the point z = 0 + 0i

    w = z / (z+i) = 0/(0+i) = 0
    gives you the point w = 0, which is outside of your circle from part (a) so you shade outside of the circle.
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    (Original post by P____P)
    Zacken
    Exam solutions did this, and he gave an explanation at the end by saying that you have to place some restrictions on the function so you can get the half line (and not the full line).

    Question: surely [tex] y<-2 [\tex] shoudl be restriction too, no?
    But x < -3 implies y < -2. I mean, sure, if you want to write it, go on. But it's implied by x < 3 and you don't need to write y < -2.
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    (Original post by Rkai01)
    Attachment 545469
    Can someone please assist on 2c it got messy when trying to mod it.
    Name:  20160607_194109[1].jpg
Views: 184
Size:  502.0 KB
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    (Original post by Cpj16)
    Hey have you guys done the review 1 ex in the FP2 book.
    I m struggling with q47 pg70- could that come up tomorrow
    If it does come up, it's the last question and would be a good few marks. I could see something like this appearing, but it would be a show (maybe show in the form f(\theta) as there are a few ways to go with the question and how you deal with fractions in it

    Kinda made me happy when I got this though, I never feel strong with using multiples of angles in trig which are not double angles.
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    June 2014 question 6b, how are you meant to know to just put all your values into the equation they've given you to find k? It seems odd when the question asks you to show the equation, not to use it?

    The mark scheme just plugs in their values and things cancel giving 25, does that count as showing an equation?
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    (Original post by somevirtualguy)
    June 2014 question 6b, how are you meant to know to just put all your values into the equation they've given you to find k? It seems odd when the question asks you to show the equation, not to use it?

    The mark scheme just plugs in their values and things cancel giving 25, does that count as showing an equation?
    the point is that you use u and v to eliminate x.

    **right i see what you mean. so the mark scheme has assumed that the circle given is THE combination of U and V that eliminates x, and plugged in U and V to subsequently show it. whereas the model answers do not 'assume' that the given circle equation will cancel X and it chugs through the algebra which will 100% remove x, thus when everything cancels out you get the given curve. basically the mark shame way is the clever sneaky way bc it relies on the fact that the equation which removes x is unique so you don't have the worry about there being any other curves Z could be mapped to. i hope that makes sense. thus both methods are valid.
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    (Original post by somevirtualguy)
    June 2014 question 6b, how are you meant to know to just put all your values into the equation they've given you to find k? It seems odd when the question asks you to show the equation, not to use it?

    The mark scheme just plugs in their values and things cancel giving 25, does that count as showing an equation?
    I just did the question from scratch not sure if you'd get the marks though but I still get the answer. And takes a lot longer.
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    (Original post by Gilo98)
    the point is that you use u and v to eliminate x
    Yes but how are you meant to know to plug your values into what they've given you rather than derviving the equation which you normally have to do?
    • Thread Starter
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    I've chosen not to revise much today, just one past paper. I will go through everything briefly before sleep, though I won't be on TSR for the rest of the day.

    Good luck to all of you taking FP2 tomorrow!
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    (Original post by somevirtualguy)
    yes but how are you meant to know to plug your values into what they've given you rather than derviving the equation which you normally have to do?
    see edit in original reply fam
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    Name:  20160607_200456[1].jpg
Views: 184
Size:  519.7 KB
    (Original post by Rkai01)
    Attachment 545471
    11c I got all wrong can someone post a solution to this, thanks.
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    (Original post by khaleesi98)
    They're just two variables, in theory you could use any letter you want ☺️
    Thank you!
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    (Original post by ImJared)
    No one else did one, so here is what I got.

    Hey does anyone know what to do for the lst part, part d? I know its easy but yh
 
 
 
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