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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    I can't do 2c. How do you choose the value of x to use?
    http://www.madasmaths.com/archive/iy...apers/c4_u.pdf

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    (Original post by eftio.gea)
    Guys can anyone help please, idk what to do with the sin(theta) Attachment 555579
    bump, tried this and cant do part a, someone show method pls!
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    (Original post by Don Pedro K.)
    SeanFM For the last part of this question, given that <ABC = 90°, could you not say that AB.BC = 0?

    Attachment 555583

    This is what the MS has done :/

    Attachment 555583555587
    I can't deal with vectors, sorry.

    But I'm not sure what you're saying. a.b = norm(a)norm(b)costheta. I'm not sure how you'd be able to say the LHS = 0.
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    (Original post by ccharlie97)
    I see now - I remember this question from when I did it lol!

    Well for part (b), you do,
    

A = \int_0^3 y \, \mathrm d x = ... = 27 \int_0^{\pi/2} \sin(2t)\sin(t) \, \mathrm d t

    Which is pretty easy to get!

    You don't get zero because you are integrating in parametric form! So as you integrate from  t = 0 to  \pi/2 , you can see the x coordinate goes from 3 to 0, and the y coordinate goes up and down, put is always positive!

    So over this t range of integration, it is actually considering the top part of the curve, so you will only get the area under the top part of the curve! So you'll get the same answer as using  y = \sqrt{f(x)} from the (a) using limits of 0 and 3 for x!

    Hope this helps
    Ah so it is because it is parametric it is different, thank you very much that is perfect Just what I needed

    Thank you for sticking with me through this haha :P
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    (Original post by pineneedles)
    I can't do 2c. How do you choose the value of x to use?
    http://www.madasmaths.com/archive/iy...apers/c4_u.pdf

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    If I say that 1 =  \frac{50}{50}...
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    (Original post by SeanFM)
    I can't deal with vectors, sorry.

    But I'm not sure what you're saying. a.b = norm(a)norm(b)costheta. I'm not sure how you'd be able to say the LHS = 0.
    Ah okay no problem And I just used the dot product rule? a.b = 0. In this case I let AB = a and BC = b. Apparently that doesn't work...
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    (Original post by Don Pedro K.)
    Ah okay no problem And I just used the dot product rule? a.b = 0. In this case I let AB = a and BC = b. Apparently that doesn't work...
    But I don't understand why you've done that (you've said because ABC <90 but I don't see how it follows). Maybe that is why it doesn't work.
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    Guys do you always have to simplify direction vectors lets say line L1 has points A(1,1,1) and B(7,13,25)

    so is the direction vector (6,12,24) or (1,2,4) ?????

    please clearify
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    (Original post by Abby5001)
    Attachment 555573 I tried this question and my constants were different to the mark scheme constants, however my answer for c=1/12 but the mark scheme said it was 1/4 . I don't know where I went wrong
    I also get 1/12. Which paper is this from?
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    (Original post by SeanFM)
    But I don't understand why you've done that (you've said because ABC <90 but I don't see how it follows). Maybe that is why it doesn't work.
    I just figured it out haha xD Thanks anyway !
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    (Original post by abro1089)
    bump, tried this and cant do part a, someone show method pls!
    What paper?
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    (Original post by jfahdgjdfshg)
    Guys on the June 2013 (R) you have to integrate 27sec^3(x) - doesn anyone know how to dothis?
    was it 27tanx ln sec x + tan x +c
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    (Original post by Mattematics)
    What paper?
    June 2013 withdrawn paper
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    (Original post by Mattematics)
    What paper?
    dunno just saw it as an attachment from previous post :s
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    Name:  Screen Shot 2016-06-23 at 16.55.28.png
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    advice on ho to do this
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    Can anyone help me with part June2013 (R) question 6 part d. Question is in spoiler
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    (Original post by eftio.gea)
    June 2013 withdrawn paper
    This should help

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    (Original post by Ainsleyy)
    Can anyone help me with part June2013 (R) question 6 part d. Question is in spoiler
    Spoiler:
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    I just did this Question! When you draw a diagram it shows that OB' = OB-2AB
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    (Original post by M&F)
    Guys do you always have to simplify direction vectors lets say line L1 has points A(1,1,1) and B(7,13,25)

    so is the direction vector (6,12,24) or (1,2,4) ?????

    please clearify
    Direction vector is b-a so (6,12,24)
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    (Original post by nathanbilli)
    I just did this Question! When you draw a diagram it shows that OB' = OB-2AB
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    Ah thank you! Did you only realise how to do that after you drew the diagram? I just looked at it and had no idea what to do haha
 
 
 
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