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Edexcel FP3 - 27th June, 2016 watch

  • View Poll Results: How did you find the Edexcel FP3 exam?
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    (Original post by Anon-)
    you're looking for the midpoint right? that's the sum of the first x coordinate and the second x coordinate, i.e (x1+x2)/2
    Cheers! I don't see how anyone would this of this under exam conditions. :lol:
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    (Original post by target21859)
    Don't really understand part d. Could someone explain it to me please?
    This was a weird one!

    Find attached my working. Apologies for the sun glare in the photo. I'm unsure if my reasoning is good enough. But as ever, a sketch really helps.

    Note that as a varies, there are two possible places for R. My reasoning here is that x is greater than or equal to a on one side of the ellipse, and less than or equal to negative a on the other side of the ellipse.

    See this post for extra clarity

    Name:  IMG_0060 copy 2.jpg
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    (Original post by target21859)
    When you differentiate (-1/3)(16-x^2)6(3/2) you get x(16-x^2)^(1/2) so you can see that there is an x on the outside. This x is needed so that you can integrate the expression exactly into what I just differentiated.
    I get that, but why can't you just integrate the initial expression, using u=x^n and dv=(16-x^2)^(1/2)? this also forms an expression in terms of In and I(n-2) but it's not what they are looking for.
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    (Original post by Ayman!)
    Zacken, Tom, could you guys help with 8(c) here? I can see how implicit differentiation works, but I'm not sure I understand the main mark scheme method of using summation of roots (which is supposedly not even in the UK spec?)
    The square root part cancels when you add them. See attached. Name:  image.jpg
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    (Original post by cjlh)
    Sweet man, what do you think of the department? I'm really excited for next year tbh
    Tbh I didn't get to see much of the department. I went to an post-offer visit day and I liked it. The course content seems really interesting too. Lots of maths & lots of programming

    Lecturers seem really nice too.

    I'm looking forward to it too!
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    (Original post by target21859)
    The square root part cancels when you add them. See attached. Name:  image.jpg
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Size:  495.0 KB
    I know how to do the algebra, I just wasn't sure why we wanted (a + b)/2. Cheers though!
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    (Original post by Anon-)
    I get that, but why can't you just integrate the initial expression, using u=x^n and dv=(16-x^2)^(1/2)? this also forms an expression in terms of In and I(n-2) but it's not what they are looking for.
    And how would you integrate the dv part nicely? If you had an x in front it would make life so much easier because there is an -x^2 term inside the root which differentiates to -2x outside the root by the chain rule. You would end up doing a lot more work than needed.
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    (Original post by target21859)
    Have you already done 3 A2 units?
    More, I think. I've already done C1-4, M1-3, S1-3, FP1-2 - just FP3 left.
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    (Original post by oinkk)
    This was a weird one!
    I'm still not really sure I understand.
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    (Original post by Ayman!)
    I know how to do the algebra, I just wasn't sure why we wanted (a + b)/2. Cheers though!
    Ah sorry
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    (Original post by Zacken)
    More, I think. I've already done C1-4, M1-3, S1-3, FP1-2 - just FP3 left.
    Well you're sorted then :P
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    (Original post by target21859)
    I'm still not really sure I understand.
    Find the locus as you would normally do, by eliminating your parameter \theta.

    Look at the coordinates of the point R. Note that the y coordinate must always be zero so the locus must lie on the real axis. Your x-coordinate is \frac{a}{cos\theta} where cos\theta is always less or equal to than one. Hence the x coordinate must be greater than or equal to a.

    Apply the same logic to the other side of the ellipse.
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    (Original post by target21859)
    And how would you integrate the dv part nicely? If you had an x in front it would make life so much easier because there is an -x^2 term inside the root which differentiates to -2x outside the root by the chain rule. You would end up doing a lot more work than needed.
    **** I'm a dumbass, of course, dv would be a ***** to work out. I thought it would integrate to -1/3x*(16-x^2)^(3/2), gg
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    (Original post by Anon-)
    **** I'm a dumbass, of course, dv would be a ***** to work out. I thought it would integrate to -1/3x*(16-x^2)^(3/2), gg
    Haha Glad you understand now.
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    (Original post by oinkk)
    Find the locus as you would normally do, by eliminating your parameter \theta.

    Look at the coordinates of the point R. Note that the y coordinate must always be zero so the locus must lie on the real axis. Your x-coordinate is \frac{a}{cos\theta} where cos\theta is always less or equal to than one. Hence the x coordinate must be greater than or equal to a.

    Apply the same logic to the other side of the ellipse.
    Yeah makes sense now thanks!
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    (Original post by target21859)
    Don't really understand part d. Could someone explain it to me please?
    Think you're complicating it more than it needs to be.

    You know that R is the point \left(\frac{a}{\cos \theta} , 0\right).

    Since -1 \leq \cos \theta \leq 1 then that means that \frac{1}{\cos \theta} \geq 1 or \frac{1}{\cos \theta} \leq -1.

    So that means that \frac{a}{\cos \theta} \geq a or \frac{a}{\cos \theta} \leq -a.

    But that is precisely x \geq a or x \leq -a.
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    (Original post by Zacken)
    Think you're complicating it more than it needs to be.

    You know that R is the point \left(\frac{a}{\cos \theta} , 0\right).

    Since -1 \leq \cos \theta \leq 1 then that means that \frac{1}{\cos \theta} \geq 1 or \frac{1}{\cos \theta} \leq -1.

    So that means that \frac{a}{\cos \theta} \geq a or \frac{a}{\cos \theta} \leq -a.

    But that is precisely x \geq a or x \leq -a.
    Yeah that explained it really well, cheers.
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    Does anyone have a copy of the FP3 June 13 withdrawn paper ?
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    (Original post by Heieieirifckfj)
    Does anyone have a copy of the FP3 June 13 withdrawn paper ?
    I don't there was one but I could be wrong.
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    Is reflections in planes and lines in vectors on the syllabus? I've seen questions of it in June 2014 IAL etc. but it's not specifically in the spec or in the textbook?
 
 
 
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