Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

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

    Offline

    4
    ReputationRep:
    (Original post by oinkk)
    What happens when we're finding a Maclaurin expansion of a function and when it comes to evaluating a particular part of it, we end up doing 1/0?
    Do you have a specific question?
    Offline

    3
    ReputationRep:
    (Original post by Craig1998)
    Do you have a specific question?
    Nah. I was just considering that scenario. Take y = cosec x for instance.
    Offline

    22
    ReputationRep:
    (Original post by oinkk)
    What happens when we're finding a Maclaurin expansion of a function and when it comes to evaluating a particular part of it, we end up doing 1/0?
    Not possible. For a MacLaurin expansion to be valid (i.e: for it to exist) you require all derivatives to be defined. i.e: if you get get 1/0, it means you're working out the Maclaurin series of a function that doesn't have a MacLaurin series. (like ln x)
    Offline

    2
    ReputationRep:
    (Original post by oinkk)
    Nah. I was just considering that scenario. Take y = cosec x for instance.
    You won't get 1/0, but you could a term which is 0, in which case you skip it and move to the next one.
    Offline

    2
    ReputationRep:
    (Original post by TheMarshmallows)
    if its a "show that" question, then you can use the answer whilst solving it (to show that it is true). Generally using u+iv will work always, doing it with modulus' is usually less algebra/shorter but isn't always possible (and takes some thinking).

    For example in the question we were talking about before, arg(z) = pi/4 is the half line y=x, so you can rewrite z = x + iy into z = x + ix (or z = y + iy if you prefer, it doesnt matter).

    Then subbing into W = (Z+1)/(z+i) we get W = (x + ix + 1)/(x + ix + i)
    grouping real and imaginary gives W = ( (x+1) + ix ) / (x + i(x+1) )

    Looking at the answer, they want you to show that mod(W) = 1 so mod(our W) = 1

    mod( (x+1) +ix) / (x + i(x+1)) ) = 1

    mod((x+1) +ix) / mod(x+ i(x+1)) = 1

    Remembering that mod(a + ib) = sqrt(a^2 + b^2) gives us

    sqrt( (x+1)^2 + x^2) / sqrt( x^2 + (x+1)^2 ) = 1

    which is clearly true because they're the same thing.
    Thanks. What do you think I should use in the exam, u + vi or x+yi. With u+vi you have to take long steps but its more straightfoward I guess but x+yi is easier but I might not get it . I also take long time to finish other questions so i dont know...
    Offline

    3
    ReputationRep:
    (Original post by Zacken)
    Not possible. For a MacLaurin expansion to be valid (i.e: for it to exist) you require all derivatives to be defined. i.e: if you get get 1/0, it means you're working out the Maclaurin series of a function that doesn't have a MacLaurin series. (like ln x)
    Understood! Cheers.
    Offline

    12
    ReputationRep:
    Hey have you guys done the review 1 ex in the FP2 book.
    I m struggling with q47 pg70- could that come up tomorrow
    Offline

    2
    ReputationRep:
    (Original post by fpmaniac)
    Thanks. What do you think I should use in the exam, u + vi or x+yi. With u+vi you have to take long steps but its more straightfoward I guess but x+yi is easier but I might not get it . I also take long time to finish other questions so i dont know...
    I'd stick with what you know and what you feel comfortable doing rather than trying to change methods now.
    Offline

    1
    ReputationRep:
    (Original post by Inges)
    http://imgur.com/xKBlDyE

    Hope this helps, it really brain teasing I know!
    (Original post by edothero)
    Product rule with u = \dfrac{1}{x} and v = \dfrac{dy}{dt}

    \dfrac{d^{2}y}{dx^{2}} = uv' + u'v
    Thank you both very much!
    Offline

    2
    ReputationRep:
    (Original post by Anon-)
    guys, could someone please explain the questions where they ask you to shade the area after a transformation in the complex plane? June 2009 6b or June 2015 5b for example
    bump
    Offline

    3
    ReputationRep:
    (Original post by economicss)
    Please could anyone do a worked solution for question 12a https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf I've tried it by substituting in for lambda plus lambda i when x is greater than 0 but for some reason I can't get it to work but I thought the method was correct? Thanks
    do you have any similar questions? i need to practice these ones!
    thanks
    Offline

    2
    ReputationRep:
    (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
    I'm almost certain that won't come up.
    Offline

    4
    ReputationRep:
    (Original post by oinkk)
    Nah. I was just considering that scenario. Take y = cosec x for instance.
    I think this is the reason we are given the expansion for ln(1+x) rather than lnx. It just wouldnt work.
    Online

    19
    ReputationRep:
    (Original post by Anon-)
    bump
    Test out a point and see what it is transformed to, if it is in the circle then you shade inside, if it is outside the circle, you shade outside. If that doesn't satisfy you, just look through your calculations and mentally replace the = with the appropriate inequalities and see what inequality you're left with at the end.
    edit: Ah I see you were talking generally, but in practice it's usually circles..
    Offline

    6
    ReputationRep:
    Name:  image.jpg
Views: 129
Size:  436.9 KB
    Can someone please assist on 2c it got messy when trying to mod it.
    Offline

    2
    ReputationRep:
    (Original post by Anon-)
    bump
    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.
    Offline

    6
    ReputationRep:
    Name:  image.jpg
Views: 140
Size:  498.9 KB
    11c I got all wrong can someone post a solution to this, thanks.
    Offline

    17
    ReputationRep:
    (Original post by TheMarshmallows)
    I'm almost certain that won't come up.
    In fairness, it was on a past edexcel exam (old spec FP2 I believe).

    These can often become easier if you multiply through by e^-0.5iø to get something with cos0.5ø on the bottom; there on it is but algebraic manipulation and using trigonometric formulae.


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (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.
    yeah that's what I thought, thanks!
    Offline

    5
    ReputationRep:
    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
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 24, 2017
Do I go to The Streets tomorrow night?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.