Hey there! Sign in to join this conversationNew here? Join for free

Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

Announcements
    Offline

    3
    ReputationRep:
    What is the essential stuff you need to memorize? Mainly trig wise ie integral of sin/cos etc
    Offline

    1
    ReputationRep:
    (Original post by skeptical_john)
    What is the essential stuff you need to memorize? Mainly trig wise ie integral of sin/cos etc
    here are 2 helpful links for the formulas you are not given in the booklet and need to know (you need to know them from not just c4 but also previous units)
    https://www.cgpbooks.co.uk/interacti...20Edexcel%20C3
    http://www.mei.org.uk/files/pdf/formulae.pdf
    Offline

    14
    ReputationRep:
    Does anybody know if we need to learn how to derive the equation a.b = a1b1 + a2b2 + a3b3?
    Offline

    2
    ReputationRep:
    I really need some help for differential equations! The first step is to rearrange it, but I struggle to know which numbers/letters to put on which side? Like obviously an x would go on the same side as the dx, but I never know what to do with constants? Please someone help I messed up c3 so badly!
    Offline

    2
    ReputationRep:
    (Original post by psychemma)
    I really need some help for differential equations! The first step is to rearrange it, but I struggle to know which numbers/letters to put on which side? Like obviously an x would go on the same side as the dx, but I never know what to do with constants? Please someone help I messed up c3 so badly!
    You have to get both sides into a form that you know you can integrate, that's the only condition. You can always move stuff around after you've integrated it to get it into the required form, but the bottom line is multiply and divide in order to get a form on both sides that you can integrate. If you have an example I can go through it for you

    (Original post by BrainJuice)
    Does anybody know if we need to learn how to derive the equation a.b = a1b1 + a2b2 + a3b3?
    No that's beyond the spec. If you have questions like that it's best to read the specification for the unit yourself to make sure:

    https://qualifications.pearson.com/c...hs_Issue_3.pdf
    Offline

    0
    ReputationRep:
    Name:  13499994_1347159238644068_1927852354_o.jpg
Views: 127
Size:  172.3 KBAttachment 554818554820
    Can anyone please explain how you do this? It's from Solomon I and according to the mark scheme the root 4-x^2 on the bottom turns into 2cosu when you substitute x=2sinu , and dx/du is also 2cosu so in the end you only have to integrate the number 1. I can't get the bottom line to be 2cosu though..
    Attached Images
     
    Offline

    19
    ReputationRep:
    (Original post by celessi)
    Name:  13499994_1347159238644068_1927852354_o.jpg
Views: 127
Size:  172.3 KBAttachment 554818554820
    Can anyone please explain how you do this? It's from Solomon I and according to the mark scheme the root 4-x^2 on the bottom turns into 2cosu when you substitute x=2sinu , and dx/du is also 2cosu so in the end you only have to integrate the number 1. I can't get the bottom line to be 2cosu though..
    root(4-(2sinu)^2) = root(4 - 4sin^2u) = root(4)root(1-sin^2u)
    Offline

    2
    ReputationRep:
    (Original post by celessi)
    Name:  13499994_1347159238644068_1927852354_o.jpg
Views: 127
Size:  172.3 KBAttachment 554818554820
    Can anyone please explain how you do this? It's from Solomon I and according to the mark scheme the root 4-x^2 on the bottom turns into 2cosu when you substitute x=2sinu , and dx/du is also 2cosu so in the end you only have to integrate the number 1. I can't get the bottom line to be 2cosu though..
    the bottom bit comes from subbing in 2sinu for x
    so we get (4-4sin^2 u)^1/2
    take out 4 as a common factor
    so we get (4(1-sin^2 u))^1/2
    but 1 -sin^2 u = cos^2
    We can then sqrt to get their answer
    hope this helps
    Offline

    0
    ReputationRep:
    (Original post by candol)
    the bottom bit comes from subbing in 2sinu for x
    so we get (4-4sin^2 u)^1/2
    take out 4 as a common factor
    so we get (4(1-sin^2 u))^1/2
    but 1 -sin^2 u = cos^2
    We can then sqrt to get their answer
    hope this helps
    Thanks!!
    • Community Assistant
    Offline

    18
    ReputationRep:
    (Original post by metrize)
    I don't know if what you did is right or not, ask Zacken or SeanFM
    how do you tag people?
    Offline

    11
    ReputationRep:
    So I manage to fly through the first Gold paper for C4, but in the real exam...
    Offline

    2
    ReputationRep:
    (Original post by particlestudent)
    So I manage to fly through the first Gold paper for C4, but in the real exam...
    Hahaha isn't it so. Could do a C3 paper in forty minutes, come yesterday I was struggling
    Offline

    3
    ReputationRep:
    Can anyone explain Question 4b on this paper, June 2013R:

    https://a086a5a2f39bda93734c56a63fab...%20Edexcel.pdf

    Hate these annoying 2/3 markers, but better to be ready for them in case they pop up
    Offline

    19
    ReputationRep:
    (Original post by Dohaeris)
    Can anyone explain Question 4b on this paper, June 2013R:

    https://a086a5a2f39bda93734c56a63fab...%20Edexcel.pdf

    Hate these annoying 2/3 markers, but better to be ready for them in case they pop up
    You want an x with |x| < 8/9 that allows you to nicely estimate (7100)^(1/3). You note that since |x| < 8/9 you are restricted to 0 < 8-9x < 16. So you're certainly not going to be able to put in 7100 or 71 or 710, but why not 7.1? Indeed you get 8 - 9x =7.1 when x = 0.1, and (7100)^(1/3) = 7.1^(1/3) * 1000^(1/3) = 10*(7.1^(1/3)) so you can easily rearrange for an estimate of 7100^(1/3) once you've plugged in x = 0.1
    Offline

    11
    ReputationRep:
    (Original post by cjlh)
    Hahaha isn't it so. Could do a C3 paper in forty minutes, come yesterday I was struggling
    Same here. I've done better in C4 in all my mocks though in comparison to C3... I just struggle to concentrate in the real exam
    Offline

    3
    ReputationRep:
    (Original post by 13 1 20 8 42)
    You want an x with |x| < 8/9 that allows you to nicely estimate (7100)^(1/3). You note that since |x| < 8/9 you are restricted to 0 < 8-9x < 16. So you're certainly not going to be able to put in 7100 or 71 or 710, but why not 7.1? Indeed you get 8 - 9x =7.1 when x = 0.1, and (7100)^(1/3) = 7.1^(1/3) * 1000^(1/3) = 10*(7.1^(1/3)) so you can easily rearrange for an estimate of 7100^(1/3) once you've plugged in x = 0.1
    Thanks for the quick reply, a couple questions:

    Where did that 16 come from, in the 0< 8-9x< 16?

    Also, why go for 0.1? Do we just shuffle around the decimal point until we get something within the range, like 7.1 from 7100 (after dividing ofc), then equal that to the expression given, like 8-9x, and solve to get 0.1?
    Offline

    2
    ReputationRep:
    Hey. Does anyone have any tips for vectors questions involving shapes. I always struggle to draw a diagram, like my vertices will always be in the wrong place so I end up using the wrong numbers in my calculations.

    Also what do you do for those questions that ask you to find the shortest distance between a line and position vector?

    Thanks

    This is what I was a bit stuck on for example
    Attached Images
     
    Offline

    19
    ReputationRep:
    (Original post by Dohaeris)
    Thanks for the quick reply, a couple questions:

    Where did that 16 come from, in the 0< 8-9x< 16?

    Also, why go for 0.1? Do we just shuffle around the decimal point until we get something within the range, like 7.1 from 7100 (after dividing ofc), then equal that to the expression given, like 8-9x, and solve to get 0.1?
    |x| < 8/9 means -8/9 < x < 8/9, so -9x can be at most -9(-8/9) = 8, which means 8 - 9x can be at most 8 + 8 = 16
    Well you'd kind of like to use something which goes into 7100 nicely, hence 7100, 710 71, 7.1, i.e. those which can be paired with multiples of 10, are nice candidates, and then we see that an approximation for 7.1 can be easily obtained since 7.1 is in the desired range and then yes we "solve" to get x = 0.1. I believe we could also go for 8 - 9x = 0.71 but this is much less clean.
    Offline

    3
    ReputationRep:
    (Original post by 13 1 20 8 42)
    |x| < 8/9 means -8/9 < x < 8/9, so -9x can be at most -9(-8/9) = 8, which means 8 - 9x can be at most 8 + 8 = 16
    Well you'd kind of like to use something which goes into 7100 nicely, hence 7100, 710 71, 7.1, i.e. those which can be paired with multiples of 10, are nice candidates, and then we see that an approximation for 7.1 can be easily obtained since 7.1 is in the desired range and then yes we "solve" to get x = 0.1. I believe we could also go for 8 - 9x = 0.71 but this is much less clean.
    Great explanation, thank you
    Offline

    2
    ReputationRep:
    Have sketching parametric curves ever come up?
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    How are your GCSEs going so far?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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