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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by BBeyond)
    Would you get the marks for this I never see it mentioned on the examiner reports or markschemes?
    Definitely. I've used the dot product in M2 and it's not taught till C4. It'll be awarded marks as long as it's correct mathematics.
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    Hi all

    http://content.yudu.com/Library/A1uzcj/StudentBookCoreC3and/resources/

    Here's a nice C3 + C4 textbook if you're running out of things to do. Questions are slightly harder than what you might expect in the exam.

    Have fun

    EDIT: Actually many of the questions are about the same difficulty as the exam
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    How do you integrate this?
    Thanks for the help!Name:  ImageUploadedByStudent Room1459439703.031749.jpg
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    (Original post by Glavien)
    How do you integrate this?
    Thanks for the help!Name:  ImageUploadedByStudent Room1459439703.031749.jpg
Views: 96
Size:  67.7 KB


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    Substitution
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    Hey guys, this is a really dum dum question but what does it mean, with regards to the dot product rule, when two directional vectors are parallel. Also does a and b in the dot product rule HAVE to be directional vectors because I think I saw a question where they used a position vector 0.O... i was probably imagining it.
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    Tfw I spent 25 mins trying to work out why I got a question wrong and it turned out that i did 2*2=10 in the second step.

    The sad thing is i went back through the question multiple times trying to find where I went wrong and made the mistake every time.

    There's no hope for me lads.

    On another note, i hope i wrap my head around vectors before school restarts. I won't have any time to revise when school begins since I'll have 5 ISAs to do.
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    (Original post by Someboady)
    Hey guys, this is a really dum dum question but what does it mean, with regards to the dot product rule, when two directional vectors are parallel. Also does a and b in the dot product rule HAVE to be directional vectors because I think I saw a question where they used a position vector 0.O... i was probably imagining it.
    It means a.b is just the product of the moduli/lengths of the vectors.
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    (Original post by 13 1 20 8 42)
    It means a.b is just the product of the moduli/lengths of the vectors.
    Thanks, and do a and b have to be directional vectors or can they be position vectors?
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    (Original post by Someboady)
    Thanks, and do a and b have to be directional vectors or can they be position vectors?
    They will be position vectors if the angle between them you are considering (in a.b. = |a||b|cos(theta)) is at O
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    (Original post by xs4)
    Substitution
    What shall I substitute?


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    (Original post by Glavien)
    What shall I substitute?


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    x = sec u - 1
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    (Original post by Zacken)
    x = sec u - 1
    XxKingSniprxX this should help
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    (Original post by Zacken)
    x = sec u - 1
    How did you figure out that was what you were meant to substitute? Why that particular trig function?


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    (Original post by Zacken)
    x = sec u - 1
    I get down to \int \sec u \tan^2 u \, \mathrm{d}u

    Not sure how to integrate that?

    Btw how could I get a space between the second u and the du? Nvm, comma ;D
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    (Original post by Glavien)
    How did you figure out that was what you were meant to substitute? Why that particular trig function?


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    x^2 + 2x \equiv (x+1)^2 - 1 make the sub u = x+1 to get \int \sqrt{u^2 - 1} which is just begging for a u = \sec v sub, so I combined them into one and gave it to you.

    Be warned that this is an incredibly tedious and boring integral whose antiderivative is supremely ugly and will likely take you an entire line to write.
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    (Original post by Student403)
    I get down to \int \sec u \tan^2 u \mathrm{d}u

    Not sure how to integrate that? [/tex]
    \sec^3 u + \sec u then the latter is standard and IBP and/or reduction formulae is needed for the first part.

    Btw how could I get a space between the second u and the du?
    whatever \, \mathrm{d}u
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    (Original post by Zacken)
    x^2 + 2x \equiv (x+1)^2 - 1 make the sub u = x+1 to get \int \sqrt{u^2 - 1} which is just begging for a u = \sec v sub, so I combined them into one and gave it to you.

    Be warned that this is an incredibly tedious and boring integral whose antiderivative is supremely ugly and will likely take you an entire line to write.
    Yeah I put the thing in to WA earlier and I saw a sinh and was like wuuuuut!?


    (Original post by Zacken)
    \sec^3 u + \sec u then the latter is standard and IBP and/or reduction formulae is needed for the first part.



    whatever \, \mathrm{d}u
    Thanks
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    (Original post by Zacken)
    x^2 + 2x \equiv (x+1)^2 - 1 make the sub u = x+1 to get \int \sqrt{u^2 - 1} which is just begging for a u = \sec v sub, so I combined them into one and gave it to you.

    Be warned that this is an incredibly tedious and boring integral whose antiderivative is supremely ugly and will likely take you an entire line to write.
    They won't give an integral like this in a C4 paper, right?


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    (Original post by Glavien)
    They won't give an integral like this in a C4 paper, right?


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    Definitely not, I'm not sure where you've managed to dig it up.
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    (Original post by Zacken)
    \sec^3 u + \sec u
    Should that be minus or am I high?
 
 
 
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