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The Last Minute C3 Questions Thread!! Watch

  • View Poll Results: How confident are you for your C3 exam?
    Very confident!
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    I'm OKAY... :bebored:
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    Hey guys! :wavey:

    Here is the final stretch to the A-Level Core 3 exams across multiple boards. We have decided to make this thread for your last minute revision that you need to secure whatever grade you're aiming for

    If you have any C3 questions, please post them here and someone will be able to help you out!

    When posting a question, make sure to post your attempt at the question, if any, so that you may get a quicker response if you've made a mistake somewhere.

    Feel free to tag me in any questions you post so I can help as soon as possible if I'm around

    In similar fashion, here is the Core 4 thread.

    After-Exam Threads
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    Tagging some people who might be interested.
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    Awesome!

    I'm dreading these A2 exams
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    (Original post by notnek)
    Tagging some people who might be interested.
    thanks for tag, Im hoping to do really well in my other modules instead so it can can average out in case c3+c4 drops me :rofl:
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    Will definitely find this useful, thanks for starting this

    (Original post by RDKGames)
    Hey guys! :wavey:

    Here is the final stretch to the A-Level Core 3 exams across multiple boards. We have decided to make this thread for your last minute revision that you need to secure whatever grade you're aiming for

    If you have any C3 questions, please post them here and someone will be able to help you out!

    When posting a question, make sure to post your attempt at the question, if any, so that you may get a quicker response if you've made a mistake somewhere.

    Feel free to tag me in any questions you post so I can help as soon as possible if I'm around

    In similar fashion, a thread for Core 4 will soon follow.

    After-Exam Threads
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    I am integrating x^3/((6-x^2)^1/2) dx

    using u = 6 - x^2 to substitute...

    My teacher has said it is best to do substitute 'u' in with only one step (so that it does not contain 'x' and 'u' in the integral at the same time.)

    I have found du/dx = -2x

    du= -2x dx

    dx = -1/2x du

    x^2 = 6-u

    when I square root this to find x, and sub into -1/2x du, how do I know whether to take the positive or negative root? When I worked this out differently by replacing the 'x's with 'u' as I went along (not in one step) I did not have this problem as the xs cancelled...

    Hopefully this makes sense?

    Thank you in advance
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    (Original post by a.myhall)
    I am integrating x^3/((6-x^2)^1/2) dx

    using u = 6 - x^2 to substitute...

    My teacher has said it is best to do substitute 'u' in with only one step (so that it does not contain 'x' and 'u' in the integral at the same time.)

    I have found du/dx = -2x

    du= -2x dx

    dx = -1/2x du

    x^2 = 6-u

    when I square root this to find x, and sub into -1/2x du, how do I know whether to take the positive or negative root? When I worked this out differently by replacing the 'x's with 'u' as I went along (not in one step) I did not have this problem as the xs cancelled...

    Hopefully this makes sense?

    Thank you in advance
    Subbing in for dx first leaves you with x^2 in the integral which you can easily be replaced by 6-u

    If you take the square root, then you need to take both negative and positive answers (unless your domain of x is positive or negative) - just seems like an awkward way to approach this type of question, I wouldn't suggest it.
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    (Original post by RDKGames)
    Subbing in for dx first leaves you with x^2 in the integral which you can easily be replaced by 6-u

    If you take the square root, then you need to take both negative and positive answers (unless your domain of x is positive or negative) - just seems like an awkward way to approach this type of question, I wouldn't suggest it.

    did this not leave you with x and u in parts of the integral at the same time?

    My teacher said it was acceptable to do this, and I wouldn't drop marks but it isn't pure. Do you think it's ok to do that, or would I lose marks?
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    (Original post by a.myhall)
    did this not leave you with x and u in parts of the integral at the same time?

    My teacher said it was acceptable to do this, and I wouldn't drop marks but it isn't pure. Do you think it's ok to do that, or would I lose marks?
    To answer your former question, of course not.

    \displaystyle I=\int \frac{x^3}{\sqrt{6-x^2}}.dx and say u=6-x^2 \Rightarrow du=-2x .dx \Rightarrow dx= \frac{du}{-2x}

    So \displaystyle I=\int \frac{x^3}{-2xu^{1/2}}.du=-\frac{1}{2} \int \frac{x^2}{u^{1/2}}.du

    Since u=6-x^2 \Rightarrow x^2=6-u so \displaystyle I=-\frac{1}{2}\int \frac{6-u}{u^{1/2}}.du

    So no, it doesn't leave me with x's and u's together in one and cleans much nicely than square rooting and considering two separate cases.

    To answer your latter question, say x=\pm \sqrt{6-u} so then dx=\frac{du}{\pm 2\sqrt{6-u}}

    so then \displaystyle I=\int \frac{x^3}{\pm 2\sqrt{u} \sqrt{6-u}} .du which just seems more painful to deal with, but doable. I wouldn't recommend it but it's up to you.
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    Where the C4 one at tho
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    (Original post by notnek)
    Tagging some people who might be interested.
    Thanks for tagging me

    My problem with maths is that i make one stupid error such as a sign error or not reading the ques properly which means i lose marks for the whole question.

    Did the 2016 C4 paper carefully the other day but ran out of time so i didnt get to even attempt q8 or 9 . I need help managing my time in the exam.
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    (Original post by Super199)
    Where the C4 one at tho
    What exam board u on?
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    (Original post by Super199)
    Where the C4 one at tho
    Up on Friday.
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    integrate 2cosec^2(2x) between pi/4 and pi/6. As cot is not defined for pi/2, what do I do here?

    (-cot(pi/2)) + (cotpi/3)


    As the final answer is just (root3)/3, have they reached this answer by just ignoring the -cotpi/2?
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    (Original post by a.myhall)
    integrate 2cosec^2(2x) between pi/4 and pi/6. As cot is not defined for pi/2, what do I do here?

    (-cot(pi/2)) + (cotpi/3)


    As the final answer is just (root3)/3, have they reached this answer by just ignoring the -cotpi/2?
    \cot(\frac{\pi}{2})=0 since \cos(\frac{\pi}{2})=0

    If you used the reciprocal of tan, you cannot use that here.
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    Hey, I can never get my head around rates of change… could someone briefly explain it for me? I’ve tried watching videos and I sometimes get it but could someone show me with an explanation of what’s actually going on so I understand what I’m doing and not just repeating a process… would greatly appreciate it
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    (Original post by zayn008)
    Hey, I can never get my head around rates of change… could someone briefly explain it for me? I’ve tried watching videos and I sometimes get it but could someone show me with an explanation of what’s actually going on so I understand what I’m doing and not just repeating a process… would greatly appreciate it
    Do you have a specific question you want to go through in detail? Might be easier to explain what is happening when and why.
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    can I get a... BONELESS PIZZA?
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    does anyone know all the things we need to remember which aren't in the formula booklet? Not sure if I'm missing some?? ie. with integrals, identities etc.

    Thank you if you do.
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    (Original post by RDKGames)
    Do you have a specific question you want to go through in detail? Might be easier to explain what is happening when and why.
    Hi! Thanks for replying,

    So question 5 from here was tough http://mei.org.uk/files/papers/C3_2014_June.pdf

    If they give it in the format of say find dy/dt and give the value of dy/dx and an equation for x in terms of T I can usually do it as it's just a matter of differentiating and cancelling out but I can never do the worded ones and i think thats because i dont actually know whats going on, i just know the technique to getting the answer which doesn't work well in a worded scenario
 
 
 
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