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AQA A2 MFP3 Further Pure 3 – 18th May 2016 [Exam Discussion Thread] Watch

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    (Original post by IrrationalRoot)
    I messed up my Cambridge application just by making silly mistakes, doesn't get worse than that lol.
    I'd rather have messed up my application and not have gotten an offer than missing my offer from silly mistakes after having worked my arsenic off for UMS interview prep and A2 grades tbh.
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    Can someone remember what Q5 was? And post the question here please? Cheers
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    (Original post by C0balt)
    I'd rather have messed up my application and not have gotten an offer than missing my offer from silly mistakes after having worked my arsenic off for UMS interview prep and A2 grades tbh.
    Are you taking step?


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    (Original post by jjsnyder)
    Are you taking step?


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    Thankfully not!
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    (Original post by C0balt)
    Thankfully not!
    Hmm okay, keep focusing on getting high ums scores in your other exams then. You haven't thrown away your offer after one exam today, I have little doubt that you still have a good chance at an A* in FM.


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    (Original post by jjsnyder)
    Hmm okay, keep focusing on getting high ums scores in your other exams then. You haven't thrown away your offer after one exam today, I have little doubt that you still have a good chance at an A* in FM.


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    Yeah, not too good with getting high ums in FM because I run out of time and leave my silly mistakes behind! But I'm counting on 600/600 in chemistry too lol
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    (Original post by C0balt)
    I'd rather have messed up my application and not have gotten an offer than missing my offer from silly mistakes after having worked my arsenic off for UMS interview prep and A2 grades tbh.
    I did 'work my arsenic off' for UMS, interview prep, STEP prep and lots more than that, then got let down by silly mistakes. Didn't even have a chance to make the offer which I could probably make. I would rather have the chance tbh.
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    (Original post by jjsnyder)
    There's death, heard that's slightly worse #firstworldproblems


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    Lol I wasn't saying it was worse than everything in general XD.
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    (Original post by C0balt)
    Yeah, not too good with getting high ums in FM because I run out of time and leave my silly mistakes behind! But I'm counting on 600/600 in chemistry too lol
    I will applaud you for that don't know how you do it! I am settling with an A this year in Chemistry, I don't see much point in wasting time working for the A*.


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    (Original post by jjsnyder)
    I will applaud you for that don't know how you do it! I am settling with an A this year in Chemistry, I don't see much point in wasting time working for the A*.


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    Well unit 5 will be a little difficult but fingers crossed! You're a mathmo, you don't need chemistry! Instead you've got that evil STEP :afraid:
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    (Original post by jjsnyder)
    I will applaud you for that don't know how you do it! I am settling with an A this year in Chemistry, I don't see much point in wasting time working for the A*.


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    I don't think there's any point in any mathmos going for the third A* tbh.
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    (Original post by C0balt)
    Well unit 5 will be a little difficult but fingers crossed! You're a mathmo, you don't need chemistry! Instead you've got that evil STEP :afraid:
    STEP is a beautiful but excruciatingly painful exam which I am not very good at :getscoat:


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    (Original post by IrrationalRoot)
    I don't think there's any point in any mathmos going for the third A* tbh.
    No, I really don't see the point in wasting time I could put towards step.


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    (Original post by jjsnyder)
    STEP is a beautiful but excruciatingly painful exam which I am not very good at :getscoat:


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    That's not true! I've seen your scores in some of your mocks, impressive stuff .
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    (Original post by Anon123hahaha)
    Can someone remember what Q5 was? And post the question here please? Cheers
    Hi there, I can't remember the question exactly but I believe this is roughly what it was:

    Question 5 - Part A

    Express \dfrac{1}{(x+1)(x+2)} in the form \dfrac{A}{x+1} + \dfrac{B}{x+2}.  \ \ [1]

    Solution

    If we multiply both sides by (x+1)(x+2), we get 1 = A(x+2) + B(x+1). By substituting in x = -1 and x = -2, we can find the values of the constants A and B. You could also do this by inspection but it's always better to show a method of some sort.

    The answer is: \dfrac{1}{(x+1)(x+2)} = \dfrac{1}{x+1} - \dfrac{1}{x+2}


    Question 5 - Part B

    Using the substitution u = \dfrac{dy}{dx}, solve the differential equation \dfrac{d^2y}{dx^2} + \dfrac{1}{(x+1)(x+2)}\dfrac{dy}{  dx} = \dfrac{x+2}{x+1}, where y = 1 and \dfrac{dy}{dx}=4 when x = 0.

    Give your answer in the form  y = f(x).  \ \ [11]

    Solution

    Noting that \dfrac{du}{dx} = \dfrac{d^2y}{dx^2}, we can rewrite the differential equation as:

    \dfrac{du}{dx} + \dfrac{1}{(x+1)(x+2)}u = \dfrac{x+2}{x+1}.

    The original second order differential equation has now been transformed into a first order differential equation of the form \dfrac{du}{dx} + Pu = Q, which can be solved analytically.

    This differential equation still looks quite tricky to solve, but that is where our answer to part a) comes in. If we rewrite it again as \dfrac{du}{dx} + \left(\dfrac{1}{x+1} - \dfrac{1}{x+2}\right)u = \dfrac{x+2}{x+1}, we can now find an integrating factor very easily:

     I = e^{\displaystyle\int\left(\frac{  1}{x+1} - \frac{1}{x+2}\right) dx}\ \ = e^{\displaystyle \bigg(\ln(x+1) -\ln(x+2)\bigg)} \ \ =e^{\displaystyle\ln \left(\frac {x+1}{x+2}\right)}

    \therefore  I = \dfrac {x+1}{x+2}

    Multiplying throughout by the integrating factor, our first order differential equation becomes:

    \dfrac{x+1}{x+2}\dfrac{du}{dx} + \dfrac{1}{(x+2)^2}u = 1 \ \ \Rightarrow \ \ \dfrac{d}{dx}\bigg(\dfrac{x+1}{x  +2}u\bigg) = 1

    If we then integrate both sides of this expression with respect to x, we get \dfrac{x+1}{x+2}u = x + C which can be rearranged to u = \dfrac{(x + C)(x+2)}{x+1}.

    Now, going back to the information given in the question, we know that u = \dfrac{dy}{dx}, and \dfrac{dy}{dx}=4 when x = 0. Using these bits of information, and our expression for u, we can find the value of C. It turns out that C=2, so:

    \dfrac{dy}{dx}=\dfrac{(x + 2)^2}{x+1}

    To find an expression for y, we need to integrate again. In its current form, the RHS of the above equation cannot be integrated. Using algebraic long division, we can rewrite it in a form that can:

    \dfrac{dy}{dx}= x + 3 + \dfrac{1}{x+1}\ \ \Rightarrow \ \ y = \dfrac{1}{2}x^2 + 3x + \ln(x+1) + D

    Using the boundary conditions y = 1 when x = 0, it can be shown that D=1. The solution of the differential equation in the form  y = f(x), subject to the boundary conditions given, is therefore:

    y = \dfrac{1}{2}x^2 + 3x + \ln(x+1) + 1

    Hope that helps anyone who is unsure

    EDIT: You could also use a substitution to perform the final integration. In which case, your answer will be in a different form to mine.
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    (Original post by sam_97)
    Hi there, I can't remember the question exactly but I believe this is roughly what it was:

    5)a) Express \dfrac{1}{(x+1)(x+2)} in the form \dfrac{A}{x+1} + \dfrac{B}{x+2}.  \ \ [1]

    Solution

    If we multiply both sides by (x+1)(x+2), we get 1 = A(x+2) + B(x+1). By substituting in x = -1 and x = -2, we can find the values of the constants A and B. You could also do this by inspection but it's always better to show a method of some sort.

    The answer is \dfrac{1}{(x+1)(x+2)} = \dfrac{1}{x+1} - \dfrac{1}{x+2}


    5)b) Using the substitution u = \dfrac{dy}{dx}, solve the differential equation \dfrac{d^2y}{dx^2} + \dfrac{1}{(x+1)(x+2)}\dfrac{dy}{  dx} = \dfrac{x+2}{x+1}, where \dfrac{dy}{dx}=4 when x = 0, and y = 1 when x = 0. Give your answer in the form  y = f(x).  \ \ [11]

    Solution

    Noting that \dfrac{du}{dx} = \dfrac{d^2y}{dx^2}, we can rewrite the differential equation as \dfrac{du}{dx} + \dfrac{1}{(x+1)(x+2)}u = \dfrac{x+2}{x+1}. The original second order differential equation has now been transformed into a first order differential equation of the form \dfrac{du}{dx} + Pu = Q, which can be solved analytically.

    This differential equation still looks quite tricky to solve, but that is where our answer to part a) comes in. If we rewrite it again as \dfrac{du}{dx} + \left(\dfrac{1}{x+1} - \dfrac{1}{x+2}\right)u = \dfrac{x+2}{x+1}, we can now find an integrating factor very easily:

     I = e^{\displaystyle\int\left(\frac{  1}{x+1} - \frac{1}{x+2}\right) dx}\ \ = e^{\displaystyle \bigg(\ln(x+1) -\ln(x+2)\bigg)} \ \ =e^{\displaystyle\ln \left(\frac {x+1}{x+2}\right)}

    \therefore  I = \dfrac {x+1}{x+2}

    Multiplying throughout by the integrating factor, our first order differential equation becomes:

    \dfrac{x+1}{x+2}\dfrac{du}{dx} + \dfrac{1}{(x+2)^2}u = 1 \ \ \Rightarrow \ \ \dfrac{d}{dx}\bigg(\dfrac{x+1}{x  +2}u\bigg) = 1

    If we then integrate both sides of this expression with respect to x, we get \dfrac{x+1}{x+2}u = x + C which can be rearranged as u = \dfrac{(x + C)(x+2)}{x+1}.

    Now, going back to the information given in the question, we know that u = \dfrac{dy}{dx}, and \dfrac{dy}{dx}=4 when x = 0. Using these bits of information, and our expression for u, we can find the value of C. It turns out that C=2, so:

    \dfrac{dy}{dx}=\dfrac{(x + 2)^2}{x+1}

    To find an expression for y, we need to integrate again. In its current form, the RHS of the above equation cannot be integrated. Using algebraic long division, we can rewrite it in a form that can:

    \dfrac{dy}{dx}= x + 3 + \dfrac{1}{x+1}\ \ \Rightarrow \ \ y = \dfrac{1}{2}x^2 + 3x + \ln(x+1) + D

    Using the boundary conditions y = 1 when x = 0, it can be shown that D=1. The solution of the differential equation in the form  y = f(x), subject to the boundary conditions given, is therefore:

    y = \dfrac{1}{2}x^2 + 3x + \ln(x+1) + 1

    Hope that helps anyone who is unsure
    In the last few step I integrated by substitution but then I end up with a different c but I think it simplifies down to the same amswear do you think I would still get full marks on that question
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    (Original post by IrrationalRoot)
    That's not true! I've seen your scores in some of your mocks, impressive stuff .
    I fluctuate a lot I guess, if the pure section on the paper is bad I won't meet my offer, if it is alright then hopefully I will.
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    (Original post by Hjyu1)
    In the last few step I integrated by substitution but then I end up with a different c but I think it simplifies down to the same amswear do you think I would still get full marks on that question
    It sounds like your answer is the same as mine, written in a different form. In which case, you'll definitely get full marks
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    (Original post by Hjyu1)
    In the last few step I integrated by substitution but then I end up with a different c but I think it simplifies down to the same amswear do you think I would still get full marks on that question
    I did the same thing, v=x+1 leads to  y =  \frac{(x+1)^2}{2} + 2(x+1) + ln(x+1) - \frac{3}{2} which should get full marks still.
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    (Original post by jjsnyder)
    I did the same thing, v=x+1 leads to  frac{(x+1)^2}{2} + 2(x+1) + ln(x+1) - frac{3}{2} which should get full marks still.
    Yeah that's it I got -3/2 for c so yeah hopefully the marker sees what I've done
 
 
 
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