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AQA Core 4 MPC4 -- 16th June 2011

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    (Original post by lamin.sessay)
    Dear Readers,

    Can you please explain in detail how you did the final 9 mark question? I am totally bemused. I tried to solve it repeatedly but to no avail.

    Many thanks,
    Lamin Sessay
    Yes lamin Pwd 2k11 BRAAAAP
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    (Original post by bish93)
    i couldn't get it to equal 0 either :/
    another exam board cock up? or is that just wisful thinking cos i couldnt do it? :P
    Well do you think our Constant of 2/e (which nearly everyone ive spoke to about the exam said they got it as!!) was incorrect? in which case i would say the possibility of an exam board cock up is very likely !!
    I differentiated it twice, got the exact same thing both times, and i dont see how it could be anything else .
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    I thought this paper was the hardest core 4 paper I've ever seen in my life, I resat Core 3 on Monday and it was really nice, and Further Maths was a nice paper too, I had high hopes for this, I felt like ripping it up!
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    (Original post by charlotte7474)
    Hiya, after differentiating it, i got C = 2/e, but then when you have to prove it was a stationary point, i couldnt get it to = 0.
    Quite a lot of people in my class couldnt either, did anyone actually get it to equal 0??!
    2y + e^(2x)y² = x² + C

    so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

    (dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

    dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    Stationary point when (1,1/e)

    2(1/e) + e²(1/e²) = 1² + C
    2/e + 1 = 1 + C
    2/e = C

    Prove that it's a stationary point or something it said,

    So dy/dx = 0

    0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    0 = (2x - 2e^(2x)y²)

    0 = 2 - 2e²(1/e²)

    0 = 2 - 2

    0 = 0 Proven.
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    (Original post by charlotte7474)
    Hiya, after differentiating it, i got C = 2/e, but then when you have to prove it was a stationary point, i couldnt get it to = 0.
    Quite a lot of people in my class couldnt either, did anyone actually get it to equal 0??!
    2y + e^(2x)y² = x² + C

    so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

    (dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

    dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    Stationary point when (1,1/e)

    2(1/e) + e²(1/e²) = 1² + C
    2/e + 1 = 1 + C
    2/e = C

    Prove that it's a stationary point or something it said,

    So dy/dx = 0

    0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    0 = (2x - 2e^(2x)y²)

    0 = 2 - 2e²(1/e²)

    0 = 2 - 2

    0 = 0 Proven.
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    (Original post by Crowaled)
    2y + e^(2x)y² = x² + C

    so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

    (dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

    dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    Stationary point when (1,1/e)

    2(1/e) + e²(1/e²) = 1² + C
    2/e + 1 = 1 + C
    2/e = C

    Prove that it's a stationary point or something it said,

    So dy/dx = 0

    0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    0 = (2x - 2e^(2x)y²)

    0 = 2 - 2e²(1/e²)

    0 = 2 - 2

    0 = 0 Proven.
    yeah i did that
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    (Original post by bish93)
    i couldn't get it to equal 0 either :/
    another exam board cock up? or is that just wisful thinking cos i couldnt do it? :P
    Wishful thinking, I'm afraid, it was right. The constant was 2/e though.
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    (Original post by Crowaled)
    2y + e^(2x)y² = x² + C

    so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

    (dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

    dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    Stationary point when (1,1/e)

    2(1/e) + e²(1/e²) = 1² + C
    2/e + 1 = 1 + C
    2/e = C

    Prove that it's a stationary point or something it said,

    So dy/dx = 0

    0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    0 = (2x - 2e^(2x)y²)

    0 = 2 - 2e²(1/e²)

    0 = 2 - 2

    0 = 0 Proven.

    2e²(1/e²) = 2?

    oh crap yeah ofcourse it does,
    thanks for clearing that up !
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    As I said earlier, I'm quite sure that the year is 2016. I came across a question in a past paper similar to this one and I also didn't get it but I did once my teacher explained. He said to think of it like this - when a baby is 3 months old, it's in its first year. You don't round down and say it's 0.
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    (Original post by Crowaled)
    2y + e^(2x)y² = x² + C

    so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

    (dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

    dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    Stationary point when (1,1/e)

    2(1/e) + e²(1/e²) = 1² + C
    2/e + 1 = 1 + C
    2/e = C

    Prove that it's a stationary point or something it said,

    So dy/dx = 0

    0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    0 = (2x - 2e^(2x)y²)

    0 = 2 - 2e²(1/e²)

    0 = 2 - 2

    0 = 0 Proven.
    Think I got that. I was unsure about the differentiating, but i got the stationary point equal to 0, so I guess I did do it correctly!
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    can any1 just remember the last question? I want to do it now again!!
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    (Original post by Crowaled)
    2y + e^(2x)y² = x² + C

    so 2(dy/dx) + 2e^(2x)y² + 2e^(2x)y(dy/dx) = 2x

    (dy/dx)(2e^(2x)y + 2) = 2x - 2e^(2x)y²

    dy/dx = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    Stationary point when (1,1/e)

    2(1/e) + e²(1/e²) = 1² + C
    2/e + 1 = 1 + C
    2/e = C

    Prove that it's a stationary point or something it said,

    So dy/dx = 0

    0 = (2x - 2e^(2x)y²)/(2e^(2x)y + 2)

    0 = (2x - 2e^(2x)y²)

    0 = 2 - 2e²(1/e²)

    0 = 2 - 2

    0 = 0 Proven.
    I was just wondering whether you thought my method was right for proving it was a stationary point. I got dy/dx on its own and then subsituted the given values for x and y (1,1/e) and this equalled zero which shows that dy/dx=0, therefore it is a SP. Would that still get the marks
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    how many marks will i get for the question where we had to solve A, B and C if i only got 2 of those values correct?
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    (Original post by charlotte7474)
    Hiya, after differentiating it, i got C = 2/e, but then when you have to prove it was a stationary point, i couldnt get it to = 0.
    Quite a lot of people in my class couldnt either, did anyone actually get it to equal 0??!
    yh i did thats how i am confident i was correct but just wondered still or perhaps i completely messed up and i am wrong :P
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    "I was just wondering whether you thought my method was right for proving it was a stationary point. I got dy/dx on its own and then subsituted the given values for x and y (1,1/e) and this equalled zero which shows that dy/dx=0, therefore it is a SP. Would that still get the marks "

    That's what I did ^
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    (Original post by harky 1)
    Messed that up then =( u think the grade boundries will get lowered?
    I would think so, because the people I talked to said it was difficult for them. Then again, there are always plenty of people who say that. We'll just have to wait and see, and hope for the best!
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    The year is 2015...
    When is the year when it will first exceed 100,000... Of course 2015
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    (Original post by haaroon321)
    I was just wondering whether you thought my method was right for proving it was a stationary point. I got dy/dx on its own and then subsituted the given values for x and y (1,1/e) and this equalled zero which shows that dy/dx=0, therefore it is a SP. Would that still get the marks
    That's exactly what I did, it's the way I've always proved a stationary point in the past, and got the marks!

    I guess theres more than one way of proving it, I guess our way was simpler.
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    The answer to that question was 2015.... 1st of August 2015!!
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    (Original post by Signed Solution)
    As I said earlier, I'm quite sure that the year is 2016. I came across a question in a past paper similar to this one and I also didn't get it but I did once my teacher explained. He said to think of it like this - when a baby is 3 months old, it's in its first year. You don't round down and say it's 0.
    I hope you're right.

    I put 2016 because I thought about those questions when they ask if I have x items and each box holds x how many boxes do you need. If you get and answer like 4.3 then you say 5 because 3 won't be enough.
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    To be honest, it seems that half have put 2015 and half have put 2016 so they'll probably just accept either because the question was difficult to understand. At least I hope they do

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