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P4 Edexcel 19th Jan 2004 watch

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    HI.has anyone done p4 today??how do you feel?could anyone drop down some answer for the paper.thanks
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    (Original post by boss-dior)
    HI.has anyone done p4 today??how do you feel?could anyone drop down some answer for the paper.thanks
    I thought it was ok. But, I am not too keen on swapping answers because it makes me nervous.
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    I thought it was pretty straight forward. Nothing unusual and pretty routine questions
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    hi.. I certainly don't expect you to post any answers, but could you possibly suggest which topics you would advise to revise for this exam as I'm doin revision at the moment for it on friday. If you could let me know I could focus on these areas. Thankyou very much anyone who can help.
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    Hmm I definitely didn't think it was easy, I thought the questions were harder than normal in places, and there was no 'show' question except for q1 i think.

    My answers are:

    2) (The NR one) third approximation was 0.8260

    3) (Complex numbers one) zw^2 = 8root3i + 8. angle BOC = 90 degrees.

    4) (1st order ODE) . y = x^-3e^-x (xe^x - e^x + C).

    Solve under boundry counditions to get y = 1/8(1+1/e) when x = 2.

    5) (inequalities). Solutions: x= 2 - root(2) and 4 - root(2).

    6) (2nd ODE.) Solved to get y = e^-2x(Acosx + Bsinx) + sin 2x - 8cos 2x.

    When x is large can approximate by 16 sin^2 (x) + sin 2x - 8.

    7) (Polar coords). Integral = 3pi/16 - 9root3/32. (could be wrong, rushed this at end). Distance between lines = 2/3 . root(6).

    What do other people think?
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    (Original post by theone)
    Hmm I definitely didn't think it was easy, I thought the questions were harder than normal in places, and there was no 'show' question except for q1 i think.

    My answers are:

    2) (The NR one) third approximation was 0.8260

    3) (Complex numbers one) zw^2 = 8root3i + 8. angle BOC = 90 degrees.

    4) (1st order ODE) . y = x^3e^-x (xe^x - e^x + C).

    Solve under boundry counditions to get y = 1/8(1+1/e) when x = 2.

    5) (inequalities). Solutions: x= 2 - root(2) and 4 - root(2).

    6) (2nd ODE.) Solved to get y = e^-2x(Acosx + Bsinx) + sin 2x - 8cos 2x.

    When x is large can approximate by 16 sin^2 (x) + sin 2x - 8.

    7) (Polar coords). Integral = 3pi/16 - 9root3/32. (could be wrong, rushed this at end). Distance between lines = 2/3 . root(6).

    What do other people think?
    Agree on all - however my mate thought (and i think hes right) that the sine approximation should have been better done by cosidering sin(2x-a) where a is arctan something - also 1st order ode should y = x^-3... presume thats what you meant as you got the same answer as me for the next part
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    (Original post by It'sPhil...)
    Agree on all - however my mate thought (and i think hes right) that the sine approximation should have been better done by cosidering sin(2x-a) where a is arctan something - also 1st order ode should y = x^-3... presume thats what you meant as you got the same answer as me for the next part
    Incidentally, how did you do the polar coords one (the final question).

    I did it quite wierdly:

    d(3sinxcos2x)/dx = 0.

    Therefore 3cos2xcosx - 6sinxsin2x = 0.

    Therefore 1/2 = tanx tan 2x. Therefore 1/2 = 2tan^2(x)/(1-tan^2(x)). so tan(x) = +/- 1/root5. Take positive: cos(x) = root(5/6). So cos2x = 2/3. So r = 2. So the distance between the point and the initial line = 2 sinx = 2/root(6). So total distance = 4/root(6) = 2/3 . root(6).
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    (Original post by theone)
    Incidentally, how did you do the polar coords one (the final question).

    I did it quite wierdly:

    d(3sinxcos2x)/dx = 0.

    Therefore 3cos2xcosx - 6sinxsin2x = 0.

    Therefore 1/2 = tanx tan 2x. Therefore 1/2 = 2tan^2(x)/(1-tan^2(x)). so tan(x) = +/- 1/root5. Take positive: cos(x) = root(5/6). So cos2x = 2/3. So r = 2. So the distance between the point and the initial line = 2 sinx = 2/root(6). So total distance = 4/root(6) = 2/3 . root(6).
    not quite like that -

    i had 3cos2xcosx - 6sinxsin2x = 0.
    => cosx(co2x - 4sin^2x) = 0
    cosx = 0 gives minima, so
    3cos2x - 2 = 0 => cos2x = 2/3 => r = 2
    also 1 - 6sin^2x = 0 so sinx = 1/root6 and the rest is the same as yours

    I spent quite a long time on the first 3 or so q's then the miiddle to late ones were fine and i finished with loads of time to spare. I also spent a while deciding whether it was quicker to find zw^2 then find mod and arg or to simply use lzw^2l = lzllwllwl and arg(zw^2) = 2argw + argz. In the end i just found zw^2 which was the right choice cos i had to sketch it in the next part
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    Incidentally, my method has one flaw in it accounting for the fact that I've forgotten to include possibilities where cos(x) = 0 or cos(2x) = 0. The only possibility is that cos 2x = 0 and x = pi /4 +2npi. But if this is the case the RHS is positive, so it's not possible. (We can't have cos(x) = 0 since -pi/4 <= x <= pi/4).
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    the inequality one,I got 2-root2<x<2+root2....
    anyone got the same or just me?/
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    (Original post by boss-dior)
    the inequality one,I got 2-root2<x<2+root2....
    anyone got the same or just me?/
    Sorry but 2 + root 2 isn't a solution, just plug it into the equation and then you might figure out why:

    (The equation was |(x-2)(x-4)| = 6 - 2x.)
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    (Original post by boss-dior)
    the inequality one,I got 2-root2<x<2+root2....
    anyone got the same or just me?/
    If you drew a graph,

    you'd see that for the equation
    +(x-2)(x-4) = 6 - 2x
    the root had to be less than two,

    and the root for
    -(x-2)(x-4) = 6 - 2x
    had to be less than 4.

    I found today's paper rather difficult.

    J.
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    (Original post by theone)
    Hmm I definitely didn't think it was easy, I thought the questions were harder than normal in places, and there was no 'show' question except for q1 i think.

    My answers are:

    2) (The NR one) third approximation was 0.8260

    3) (Complex numbers one) zw^2 = 8root3i + 8. angle BOC = 90 degrees.

    4) (1st order ODE) . y = x^-3e^-x (xe^x - e^x + C).

    Solve under boundry counditions to get y = 1/8(1+1/e) when x = 2.

    5) (inequalities). Solutions: x= 2 - root(2) and 4 - root(2).

    6) (2nd ODE.) Solved to get y = e^-2x(Acosx + Bsinx) + sin 2x - 8cos 2x.

    When x is large can approximate by 16 sin^2 (x) + sin 2x - 8.

    7) (Polar coords). Integral = 3pi/16 - 9root3/32. (could be wrong, rushed this at end). Distance between lines = 2/3 . root(6).

    What do other people think?
    I agree on those apart from Newton raphson, I think I got 0.8177. I am glad you got 3pi/16 - 9root3/32 for the integral, I was worried that I messed that up for some reason. I was rushing towards the end.


    Also, I agree that the paper was harder than usual. When doing past papers I found that I could finish a paper quite quickly. I found this one quite intensive.
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    (Original post by mikesgt2)
    I agree on those apart from Newton raphson, I think I got 0.8177. I am glad you got 3pi/16 - 9root3/32 for the integral, I was worried that I messed that up for some reason. I was rushing towards the end.


    Also, I agree that the paper was harder than usual. When doing past papers I found that I could finish a paper quite quickly. I found this one quite intensive.


    I got 0.8177 as well.....quite happy.....
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    (Original post by hornblower)
    If you drew a graph,

    you'd see that for the equation
    +(x-2)(x-4) = 6 - 2x
    the root had to be less than two,

    and the root for
    -(x-2)(x-4) = 6 - 2x
    had to be less than 4.

    I found today's paper rather difficult.

    J.


    if I got the roots wrong,will I get follow through marks for it in part c.???
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    (Original post by boss-dior)
    if I got the roots wrong,will I get follow through marks for it in part c.???
    Yes, although you'd lose A (Accuracy) marks you'd get M (method) marks and possibly - it depends on the type of question - B (don't ask!) marks.

    This is because it is unfair to penalise you more than once for a single error.

    J.
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    I agree with your third approximation for the Newton Raphson one, i did it wrong in the exam.
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    (Original post by theone)
    I agree with your third approximation for the Newton Raphson one, i did it wrong in the exam.
    I almost did but went back and checked my number crunching. I hate questions like that - like linear interpolation, iteration, and most of S1 where its just typing numbers in a calculator - and i usually do it wrong somewhere
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    (Original post by It'sPhil...)
    I almost did but went back and checked my number crunching. I hate questions like that - like linear interpolation, iteration, and most of S1 where its just typing numbers in a calculator - and i usually do it wrong somewhere
    Ah well, I'm not too worried just yet, I should get pretty much full marks, except lose 3 on the NR one for getting the 3rd approx wrong (got the 2nd right though ) and 2-3 for not considering the cos(x) or cos(2x) = 0 solution for the polar one (although these are trivial, so if I get a dosy examiner they may not notice ). Looks like everything is pretty much good, so I'm ok
 
 
 

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