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    (Original post by zmai)
    Thankyou!! I forgot to plug x values back in to find y values for question 4.

    Would it be okay to write 1/2ln(root3) ? Did they specify the format?


    Posted from TSR Mobile
    I think that this would be okay because the question never specified what type of numbers "a" and "b" needed to be (i.e. rational, integer etc.) when in the form a ln(b).
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    Anyone have any idea of what grade boundaries will be?

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    (Original post by a10)
    but cos x cant be zero the limits of x where greater than zero and less than pi? therefore u can divide by cos?
    ...Forgot about the limits.
    EDIT: Cos(pi/2)=0
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    How many marks will I lose on the 9 marker for making an error at the very beginning of -k/root R rather than just k/root R? Absolutely hated that exam!
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    (Original post by Vip3rgt9)
    ...Forgot about the limits.
    That's for x.. Pi/2 gives you cosx = 0
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    (Original post by h2shin)
    Stationary points for the trig: (pi/2, 1) (pi/6, 1.5) (5pi/6, 1.5)
    Any agreements?

    Oh and gradient to left was negative and gradient to right was positive so I put minimum but can't be too sure...
    Wow I feel stupid, I worked out gradient to left is negative, and right was positive and wrote that was a turning point!
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    For 8i), the differential equation.

    I got the correct answer but didn't leave it as \[r = {(4.86t + 2.7)^{\frac{2}{3}}}\]

    Instead, when I got to \[{r^{\frac{3}{2}}}\ = (4.86t + 2.7)
    I took the root by power 3/2 of the things on the right hand side so I got

    r = 2.87t + 1.94

    Will that still be okay?
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    (Original post by JASApplications)
    For 8i), the differential equation.

    I got the correct answer but didn't leave it as \[r = {(4.86t + 2.7)^{\frac{2}{3}}}\]

    Instead, when I got to \[{r^{\frac{3}{2}}}\ = (4.86t + 2.7)
    I took the root by power 3/2 of the things on the right hand side so I got

    r = 2.87t + 1.94

    Will that still be okay?
    for some reason in the bracket I got 27 rather than 2.7 :s
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    (Original post by JASApplications)
    For 8i), the differential equation.

    I got the correct answer but didn't leave it as \[r = {(4.86t + 2.7)^{\frac{2}{3}}}\]

    Instead, when I got to \[{r^{\frac{3}{2}}}\ = (4.86t + 2.7)
    I took the root by power 3/2 of the things on the right hand side so I got

    r = 2.87t + 1.94

    Will that still be okay?
    I think you will have dropped one or two marks because if you multiplied (4.86t+2.7)^2/3 it wouldn't give you that?


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    (Original post by JakePreedy)
    I think that this would be okay because the question never specified what type of numbers "a" and "b" needed to be (i.e. rational, integer etc.) when in the form a ln(b).
    Brilliant thankyou!


    Posted from TSR Mobile
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    would you lose marks for leaving ur intergrated answer in terms of u, I got the right answer as most people but I just didn't sub back what u was? I hope they aren't that harsh :L
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    (Original post by a10)
    would you lose marks for leaving ur intergrated answer in terms of u, I got the right answer as most people but I just didn't sub back what u was? I hope they aren't that harsh :L
    Actually, why do you have to sub it back in anyway? You've stated what u equals.
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    (Original post by Vip3rgt9)
    ...Forgot about the limits.
    EDIT: Cos(pi/2)=0
    ****, so iv lost 2 marks for not including the other two then
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    (Original post by a10)
    can someone confirm this:

    for the tan question my final integral was


    -2ln(2- sec^2(x))

    and I subbed in values wrong for the first part so I ended up with

     (-2 ln 2/3) + 2ln(-6+4root 3)

    can someone confirm whether they got the -6+4 root 3 part?
    Did exactly this. Integrated it wrong

    Dunno how many marks we'll get, maybe 1 for 'putting limits in correctly'
    and optimistically maybe 1 for 'using log rules correctly'
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    (Original post by zmai)
    Thankyou!! I forgot to plug x values back in to find y values for question 4.

    Would it be okay to write 1/2ln(root3) ? Did they specify the format?


    Posted from TSR Mobile
    I think you had to find the y-values but I doubt you'd lose more than 1 mark after finding the correct x-values.

    I don't think they specified a format other than alnb so I think that would be fine
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    A* grade boundary anyone?
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    (Original post by zmai)
    I think you will have dropped one or two marks because if you multiplied (4.86t+2.7)^2/3 it wouldn't give you that?


    Posted from TSR Mobile
    It does. 2.7^2/3 = 1.94 and 4.86^2/3 = 2.87
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    Really struggled, it was so difficult! Everyone seems to be finding it okay gah
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    (Original post by Benjy100)
    My money is on 62-63
    How many marks do you think you got Ben?
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    (Original post by alastair999100)
    Did exactly this. Integrated it wrong

    Dunno how many marks we'll get, maybe 1 for 'putting limits in correctly'
    and optimistically maybe 1 for 'using log rules correctly'
    why is the integral wrong? I differentiated it and got the tan2x fraction?
 
 
 
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