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    (Original post by Maid Marian)
    That's what I did, but apparently it's not right and I didn't think it could be right either because it was worth 5 marks
    It is just 0.5ln(sec2x).
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    (Original post by SJ12345)
    Did anyone get 0.136 on any part of question 10?
    That was the approximation yeah
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    (Original post by Benjy100)
    This is effectively a preliminary markscheme until Mr M posts his full definitive solutions. Some of my answers may be wrong. Spoilered incase you don't want to see them

    Spoiler:
    Show
    1. \frac{4}{{x + 2}} - \frac{3}{{x - 1}} + \frac{2}{{{{(x - 1)}^2}}}

    2. \frac{1}{9}{x^9}\ln 3x - \frac{1}{{81}}{x^9} + k

    3. Skew. Solved the first two equations for \lambda  = 3 and \mu  = 8 and then substituted into the third equation for 18=-38 which is clearly untrue, thus the equations are inconsistent and the lines are skew.

    4. \frac{{dy}}{{dx}} =  - 2\sin 2x + 2\cos x

    Stationary points are (\frac{1}{2}\pi ,1) \left( {\frac{1}{6}\pi ,\frac{3}{2}} \right) \left( {\frac{5}{6}\pi ,\frac{3}{2}} \right)

    5. \frac{1}{{1 - \tan x}} - \frac{1}{{1 + \tan x}} = \frac{{1 + \tan x - (1 - \tan x)}}{{1 - {{\tan }^2}x}} = \frac{{2\tan x}}{{1 - {{\tan }^2}x}} = \tan 2x

    Integral was \frac{1}{4}\ln 3

    6. \ln \left| {1 + \ln x} \right| + \frac{1}{{1 + \ln x}} + k

    7. \left| {AB} \right| = \sqrt {91} \left| {AC} \right| = 3\sqrt 3

    Angle BAC = 171.3 degrees (1 d.p.)

    I messed up on the volume of the pyramid

    8. \frac{{dr}}{{dt}} = \frac{k}{{\sqrt r }}

    \[r = {(4.86t + 2.7)^{\frac{2}{3}}}\]

    Letting t=0, finding r and substituting gives V = 30.5 cm^3 (3 s.f.)

    9. \[\frac{{dy}}{{dx}} = \frac{{2 - 2{t^{ - 3}}}}{{ - {t^{ - 2}}}} = \frac{{2{t^3} - 2}}{{ - t}} = \frac{{2 - 2{t^3}}}{t}\]

    Solving dy/dx yields t=1 which corresponds to the point (0,3)

    When t>1 x<0 and dy/dx is negative
    When t<1 x>0 and dy/dx is positive

    Hence it is a minimum

    10. Show.

    Let x=0.1 for 0.136

    \[ - \frac{1}{{{x^2}}}(1 + \frac{3}{x} + \frac{6}{{{x^2}}}) =  - \frac{1}{{{x^2}}} - \frac{3}{{{x^3}}} - \frac{6}{{{x^4}}}\]

    Same substitution would be unsuitable as the expansion here is valid for \left| { - \frac{1}{x}} \right| &lt; 1\] which is the same as \left| x \right| &gt; 1\]
    Thanks for the MS

    For #3 how did you show they are not parallel? I just stated they are not factors
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    (Original post by printergirl)
    i did

    but i can tell you now we're wrong
    t is 1. You just had to substitute this back into your parametric equations to get your maximum point as (0, 3) - or negative 3, I can't remember!
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    Do you think i can drop 13 marks and still get an A* (I have 93% in C3)?
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    It's so unfair, I'd much rather have an easier paper and actually give students the chance to show their talent.
    That paper was so bad, the difference between the top and bottom achievers won't be very much because there were so 'few' standard questions. I don't mind them being hard, just that they are fairly standard. Question 10 WTF?
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    (Original post by a10)
    I totally agree with this....
    I picked up mine cheap second hand on ebay. Not much more than a scientific calculator.

    Posted from TSR Mobile
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    (Original post by Kirity)
    Thanks for the MS

    For #3 how did you show they are not parallel? I just stated they are not factors
    I just found the two direction vectors and showed that they are not multiples of each other and so they are not parallel.
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    (Original post by JoshL123)
    Did anyone get t= -1? Fml :/
    Yes until I realised I multiplied out my brackets wrong (sorry :/ )
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    That was an awful paper

    Does anyone have any ideas on grade boundaries?
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    angle can be 171.3 or 6.68 because its the angle between two vectors and it doesn't affect the area of the base because sin(x)=sin(180-x)
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    (Original post by Kirity)
    Thanks for the MS

    For #3 how did you show they are not parallel? I just stated they are not factors
    Just show that the vectors are not scalar multiples of each other. Which I think that's what you did.

    Posted from TSR Mobile
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    for my differential equation I got r^(3/2)=(81/25)t+(9/5) and I got 30.5 3.s.f for the second bit, is this correct?
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    (Original post by ttylerbevann)
    Yes until I realised I multiplied out my brackets wrong (sorry :/ )
    So t wasn't -1 ;/
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    (Original post by eggfriedrice)
    Isnt it slightly unfair people can plot the graphs using theirs calculators during the exam? :| "pay £80 and get a slight advantage. " Meh.

    I only paid £25 for my graphical, but it forgot to even use it to help me with that parametric question!
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    I just can't believe I forgot what inversely proportional meant :facepalm::facepalm::facepalm::facepalm::facepalm::facepalm:

    That's going to haunt me forever. 9 marks down the drain.

    :banghead:
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    (Original post by samthegooner)
    It's so unfair, I'd much rather have an easier paper and actually give students the chance to show their talent.
    That paper was so bad, the difference between the top and bottom achievers won't be very much because there were so 'few' standard questions. I don't mind them being hard, just that they are fairly standard. Question 10 WTF?
    That is so true! Sums up the whole exam! :/
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    A few of my answers:

    For the first stationary points one, x was pi/6 and 5pi/6
    Lines were skew
    Volume of pyramid was around 9 I think? Length if ad x ac x ab x sin(bac)
    I got the air in the balloon as 35..did anyone else?
    Parametric gave t=1 for the stationary which gave 0,3 I think
    For the last part x=0.1 wasn't valid for modulus x less than 1,hence it not being accurate.
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    I also got t=-1? I dont understand how it isnt?
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    (Original post by Chrissii)
    I only paid £25 for my graphical, but it forgot to even use it to help me with that parametric question!
    In my opinion that is still cheating... why should someone have an advantage just because they bought high tech. equipment? Casio fx-85GT all the way! :P
 
 
 
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