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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?
    yeah i got them apart from the first one.... oh well
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    (Original post by ihatepeople37)
    when you had to integrate x^8 ln3x ... i put u as ln3x but what is that differentiated please???
    its 3/3x .. f'(x)/f(x)
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    (Original post by 94Singh)
    its 3/3x .. f'(x)/f(x)
    yeah and when you simplify 3/3x it becomes 1/x
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    (Original post by eggfriedrice)
    Pretty sure it was a maximum. You differentiate it again and find it's negative. (C1)
    Yeah, I got maximum, too. d*2y/dx*2 was <0, which made it a maximum. I think.
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    that paper was messed up!!!!:mad:
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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...
    how did you get the first one?
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    (Original post by ihatepeople37)
    yeah i got them apart from the first one.... oh well
    I only got two stationary points, also! :eek: Was there another one?
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    for the stationary point question was t=-1?
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    ohhh **** just realized they had to be 3 stationary points I put 1
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    (Original post by eggfriedrice)
    Pretty sure it was a maximum. You differentiate it again and find it's negative. (C1)
    Just plotted the function. It was a minimum.
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    (Original post by Dizzy in my Head)
    Yeah, I got maximum, too. d*2y/dx*2 was <0, which made it a maximum. I think.
    but it said by considering the gradient either side
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    (Original post by louise_234)
    yeah and when you simplify 3/3x it becomes 1/x
    but 1/x is integrated it is lnx not ln3x, or am i being really stupid??
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    So much volume!
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    (Original post by a10)
    ohhh **** just realized they had to be 3 stationary points I put 1
    Thats what I got as well :/ fml
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    Okay I'm petrified now, you guys all seemed to find it alright!

    I can't be the only one who thought it was horrific?
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    (Original post by louise_234)
    yeah and when you simplify 3/3x it becomes 1/x
    yeah .. they differentiate to become the same .. try putting them into your calculator if you have one that can differentiate
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    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

    *A few additions have been made to the markscheme* - namely the cartesian equation (forgot to put it in)

    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.)

    Show that AD is perpendicular to AB and AC by finding the dot product of AD and AB, and then AD and AC - and showing that both dot products equate to 0.

    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

    Cartesian equation...

    \[\begin{array}{l}

xt = 1 - t \Rightarrow t(x + 1) = 1 \Rightarrow t = \frac{1}{{x + 1}}\\

y = \frac{2}{{x + 1}} + {(x + 1)^2}

\end{array}\]

    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\]
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    (Original post by Maid Marian)
    Okay I'm petrified now, you guys all seemed to find it alright!

    I can't be the only one who thought it was horrific?
    Dont worry, so did I and most of my year. Just realised marks here and there missed too
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    (Original post by Maid Marian)
    Okay I'm petrified now, you guys all seemed to find it alright!

    I can't be the only one who thought it was horrific?
    No. I agree. I thought is was dreadful. :mad: I'm so angry with myself! I can't wait for someone to post an unofficial mark scheme.
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    I got t=1 and two stationary points ._.
 
 
 
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