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    Everyone got 5*(1/100pi) for the rate of change question right? It went to 0.0159 or something?
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    (Original post by conorw6)
    Everyone got 5*(1/100pi) for the rate of change question right? It went to 0.0159 or something?
    YES!
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    (Original post by conorw6)
    Everyone got 5*(1/100pi) for the rate of change question right? It went to 0.0159 or something?

    I simplified to \frac{1}{20\pi}
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    What were the limits for the integral on question 8??? Was it 4 and 1?? Some ppl are telling me we had to use x coordinate of P?
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    (Original post by corleone12)
    What were the limits for the integral on question 8??? Was it 4 and 1?? Some ppl are telling me we had to use x coordinate of P?
    It is 1 and 4, then you subtract it from the area of rectangle 3.
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    (Original post by RemainSilent)
    same her lol i just did something like sin x = cos x then tan came out as pi/4 and it didnt wokr out
    Wait my I got tan(x) = 1
    So my x was pi/4


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    (Original post by Primus2x)
    It is 1 and 4, then you subtract it from the area of rectangle 3.
    What was the final area ?
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    For 9iii) I was correct up until substituting in my limits and where I got e^2x, I just substituted in ln(3) as when it should have been 2ln(3) causing me to get the wrong answer. How many marks do you think I will get out of 5?

    3?
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    (Original post by corleone12)
    What was the final area ?
    I don't remember, sorry. If someone tell me what the function was, I could do it.
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    (Original post by Primus2x)
    I don't remember, sorry.
    Somehow i got 3 for the integral and 4 for rectangle over all i got area of 1
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    (Original post by corleone12)
    YES!
    yessss!
    (Original post by Primus2x)
    I simplified to \frac{1}{20\pi}
    I'm pretty sure they accept it in exact form or with suitable rounding!
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    Please vote on this poll for A boundary predictions:
    http://www.thestudentroom.co.uk/show....php?t=3399719
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    Wait, I remember now.
    (Someone seriously needs to fix latex on TSR)
    Copy the code to this site http://www.codecogs.com/latex/eqneditor.php
    

f(x)=\frac{(x-2)^2}{x}=x-4+\frac{4}{x}

\\

\int_{1}^{4} f(x) dx=\left [ \frac{x^2}{2}-4x+4\ln{x} \right ]_{1}^{4}=[8-16+8\ln2]-\left [\frac{1}{2}-4+0 \right ]

\\

=\frac{-9}{2}+8\ln{2}

\\

Area=3-\left [\frac{-9}{2}+8\ln{2}

 \right ]=\frac{15}{2}-8\ln{2}
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    (Original post by Connorbwfc)
    For arcsin(x)=arcos(x)

    Would you get any marks for just putting the answer as Root(2) / 2 ?????
    it was 2 marks so i think you will get 1. this was probably the trickiest question on the paper to get 2 marks in i think- mei always have an a* question in c3/c4 papers so this is probably it.

    i used arccos x + arcsin x= pi/2 and the equation to get root2 /2

    i think you might have to say because they are the smae values at pi/4 or something or solve it properly using an identity
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    If you can't read the equations that were parsed into that mess:

    The answer to the area bounded by f(x) and y=1 should be \frac{15}{2}-8\ln{2}

    And the integral of g(x) with respect to x between 0 and 3 is -\frac{15}{2}+8\ln{2} because if you translate y=1, x=1, x=4 by the same vector that transformed f(x) into g(x), they become y=0, x=0, and x=3 and that is equivalent to integrating g(x) between 0 and 3.
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    (Original post by Primus2x)
    If you can't read the equations that were parsed into that mess:

    The answer to the area bounded by f(x) and y=1 should be \frac{15}{2}-8\ln{2}

    And the integral of g(x) with respect to x between 0 and 3 is -\frac{15}{2}+8\ln{2} because if you translate y=1, x=1, x=4 by the same vector that transformed f(x) into g(x), they become y=0, x=0, and x=3 and that is equivalent to integrating g(x) between 0 and 3.
    you can also say area is the same but under x-axis instead of above it
    or you can say modulus area is the same
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    I think I have got between 59 and 64 on this paper. Is it still possible to get an A*?
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    53 = B grade boundary?
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    (Original post by ETRC)
    it was 2 marks so i think you will get 1. this was probably the trickiest question on the paper to get 2 marks in i think- mei always have an a* question in c3/c4 papers so this is probably it.

    i used arccos x + arcsin x= pi/2 and the equation to get root2 /2

    i think you might have to say because they are the smae values at pi/4 or something or solve it properly using an identity
    I did it like this
    

\arcsin{x}=\arccos{x}

\\

x=\sin\left ( \arccos{x} \right )

\\

x=\sqrt{1-\left ( \cos\left ( \arccos{x} \right ) \right )^2} \because \sin^2 \theta+\cos^2 \theta =1 \Rightarrow \sin \theta=\sqrt{1-\cos^2 \theta}

\\x=\sqrt{1-x^2}

x^2=1-x^2

\\

2x^2=1

\\

x^2=\frac{1}{2}

\\

x=\sqrt{\frac{1}{2}}

\\

\therefore x=\frac{\sqrt{2}}{2} \because \text{rationalise denominator}
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    I'm going to try and do a mark scheme here. Bare in mind that I'm in Year 12 too >.>

    Firstly, thanks to Barrel who actually posted the paper (page 15). here are the pages in order:
    http://www.thestudentroom.co.uk/atta...9&d=1434104177
    http://www.thestudentroom.co.uk/atta...1&d=1434104202
    http://www.thestudentroom.co.uk/atta...3&d=1434104269
    http://www.thestudentroom.co.uk/atta...5&d=1434104300


    1. \,\, y = e^{2x} \cos x 

\dfrac{dy}{dx} = e^{2x} (-\sin x) + 2e^{2x} \cos x

\dfrac{dy}{dx} = e^{2x} (2 \cos x - \sin x) \,\, [2]

\mathrm{- solve for max:}

0 = e^{2x} (2 \cos x - \sin x) 

2 \cos x = \sin x 

\tan x = 2 \,\, [2]

x = 1.1 \, \mathrm{to 2 s.f.}

\mathrm{- sub into y = e^{2x} \cos x}

y = 4.1 \, \mathrm{to 2 s.f.} 

P(1.1, 4.1) \,\, [2]


    2. \,\, \displaystyle \int \sqrt [3]{2x - 1} \, dx

\mathrm{- solve by inspection of substitution.  Substitution would be u = 2x - 1}

= \dfrac {3}{8} (2x - 1)^{\frac {4}{3}} + c \,\, [4]


    3. \,\, \displaystyle \int^2 _1 x^{3}\ln x \, dx

\mathrm{- integrate by parts:}

\dfrac {dv}{dx} = x^{3}, u = lnx

v = \dfrac {x^4}{4}, \dfrac{du}{dx} = \frac{1}{x} \,\, [2]

\mathrm{- follow through [2], and}

= 4 \ln 2 - \frac{15}{16} \,\, [1]


    4. \,\, V = \dfrac{1}{3} \pi r^{2} h, \, \dfrac{dV}{dt} = 5

\mathrm{- draw a triangle:}

\tan 45 = \dfrac{r}{h}

r = h

V = \dfrac{1}{3} \pi h^3 \,\, [2]

\dfrac{dV}{dh} = \pi h^2

\mathrm{- chain rule:}

\dfrac{dh}{dt} = \dfrac{dV}{dt} \times \dfrac{dh}{dV} = \dfrac{\frac{dV}{dt}}{\frac{dV}{  dh}}

\dfrac{dh}{dt} = \dfrac{5}{\pi h^2} \,\, [2]

h = 10, \, \dfrac{dh}{dt} = \dfrac{5}{100 \pi} = \dfrac{1}{20 \pi} \,\, [1]


    5. \,\, y^2 + 2x \ln y = x^2

\mathrm{- just put (1, 1) in - I'm not writing that out for you... [1]}

\mathrm{- differentiate implicitly:}

2y \dfrac{dy}{dx} + 2x \dfrac{1}{y} \dfrac{dy}{dx} + 2 \ln y = 2x

\dfrac{dy}{dx} 2 \left(y + \dfrac{x}{y} \right) = 2 \left(x - \ln y \right)

\dfrac{dy}{dx} = \dfrac{x - \ln y}{y + \frac{x}{y}} \,\, [3]

\mathrm{- substitute in (1, 1)}

\dfrac{dy}{dx} = \dfrac{1 - 0}{1 + 1} = \dfrac{1}{2} \, [1]


    - I wasn't completely sure of the method for the second part, this is what a friend told me the method was (I think I do it correctly). Because of the nature of these two questions, however, I'm not 100% sure you need a method if you get the answer correct. It's only 1 mark each either way ~
    6. \,\, 6 \sin ^{-1} x - \pi = 0

\sin ^{-1} x = \dfrac{\pi}{6}

x = \sin \dfrac{\pi}{6} = \dfrac {1}{2} \,\, [2]

\mathrm{method:}

\sin ^{-1} x = \cos ^{-1} x

\theta = \sin ^{-1} x, \, \theta = \cos ^{-1} x

x = \sin \theta, \, x = \cos \theta

\mathrm{- divide through (solve them simultaneously)}

1 = \tan \theta

\theta = \dfrac{\pi}{4}

x = \sin \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}} ( = \dfrac{\sqrt{2}}{2}) \,\, [2]


    7. \,\, \mathrm{- for \ the \ first \ part \ you \ just \ substitute \ f(x) \ in \ wherever \ there \ is \ an \ x \ in \ f(x) \ [2]}

\mathrm{- if you really want to see it reply to this}

\mathrm{- for the second part, I didn't know this as an actual rule, but:}

f^{-1}(x) = f(x) \,\, [1]

\mathrm{- sub (-x) into g(x), prove that g(x) = g(-x), easy. [2]}

\mathrm{- g(x) has a line of symmetry in the y-axis/is reflected in the y-axis owtte [1]}


    8. \,\, f(x) = \dfrac{(x - 2)^2}{x} = \dfrac{x^2 - 4x + 4}{x}

f(x) = x - 4 + 4x^{-1}

\mathrm{- no quotient rule for me, thanks}

f'(x) = 1 - 4x^{-2} = 1 - \dfrac{4}{x^2} \,\, [2]

f''(x) = 8x^{-3} = \dfrac{8}{x^3} \,\, [2]

\mathrm{- solve for Q:}

0 = 1 - \dfrac{4}{x^2}

\dfrac{4}{x^2} = 1

4 = x^2

x = \pm 2

x = +2 \, \mathrm{is solution for P}

x = -2 \, \mathrm{for Q}

f(-2) = -8

Q(-2, -8) \,\, [2]

f''(-2) = -1<0 \, \mathrm{so Q is a maximum} \,\, [1]

    \mathrm{- \ verify \ is \ simple \ [2]}

\mathrm{- the \ next \ part \ can \ be \ done \ two \ ways...}

\mathrm{- rectangle - integral:}

\displaystyle A = 1 \times 3 - \int ^4 _1 f(x) \, dx

\mathrm{- or as an integral, top curve - bottom curve:}

\displaystyle A = \int ^4 _1 1 - f(x) \, dx

\mathrm{- follow either through}

\mathrm{- the integration is easy using the expansion of f(x)}

A = \dfrac{15}{2} - 4 \ln 4 \,\, [4]

\mathrm{- next part...}

g(x) = f(x + 1) - 1 \, qed. \,\, [3]

\displaystyle \int ^3 _0 g(x) \, dx = 4 \ln4 - \dfrac{15}{2}

= - \mathrm{[your answer from before]} \,\, [1]

\mathrm{- I \ can't \ really \ say \ what \ the \ words \ for \ this \ answer \ are} \,\, [1]


    9. \,\, f(x) = (e^x - 2)^2 - 1

0 = (e^x - 2)^2 - 1

1 = (e^x - 2)^2

e^x - 2 = \pm 1

e^x = 1, 3 \,\, \mathrm{(ln1 \ is \ 0 \ which \ is \ for \ O)}

x = \ln 3 \,\, [2]

f'(x) = 2e^{x} (e^x - 2)

0 = 2e^{x} (e^x - 2)

e^x = 2

x = \ln 2 \,\, [3]

f(\ln 2) = -1 \,\, [1]

Q(\ln 2, -1)

\mathrm{- as I said earlier, it wants the 'enclosed' area...}

\displaystyle \int ^{\ln 3} _0 (e^x - 2)^2 - 1 \, dx

\displaystyle= \int ^{\ln 3} _0 e^{2x} - 4e^{x} + 3 \, dx

\mathrm{- follow through [4]}

A = 3 \ln 3 - 4 \,\, [1]

    \mathrm{- \ swap \ y \ and \ x \ and \ rearrange \ etc}

f^{-1}(x) = \ln {( \sqrt {x + 1} + 2)} \,\, [3]

domain: \, x \geq -1 \,\, [1]

range: \, f^{-1}(x) \geq \ln 2 \,\, [1]

\mathrm{- graph is reflection in y = x [2]}

\mathrm{- starts at (-1, ln 2) intercept at ln3, then curve}

\mathrm{- should cross with f(x) at y = x}


    Anything I missed or got wrong just ask. If you think it would be marked differently then say so too.
 
 
 
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