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    (Original post by SH0405)
    1. (1.107, 4.094) [6]
    2. (3(2x-1)(4/3))/8 + c [4]
    3. u = 2x-1, (1/16)(64ln2 - 15) [5]
    4. r=h, 1/20pi = 0.0159 [5]
    5. gradient = ½ [6]
    6. i) x = ½ [2]
    ii) x = sqrt(2)/2 [2]
    7. i) f(f(x)) = x via substituting in. f-1(x) = f(x) = (1-x)/(1+x) [3]
    ii) g(x) is even since g(-x) = g(x). Symmetrical about the y-axis. [3]
    8. i) f'(x) = 1 - 4/x2, f''(x) = 8/x3which is less than zero when x=-1 and so Q is a maximum point. [7]
    ii) Area was 3 take away integral which gave ½(15-8ln4) [6]
    iii) f(x-1)-1=g(x) so g(x) = (x2-3x)/(x+1) [3]
    iv) This is the negative of the previous answer, and yes it's negative, because it asked for the value. [2]
    9. i) x = ln3 [2]
    ii) x = ln2, y = -1 [4]
    iii) The integral came out as negative, but this question asked for the area, so the given answer must be positive; 4 - 3ln3 [5]
    iv) f-1(x) = ln(sqrt(x+1)+2), with the domain as x >= -1 and range as y >= ln2. The sketch of the graph was a reflection in the line y = x, but subject to the 'new' domain and range (i.e. not below ln2). [7]

    Please quote me if I have made a mistake...
    8.ii) I'm pretty sure answer was 1/2(15-8ln2) not ln4 😊
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    (Original post by chizz1889)
    8.ii) I'm pretty sure answer was 1/2(15-8ln2) not ln4 😊
    I got 15/2 - 8ln2... I think?

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    (Original post by _Caz_)
    I got 15/2 - 8ln2... I think?

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    I'm fairly certain it was 15/2 - 4ln2

    EDIT: Infact no it was 3ln3!!! I think, I will work it out again in a sec 😊
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    Mark scheme done

    http://www.thestudentroom.co.uk/show...5#post56931025
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    Infact yes I concur! I was getting mixed up between questions ☺️
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    (Original post by _Caz_)
    I got 15/2 - 8ln2... I think?

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    Yeah that's right 😀
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    (Original post by _Caz_)
    I got 15/2 - 8ln2... I think?

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    The answer was 15/2 - 4ln4

    which is the same as what you got
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    I forgot about the rectangle in q.8 completely worked out the area perfectly for the integral of f(x) just forgot to to minus it from the area of the rectangle. I didn't do any calculations for the rectangle either, or write out how to obtain the area. I also could work out what it was on about for g(x), so put it into my graphical calculator and go -1.95... the decimal answer. And I also messed up my graph but I did a some sort of sketch (but then wrote reflected in the line y=x. I think these were my only mistakes, that I know of.

    How many marks would I lose for this?
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    isnt there a asymptote at x=3 therefore the domain for f(x) should be ln2<=x<3
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    Does anybody have a list of 90UMS boundaries for past papers?
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    arcsinx=arccosx --> tanx=1 -->x=pie/4
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    (Original post by Duskstar)
    The answer was 15/2 - 4ln4

    which is the same as what you got
    (Original post by chizz1889)
    Yeah that's right 😀
    (Original post by chizz1889)
    I'm fairly certain it was 15/2 - 4ln2

    EDIT: Infact no it was 3ln3!!! I think, I will work it out again in a sec 😊

    to be fair there was so much natural log in that paper it's all blurred into one! Hope the exam was okay for everyone. Strange paper this year - the whole thing seemed to be on functions?! No proof really, no exponential equations, no modulus. Was quite surprised.
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    (Original post by 123BOBBY)
    arcsinx=arccosx --> tanx=1 -->x=pie/4
    I think it might be sqrt2/2 although I put pi/4 as well. :/
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    (Original post by OwenEllicott1996)
    I forgot about the rectangle in q.8 completely worked out the area perfectly for the integral of f(x) just forgot to to minus it from the area of the rectangle. I didn't do any calculations for the rectangle either, or write out how to obtain the area. I also could work out what it was on about for g(x), so put it into my graphical calculator and go -1.95... the decimal answer. And I also messed up my graph but I did a some sort of sketch (but then wrote reflected in the line y=x. I think these were my only mistakes, that I know of.

    How many marks would I lose for this?
    You probably lost 2 marks for not doing the rectangle, and then 2 marks for the g(x) question, because the intention was to literally write down you answer before * -1. On the sketch question you probably lost 1 or 2 marks on the actual sketch (out of 2) depending what qualities it had of what it was supposed to be.
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    (Original post by 123BOBBY)
    arcsinx=arccosx --> tanx=1 -->x=pie/4
    (Original post by _Caz_)
    I think it might be sqrt2/2 although I put pi/4 as well. :/
    Yep, refer to this:

    \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})
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    (Original post by Duskstar)
    Yep, refer to this:

    \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})
    do you reckon ill get a mark for getting up to the pi/4? I mean it was a two mark question so maybe not...
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    wasnt there an asymptote at x=3
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    (Original post by Duskstar)
    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 &lt; 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 = 0, 2 \,\, \mathrm{(0 \ is \ impossible)}

x = \ln 2 \,\, [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.
    e^x - 2 = \pm 1
    e^x = 0, 2 \,\, \mathrm{(0 \ is \ impossible)}

    I'm sorry if this has already been corrected, but in the solution of 9) you state that 2+1=2 and 2-1=0.
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    Grade boundries for a c anyone ?
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    (Original post by AccountName)
    e^x - 2 = \pm 1
    e^x = 0, 2 \,\, \mathrm{(0 \ is \ impossible)}

    I'm sorry if this has already been corrected, but in the solution of 9) you state that 2+1=2 and 2-1=0.
    That hasn't, thank you. I keep getting ln2 and ln3 mixed up lol - I thought I was doing the second part :doh:
 
 
 
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