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    How many marks was question 5 ? Specifically for finding the derivative?


    Posted from TSR Mobile
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    In the integration in Q8 where you had to do the area of the rectangle - the area under the graph, was the area of the rectangle 3(units^2)?
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    (Original post by WaiKon1337)
    In the integration in Q8 where you had to do the area of the rectangle - the area under the graph, was the area of the rectangle 3(units^2)?
    Yeah it was, area between x=1 x=4 y=1 and y=0 3*1=3
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    (Original post by 123BOBBY)
    arcsinx=arccosx --> tanx=1 -->x=pie/4
    Can you explain this stage of your working please?
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    guys for volume ques; if i put r = hsin 45 (stupid me :/ ) and done it but i got a differnet dV/dr would i get any follow through marks? also would you get a mark just writing dV/dt or something = 5 ?
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    (Original post by RemainSilent)
    guys for volume ques; if i put r = hsin 45 (stupid me :/ ) and done it but i got a differnet dV/dr would i get any follow through marks? also would you get a mark just writing dV/dt or something = 5 ?
    Perhaps 1 mark for writing dV/dt = 5 (PERHAPS).
    I would have thought you will lose 2 marks for putting r = hsin45 (one for the straight mistake, and one for the answer mark that followed for dV/dh).

    Not sure though - 2 or 3 out of 5.
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    is someone able to draw the graph please think mines wrong
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    (Original post by RemainSilent)
    is someone able to draw the graph please think mines wrong
    http://www.wolframalpha.com/input/?i...5E0.5+%2B+2%29
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    (Original post by SH0405)
    Can you explain this stage of your working please?
    set each side of the equation equal to theta (an angle)
    rearrange to obtain x in terms of theta
    divide through to obtain expression in terms of theta
    work out theta
    work out x using one of the 'x in terms of theta' equations
    \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 Primus2x)
    How many marks was question 5 ? Specifically for finding the derivative?


    Posted from TSR Mobile
    I think finding the derivative is worth 3 marks. 1 mark for verifying (1, 1) and 1 mark for substituting in (1, 1) and obtaining the correct answer.

    Another alternative is 1 for verify. 2 for derivative (both method). 1 method for subbing in (1, 1). 1 accuracy for answer = 1/2
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    (Original post by Duskstar)
    set each side of the equation equal to theta (an angle)
    rearrange to obtain x in terms of theta
    divide through to obtain expression in terms of theta
    work out theta
    work out x using one of the 'x in terms of theta' equations
    \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})
    I swear I didn't quote you...your working's fine .
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    (Original post by SH0405)
    I swear I didn't quote you...your working's fine .
    Oh I thought you just wanted to know how they did it lol. And they didn't do it correctly anyway, but those steps are how you get the tan bit ~ thanks
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    (Original post by Duskstar)
    Oh I thought you just wanted to know how they did it lol. And they didn't do it correctly anyway, but those steps are how you get the tan bit ~ thanks
    Yes, it was a questioning approach; 'can you show me your working (which I know is wrong...)?'
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    How many marks do you think you got today?
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    (Original post by Duskstar)
    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.
    Thanks, that makes me feel better. Bring on C4.
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    In Q 9 when you have to find the area enclosed by the curve (the bit under the x axis) and by integrating from 0 to ln3, you got 3ln3-4 or something. This gets a negative value as its under the x axis, but surely an area must be positive so it should be 4-3ln3?
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    (Original post by 123BOBBY)
    wasnt there an asymptote at x=3
    No
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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<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.
    Didn't do this exam (I do edexcel), but I like snooping on other papers. Here's how I would do 6b)

    Arcsin(x))=arccos(x)

    Cos(arcsin(x))=x

    Sin(90-arcsin(x))=x

    90-arcsin(x)=arcsin(x)

    Then solve for x like this
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    (Original post by Gome44)
    Didn't do this exam (I do edexcel), but I like snooping on other papers. Here's how I would do 6b)

    Arcsin(x))=arccos(x)

    Cos(arcsin(x))=x

    Sin(90-arcsin(x))=x

    90-arcsin(x)=arcsin(x)

    Then solve for x like this
    Neat
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    (Original post by Duskstar)
    Neat
    Your papers look so much harder than ours
 
 
 
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