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    All the functions that appeared in the paper, as follows:

    (I'll try and add questions as I remember them, though I have almost no idea of question numbers :P )

    \displaystyle f(x) = e^{2x} \cos x

    ~ differentiate and find the coordinate of the maximum (1.1, 4.1)

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

    ~ Integrate for:
    \displaystyle \frac {3}{8} (2x - 1)^{\frac {4}{3}} + c


    \displaystyle \int^2 _1 x^{3}\ln x

    ~ Integration by parts for something I can't quite remember, but someone will bother to work out or ask about...


    \displaystyle y^2 + 2x \ln y = x^2

    ~ verify point (1, 1) which means sub it in and prove it still equates, then differentiate implicitly to get the gradient at that point, with was 1/2.


    \displaystyle f(x) = \frac {1 - x}{1 + x}

    ~ prove f(f(x)) = x, which involves subbing f(x) in in place of each x, then rearranging.
    ~ hence write down the inverse of f(x), which is f(x) (I personally don't understand why this was a 'hence' question - I haven't seen a rule that would describe this before, although it's obvious there is one)

    \displaystyle g(x) = \frac {1 - x^2}{1 + x^2}

    ~ show g(x) is even - sub (-x) for x, and show that is the same as g(x). This means that g(x) has a line of symmetry in the y-axis, or x = 0.


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

    ~ you had to differentiate this twice. I found it easier to expand the top and divide through by x than use the quotient rule, but either should work and I think the answer was f'(x) = 1 - 4x^(-2), then differentiate again for f''(x) = 8x^-3 you should fin the point Q(-2, -8), then show it is a maximum because f''(x) < 0

    ~ I think this was the question where you had to integrate to find the area enclosed by the function and the line y = 1 (after proving the intersections were x = 1 and 4), which was easy enough - either use a rectangle, or take the integral of 1 - f(x). I think my answer was 15/2 - 4ln4

    translate (-1, -1) for:

    \displaystyle g(x) = \frac {x^2 - 3x}{x + 1}

    ~ you had to show this, which just meant showing f(x + 1) - 1 = g(x).
    ~ then, it asked you to write down the integral from 0 to 3 of g(x) with no further calculation, which is the integral from the second part (I don't know if they would want you to indicate that it was this value but negative), and then describe why this was the case. I guess you just talk about the translation and stuff ~

    \displaystyle f(x) = (e^x - 2)^2 - 1

    ~ I can't remember quite what the questions were here, but I think you had to differentiate it, find x-intercept at ln3, and minimum at (ln2, -1)

    I got for the inverse:
    \displaystyle f^{-1}(x) = \ln {( \sqrt {x + 1} + 2)}

    ~ domain of inverse x>= -1 range f-1(x) >= ln2
    ~ then sketch the inverse by reflecting in y = x. The easy way to do this was plot the images of Q and P, and make sure you cross the line y=x in the same place, and continue the line without changing gradient much (it was pretty much a flat line past the intersection with y=x).


    and somewhere in the paper, two parts to a question:

    \displaystyle 6 \sin ^{-1} x - \pi = 0 , x = \frac {1}{2}
    \displaystyle  \sin ^{-1} x = \cos ^{-1} x ,x = \frac {1}{\sqrt 2}
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    (Original post by Duskstar)
    All the functions that appeared in the paper, as follows:

    \displaystyle f(x) = e^{2x} \cos x
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle \int \root {3}{2x - 1} dx

    \displaystyle \int^2 _1 x^{3}\ln x
    \displaystyle y^2 + 2x \ln y = x^2
    verify point (1, 1)
    \displaystyle f(x) = \frac {1 - x}{1 + x}
    \displaystyle g(x) = \frac {1 - x^2}{1 + x^2}
    \displaystyle f(x) = \frac {(x - 2)^2}{2}
    translate (-1, -1) for:
    \displaystyle g(x) = \frac {x^2 - 3x}{x + 1}
    \displaystyle f(x) = (e^x - 2)^2 - 1
    I got for the inverse:
    \displaystyle f^{-1}(x) = \ln {\root {x + 1} + 2}

    and somewhere in the paper, two parts to a question:
    \displaystyle 6 \sin ^{-1} x - \pi = 0 (x = \frac {\pi}{6}
    \displaystyle  \sin ^{-1} x = \cos ^{-1} x (x = \frac {1}{\root 2}
    For arcsinX and arccosx I set them both equal to theta. Theta =pi/4 then x equals root2/2
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    (Original post by chizz1889)
    Ahaha safe then 😆 I got 4-3ln3 as it has to be positive right 😊
    They'll probably accept positive or negative, because conceptually you can think of it as a negative area. I solved for the negative area then wrote |Area| = 4-3ln3 as well just in case.
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    Implicit differentiation gradient = 0.5
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    (Original post by chizz1889)
    For arcsinX and arccosx I set them both equal to theta. Theta =pi/4 then x equals root2/2
    Yep, I just had some formatting trouble lol.
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    (Original post by chizz1889)
    Implicit differentiation gradient = 0.5
    I got the same ~
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    Any idea of grade boundaries?

    Was surprised to see no modulus questions.

    AND NO DODGY PROOF. YES.


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    (Original post by _Caz_)
    Any idea of grade boundaries?

    Was surprised to see no modulus questions.

    AND NO DODGY PROOF. YES.


    Posted from TSR Mobile
    I know thank f**k haha
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    (Original post by pierre225)
    Such a nice exam ahhh so annoyed at myself for losing a mark on integration! hopefully 71/72 will be enough?
    Dude I'd happily take 71/72

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    (Original post by _Caz_)
    Any idea of grade boundaries?

    Was surprised to see no modulus questions.

    AND NO DODGY PROOF. YES.


    Posted from TSR Mobile
    definitely higher than last year, since this paper was freakishly easy
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    I got so many answers wrong

    Will I lose only 1 mark for wrong answers each, as long as method was right?
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    (Original post by _Caz_)
    Any idea of grade boundaries?

    Was surprised to see no modulus questions.

    AND NO DODGY PROOF. YES.


    Posted from TSR Mobile
    Grade boundaries will probably be pretty normal.

    I personally found it very easy - it was very calculus heavy, function heavy, and you could check literally 90% of your answers with a graphical calculator.
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    (Original post by chizz1889)
    (1.11,4.09)
    How did you get that? I got lost at was it tanx=2 or something?
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    (Original post by chizz1889)
    Implicit differentiation gradient = 0.5
    Well that's one i know i got right now XD

    Why was the paper so ****ing hard?!?!?!?!
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    (Original post by JamesExtra)
    How did you get that? I got lost at was it tanx=2 or something?
    Yeah that's right but you had to have the calculator in rad mode then do arctan 2
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    (Original post by Duskstar)
    Grade boundaries will probably be pretty normal.

    I personally found it very easy - it was very calculus heavy, function heavy, and you could check literally 90% of your answers with a graphical calculator.
    Nah i reckon the grade boundaries will be a bit lower than normal, everyone i've spoken to found it pretty hard. Even my asian friend said so haha
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    (Original post by chizz1889)
    Yeah that's right but you had to have the calculator in rad mode then do arctan 2
    Ah ok I left it in degrees. I was wondering why I was getting a ridiculous answer for Y. Oh well hopefully have lost much seeing as I got up to there
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    (Original post by _Caz_)
    Dude I'd happily take 71/72

    Posted from TSR Mobile
    yeah I would too I actually did quite bad I couldn't answer loads of the questions
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    (Original post by Buses)
    definitely higher than last year, since this paper was freakishly easy
    You must be crazy if you think that paper was easy lol
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    Ok not like a maths genius or anything and I found it fairly easy compared to other papers. Only ones I got stumped on we're Q1 after tanx=2 and the arcsin=arccos. Expect the grade boundaries to be high to be honest
 
 
 
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