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    (Original post by X_IDE_sidf)
    My answers:

    1 b) (2 marks) 256
    c) (2 marks) 1.602

    2 a) (1 mark) 7
    b) (3 marks) 20.75
    c) (2 marks) 5.75

    3 a) (2 marks) \sqrt{34}
    c) (4 marks) 3x+5y-95=0

    4 a) (2 marks) 5
    b) (2 marks) f(-2)=0
    c) (4 marks) f(x)=(x+2)(3x+2)(2x-1)

    5 a) (4 marks) 16-288x+1944x^2
    b) (1 mark) 16
    c) (2 marks) \frac{7}{2}

    d) (2 marks) 936

    6 i) (3 marks) \frac{8\pi}{15}or \frac{-2\pi}{15}

    ii) (6 marks) 345.5$^{\circ}$ or 194.5$^{\circ}$

    7 b) (3 marks) 24.3

    8 ii) (4 marks) -2.19

    9 a) (2 marks) \frac{\pi x^2}{3}
    d) (5 marks) x=16.63 P= 120m
    c) (2 marks)  f''x = 0.437 > 0 \therefore is a minimum at x

    my unofficial mark scheme
    Thanks! But how come the marks don't add up?
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    (Original post by Frappé)
    Thanks! But how come the marks don't add up?
    I didn't include the proofs.
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    (Original post by X_IDE_sidf)
    I didn't include the proofs.
    Add in how many marks each of the proofs were at least and it's perfect! Good job man.
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    (Original post by X_IDE_sidf)
    My answers:

    1 b) (2 marks) 256
    c) (2 marks) 1.602

    2 a) (1 mark) 7
    b) (3 marks) 20.75
    c) (2 marks) 5.75

    3 a) (2 marks) \sqrt{34}
    c) (4 marks) 3x+5y-95=0

    4 a) (2 marks) 5
    b) (2 marks) f(-2)=0
    c) (4 marks) f(x)=(x+2)(3x+2)(2x-1)

    5 a) (4 marks) 16-288x+1944x^2
    b) (1 mark) 16
    c) (2 marks) \frac{7}{2}

    d) (2 marks) 936

    6 i) (3 marks) \frac{8\pi}{15}or \frac{-2\pi}{15}

    ii) (6 marks) 345.5$^{\circ}$ or 194.5$^{\circ}$

    7 b) (3 marks) 24.3

    8 ii) (4 marks) -2.19

    9 a) (2 marks) \frac{\pi x^2}{3}
    d) (5 marks) x=16.63 P= 120m
    c) (2 marks)  f''x = 0.437 > 0 \therefore is a minimum at x

    my unofficial mark scheme
    Not my answers. Found it on another thread. Has marks included.



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    Here is the tread where I have all the marks and all the questions, http://www.thestudentroom.co.uk/show...691&p=65157957 <-- Go there to comment on them (they are from memory/calculator history so may be slightly off.)
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    You had to put your 16.7 value for x back into the origional equation to find the perimeter.
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    That wasn't bad. They have asked a lot worse questions in the past
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    (Original post by M8ty)
    That was so easy😂
    Expect grade boundries to be high.
    You must be stupid to have not got at least 70+
    Ok somebody is feeling cocky, sorry not everyone has a mistake free brain
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    (Original post by shayefc)
    can anyone remember the total marks for Q9?
    yes it was 15 marks.. if you got all of it right, that 20UMS in the bag
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    Would I lose a mark if I didn't put + C on the integration question? I thought we didn't need it in C2?
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    Completely ran out of time
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    I got the difference between u9 and u10 negative does that matter
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    I did that too actually stressing so much
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    (Original post by Failedexamms)
    I got the difference between u9 and u10 negative does that matter
    Name:  sdafaefa.PNG
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    (Original post by musictagj)
    Guys it asked for P not x, so 17 is wrong because you had to substitute in x=16.633 and then you get 120 to the nearest metre


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    I actually laughed at everyone saying they got 17...

    CLASSSIICCC mistake, they forgot to substitute back into the original formula for P. I literally came so close to keeping my answer of 17 before realizing i didnt sub it back in, yeah the actual answer was 120.something and rounded to 120.

    For the Next part you sub this number into the second differential.

    P'' = 2000/x^3 = 0.0*something tiny but bigger than zero*

    because P'' > 0 (and P' = 0) we know its a minimum point.
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    (Original post by Failedexamms)
    I got the difference between u9 and u10 negative does that matter
    I used mod notation; it means turn the answer into a positive if its negative.
    I then wrote underneath the actual negative difference just to be sure of the mark
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    (Original post by Hangk13)
    What did everyone get when substituting their x value into the 2nd differential to show it was a minimum point? (very last question part e)

    I got a really small decimal??
    0.4 something something
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    (Original post by Tobiq)
    I actually laughed at everyone saying they got 17...

    CLASSSIICCC mistake, they forgot to substitute back into the original formula for P. I literally came so close to keeping my answer of 17 before realizing i didnt sub it back in, yeah the actual answer was 120.something and rounded to 120.

    For the Next part you sub this number into the second differential.

    P'' = 2000/x^3 = 0.0*something tiny but bigger than zero*

    because P'' > 0 (and P' = 0) we know its a minimum point.
    Hate to sound like a **** but totally agree hahaha, am resitting and was super happy at the number of year 12s looking like they had just been hit by a bus, sorry guys
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    (Original post by HannahC-H)
    0.4 something something
    it was like 1.5 x 10^-3, i remember. 0.4 is way too big but i dont think the size matters, just as long as its bigger than zero
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    (Original post by Tobiq)
    it was like 1.5 x 10^-3, i remember 0.4 way too big but i dont think the size matters, just as long as its bigger than zero
    Ah well, pretty sure I can afford to lose one mark
 
 
 
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