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2016 May 25th Edexcel Core 2 Questions and answers. [Unofficial mark scheme] 2016 Watch

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    (Original post by jenna191)
    wait wait when u integrate for area do uhave to have a +C ?
    yes, there are no limits.
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    According to this 67+/- 3 marks and that was being harsh on the markings

    I would be surprised on the grade boundaries were higher than average as it was pretty average/easy
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    (Original post by jenna191)
    wait wait when u integrate for area do uhave to have a +C ?
    Doubt it, as it disappears when you do the bounds
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    I calculated my predicted UMS score and I got 234-235 ( from the 2015 grade boundaries), is that still a B.
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    (Original post by jenna191)
    wait wait when u integrate for area do uhave to have a +C ?
    When you integrate you get = x+c
    However when you are trying to find an area for values between a and b
    b+c -(a+c) the constants don't change no matter what x is so it's safe to cancel them out instantly.
    b-a+c-c = b-a
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    (Original post by Dieselblue)
    When you integrate you get = x+c
    However when you are trying to find an area for values between a and b
    b+c -(a+c) the constants don't change no matter what x is so it's safe to cancel them out instantly.
    b-a+c-c = b-a

    The question didn't have limits at that point, part (a) was simply \int (3x-x^{\frac{3}{2}})dx
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    Im pretty sure 8i was write B in terms of A?
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    I did that too omg thought I was the only one! So the negative value is right




    QUOTE=Qweudhs56;65157369]1.b. Is wrong. The answer is -1.602. It has to be negative because in geometric sequence, the biggest term is subtract by the smallest.

    Therefore 64(3/4)^9 - 64(3/4)^8 = -1.602 (to three decimal places).

    P.S. I don't know if the markscheme would say 'ignore' signs.[/QUOTE]
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    (Original post by LuckyMN)
    I calculated my predicted UMS score and I got 234-235 ( from the 2015 grade boundaries), is that still a B.
    Its 240 for an A so not too far off, if you're lucky you might get an A
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    (Original post by X_IDE_sidf)
    PDF COPY: http://www.thestudentroom.co.uk/atta...7&d=1464189347

    1 Geometric series question, prove a=64 given S_4=175 and r=\frac{3}{4} then workout sum to infinity. Then find the difference between the 9th and 10th term

    2 Trapezium rule. y=8-2^{x-1} in the interval [0,4] with 4 trapeziums

    3 Circle centred at (7,8). Find the equation of it and of a tangent at point (10,13)

    4 where f x =6x^3+13x^2-4 find the remainder when divided by (2x+3) then factorise it fully given (x+2) is a factor.

    5 Expansion of (2-9x)^4. The using that expand (1+kx)(2-9x)^4 in the form A -232x + Bx^2 given the coefficient of x

    6 1-2\sin(\theta - \frac{\pi}{5})=0 solve for \theta and 4\cos^2 x + 7\sin x - 2 = 0

    7 This was \int (3x-x^{\frac{3}{2}}) dx and then find the limits (where it crossed the x axis.

    8 \log_3(3b+1)-\log_3(a-2)=-1, write b in terms of a then find x given 2^{2x+5}-7(2^x)=-1.

    9 Find optimum perimeter of a funny shape which comprised a rectangle, sector and a equilateral triangle, need diagram.
    Image by Cake_Chan
    Equations given, that needed proving are, y=\frac{500}{x}-\frac{x}{24}(4\pi+3\sqrt{3})
    and P=\frac{1000}{x}+\frac{x}{24}(4\  pi+36-3\sqrt{4})


    My answers:
    1 a) (2 marks) proof
    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}
    b) (3 marks) (x-7)^2+(y-8)^2=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 a) (3 marks) \frac{3}{2}x^2-\frac{2}{5}x^{\frac{5}{2}}+c
    b) (3 marks) 24.3

    8 i) (3 marks) b= \frac{3a-5}{9}
    ii) (4 marks) -2.19

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

    This is useful and was spread out over another thread so I thought I would post it here. The missed answers were all either proofs or not suited to a single number response.

    Thank you new arsey
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    feeling so much better after seening even if i did good or bad lol
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    (Original post by Randall13)
    for the first part of the logs

    I said 1/3 = 3b+1/a-2

    then 1/3a -2/3= 3b+1

    1/3a-5/3=3b

    1/3a-5/3 all divided by 3 = b

    Is this correct and how many marks would i lose if incorrect?
    Nope, there's a mistake here.
    1/3 = (3b+1)/(a-2) is right, but you mutliplied out 'a' wrong
    It should be
    3b+1 = a/3 - 2/3
    3b = a/3 - 5/3
    3b = (a-5)/3
    b = (a-5)/9
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    I did that too! The difference was negative thiugh so I hope it's right


    QUOTE=Fatmanlolololol6;65158247]I am quite pissed. I got full marks in this paper, but for question 1.b. I wrote -1.602. Would that be penalised? And does that mean I can still get full UMS?[/QUOTE]
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    (Original post by Failedexamms)
    I did that too! The difference was negative thiugh so I hope it's right


    QUOTE=Fatmanlolololol6;65158247]I am quite pissed. I got full marks in this paper, but for question 1.b. I wrote -1.602. Would that be penalised? And does that mean I can still get full UMS?
    [/QUOTE]

    The difference was positive, difference means that |u9-u10| = |u10-u9|, so the answer had to be positive, you'd get away with 1 mark though
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    For question 8i would this be a valid way to get to the answer?
    I did it this way during the exam but everyone I've spoken to seems to have done it a different way.

    log_3(3b+1)-log_3(a-2)=-1

    log_3(3b+1)+1=log_3(a-2)

    log_3(3b+1)+log_33=log_3(a-2)

    log_33(3b+1)=log_3(a-2)

    3(3b+1)=a-2

    9b+3=a-2

    9b=a-5

    b=\frac{a-5}{9}
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    (Original post by UKoE Luna)
    For question 8i would this be a valid way to get to the answer?
    I did it this way during the exam but everyone I've spoken to seems to have done it a different way.

    log_3(3b+1)-log_3(a-2)=-1

    log_3(3b+1)+1=log_3(a-2)

    log_3(3b+1)+log_33=log_3(a-2)

    log_33(3b+1)=log_3(a-2)

    3(3b+1)=a-2

    9b+3=a-2

    9b=a-5

    b=\frac{a-5}{9}
    That method looks perfectly fine, i would worry about it aha
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    (Original post by SLamb)
    That method looks perfectly fine, i would worry about it aha
    Awesome Thanks for clearing that up I wasn't quite sure if I was just allowed to remove the logs like that.
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    any idea what the grade bondaries might be
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    (Original post by Failedexamms)
    I did that too! The difference was negative thiugh so I hope it's right
    (Original post by Fatmanlolololol6)
    I am quite pissed. I got full marks in this paper, but for question 1.b. I wrote -1.602. Would that be penalised? And does that mean I can still get full UMS?
    (Original post by Failedexamms)
    I did that too omg thought I was the only one! So the negative value is right
    (Original post by Qweudhs56)
    1.b. Is wrong. The answer is -1.602. It has to be negative because in geometric sequence, the biggest term is subtract by the smallest.Therefore 64(3/4)^9 - 64(3/4)^8 = -1.602 (to three decimal places).P.S. I don't know if the markscheme would say 'ignore' signs.
    I have no idea what is meant by ' It has to be negative because in geometric sequence, the biggest term is subtract by the smallest.'. The question just asked for the difference. This would imply that the answer should be positive. However, I feel that it should be positive but they will not penalise you in the mark scheme.To put this to rest finally, I found this in jan 2006:Name:  sdafaefa.PNG
Views: 163
Size:  95.9 KB which prooves my point
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    Hi guys

    if i put 5y + 3x - 95 = 0, will i still get the mark even though it said put it as 'ax + by + c = 0'?
 
 
 
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