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

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    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?
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    (Original post by X_IDE_sidf)
    Thankyou, sorry I missed that, \log_3(3b+1)-\log_3(a-2)=-1, write b in terms of a
    What would you say I would get, I wrote a in terms of b?
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    (Original post by Olmeister)
    What would you say I would get, I wrote a in terms of b?
    see first post/mark scheme 8 i. I have included it.
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    (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?
    I would imagine the MS would say ignore signs. and full ums is usually 73-74 not 75.
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    (Original post by X_IDE_sidf)
    see first post/mark scheme 8 i. I have included it.
    No I mean I did it the wrong way around?
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    8.i is wrong. It's b=(a-5)/9
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    Indeed, I believe 8(i) is (a-5)/9.

    Other than that, all of the others seem right. I suppose it's now down to just showing enough on proof questions. Though on Q9 they were only 3 marks each...
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    (Original post by Olmeister)
    No I mean I did it the wrong way around?
    Ahh sorry, I would suspect that you would get nothing for that. maybe M1 but it depends what you wrote.
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    Which was the q that said don't use graphical or numerical methods ?
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    (Original post by X_IDE_sidf)
    1 Geometric series question, prove a=64 given S_4=175 and 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 \thetaand4\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= (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.
    Just to point out fort he first question, you knew what r was... I can't remember correctly but I think it was 3/4
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    8i is wrong.

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

     log_3{(\frac{3b+1}{a-2})} =-1

     3^{-1} =\frac{3b+1}{a-2}

     \frac{a-2}{3} =3b+1

     a-2 =9b+3

     b = \frac{a-5}{9}
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    (Original post by NotNotBatman)
    8i is wrong.

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

     log_3{(\frac{3b+1}{a-2})} =-1

     3^{-1} =\frac{3b+1}{a-2}

     \frac{a-2}{3} =3b+1

     a-2 =9b+3

     b = \frac{a-5}{9}
    (Original post by X_IDE_sidf)
    see first post/mark scheme 8 i. I have included it.
    (Original post by Waterrh72)
    8.i is wrong. It's b=(a-5)/9
    (Original post by LukeB98)
    Indeed, I believe 8(i) is (a-5)/9.

    Other than that, all of the others seem right. I suppose it's now down to just showing enough on proof questions. Though on Q9 they were only 3 marks each...
    Thankyou for pointing my miss copy out. It is minus 5 not plus.
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    (Original post by asinghj)
    Just to point out fort he first question, you knew what r was... I can't remember correctly but I think it was 3/4
    Thank you, it was \frac{3}{4}
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    (Original post by NotNotBatman)
    8i is wrong.

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

     log_3{(\frac{3b+1}{a-2})} =-1

     3^{-1} =\frac{3b+1}{a-2}

     \frac{a-2}{3} =3b+1

     a-2 =9b+3

     b = \frac{a-5}{9}
    Bro could you write :

    1/3 a -5/9 all divided by 3. Is this correct as well?
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    (Original post by Randall13)
    Bro could you write :

    1/3 a -5/9 all divided by 3. Is this correct as well?
    No, it would have to be \frac{1}{9}a -\frac{5}{9}
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    (Original post by X_IDE_sidf)

    8 i) (3 marks) b= \frac{3a-5}{9}
    ii) (4 marks) -2.19
    I got a different answer to 8i and cant see where I went wrong

    (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?
    I got the same as this
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    (Original post by NotNotBatman)
    No, it would have to be \frac{1}{9}a -\frac{5}{9}
    How many marks would i lose out of 3?
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    what if i find the limits of 7b in 7a?
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    hahahhahahah i wish we could send this to edexcel
    they tryna stop arsey but arsey left a legacy
    they aint gonna stop tsr
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    (Original post by BainesyA)
    I got a different answer to 8i and cant see where I went wrong



    I got the same as this
    8 i)

    \log_3(\frac{3b+1}{a-2}) = -1

    \frac{3b+1}{a-2} = 3^{-1}

    3b+1=\frac{a-2}{3}

    3b=\frac{a-2}{3}-1

    b=\frac{a-2}{9}-\frac{3}{9}

    b=\frac{a-5}{9}
 
 
 
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