Turn on thread page Beta
    Offline

    0
    ReputationRep:
    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?
    Offline

    12
    ReputationRep:
    (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?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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.
    Offline

    12
    ReputationRep:
    (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?
    Offline

    0
    ReputationRep:
    8.i is wrong. It's b=(a-5)/9
    Offline

    4
    ReputationRep:
    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...
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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.
    Offline

    0
    ReputationRep:
    Which was the q that said don't use graphical or numerical methods ?
    Offline

    16
    ReputationRep:
    (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
    • Community Assistant
    Online

    20
    ReputationRep:
    Community Assistant
    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}
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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}
    Offline

    3
    ReputationRep:
    (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?
    • Community Assistant
    Online

    20
    ReputationRep:
    Community Assistant
    (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}
    Offline

    2
    ReputationRep:
    (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
    Offline

    3
    ReputationRep:
    (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?
    Offline

    0
    ReputationRep:
    what if i find the limits of 7b in 7a?
    Offline

    2
    ReputationRep:
    hahahhahahah i wish we could send this to edexcel
    they tryna stop arsey but arsey left a legacy
    they aint gonna stop tsr
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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}
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 6, 2017

University open days

  • Heriot-Watt University
    School of Textiles and Design Undergraduate
    Fri, 16 Nov '18
  • University of Roehampton
    All departments Undergraduate
    Sat, 17 Nov '18
  • Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 17 Nov '18
Poll
Have you ever experienced bullying?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.