Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    2
    ReputationRep:
    (Original post by GPW333)
    What was question 6 about?
    Something to do with AP and GP - you were given Un+1 in part i) I think


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by proximity)
    Someone please send in their solution of the integration question with answer 512/15
    Do you remember the exact question?


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by AlfieH)
    My teacher has confirmed that the answer to 9iii is Pi/3a and 4Pi/3a.
    Unless i'm wrong, with Pi/3A, you're multiplying by A twice.

    The solution is 5/3 * 1/5Pi, which equals 1/3Pi but not 1/3Pi times by another A.

    Offline

    2
    ReputationRep:
    (Original post by AlfieH)
    I am gonna start an unofficial mark scheme - I think this is about correct!

    1i) 10cm (2m)

    1ii) 5.04cm (2m)

    2i) 3/10π (2m)

    2ii) 20.4cm (3m)

    3i) 27+27kx+9k^2x^2+k^3x^3 (4m)

    3ii) +/-√3 (2m)

    4i) log base 3 x^2/x+4 (2m)

    4ii) x=12 (4m)

    5a) x^4/2-x^3+2x^2-6x (3m)

    5bi) -6/a+2/a^2+4 (4m)

    5bii) 4 (1m)

    6i) k=91 (3m)

    6ii) Sum of terms = 978 (2m)

    6iii) N=38 (6m)

    7i) R=0, quotient = x^2-4x+3 (3m)

    7ii) (x+1), (x-1), (x-3) (3m)

    7iii) Prove (2m)

    7iv) 512/15 (4m)

    8i) Translation 2 units in positive x direction (2m)

    8ii) Stretch scale factor 1/9 in y direction (2m)

    8iii) Sketch, crosses y at (0,1/9) (2m)

    8iv) x=6.73 (3m)

    8v) Area =9.60 (3m)

    9i) 2π/a (1m)

    9ii) a=5/3 and k=√3/2 (3m)

    9iii) π/3a and ? (4m)
    8 iii and 9 iii were both 3 marks I think
    Offline

    1
    ReputationRep:
    Does anyone remember roughly what the questions were for these answers?
    5bi) -6/a+2/a^2+4 (4m) 8iv) x=6.73 (3m)
    Offline

    8
    ReputationRep:
    For the last part of question 9, I answered in degrees and not radians (60/a and 180 - 60/a) - the question didn't ask for radians did it so I'd still get all the marks?
    Offline

    2
    ReputationRep:
    (Original post by proximity)
    Someone please send in their solution of the integration question with answer 512/15
    Thank god, loads of people said you were meant to split it out and I didn't .
    Offline

    2
    ReputationRep:
    (Original post by CLGC98)
    How were you meant to do the integration to find the area , was it with limits of 3 to -1 or do you do -1 to 0 and 0 to 3?
    3 to -1
    Offline

    2
    ReputationRep:
    (Original post by DziNe)
    For the last part of question 9, I answered in degrees and not radians (60/a and 180 - 60/a) - the question didn't ask for radians did it so I'd still get all the marks?
    It did say radians
    Offline

    2
    ReputationRep:
    (Original post by louisa190)
    for definite?
    Yep!
    Offline

    3
    ReputationRep:
    (Original post by DziNe)
    For the last part of question 9, I answered in degrees and not radians (60/a and 180 - 60/a) - the question didn't ask for radians did it so I'd still get all the marks?
    http://static1.squarespace.com/stati...ark+scheme.pdf

    ^ Last yea's mark scheme. Question 9 on that was very similar. They state explicitly that the answer "Must be radians and not degrees" but gave a method mark if "attempted in degrees not radians".
    Offline

    2
    ReputationRep:
    Can someone go through the area under curve one please? I got the intervals as 3 and -1 and integrated right but then somehow went wrong
    Offline

    1
    ReputationRep:
    (Original post by cc262626)
    For the coeffient is equal to the constant question, did anyone else think they meant the constant k? And get answers of 1/9 and 0 😟
    I did the same but since they didn't make it clear I think they should accept both answers
    Offline

    0
    ReputationRep:
    Does anyone remember the exact geometric question where you had to find N that's greater than infinity?
    Offline

    1
    ReputationRep:
    (Original post by Alevelz)
    Does anyone remember the exact geometric question where you had to find N that's greater than infinity?
    N=38
    Offline

    0
    ReputationRep:
    No I mean the question but thanks
    Offline

    1
    ReputationRep:
    (Original post by Alevelz)
    Does anyone remember the exact geometric question where you had to find N that's greater than infinity?
    not word for word but it was:
    sum of first 20 terms of AP > sum to infinity of GP

    (for the AP a=5 d=1.5) (for the GP a=1.2 (?) r=0.9)
    Offline

    1
    ReputationRep:
    (Original post by CLGC98)
    How were you meant to do the integration to find the area , was it with limits of 3 to -1 or do you do -1 to 0 and 0 to 3?
    I think it was -1 to 3 since the the are was above the axis, you only split it if it's part below part above. That's what I think anyway (might be wrong)
    Offline

    3
    ReputationRep:
    (Original post by Alevelz)
    Does anyone remember the exact geometric question where you had to find N that's greater than infinity?


    37.27 > N

    So 38 is the smallest whole number N can take.
    Offline

    1
    ReputationRep:
    (Original post by iamDev)
    I was thinking you were right but the period isnt 2pi over a. 2pi over a is the period of a sin ax graph not a tan ax graph
    u ARE supposed to use the period of the sin or cos graph... NOT the tan graph.
    At the end of the day, the tan graph will give you many possible solutions, but the ones required by the question are the ones where the sin and cos are positive, so u work it out using the period of the sin and cosine in the quadrants where both sin and cos are positive
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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.