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

OCR (Not MEI) Core 2 Discussion Thread 17th May 2013 + Jan 13 Paper and MS watch

    Offline

    0
    ReputationRep:
    I got 37/30, though I don't know anyone else who did
    Offline

    16
    ReputationRep:
    This exam went amazingly well for me after a pretty shoddy c1 on Monday! I wrote down my answers on a scrap bit of paper so il post them up in a while.


    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    (Original post by Ellie_Rosa)
    I got 37/30, though I don't know anyone else who did
    that's correct.
    Offline

    16
    ReputationRep:
    C2.

    1)6.39

    2)i)106.3,253.7
    ii) 71.6, 251.6

    3) 64 + 960x +6000x^2
    ii) c=-11

    4) 4/5x^4 -3x^2 + x + c
    ii) -12x^-2 +c
    iii)a=2

    5i)62.2
    ii)34.0

    6)i)963
    ii) n=<17.2 therefore greatest n=17

    7i) shown
    ii)37/30

    8)y=1 , y=4
    ii) interger then fraction
    iii) shown

    9)i) 15
    ii)x=-1/2, x=-1, x=3/2
    iii) x= 2/3pi, pi, 4/3pi

    i THINK that's all correct
    Offline

    0
    ReputationRep:
    Logs 'show that' (question 8), I did: 7 'show that' question:
    y=y
    a^x=4b^x
    log2(a^x)=log2(4)+log2(b^x)
    x(log2(a)-log2(b)=2
    ab=2 therefore b=2/a, sub this in to get
    x(log2(a)-log2(2/a))=2
    x(log2(a)-(log2(2)-log2(a))=2
    x(log2(a)-1+log2(a))=2
    x(2log2(a)-a)=2
    Therefore x= 2/(2log2(a)-1)
    Hope this helps! (:
    Offline

    0
    ReputationRep:
    (Original post by Goods)
    C2.

    1)6.39

    2)i)106.3,253.7
    ii) 71.6, 251.6

    3) 64 + 960x +6000x^2
    ii) c=-11

    4) 4/5x^4 -3x^2 + x + c
    ii) -12x^-2 +c
    iii)a=2

    5i)62.2
    ii)34.0

    6)i)963
    ii) n=<17.2 therefore greatest n=17

    7i) shown
    ii)37/30

    8)y=1 , y=4
    ii) interger then fraction
    iii) shown

    9)i) 15
    ii)x=-1/2, x=-1, x=3/2
    iii) x= 2/3pi, pi, 4/3pi

    i THINK that's all correct
    Hi, I got all of them, but on 9ii, you were only meant to factorise completley, not find the solutions too (: Can you remember what question 5 is??
    Offline

    0
    ReputationRep:
    (Original post by _JC95)
    I tried to do the tangent area under the graph question 2 ways...

    First I got 37/30 by subtracting the area of the right angled triangle.

    Then, to check, i tried integrating. (x^3/2 - 1) - (3x-5) and got a completely different answer of 1.9

    Surely both methods are valid and should both produce the exact same answer??

    In the end I stuck with the original 37/30, the rest of the paper was pretty fair.


    Posted from TSR Mobile
    In theory both methods are valid, but when you integrate (x^3/2 - 1) - (3x-5) using the limits x=4 and x=1, the area between the tangent (y=3x-5) and the x-axis includes a little chunk below the x-axis within these limits.
    That's why you get a different answer to integrating separately and using the limits x=4 and x=1 for (x^3/2 - 1), and x=4 and x=5/3 for (3x-5), then subtracting one from the other.
    Offline

    16
    ReputationRep:
    (Original post by Hannburger)
    Hi, I got all of them, but on 9ii, you were only meant to factorise completley, not find the solutions too (: Can you remember what question 5 is??
    Ehh had to factories to solve so ill get them

    It was a arc of 0.8 rad radius 16 minus a triangle with a side as the radius and a side of 7. Find area then perimeter of arc minus the triangle


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by stefanconstant)
    Well 37/30
    is the right answer so you're fine..
    Did you integrate it right?
    I believe so:s have a go?


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by Goods)
    Ehh had to factories to solve so ill get them

    It was a arc of 0.8 rad radius 16 minus a triangle with a side as the radius and a side of 7. Find area then perimeter of arc minus the triangle


    Posted from TSR Mobile
    Ah yeah, thanks, got those answers too yey! (:
    Offline

    0
    ReputationRep:
    (Original post by _JC95)
    I believe so:s have a go?


    Posted from TSR Mobile
    Integrated is 2/5x^5/2 -3/2x^2 +4x?
    Offline

    0
    ReputationRep:
    Did anyone else find the equation of the tangent and theb integrate between the two points and take that from the area under the curve?? Everyone seems to have used a triangle

    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    wat was question 2 van someone tell me... please thanks
    Offline

    0
    ReputationRep:
    (Original post by stefanconstant)
    Integrated is 2/5x^5/2 -3/2x^2 +4x?
    yeah...


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    yeah I got 1.9 too lol...
    I put 37/30
    they should both be the same unless we're missing something
    Offline

    0
    ReputationRep:
    What do people think roughly grade boundary will be for an A?
    Also for the area under curve question would I get some marks for workings as i worked out equation of tangent and area of small triangle but got complete wrong answer?
    thanks
    Offline

    16
    ReputationRep:
    57? (80% of 72)
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    Answers to Summer paper are here:

    http://www.thestudentroom.co.uk/show....php?t=2354368

    I'll be answering questions in a while.
    Offline

    0
    ReputationRep:
    What were people's working for 5ii) the question on finding the perimeter?
    Offline

    0
    ReputationRep:
    On the last part of the last question, I made a silly mistake. I got 2 of the 3 required answered correct, but then ended up with another 2 obviously wrong answers (I don't know what the hell I was thinking, but I put it down to rushing the last question, as I had skipped question 5 and wanted to get back to it). How many marks (out of 4) could I expect to drop for this mistake? I was thinking 1 or 2, but I don't know. I'm kicking myself, because it was probably one of the easiest questions on the paper.
    Another thing, some people have been telling me that the UMS at AS level isn't particularly significant, which becomes more important at A2. For instance, to get an A* overall, you need an average of 90UMS at A2, and an average of 80UMS or more across AS and A2. This would mean that if you got an avrg of 90UMS at A2, 70UMS at AS would suffice to get the A* (90+70=160/2=80). What are your thoughts on this?
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

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.