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

    0
    ReputationRep:
    (Original post by Buymoria)
    Just did the June 2007 FP1 paper (grade boundary for an A was 54) and was far far easier
    Lets just hope we get really low grade boundaries XD
    Offline

    2
    ReputationRep:
    how did people find k in the question which was something like x^2 + 2x +11 = k(2x-1)?
    Offline

    2
    ReputationRep:
    What was the question number 10? about the second tangent?
    Offline

    0
    ReputationRep:
    Question 1?! Please someone help
    Offline

    0
    ReputationRep:
    How many marks was question 8 worth? It was the complete the square question.
    Offline

    1
    ReputationRep:
    By getting an A in this module, do people mean above 80 ums?
    Offline

    7
    ReputationRep:
    (Original post by student199919)
    how did people find k in the question which was something like x^2 + 2x +11 = k(2x-1)?
    Expand the brackets on the right to get x^2 +2x +11= 2kx-k

    Move to one side to get = 0, so x^2 +(2- 2k)x+11+k= 0

    Sub the values into b^2-4ac > 0
    a=1, b=2-2k, c=11+k

    Then solve to get k^2-3k-10 = 0, factorise to get (k-5)(k+2)>0,

    Draw a quadratic graph that intercepts x axis at k=5 + k=-2, so k<-2 and k>5

    Hope this helps
    Offline

    0
    ReputationRep:
    (Original post by DanMcGrathKB)
    Question 1?! Please someone help
    I think it was 2x^2-9
    Offline

    0
    ReputationRep:
    Anyone got kik? NEED TO DISCUSS ANSWERS
    Offline

    0
    ReputationRep:
    hoping c2 is better 🙌🏽
    Offline

    2
    ReputationRep:
    (Original post by liziepie)
    Expand the brackets on the right to get x^2 +2x +11= 2kx-k

    Move to one side to get = 0, so x^2 +(2- 2k)x+11+k= 0

    Sub the values into b^2-4ac > 0
    a=1, b=2-2k, c=11+k

    Then solve to get k^2-3k-10 = 0, factorise to get (k-5)(k+2)>0,

    Draw a quadratic graph that intercepts x axis at k=5 + k=-2, so k<-2 and k>5

    Hope this helps

    Thank you! i followed all those steps but i got 2 surds as my answers rather than -2 and 5
    Offline

    17
    ReputationRep:
    (Original post by student199919)
    how did people find k in the question which was something like x^2 + 2x +11 = k(2x-1)?
    That was one of the questions I didn't think was too bad but that I most likely got wrong anyway. Basically put everything on one side so that would be:
    X^2 +2X -2kx +11 +K=0 (think I went wrong here put +2KX and -K aka simple adding and subtraction error :/)
    You then you the discriminant and I believe the question specified that there would be 2 real roots so b^2-4ac>0.
    Substitute in the values of a, b and c:
    (2+2K)^2 -4x1x(11+K)>0
    4K^2 +8K +4 -44-4K>0
    4K^2+4K-44>0
    Now what I did here was divide by 4 which should be correct.
    K^2 +K -11>0
    So as this is a positive K squared graph, I said there would be 2 separate inequalities as the graph would dip below the x-axis therefore 2 distinct areas where K is greater than 0.
    I also used completing the square to solve this which may not have been a good idea...anyway I got:
    (k+0.5)^2 -0.25-11>0
    K>-0.5+11.25^0.5
    K<-0.5-11.25^0.5
    (^0.5 slightly award to write compared to square root but I don't know how to express that in the forum's text).
    So yeah don't think this is right and I'm pretty sure I made a stupid error in putting 10 instead of 11 or something so I would have lost the one of the method marks I may have got for that last step....
    Offline

    7
    ReputationRep:
    Thought was harder than usual but alright overall, managed to get the last question 👍
    Offline

    2
    ReputationRep:
    what are the predictions for grade boundaries?
    Offline

    10
    ReputationRep:
    How many marks do you think I will get for the last question I got 18root27/12 which is equivalent to 27 but will I lose marks for not having it as a whole number? Thanks
    Offline

    17
    ReputationRep:
    One the last question I got as far as a=8x^3 but after that I am sure my method was the incorrect way of getting to the answer, hence my answer was definitely wrong, being a=216 XD.
    Offline

    2
    ReputationRep:
    (Original post by black1blade)
    That was one of the questions I didn't think was too bad but that I most likely got wrong anyway. Basically put everything on one side so that would be:
    X^2 +2X -2kx +11 +K=0 (think I went wrong here put +2KX and -K aka simple adding and subtraction error :/)
    You then you the discriminant and I believe the question specified that there would be 2 real roots so b^2-4ac>0.
    Substitute in the values of a, b and c:
    (2+2K)^2 -4x1x(11+K)>0
    4K^2 +8K +4 -44-4K>0
    4K^2+4K-44>0
    Now what I did here was divide by 4 which should be correct.
    K^2 +K -11>0
    So as this is a positive K squared graph, I said there would be 2 separate inequalities as the graph would dip below the x-axis therefore 2 distinct areas where K is greater than 0.
    I also used completing the square to solve this which may not have been a good idea...anyway I got:
    (k+0.5)^2 -0.25-11>0
    K>-0.5+11.25^0.5
    K<-0.5-11.25^0.5
    (^0.5 slightly award to write compared to square root but I don't know how to express that in the forum's text).
    So yeah don't think this is right and I'm pretty sure I made a stupid error in putting 10 instead of 11 or something so I would have lost the one of the method marks I may have got for that last step....
    Thank you. I was going on the right lines but I simplified -8k - -4k as -12k rather than -4k which is annoying
    Offline

    2
    ReputationRep:
    (Original post by Bobby21231)
    How many marks do you think I will get for the last question I got 18root27/12 which is equivalent to 27 but will I lose marks for not having it as a whole number? Thanks
    did it ask to give a as an integer?
    Offline

    0
    ReputationRep:
    Any mark-schemes yet?
    Offline

    1
    ReputationRep:
    No the question told us that "a is a positive constant"
 
 
 
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.