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    (Original post by Badman1231)
    Its the same thing Stop worrying!
    haha it is kinda worrying

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    (Original post by Rubyturner94)
    for rearranging tk make r the subject, in the denomenator everyone put pi(a+b) i put pia+pib

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    Yeah, like everyone else is saying, it's the same thing - all you've done is expand the brackets. They only prefer pi(a+b) because it's a bit neater and simpler I guess, but you're fine.
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    (Original post by Deceived)
    (r^2)(a+b) = 3V/pi

    (r^2) = 3V/pi/(a+b)

    ^ This is the wrong step.

    r2(a+b)=3V/
    π

    r2=(3V/π)/(a+b)

    r2=(3V/π)*(1/(a+b))

    r2=3V/(π*(a+b))

    I hope this makes sense. Basically, dividing by something is the same as multiplying by the reciprocal.

    yeah thanks, should get a mark. silly mistake. dont think i messed up on the rest
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    one last time guys hahah
    is this ok?
    i did it a long way kind of


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    (Original post by Rubyturner94)
    one last time guys hahah
    is this ok?
    i did it a long way kind of


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    Yeah that's correct.

    I'm so mad at myself for putting ± before the square root. Next time I'm going to read the damn question. Hopefully, that was my only silly mistake.
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    (Original post by Badman1231)
    Are well done mate. I did this method but instead of putting (2+k)(2-k) I put (2-k)(2-k) which got my final awnser to
    K^2-12k
    K(K-12) therfore K=0 or 12
    I think I meant to write (2+k)(2+k), I went back to correct what I wrote when I typed it up but forgot that second bracket, oops
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    Yes! haha
    Forget about C1 now lets focus on C2!
    (Original post by Rubyturner94)
    one last time guys hahah
    is this ok?
    i did it a long way kind of


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    (Original post by Deceived)
    Damn. That actually works. Spent ~5 minutes trying to do that but for some reason I couldn't get it to work so I used the differentiation method. Way to go for using that method. I assume that's the method OCR want their candidates to use as chain rule isn't on the C1 syllabus. Do you think I'll still receive full marks for this question even though I used differentiation? I got the same answer (k=0 or 4).

    Thanks
    I confused myself trying to work it out again!

    I think it should be fully credited, as you ended up with the right answer - I mean the Jan11 question was a show that question so it should have been marked differently
    It's a bit like for the question where you had to find the y coordinate for the minimum point(?) in Section A, I read -7 off the completing the square intercept, then checked it by differentiating the original expression to find the y value from the x value when dy/dx =0 if that makes sense
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    (Original post by LRJ)
    I confused myself trying to work it out again!

    I think it should be fully credited, as you ended up with the right answer - I mean the Jan11 question was a show that question so it should have been marked differently
    It's a bit like for the question where you had to find the y coordinate for the minimum point(?) in Section A, I read -7 off the completing the square intercept, then checked it by differentiating the original expression to find the y value from the x value when dy/dx =0 if that makes sense

    Thanks. I pray it does end up getting full credit. Yes, I did the exact same thing for the minimum question. Differentiation is so useful haha. Best of luck with C2
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    (Original post by KanKan)
    Omg, I actually got all this, untill this point
    then somehow ended up with x = +-(-1)^1/2
    .. which can't be right because you cant root a negative (in real terms)

    How many marks do you reckon I picked up then?
    I'm not sure, how many marks was it out of?

    I'm guessing here, but there should have been a mark for eliminating y, finding the quadratic expression involving k's, a mark for showing that b^2 -4ac = 0 for a tangent, a mark for the right answer, and perhaps another method mark somewhere in there?
    -and for the differentiation method I'm not too sure xD
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    Hi can someone confirm the workings for question 2.) section A i seem to have got Integers and I'm not sure how :/
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    (Original post by vincentky)
    Hi can someone confirm the workings for question 2.) section A i seem to have got Integers and I'm not sure how :/
    same!
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    I think I need to stop reading this thread and calculating how many marks I lost. I've lost about 8-10 marks on stupid human errors and probably more. Hopefully I'll make up for the lost marks in C2.

    Can anyone comment on how hard this paper was to previous past papers? I hope the grade boundaries for an A is around 57.
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    (Original post by vincentky)
    Hi can someone confirm the workings for question 2.) section A i seem to have got Integers and I'm not sure how :/
    What was question 2 again?
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    (Original post by Deceived)
    What was question 2 again?
    Something to do with the intersection between two lines?
    I think I got one of them to be... y=-1/3x+1/3 when I rearranged for Y. then equated it to the other lines y=3x+2 and attempted to solve..
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    (Original post by vincentky)
    Something to do with the intersection between two lines?
    I think I got one of them to be... y=-1/3x+1/3 when I rearranged for Y. then equated it to the other lines y=3x+2 and attempted to solve..
    Oh yeah, that's right.

    One equation was x+3y=1 and the other was y=3x-2 I think

    3y=-x+1
    y=-(1/3)x+(1/3)

    3x-2=-(1/3)x+(1/3)
    (10/3)x-(7/3)=0
    (10/3)x=(7/3)
    10x=7
    x=7/10

    y=3x-2
    y=(3*(7/10))-2
    y=(21/10)-2
    y=1/10

    Point of intersection = (7/10,1/10)

    I'm fairly sure that was the answer I got.

    I hope this helps.
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    (Original post by Deceived)
    Oh yeah, that's right.

    One equation was x+3y=1 and the other was y=3x-2 I think

    3y=-x+1
    y=-(1/3)x+(1/3)

    3x-2=-(1/3)x+(1/3)
    (10/3)x-(7/3)=0
    (10/3)x=(7/3)
    10x=7
    x=7/10

    y=3x-2
    y=(3*(7/10))-2
    y=(21/10)-2
    y=1/10

    Point of intersection = (7/10,1/10)

    I'm fairly sure that was the answer I got.

    I hope this helps.
    I did everything correct up until the part 10/3x =7/3 i couldn't remember the equations of the lines exact. but i didn't multiply the numerators by 3 i just multiplied 7/3 by the inverse of 10/3 and got 21/30 but I did not simplify and used the x val to find y. do you think I would still get credibility?
    and thanks btw
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    (Original post by krishparscrotam)
    My answers were as follows:
    8. 3(x-2)^2 -7 and minimum y value was -7
    Hi, did they ask for the minimum y value, or the turning point. I put (2,-7). Am i wrong?
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    (Original post by vincentky)
    I did everything correct up until the part 10/3x =7/3 i couldn't remember the equations of the lines exact. but i didn't multiply the numerators by 3 i just multiplied 7/3 by the inverse of 10/3 and got 21/30 but I did not simplify and used the x val to find y. do you think I would still get credibility?
    You'd definitely get a mark for substituting out y. You may receive an additional mark but I guess it really depends on how many marks it was out of. I assume 1 mark is rewarded for finding the y coordinate at the end. So if it was a 3 mark question, you're probably looking at receiving 2 marks because 21/30=7/10 so you shouldn't lose an A1 mark. So 2/3 marks probably.

    Hope the exam went well for you and best of luck with C2.
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    (Original post by Studenthsr)
    Hi, did they ask for the minimum y value, or the turning point. I put (2,-7). Am i wrong?
    They asked for the minimum y value. I assume you'll receive full credit because you put (2,-7).
 
 
 
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