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    (Original post by julesaquilina)
    I got k is less than -2 and greater than 5? And many others got that as well
    cant be as y=x+5 would intercept the curve
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    0<r<√45 - 5 i got this too- anyone else?
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    (Original post by Gregis)
    The line equation was y=2x^2, then the x-coordinate of A=5 and the x-coordinate of B=5+h and you had to prove the gradient was 20+2h.

    I think you had to find the midpoint of the x-coordinates and then put that into the dy/dx of the equation.
    i found y values and did y2-y1 thing, luckily it was a proof so check your answer haha
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    (Original post by AlfieH)
    Wasn't it something like y=4x^2+a/x+5 and y has a stationary point at which the y coordinate = 32.

    I differentiated to find 8x-a/x^2 and then equated to 0. Then said that 8x^3-a=0 therefore 8x^3=a.

    Plug that back into the original equation to get 32=4x^2+8x^2+5 (32 is the y coordinate we were given).

    Solving that to get x= 0 and + or - 3/2. Then plugging that back in to the original equation gave a=0, not possible or a negative solution when it asked for the positive solution so solving that eventually gave 27.

    I hope that makes sense? Think it's correct (I BLOODY HOPE SO!)
    Yes that's what i got
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    Did anyone else find the exam hard?
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    What will the grade boundaries be guys? Jan 2011 was 54 for an A and that was much easier than this paper. I got 67 on 2011 paper. This paper was a wreck.
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    (Original post by Ozil5)
    Can someone make an online Google document thing so we can all input our answers to make the mock MS
    https://docs.google.com/document/d/1...it?usp=sharing
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    (Original post by AlfieH)
    Can anybody remember the indices questions? I remember the first one was (2^5 divided by 2^7) but there was something else in that question.

    Also - what about the 2nd question, I got -¼+or-¾root5

    It was 5 x 4^2/3 + 3 or something x 16^1/3
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    (Original post by Gregis)
    The line equation was y=2x^2, then the x-coordinate of A=5 and the x-coordinate of B=5+h and you had to prove the gradient was 20+2h.

    I think you had to find the midpoint of the x-coordinates and then put that into the dy/dx of the equation.
    Did you do that?! Exactly what I did! What about the next question, the 1 marker where it said talk about the gradient or something?
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    guys are we talking OCR or OCR MEI - seems to be a mix?
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    (Original post by AlfieH)
    I think I remember now wasn't it prove the gradient was 4h+10 ?

    I did it different by finding the midpoint AB, then plugging that into the dy/dx of 2x^2 which then gave me 4x therefore 4(h+2.5h) which still gave me 4h+10..

    Reckon I'll still get the 3 marks?
    I definitely remember there being a 20 in there, as I messed up expanding the brackets at first :P I did it like this:

    If y=2x^2, then y for A would be 2x5^2, 2x25 = 50

    y for B = 2(h+5)^2 = 2(h+5)(h+5) = 2(h^2+10h+25) = 2h^2+20h+50

    so A = (5,50) + B = (5+h, 2h^2+20h+50)

    y1-y2/x1-x2 = 50-(2h^2+20h+50) / 5 -(5+h)

    = -2h^2-20h / -h

    = 20+2h (I think :P)

    hey man, if you did it your way and it was right, then you should get the marks.
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    Here our my answers for the questions I remember for people to compare to. This was the hardest paper I have ever done.
    Surds question -1/4 +3/4root5
    Circles tangents first one 2x-20 and second one 2x+10 and radius r<2 (but I'm sure this is wrong)
    Substitution quadratic 1/16 and 81
    Last question 18root27/12 this is equivalent to 27 hopefully I will still get full marks
    Indices 2^-6 and 2^13/3
    Discriminant k< something or k>5
    Transformation (x-2)^2(3-x) and stretch parallel to the y axis by scale factor 1/2 (don't think these are right)
    Complete the square -2(x-3) +22 (3,22)
    Coefficient of x^3 is -10
    Y coordinate of normal 50.25
    I'll add more when I remmber them
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    (Original post by Btec certified)
    What did u get liark? Get coordinates of (10,0) for A aye?
    I get it now, yeah. Centre C was midpoint of the two points so use that to find the other point and sub in to the equation of a line with a gradient 2.

    Think I'll get a mark for using gradient 2 but that's all? Or also one for using equation of the line? It was 3 marks overall.
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    (Original post by nobodycarescarla)
    0<r<√45 - 5 i got this too- anyone else?
    Same!! Only I wrote it as 0<r<(3root5-5)
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    could someone please go through the answer to question 9 and how they achieved it? that question destroyed me but i think i may get some method marks.
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    Did anyone else get a=27?! Oh my god that seemed so much harder than any of the other past papers. Good luck everyone!!
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    Does pencil not show up on the test by any chance? I switched between pencil and pen constantly during the test.
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    (Original post by AlfieH)
    Did you do that?! Exactly what I did! What about the next question, the 1 marker where it said talk about the gradient or something?
    I said that at the point (could be B? The one with the h) i said the one with the h would have a steeper gradient than at point A but i wasnt sure
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    Well that didn't work.
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    Has anyone made an unofficial mark scheme yet? I know it's a bit soon 😅
 
 
 
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