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    (Original post by KevinGu123456789)
    For the question about the circle being wholly inside the other, did anyone else get 0<r<√45 - 5?
    Most people got 0<r<2, but I thought that this may not be right
    I got this, as did most other people at my school who managed to answer it.
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    (Original post by KevinGu123456789)
    Does anyone remember the actual last question though?
    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!)
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    I got y=2x-20 and y=2x+10 for tangents of circles
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    Yep. Deffo just got nearly every question wrong.
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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
    Yep, myself and at least 4 others got the same answer for the second question
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    How did people get 27? I got 27/12 fek
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    (Original post by Dmitry_Mednov)
    It is wrong as if you use k as -5 or -6 it will not intersect the curve, I checked it at the end.
    He's right, you're wrong. b^2-4ac when K<-7.5 was a number less than zero. Above -7.5 it gave a positive number, so when K<-7.5 ether are no real roots to the equation and the lines do not cross, you probably subbed it in wrong.

    That paper wasn't hard.
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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
    Yeah that's right
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    Can someone explain how they did the last question?
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    (Original post by duncant)
    I got y=2x-20 and y=2x+10 for tangents of circles
    same man
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    (Original post by KevinGu123456789)
    For the question about the circle being wholly inside the other, did anyone else get 0<r<√45 - 5?
    Most people got 0<r<2, but I thought that this may not be right
    I got that! I said the radius was 3√5 whereas lots thought it was 2√5.

    Thing is on that question the distance from Centre to 0,0 was 5 therefore the radius couldn't have been 2√5 so I quickly changed that!

    I got 0<r<3√5-5 (same as you).
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    (Original post by nobodycarescarla)
    How did people get 27? I got 27/12 fek
    8(27/12)^3/2 simplifies to 27
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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
    I got that as well😊
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    With the indice questions, was it 2^-6 for part (i) and 2^13/3 for (ii)?
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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!)
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    i think the grade boundaries will be in the 50s for an A - i found january 2011 a lot easier and it was 54 for an A
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    A=27 for last question. Y=1/16 and y=8 for hidden quadratic. K less than -2 but great than 5 for question 9. Y=(x-2)^2 (5-x) for translation.
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    (Original post by JackA123)
    With the indice questions, was it 2^-6 for part (i) and 2^13/3 for (ii)?
    I got 2^-6 but was worrying cos all I remember from the exam is 2^5 divided by 2^7 which gives ¼ so there must've been something else I can't remember.

    Mathematically speaking should be a 2^-4...
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    question 9 was x^2 + 2x + 11= k(2x-1), find k

    and 11 was y= 4x^2 + a/x + 5, find a, also when dy/dx=o, y=32
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    (Original post by duncant)
    A=27 for last question. Y=1/16 and y=8 for hidden quadratic. K less than -2 but great than 5 for question 9. Y=(x-2)^2 (5-x) for translation.
    (x-2)^2 (5-x)?

    I put x^2(5-x) If it was (x-2)^2 it would've also resulted in a stretch.
 
 
 
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