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    (Original post by kas69)
    Easy paper if u did most/all the other past papers! Hard questions were the minimum length, cylinder question and finding the equation after vector 4,2...
    If u want to no any of the methods msg me...
    For 3a did you get gradient as -7 or -10
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    Just thought id correct my previous answer - you are correct that it is (x+2) when x=-2. I had to try and remember my answers...
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    What do you lots think the grade boundaries will be for an A?
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    (Original post by Heffalump .)
    x^2-x+4
    Oh yeah, I got that
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    (Original post by student0042)
    (y+3) = 5/3(x+2)
    3y + 9 = 5x + 10
    5x - 3y + 1 = 0

    Yes, working through that you are right, I will change my answer. Thanks.
    got a bit scared then! thank you so much, i think i am now certain of my mark (or at least a confidence interval within which it definitely lies). will make results day less of a shock.

    also, in the integration question, i must have added something up wrong because i got 107/5 for the first part, not 108/5. this means that the next part of the question (54-107/5= 163/5) was also wrong, as it was carried forward. how many marks would i lose in total? do i get penalised twice for that one mistake, or can i still the full 3 marks for the second part of calculating the area, even though it is wrong? thank you
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    (Original post by SunDun111)
    For 3a did you get gradient as -7 or -10
    -10 i think
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    (Original post by Heffalump .)
    x^2-x+4
    Got that!!!
    Lets hope it is correct

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    (Original post by Heffalump .)
    I put
    -5x + 3y -1 =0
    It's the same thing, but did the question specifically ask for it to be in the form ax - by + r = 0 ?
    That should also be correct. I actually wrote it in both formats to be on the safe side, though it didn't ask for it to be written in any specific order format so long as all of the parts are on one side of the equation.
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    (Original post by SunDun111)
    For 3a did you get gradient as -7 or -10
    I got -10 as the gradient

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    (Original post by ProNoob)
    I got -10 as the gradient

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    So did I but my teacher got -7? How what was the dy/dx equation again
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    (Original post by tanyapotter)
    got a bit scared then! thank you so much, i think i am now certain of my mark (or at least a confidence interval within which it definitely lies). will make results day less of a shock.

    also, in the integration question, i must have added something up wrong because i got 107/5 for the first part, not 108/5. this means that the next part of the question (54-107/5= 163/5) was also wrong, as it was carried forward. how many marks would i lose in total? do i get penalised twice for that one mistake, or can i still the full 3 marks for the second part of calculating the area, even though it is wrong? thank you
    I made a mistake in the first bit too, so maybe -1 or -2 marks there and in the second bit I think -1 mark as you didn't get the final correct answer. (I presume one for 54 is area of trapezium, 1 for follow through? and one for correct answer.)
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    Question 1

    (a) -3/5
    (b)(i)perp. gradient = 5/3
    in the form mx + ny + p = 0
    5x - 3y + 1 = 0

    Question 2

    7+root15
    n=7 (Had to be stated separately)

    Question 3

    a) y-6 = -10(x+1)
    y = -10x - 4

    bi) 108/5 or 21.6
    bii) 162/5 or 32.4

    Question 4

    a) (x+1)^2 + (y-3)^2 = 50
    bi) C is (-1, 3)
    bii) r = √50 = 5√2
    c) k= 8, -2
    d) min. distance = √49 = 7

    Question 5

    a) p = 3/2, q = -¼
    bi) (-3/2, -1/4)
    bii) x=-3/2
    c) y = x^2 - x + 4

    Question 6

    ai) Surface Area = 2πrh
    48π = 2πrh - 2πr^2
    h = 24/r - r/2

    aii) V = πr^2h
    V = πr^2 * (24/r - r/2)
    V = 24πr - (π/2 * r^3)

    bi) 24π - (3πr^2) / 2
    bii) 24π - (3πr^2) / 2 = 0 when r = ±4
    r = 4 as positive value
    d^2V/dr^2 = 3πr
    when r = 4, second derivative = -12π, therefore maximum

    Question 7

    a) draw x^2(x-3) where line touches at 0 and crosses at 3
    bi) Remainder = 36
    bii) R = 0 therefore is a factor
    biii) (x-2)(x^2-5x+10)
    biv) b^2 - 4ac < 0 therefore no real roots for (x^2-5x+10)
    (x-2) only root when x = 2

    Question 8

    a) show that = x^2 + 3(k-2)x -13-k=0 by setting two equations equal to each other
    b) b^2 - 4ac < 0 therefore 9k^2 -32k -16 < 0
    c) -4/9 < k <4
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    (Original post by SunDun111)
    So did I but my teacher got -7? How what was the dy/dx equation again
    I have forgotten sorry

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    (Original post by SunDun111)
    So did I but my teacher got -7? How what was the dy/dx equation again
    Remember my dy/dx being 4x^3 + 6x so substituting in -1 gives you -10. Your teacher is wrong judging by the fact everyone else got -10.
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    (Original post by Lobster37)
    my exact words were 'x=2 is only real root' would i get the mark for this. Also are there any oher statements you can think of that we should have written?
    Yeah I think you will get a mark for that. I think you get marks for saying dy/dx =0 when talking about a stationary point
    And d2y/dx2 is less than zero so it was a maximum point
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    (Original post by JackC3001)
    That should also be correct. I actually wrote it in both formats to be on the safe side, though it didn't ask for it to be written in any specific order format so long as all of the parts are on one side of the equation.
    I always assume that when it wants it in the form ax + ... = 0, that a should be positive. Someone did say earlier that in the mark scheme it says condone -ax + ... =0.
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    (Original post by SunDun111)
    For 3a did you get gradient as -7 or -10
    -10 i think.
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    (Original post by student0042)
    I always assume that when it wants it in the form ax + ... = 0, that a should be positive. Someone did say earlier that in the mark scheme it says condone -ax + ... =0.
    It definitely didn't ask for a certain format other than grouping the terms, so judging by other mark schemes, they will allow either.
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    (Original post by SunDun111)
    So did I but my teacher got -7? How what was the dy/dx equation again
    You got -10 by using the dy/dx method.

    If you used the method of y-y/x-x to get the gradient, you got -7 or something. I don't think you were supposed to se that method though.
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    (Original post by aoxa)
    You got -10 by using the dy/dx method.

    If you used the method of y-y/x-x to get the gradient, you got -7 or something. I don't think you were supposed to se that method though.
    What was the original equation of the curve can you remember?
 
 
 
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