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    (Original post by student0042)
    1a. -3/5 [2]
    1b. 4x-3y+1=0 (I thought it was this form) [3]
    1c. A(9,-4) [3?]
    2. 7 + √15 [5]
    3a. y-6=-10(x+1) [4]
    3bi. 108/5 [5]
    3bii. 162/5 [3]
    4a. (x+1)^2 + (y-3)^2 = 50 [2?]
    4bi. C(-1,3) [1]
    4bii. 5√2 [2]
    4c. k=-8, 2 [2]
    4d. minimum distance =7 [2]
    5a. p = 3/2, q = -¼ [3]
    5bi. (-3/2, -¼) [2]
    5bii. x=-3/2 [1]
    5c. y = x^2 - x + 4 [3?]
    6ai. h = 24/r -r/2 [3]
    6aii. show V = 24πr -π/2 r^3 [3]
    6bi. 24π -3π/2 r^2 [2]
    6bii. =-12π therefore max (r=+4) [4]
    7a. draw x^2(x-2) [3]
    7bi. R = 36 [2]
    7bii. R= 0 therefore root [2]
    7biii.(x-2)(x^2-5x+10) [2]
    7biv. x=2 [3]
    8a. Show that x^2 + 3(k-2)x -13-k=0 [1]
    8b. 9k^2 -32k -16 < 0 [3]
    8c. -4/9 < k <4 [4]

    Edit: added marks, not sure about a few of them.
    if this is right then ive probably only lost around 3-4 marks
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    (Original post by SaadMuhammad)
    I think my paper did not go as well as I was expecting. I have definitely lost 2 marks on the question where you had to work out the shortest distance between the centre and QR. By the looks of other answers on TSR, I think I have lost the last question, which is out of 4 I think. I believe I may have made a mistake on the integration question - though I am unsure about this; hopefully full marks on this question. So in total, I definitely lost 6 and maybe another 4-5 just to be on the safe side.

    In the end, a 60/75 is still 80% - an A I hope! I will work harder for C2 and Decision and should get higher marks to compensate for this paper.

    Good luck for your other exams!.
    You too 😄
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    (Original post by FS492)
    Its not completing the square though? [+] value is taken away from x and +[y] is added to integer?
    You have to complete the square to find the new equation. Completing the square gives you the original vertex and then you can slot in the numbers for the translation on the vertex and multiply out.

    You can't simply add 4 to the y-intercept of 2 because the x-coordinate moves too and if you imagine the quadratic on a graph and moving it along -2 in the x-direction , this will change where it crosses the y-axis. And then you apply the translation in the y-direction and it changes again. So the y-intercept is the new position when translated by x + the translation on y
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    (Original post by Lobster37)
    -2***
    Yes, X=-2
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    (Original post by AbbiC)
    Can someone do an unofficial mark scheme for this?
    http://www.thestudentroom.co.uk/show....php?t=3322639
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    Yyaaaaaay i got that too!!!
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    (Original post by RhyaCeri)
    if this is right then ive probably only lost around 3-4 marks
    in the first one, shouldn't it be 5x, not 4x? or can i not remember properly?
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    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...
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    I got x^2 + 3x + 6 when I applied the vector (minus 2 and add 4)

    (x+1.5)^2 - 0.25

    (x+ (1.5-2))^2 - (0.25+4)

    when you multiply out its x^2 + 3x + 6
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    (Original post by student0042)
    1a. -3/5 [2]
    1b. 4x-3y+1=0 (I thought it was this form) [3]
    1c. A(9,-4) [3?]
    2. 7 + √15 [5]
    3a. y-6=-10(x+1) [4]
    3bi. 108/5 [5]
    3bii. 162/5 [3]
    4a. (x+1)^2 + (y-3)^2 = 50 [2?]
    4bi. C(-1,3) [1]
    4bii. 5√2 [2]
    4c. k=-8, 2 [2]
    4d. minimum distance =7 [2]
    5a. p = 3/2, q = -¼ [3]
    5bi. (-3/2, -¼) [2]
    5bii. x=-3/2 [1]
    5c. y = x^2 - x + 4 [3?]
    6ai. h = 24/r -r/2 [3]
    6aii. show V = 24πr -π/2 r^3 [3]
    6bi. 24π -3π/2 r^2 [2]
    6bii. =-12π therefore max (r=+4) [4]
    7a. draw x^2(x-2) [3]
    7bi. R = 36 [2]
    7bii. R= 0 therefore root [2]
    7biii.(x-2)(x^2-5x+10) [2]
    7biv. x=2 [3]
    8a. Show that x^2 + 3(k-2)x -13-k=0 [1]
    8b. 9k^2 -32k -16 < 0 [3]
    8c. -4/9 < k <4 [4]

    Edit: added marks, not sure about a few of them.
    Was gradient -10 how it was -7?
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    (Original post by RhyaCeri)
    if this is right then ive probably only lost around 3-4 marks
    Even though I've put 162/5 for 6, in the exam I put the wrong answer so I think I've lost around 3-4 marks too. This should still be 95+ UMS however. (hopefully.)
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    (Original post by tanyapotter)
    in the first one, shouldn't it be 5x, not 4x? or can i not remember properly?
    its 5x? i think, im not sure, pretty sure im gonna have nightmares about that cylinder question tonight though
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    (Original post by student0042)
    Even though I've put 162/5 for 6, in the exam I put the wrong answer so I think I've lost around 3-4 marks too. This should still be 95+ UMS however. (hopefully.)
    i sat the exam last year aswell, so this paper was so much better in comparison
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    (Original post by emzchalker98)
    I got that!
    Btw that is wrong lol
    It is 32.4 or 162/5

    Posted from TSR Mobile
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    would I get the marks if i wrote 1.5 instead of 3/2 when it said rational number?
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    (Original post by tanyapotter)
    in the first one, shouldn't it be 5x, not 4x? or can i not remember properly?
    (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.
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    (Original post by student0042)
    I'm not certain that i'm right, just the answers I got.
    Got the same answers. Your co-ordinates of A match mine and 32.4 is definitely right. Many others seemingly got that too.
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    damn i know.i got every question right except the radius of cylinder =4 i put 4pi fml
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    For the translation I think I got x^2+3x+4
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    (Original post by Urvs)
    would I get the marks if i wrote 1.5 instead of 3/2 when it said rational number?
    You should get marks as they are the same value.
 
 
 
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