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    willl You lose marks for not keeping it in
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    (Original post by Signorina)
    Haha, I think I remember writing (x+2) was a factor... but a few people have said it was (x-2) so not sure:confused:
    come to think of it im fairly sure i put x=-2
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    (Original post by Josh2345)
    Like did u differentiate v?
    24-((r^2)/2) ??
    Well V = 24πr - πr^3/2

    So when you differentiated this, you got 24pi - (3πr^2)/2

    You then rearranged to get r=4 at the stationary point, then used the second differential which was -3πr, and then subbed r=4 in to find whether it was a max/min point.
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    ra) 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

    for drawing the cubic i did a positive cubic graph, points were 0,3 but i drawedd the line going through 0 and touching x axis?
    how many marks out of 3 wouldi get for this>?

    for biii i got b as -x i didnt expand brackets and got c as 10 how many marks would i lose for this, i think it was 2 marka???

    and for biv i used b^2-4ac sed it was less then 0 and sed only real root is x=-2 how many marks would i get forthis? i think this was 3 marks? but i got the value of discrimiant wrong? thanks!
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    I got h=(48pi - pir^2)/2pir
    I didnt simplify it to 24/r -r/2
    It was worth 3 marks, how many marks do you think I lost for doing that?
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    If you reexpand out the bracket for Q7, you only get the origional equation x^2(x-3)-20 if the factor is (x+2) not (x-2)
    Therefore the only root to the equation is x=-2
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    Someone mentioned I was wrong on here when I said the graph sketch was actually worth 2 not 3, I did double check that before paper end and I just had it confirmed by a teacher so it was worth 2 marks, not 3. I was also surprised but I did check.
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    (Original post by Heffalump .)
    I got h=(48pi - pir^2)/2pir
    I didnt simplify it to 24/r -r/2
    It was worth 3 marks, how many marks do you think I lost for doing that?
    Did anyone get 12pi and therefore its a minimum
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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. 2√5 [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.
    Yay I got all of those. Hoping for full marks or near enough... :crossedf:
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    (Original post by trad1998)
    Cylinder question I messed up the first 5 marks. But the second part of it was okay.


    Posted from TSR Mobile
    Same I think
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    (Original post by omarharoun3)
    Did anyone get 12pi and therefore its a minimum
    It was -12pi therefore maximum
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    (Original post by Heffalump .)
    I got h=(48pi - pir^2)/2pir
    I didnt simplify it to 24/r -r/2
    It was worth 3 marks, how many marks do you think I lost for doing that?
    I don't think you needed to simplify it, at least until the second part where it it was a show that V equals... so I can't imagine you'd lose any, and if you did, it'd be one at most.
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    How many marks was the Circle K question where k was 8 and -2?

    Also, do you know what the marks would be for
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    (Original post by B82)
    How many marks was the Circle K question where k was 8 and -2?

    Also, do you know what the marks would be for
    3
    1 mark: substituting k and 4 into the circle equation
    2 mark: expanding out correctly
    3 mark: factorising and obtaining the right values from this
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    (Original post by Jupers)
    Damnnn, for the chord i left it as root49, it was only 2 marks do so i might not even lose one i guess
    Still correct They didn't specify it had to be in it's simplest form or anything I suppose
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    (Original post by Smileyface97)
    Yay I got all of those. Hoping for full marks or near enough... :crossedf:
    i got these but i got 5root2? and im not sure about the first equation of the line?
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    http://www.thestudentroom.co.uk/show....php?t=3322885 can you guys answer the poll on the difficulty of the exam, it would be interesting to know what everyone thought of it
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    (Original post by Signorina)
    Haha, I think I remember writing (x+2) was a factor... but a few people have said it was (x-2) so not sure:confused:
    (x+2) was a factor, which meant that x=-2 is a root. Maybe people have got confused there?
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    (Original post by student0042)
    Yes. which with d^2V/dr^2 implies max.
    Oh thank goodness XD I went back to it at the end and got that in the last 3 minutes or something
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    ra) 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

    for drawing the cubic i did a positive cubic graph, points were 0,3 but i drawedd the line going through 0 and touching x axis?
    how many marks out of 3 wouldi get for this>?

    for biii i got b as -x i didnt expand brackets and got c as 10 how many marks would i lose for this, i think it was 2 marka???

    and for biv i used b^2-4ac sed it was less then 0 and sed only real root is x=-2 how many marks would i get forthis? i think this was 3 marks? but i got the value of discrimiant wrong? thanks!
 
 
 
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