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    I found it good thanks. The cylinder question involved you knowing the area of a circle equation and the circumference equation in terms of pi.
    It was an optimisation question.

    BTW the graph bounced off 0 under the x axis and then went down and then up to 3
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    (Original post by Moe21)
    anyone know how we were supposed to sketch the graph?
    I sketched a cubic with points at x-intercepts at -3, 0, 3
    I think it was meant to be a double root at 0 and passes through x at 3... I think


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    I think it will be around 58(+ or-)1
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    (Original post by Flangastic)
    The formula for the area of a cylinder was in the formulae booklet...guessing no one thought of looking in there, during a core 1 paper 😂


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    I could only see area of a cone not a cylinder. Luckily I remembered what it was anyway aha:')
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    (Original post by DexterMabel123)
    i got 21.6 for the integration question anyone get this?
    I got that!
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    (Original post by Lizzie33)
    I think it was meant to be a double root at 0 and passes through x at 3... I think


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    yeah, what i did was the graph starts at the bottom left hand side quadrant, touches the origin, goes back down, has a minimum at the bottom right quadrant, then goes back up crossing the x axis at +3
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    For the fourth question (with the equations of circles) what values did everyone get for k????
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    I found it ridiculously easy
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    In the circle eq it was =d so diameter
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    (Original post by Heffalump .)
    There was a double '0' root. so it started under the x axis and to the left of the y axis, the went up to the origin, touched it, went back down, then went back up to '3' on the a axis and passed above to x axis to the right of the y axis.
    would i lose all marks if i sketched 3 roots?
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    Pretty sure i got full marks. Wasn't that bad. If you didnt know the area of a cylinder equation, you could use the fact that the area of a circle is 2Pi * r and then multiply it by the height h.
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    (Original post by Anam)
    i don't understand what you mean! i did it so that the curve touched the 3 on x axis and went up and down towards 0
    As it's a positive cubic equation it starts bottom left and ends up top right, it has a single root at 3, so it goes through, and a double root at 0, so it touches
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    (Original post by Anam)
    i don't understand what you mean! i did it so that the curve touched the 3 on x axis and went up and down towards 0
    I think that's wrong I'm afraid, there was definitely a double root at 0, not 3
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    (Original post by tanyapotter)
    yeah, what i did was the graph starts at the bottom left hand side quadrant, touches the origin, goes back down, has a minimum at the bottom right quadrant, then goes back up crossing the x axis at +3
    yeah this is right
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    (Original post by Moe21)
    would i lose all marks if i sketched 3 roots?
    What do you mean by 3 roots? it depends on where they were, it was 3 marks, but if you plotted on of the roots correctly (the root 3) then you would have only lost two I think...
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    I thought it was very good actually. Did anyone get 7 for re shortest distance between C and the chord RQ or something??
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    (Original post by tadit)
    As it's a positive cubic equation it starts bottom left and ends up top right, it has a single root at 3, so it goes through, and a double root at 0, so it touches
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    Aw man! i lost those 3 marks then but thanks for explaining
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    (Original post by RhysKirkwood)
    I thought it was very good actually. Did anyone get 7 for re shortest distance between C and the chord RQ or something??
    Yeah that's correct.

    For anyone that wants to know how:

    The radius of the circle was 5sqrt(2) and the chord length was 2.

    You can turn this into a right angled triangle with the hypotenuse being the radius, and another side being length 1 (half chord). Then you can work out the distance between the chord and centre using Pythagoras.
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    Think this one was just 2 marks if I am not mistaken.


    (Original post by Anam)
    Aw man! i lost those 3 marks then but thanks for explaining
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    You put the co ordinates of x2 x1 and y2 and y1 into
    The gradient equation
    Y2-y1 ÷ x2 - x1
    Then multiply both top and bottom by x2 +x1
    To get the denominator and numerator which cancel divides to get a simplified form of something like

    7- square root of something
    Forgot the digits sorry :/
 
 
 
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