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    (Original post by ella_chloe)
    that was dy by dx i think? and it had -2a and a was 13 or something idk
    it was a cubic i thought?
    yeah I said it was the gradient function i.e dy/dx
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    if an A was say 62, what mark will be required for 100 UMS? 71 or 72?
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    Anyone else get 20 root 5 for the last one?
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    (Original post by ccfctom)
    if an A was say 62, what mark will be required for 100 UMS? 71 or 72?
    Usually yes, maximum 3 mark lost.
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    (Original post by eddy29)
    what was the question involving k?
    It was the one where the first part was draw the graph of the quadratic function they gave you. The second part was find where y > 0 which was either side of where the graph crossed the x axis, x<-1 or x>3/2. Then it was whatever the quadratic was = k has no real roots, find the set of values for k. Could either do this by finding the vertex of the graph or by the discriminant of the quadratic, ended up with k<-25/8
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    (Original post by Klaxoii)
    Anyone else get 20 root 5 for the last one?
    40 seems to be what most people agree on.
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    so like for the last question, i just wrote "the area is 40" - do you need to specify 40 square units? Similarly I said the radius of circle was 5, not 5 units.
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    wait so the question about the stretch in the x-axis was not scale factor 3? how can it be 1/3 ?
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    (Original post by chloe-jessica)
    40 seems to be what most people agree on.
    Yeah I realise where I went wrong now, it was only 4 marks, so hopefully I should pick 2 or 3 method marks up
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    (Original post by Parallex)
    can anyone remember the find a question

    just wanted to check something
    I think I got a=13 for that one..?
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    (Original post by zintanax)
    so like for the last question, i just wrote "the area is 40" - do you need to specify 40 square units? Similarly I said the radius of circle was 5, not 5 units.
    Technically it should be 40 square units but I don't think you will be penalised for this.
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    (Original post by Gcselong)
    wait so the question about the stretch in the x-axis was not scale factor 3? how can it be 1/3 ?
    Y=-1/x => y=-1/3x.
    This could be done by either multiplying the x by 3, which is equivalent to y=f(ax) so would be scale factor 1/3 in x direction. Alternatively you could multiply the whole original function by 1/3 which is equivalent to y=af(x) so would be scale factor 1/3 in y direction.
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    (Original post by mbags)
    I think I got a=13 for that one..?
    Yeah I got that
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    HEY can anyone tell me how they split your grade between the three papers? Is it like added all three up so i have to almost get like a certain amount of marks over all three in total, or what? how many marks do I need in All three to get an A?? thanks xx
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    (Original post by Simonium1010)
    A few definite correct answers:
    1) first question was 4(√ 3)+4
    2) indices:
    i) 5^8
    ii) 5^(-1/4)
    iii) 5^(9/2)
    3) with two simultaneous equations, solutions were:
    x=14/3, y=-13/3, and
    x=2, y=1
    4) The quadratic with unknown 'a', a=13
    5) transformations:
    i) 1/(x-2)^.... (Can't remember the denominator power - it was the '-2' that's important)
    ii) the one with denominator '3x' was a stretch scale factor 1/3 parallel to the x-axis (in the x direction) OR parallel to the y axis (in the y axis) as both do the same transformation in this example.
    X DIRECTION AND Y DIRECTION BOTH RIGHT!
    6) with another quadratic curve question, with something like 2x^2-x-3=0;
    The curve intercepted the x axis at -1 and 3/2, and the y axis at -3.
    When it asked for the values of k, k<-25/8
    7) with the curve y=x^(-1/3) and at the point x=-8, the gradient was -1/48
    8) the circle question: centre (5,-2), radius 5.
    The then triangle PQR later in that question had area 40.
    Sorry if they aren't in their test order, and good luck everyone doing C2 tomorrow!
    Doing c2 tomorrow????????
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    (Original post by chloe-jessica)
    Y=-1/x => y=-1/3x.
    This could be done by either multiplying the x by 3, which is equivalent to y=f(ax) so would be scale factor 1/3 in x direction. Alternatively you could multiply the whole original function by 1/3 which is equivalent to y=af(x) so would be scale factor 1/3 in y direction.
    So what would this be?
    (iii) The curve y =x^1/2 is stretched by a scale factor of 5 parallel to the x-axis. State the equation of thetransformed curve.
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    (Original post by chloe-jessica)
    Can't remember the question order so sorry if it's wrong (it's definitely wrong). From what I can remember:
    1) 4 + 4root3
    2) can't really describe the graph, if you know what y=1/x looks like then reflect in either axis.
    Translation of 2 in x was -1/(x-2)
    The translation they wanted you to state was a stretch of 1/3 in either axis.
    3) 5^8
    5^(-1/4)
    5^(9/2)
    4) was this the one with k?
    Factorise the quadratic, draw graph. For y>0, x<-1 or x>3/2
    K < -25/8
    5) differentiation? Can't remember the answer to this, was reasonably straight forward though if you expanded brackets.
    6) simultaneous equation - x=1, y=2 or x=14/3, y=-13/3? Not sure about signs
    7) differentiate x^(-1/3)? Should get out -1/48 when you put x in.
    8) coordinate geometry? Line was something like 6x-8y-29=0
    9) a=13
    Second derivative was positive therefore min points
    X=1/3
    10) centre (5,-2) radius 5
    Prove tangent
    Area of triangle = 40
    I've just realised how much I really cannot remember the order. These are my answers which seem to be generally accepted within my school and by various people on here. They may not be right.
    For question 6 it was y=5-2x so your first x and y solutions arent valid.

    I got X=2 y =1
    And x=14/3 y=-13/3
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    (Original post by Brian luiz)
    For question 6 it was y=5-2x so your first x and y solutions arent valid.

    I got X=2 y =1
    And x=14/3 y=-13/3
    Couldn't remember which way the x and y were. 2 and 1 were definitely right, though I probably got them mixed up when I was trying to remember what I wrote to post this. I'll change it now, thanks.
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    (Original post by Gcselong)
    So what would this be?
    (iii) The curve y =x^1/2 is stretched by a scale factor of 5 parallel to the x-axis. State the equation of thetransformed curve.
    Y=(x/5)^1/2
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    (Original post by Parallex)
    find a given that x=4 was a stationary point or something

    like can anyone remember f(x)
    I think it was 2x^3-13x^2+8x+8 (I think, also assuming that a=13 which I'm fairly sure it does)
 
 
 
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