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    (Original post by DomRusky)
    9)i) Positive quadratic curve (smiley face, not sad face). Intercepts y at (0,-6) and x at (-3/2,0) and (2,0)
    I (think) i got it the other way round, (-2,0) and (3/2,0)?!
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    (Original post by Big-Daddy)
    To be precise it specified "the coordinates of the point". Point, singular. Hence only one answer.

    The coordinates weren't (4,11) - they were (-2,-27). That was an example, right?

    Yes, the assumed coordinates for question 10 are (-2,-27). Sorry for the confusion. I was showing an example.
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    (Original post by DomRusky)
    Ok, I thought this was a really easy exam, but that's my own opinion- don't bite me for it!

    I brought my answers out of the exam on scrap paper, so I'll do my best to translate. I'm pretty confident they're all right, if I'm wrong correct me, but I went through with teachers and other students afterwards and these seem to be correct answers. If anybody did get similar, shout up!

    1)i) 12 rt.5
    ii) 4 rt.5
    iii) 5 rt.5

    2) x^3 = 1/8 and -1 ... therefore x = 1/2 and -1

    3) f'(x) = -(12x^-3) +2
    f''(x) = 36x^-4

    4)i) 3(x+3/2)^2 + 13/4 [Because 3(3/2)^2 = 3(9/4) = (27/4) .. so then take that from the ten to give 13/4]
    ii) Vertex is (-3/2, 13/4)
    iii) b^2 - 4ac = 9^2 - 120 = -39

    5)i) Basically the same graph as 1/x^2... Curves symmetrical and never touch any of the axes
    ii) Stretch in Scale Factor 1/2 in y (vertical) direction

    6)i) x^2 + (y+4)^2 = 40 Centre is (0, -4) Radius is rt.40 = 2 rt.10
    ii) B co-ordinate is (-2, -10)

    7)i) x < -1/8
    ii) 0 </= x </= 5 ( </= means greater than or equal to! solved this one by drawing out the curve.. )

    8) m of perpendicular is 1/3, so m of line is -3.. put into formula to get 3x + y -1/2 = 0 .. x2 to get into integers, 6x + 2y - 1 = 0

    9)i) Positive quadratic curve (smiley face, not sad face). Intercepts y at (0,-6) and x at (-3/2,0) and (2,0)
    ii) Vertex at x=1/4 so function is decreasing for x < 1/4
    iii) Submit into formula for curve, use quadratic formula to solve, find that points P and Q are (-2,4) and (2.5,4) so distance is 4.5

    10)i) Solve to find that k = -5
    dy/dx = -3x^2 - 6x + 4 - k
    0 = -3(-3)^2 - 6(-3) + 4 - k
    k = -27 + 18 + 4 = -5

    ii) d2y/dx2 = -6x - 6 ... submit -3 in to get 12, 12>0 therefore it's a minimum point
    iii) This one was really long winded for 5 marks, basically you had to put the formulas together, solve the quadratic to find the two points on the cubic curve that satisfied a gradient of 9, you then put those two points (0 and -2) into the y=9x-9, and see which works. 0 doesn't work, but -2 does, so A = (-2, -27)

    Hope this helps! Please point out any mistakes you think I've made, would be awesome if someone could scan a copy of the question sheet too
    Hey,
    Do you remember the exact question for 7.i)???
    Thanks
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    (Original post by Big-Daddy)
    Given that Mr M is posting answers at some point tonight, discussing solutions seems a little fruitless.

    However, if anyone wants to post a question fully I can post my answer to it. Does anyone remember the two inequalities (questions not answers), or the 7-mark Question 8?
    3-8x>4,
    and
    (2x-4)(x-3)<=12

    I got x<-1/8 for the first one but flopped on the 2nd one. I did expand, and brang 12 to the other side so it would be 2x^2-10x<=0 and my mind went blank from there -.-
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    Thought i would upload the exam paper for you guys, in the attachments hopefully...
    Attached Images
      
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    (Original post by TwlilightLoz)
    what did guys put for the question where you had to state what happens to graph as it transforms from 2/x^2 to 1/x^2. I wrote that it stretches parallel in the x axis by 0.5
    I put it was parallel to the y-axis instead.
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    (Original post by random_kid)
    Hey,
    Do you remember the exact question for 7.i)???
    Thanks
    Solve

    3 - 8x > 4

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    so what do we think grade boundaries wise ?
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    (Original post by TwlilightLoz)
    what did guys put for the question where you had to state what happens to graph as it transforms from 2/x^2 to 1/x^2. I wrote that it stretches parallel in the x axis by 0.5

    Crap I put -0.5
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    (Original post by daisyrchds)
    I put it was parallel to the y-axis instead.
    It was the y axis.
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    (Original post by roar96)
    Ahhh... (Y^3)^2 .... Darn... I was quite sure it would have to be cubed.... How stupid of me
    Posted from TSR Mobile
    im sure you did fine, relax
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    The stretch was as such:


    "Stretch by a scale factor of 1/2, parallel to the y axis".
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    (Original post by Jackaoliver)
    Thought i would upload the exam paper for you guys, in the attachments hopefully...
    ah thank you!!
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    (Original post by daisyrchds)
    I (think) i got it the other way round, (-2,0) and (3/2,0)?!
    I hate to be that guy but I think that's wrong, cos that'd factorise to (x+2)(2x-3) and then you'd get 2x^2 + x - 6 .. whereas it's -x not +.. probably get marks for everything else though!
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    (Original post by TwlilightLoz)
    3-8x>4,
    and
    (2x-4)(x-3)<=12

    I got x<-1/8 for the first one but flopped on the 2nd one. I did expand, and brang 12 to the other side so it would be 2x^2-10x<=0 and my mind went blank from there -.-
    I couldnt explain the second question better than this link:
    http://www.wolframalpha.com/input/?i...-3%29%3C%3D12+

    The answer is 0<=x=<5
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    (Original post by TwlilightLoz)
    3-8x>4,
    and
    (2x-4)(x-3)<=12

    I got x<-1/8 for the first one but flopped on the 2nd one. I did expand, and brang 12 to the other side so it would be 2x^2-10x<=0 and my mind went blank from there -.-
    You find the two roots, which are 0 and 5 (since it turns to x(2x-10)<= 0)

    Since it's under 0, x is a value been 0 and 5 (including both)
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    (Original post by TwlilightLoz)
    3-8x>4,
    and
    (2x-4)(x-3)<=12

    I got x<-1/8 for the first one but flopped on the 2nd one. I did expand, and brang 12 to the other side so it would be 2x^2-10x<=0 and my mind went blank from there -.-
    For second one you then do x(2x-10)<=0
    So x is 0 for one solution, x is 5 for the other
    0<=x<=5

    Posted from TSR Mobile
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    (Original post by BankOfPigs)
    I don't think they can ask you to solve cubics in core 1, thats restricted to c2 with factor theorem. Unless they already give you the factors.

    For those questions they're always in a form that can easily turn into a quadratic.
    I solved the cubic (x^3+3x^2-4=0, (x+2)^2*(x-1)=0). It was clear from this solution that x=-2 was the answer as we're looking for a tangent (repeated root). However there was another way of doing this and that's the one expected at C1.
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    (Original post by DomRusky)
    I hate to be that guy but I think that's wrong, cos that'd factorise to (x+2)(2x-3) and then you'd get 2x^2 + x - 6 .. whereas it's -x not +.. probably get marks for everything else though!
    argh i hate you i got it right in the first place then decided to change it in the last minute, forgetting it was 2x! gutted, oh well
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    how many marks do you think you would lose for the last one if you find x=0 and x=-2 , but then only use x=0 as your solution?? thanks
 
 
 
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