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    (Original post by Big-Daddy)
    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.
    Would we still get 5/5 marks for the question if we solved it as a cubic and still obtained (-2,-27) though?
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    (Original post by nsoods)
    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
    probably only 2 of the 5, cos you do a lot with re-arranging the formulas and solving the quadratic, so you'll only drop marks for not finding the distance and not using the right one.
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    I got k=-5 but messed up on the last question what do u guys think the grade boundaries will be like
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    Just remembered the completing the square answer: p=3/2, q=13/4. So the vertex was (-3/2,13/4).

    (Original post by Jack_King)
    Would we still get 5/5 marks for the question if we solved it as a cubic and still obtained (-2,-27) though?
    I would be disgusted if we didn't.

    However I wrote down both ways in the end At first I got a cubic and tried every method I could to solve it, but then realized that a) it was unsolvable and b) I had ****ed up my differentiation to find k earlier (it took me a good 15 minutes of effort on the cubic before I realized it was unsolvable), so then I went back and redid the rest of that question (getting k=1, second derivative=12>0 thus minimum) by the gradient method (don't ask why), then at the end it occurred to me to try with the cubic again and the answer was obvious the second time So I ended up with both methods on paper. But you should get the marks.
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    (Original post by a10)
    you would lose method marks because its not used in the C1 syllabus so it would not be fair if they gave full marks for a method not taught in C1.
    Well that is rather stupid in my opinion, you've obtained the correct answer but no you're not allowed the marks, it's not like the question stated : By only using differentiation find the coordinates of A..
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    (Original post by a10)
    you would lose method marks because its not used in the C1 syllabus so it would not be fair if they gave full marks for a method not taught in C1.
    Sorry buddy, you're wrong here. I've asked Mr M about using implicit differentiation for the gradient at a point on a circle, and he said it was fine. That's a C3 technique (well not really but you never learn it before you learn the chain rule). Factorisation is C2 if not general knowledge. You'll definitely get the marks for using it.
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    (Original post by DomRusky)
    probably only 2 of the 5, cos you do a lot with re-arranging the formulas and solving the quadratic, so you'll only drop marks for not finding the distance and not using the right one.
    thanks Dom, you seam like a great guy. Best of luck with everything
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    (Original post by Jack_King)
    Well that is rather stupid in my opinion, you've obtained the correct answer but no you're not allowed the marks, it's not like the question stated : By only using differentiation find the coordinates of A..
    Don't worry. If you landed up with (-2,-27) you'll get full marks.
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    (Original post by Big-Daddy)
    Just remembered the completing the square answer: p=3/2, q=13/4. So the vertex was (-3/2,13/4).



    I would be disgusted if we didn't.

    However I wrote down both ways in the end At first I got a cubic and tried every method I could to solve it, but then realized that a) it was unsolvable and b) I had ****ed up my differentiation to find k earlier (it took me a good 15 minutes of effort on the cubic before I realized it was unsolvable), so then I went back and redid the rest of that question (getting k=1, second derivative=12>0 thus minimum) by the gradient method (don't ask why), then at the end it occurred to me to try with the cubic again and the answer was obvious the second time So I ended up with both methods on paper. But you should get the marks.
    Okay excellent stuff, you've definitely got the marks then for that question! It'd be ridiculous if I lost marks for it, after all the correct answer was stated at the end with a correct method, first thing that came to my mind when I saw the question was solve it via the remainder theorem, none of this differentiation bull**** haha.
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    Here's the unofficial markscheme, fairly sure it's all correct:
    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)
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    (Original post by Big-Daddy)
    Sorry buddy, you're wrong here. I've asked Mr M about using implicit differentiation for the gradient at a point on a circle, and he said it was fine. That's a C3 technique (well not really but you never learn it before you learn the chain rule). Factorisation is C2 if not general knowledge. You'll definitely get the marks for using it.
    you sure about that bro? I don't think so. It states on the mark schemes the correct method and if its not on there you'll simply lose those method marks but still gain an answer mark. It wouldn't be fair on first year students if people got awarded full marks for using a method not taught in C1.
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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
    As far as I can tell, my answers are all the same as yours! I found it easy too, but that's mainly because I'm sitting core 3 and 4 in a few weeks time so core 1 just couldn't be difficult! Thanks for putting your answers on
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    (Original post by DomRusky)
    If you put the answer in the wrong box, don't worry. My teacher's an exam marker and said that it's just to help set it out. You'll get the marks for the working, even if it's in another question's box
    Awesome, thank you
    Thanks for the mark scheme aswell
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    anyone estimate the grade boundaries? is 61/72 a relatively high one?
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    (Original post by joostan)
    I couldn't find a thread for this so I figured I'd make you guys one.
    Good Luck with your exams (If you're reading this before the exam starts.)- I'm sure it went well.
    To the best of my knowledge there aren't any rules regarding discussion of OCR Exams so talk away
    If you want to make an “unofficial mark scheme” , just quote me and I’ll set up links in the OP for you all to have a look at.

    EDIT: MrM. Will be Providing solutions at some point in the not too distant future - I'll put a link in to wherever he puts them

    Just some people's thoughts: - If you can remember a question, post it and I'll answer.

    Q1:

    Q10:
    Unofficial markscheme in above post.
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    (Original post by a10)
    im sure you did fine, relax
    How many marks do you reckon i'll get for putting y=x^2 doing that to get something to the power of 4, then putting z=y^2 and then getting a quadratic to work out x from there? I know it was stupid because it got the wrong answer and wouldnt actually end up calculating it for ^6, but i did use the substitution method and stuff...

    Posted from TSR Mobile
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    (Original post by nsoods)
    thanks Dom, you seam like a great guy. Best of luck with everything
    Just tryin to help and thanks man! you too
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    Answers are here:

    http://www.thestudentroom.co.uk/show....php?t=2350074
    • Community Assistant
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    Answers are here:

    http://www.thestudentroom.co.uk/show....php?t=2350074
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    (Original post by Majeue)
    ii) B co-ordinate is (-2, -10)
    Do you remember the question for either this, or Question 8?
 
 
 
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