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    I did not calculate x, i used the value from the graph... Not sure if it was question 7... The one with the area between curve and tangent

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    (Original post by roar96)
    I did not calculate x, i used the value from the graph... Not sure if it was question 7... The one with the area between curve and tangent

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    I don't actually know what you are talking about. 7 was a y value and this was used in the solution.
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    Q7 part ii, to calculate the area under the tangent, i integrated between one correct point and one incorrect point, (x value), and took this away from the area under the curve.

    Sorry if I'm not explaining it very well...

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    (Original post by roar96)
    Q7 part ii, to calculate the area under the tangent, i integrated between one correct point and one incorrect point, (x value), and took this away from the area under the curve.

    Sorry if i'm not explaining it very well...

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    You will probably get a couple of marks for that.
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    I put N less than or equals to not just N equals 17 as it asked for the greatest value so lost one mark?
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    (Original post by stefanconstant)
    I put N less than or equals to not just N equals 17 as it asked for the greatest value so lost one mark?
    yes I expect so
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    (Original post by Mr M)
    yes I expect so

    Do you think the examiners would be pedantic if I wrote '34cm' instead of '34.0cm', as in previous mark-schemes they have generally allowed answers such as '52' instead of '52.0'?
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    (Original post by Zakee)
    Do you think the examiners would be pedantic if I wrote '34cm' instead of '34.0cm', as in previous mark-schemes they have generally allowed answers such as '52' instead of '52.0'?
    Entirely depends on the mark scheme. It may be penalised, it may not. I don't understand why people find it so hard to follow the instructions on the paper and give all non-exact answers to 3 s.f.?
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    (Original post by Mr M)
    Mr M's OCR (not OCR MEI) Core 2 Answers May 2013


    1. 6.39 (4 marks)


    2. (i) 254 and 106 degrees (3 marks)

    (ii) 71.6 and 252 degrees (3 marks)


    3. (i) 64+960x+6000x^2 (4 marks)

    (ii) c = - 11 (3 marks)


    4. (a) \frac{5}{4}x^4 -3x^2+x+k (3 marks)

    (b) (i) -12 x^{-2} + k (2 marks)

    (ii) a = 2 (3 marks)


    5. (i) 62.2 (4 marks)

    (ii) 34.0 (4 marks)


    6. (i) 963 (3 marks)

    (ii) 17 (6 marks)


    7. (i) Show ... (4 marks)

    (ii) \frac{37}{30} (5 marks)


    8. (i) (a) (0, 1) (1 mark)

    b) (0, 4) (1 mark)

    c) Any value where a > 1 and any value where 0 < b < 1 (2 marks)

    (ii) Show ... (5 marks)


    9. (i) 15 (2 marks)

    (ii) Show f(-0.5)=0 and f(x)=(2x+1)(2x-3)(x+1) (6 marks)

    (iii) \theta = \frac{2\pi}{3} and \theta =\frac{4\pi}{3} and \theta=\pi (4 marks)


    I'll be around a bit this afternoon and then part of this evening to answer questions. Please post in this thread rather than sending me private messages.
    thanks for everything sir !!!
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    I think I did relatively well! But could someone please remind me what questions 5 and 6 were about because those answers don't look familiar and I can't seem to remember the questions.


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    For the logs question, i was being stupid and completely missed out the fact that the had gave as the equation ab=2 so i tried showing it by making things up, which i shouldn't have.However this is what i put for the logs question:
    Loga^2X = Log4b^X
    2Xloga = Log4 + Logb^X , sub in b=2
    2Xloga = 2 + Log2^X
    2Xloga - Log2^X = 2
    2XLoga - XLog2 = 2
    2XLoga - X1 = 2
    X (2Loga - 1) = 2
    X = 2/(2Loga - 1)

    I know this is me being really stupid . However Mr M can you please tell me if i would get any marks for putting that as my working. (BTW i put Log here but on my script i wrote Log2 ). Also thanks for the markscheme Mr M.
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    (Original post by Asad96)
    For the logs question, i was being stupid and completely missed out the fact that the had gave as the equation ab=2 so i tried showing it by making things up, which i shouldn't have.However this is what i put for the logs question:
    Loga^2X = Log4b^X
    2Xloga = Log4 + Logb^X , sub in b=2
    2Xloga = 2 + Log2^X
    2Xloga - Log2^X = 2
    2XLoga - XLog2 = 2
    2XLoga - X1 = 2
    X (2Loga - 1) = 2
    X = 2/(2Loga - 1)

    I know this is me being really stupid . However Mr M can you please tell me if i would get any marks for putting that as my working. (BTW i put Log here but on my script i wrote Log2 ). Also thanks for the markscheme Mr M.
    The first line is wrong so you can't expect much credit. You would probably pick up a mark somewhere there for applying laws of logarithms correctly.
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    (Original post by Mr M)
    The first line is wrong so you can't expect much credit. You would probably pick up a mark somewhere there for applying laws of logarithms correctly.
    Thank you
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    (Original post by Mr M)
    That will be ok.
    Thank you.
    And sorry, I did not realise my last post had submitted.
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    Mr M,

    For the question about finding the shaded area, I remember integrating the curve but I needed to find the area of the triangle (formed by the line and the x-axis) but I didn't know the equation of the line so I differentiated at the point of intersection and then plugged the values in to find the gradient at the point, then I found the gradient of the line so I could find the x-intercept which I found to be something like 1.6recurring and then I found the area of the triangle and subtracted it from the area underneath the curve.

    I'm scared as to if I'll even get any marks because Differentiation isn't part of C2 is it?

    Thanks for the answers though! I really appreciate you taking your time to answer questions.


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    (Original post by iDraw)
    Mr M,

    For the question about finding the shaded area, I remember integrating the curve but I needed to find the area of the triangle (formed by the line and the x-axis) but I didn't know the equation of the line so I differentiated at the point of intersection and then plugged the values in to find the gradient at the point, then I found the gradient of the line so I could find the x-intercept which I found to be something like 1.6recurring and then I found the area of the triangle and subtracted it from the area underneath the curve.

    I'm scared as to if I'll even get any marks because Differentiation isn't part of C2 is it?

    Thanks for the answers though! I really appreciate you taking your time to answer questions.


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    That's the correct method. The C2 specification states that knowledge of C1 may be required. This continues at A2 - C4 assumes knowledge of C1, C2 and C3.
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    (Original post by Mr M)
    That's the correct method. The C2 specification states that knowledge of C1 may be required. This continues at A2 - C4 assumes knowledge of C1, C2 and C3.
    Yay thank you! Oh my gosh, what a relief!

    Also, for the last question, I got two values of \theta, where \theta = \frac{2\pi}{3} and \theta =\frac{4\pi}{3} but for the last one, I got something like cos \theta = \frac{3}{2} and so I didn't get a third value for \theta as its impossible:confused: How many marks will I drop?
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    (Original post by iDraw)
    Yay thank you! Oh my gosh, what a relief!

    Also, for the last question, I got two values of \theta, where \theta = \frac{2\pi}{3} and \theta =\frac{4\pi}{3} but for the last one, I got something like cos \theta = \frac{3}{2} and so I didn't get a third value for \theta as its impossible:confused: How many marks will I drop?
    Drop one mark.

    You seem to have forgotten that \cos \theta = -1 was a solution.
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    Any opinions on the S1 exam on Friday Mr M? Do you think the people saying that the tests have been gradually getting harder in S1 are right ? With the Jan 13 test (68% for an A) being an Omen of the most Devious statistics paper ever conceived on Friday?
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    (Original post by dakenSoren)
    Any opinions on the S1 exam on Friday Mr M? Do you think the people saying that the tests have been gradually getting harder in S1 are right ? With the Jan 13 test (68% for an A) being an Omen of the most Devious statistics paper ever conceived on Friday?
    I don't do S1 so have no idea but wouldn't believe any conspiracy theories about an increase in difficulty.
 
 
 
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