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    (Original post by missypink<3)
    Shouldn't 9(iii) be 5pie/3 not 4pie/3?
    Also how did you get the answer to 7(ii). I got 1.9
    2 - \frac{2}{3} \neq \frac{5}{3}

    7ii) Find the equation of the tangent, then where the tangent meets the x axis. Now calculate the area of the triangle and subtract this from the value of the integral provided previously.
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    (Original post by Mr M)
    Of course.
    For 7ii) :
    I wrote, Area = 37/30 = 1.23 (3sf)
    then underlined the decimal, does this lose a mark?
    Thanks in advance
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    (Original post by missypink<3)
    I put b as -2 at it was y=4b^x and it crossed the y-axis at (0,4) which is only possible is b is negative.
    That's wrong - sorry.

    Edit: I should explain why. Try letting x=\frac{1}{2} for example.
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    (Original post by frankdatank)
    For 7ii) :
    I wrote, Area = 37/30 = 1.23 (3sf)
    then underlined the decimal, does this lose a mark?
    Thanks in advance
    That will be ok.
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    (Original post by frankdatank)
    For 7ii) :
    I wrote, Area = 37/30 = 1.23 (3sf)
    then underlined the decimal, does this lose a mark?
    Thanks in advance
    Yeah, Mr M said as long as you state 37/30 it's fine.


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    67/72 roughly what ums for this paper?


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    (Original post by Mr M)
    2 - \frac{2}{3} \neq \frac{5}{3}

    7ii) Find the equation of the tangent, then where the tangent meets the x axis. Now calculate the area of the triangle and subtract this from the value of the integral provided previously.
    If cosx= -1 then x= 2pie/3
    Cosine is negative in the 4th quadrant in the |CAST diagram. Therefore surely you would do pie + 2pie/3 in order to get the value of 5pie/3?

    Also for question 7(ii).
    I correctly found the equation of the tangent and then integrated the line between 4 and 0. I then subtracted this value from the value found in 7(i).
    How many marks would I get?
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    (Original post by germankid101)
    For the perimeter question 5(ii) I calculated the length BD using the formula 2rsin(θ/2) which gave me 12.461....
    I then added this to 12.8 and 9 to give an answer of 34.3cm (3sf).
    How many marks would I lose for this and why isn't this correct?

    (Original post by Mr M)
    I have no idea where that formula appeared from! Drop 2 marks.
    I'm not sure whether you're interested but I suspect the formula works out the length of BC. You know for triangle BAC, and another triangle BDC with D on the circumference of the circle, angle BDC will be half angle BAC, or half theta. From the extended sine rule, the circumcircle of BCD will be the circle with centre A and radius 16 so:

    BC/ sin(half theta) = 2R, so BC = 2Rsin(half theta).

    However, germankid has made the mistake that D is not on the circumference of the circle.
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    (Original post by missypink<3)
    If cosx= -1 then x= 2pie/3
    Cosine is negative in the 4th quadrant in the |CAST diagram. Therefore surely you would do pie + 2pie/3 in order to get the value of 5pie/3?

    Also for question 7(ii).
    I correctly found the equation of the tangent and then integrated the line between 4 and 0. I then subtracted this value from the value found in 7(i).
    How many marks would I get?
    Don't bother with that CAST rubbish. It is just confusing you.

    \cos x = \cos (2 \pi - x)

    7ii) drop 3 marks
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    Hi Mr M,

    Q1 - I worked out the wrong value value of h (stupid I know) but apart from that my method going forward was fine
    Q7ii - I differentiated the equation but made a silly mistake and worked out m as 6 instead of 3. From then on however, my method was fine.
    How many marks would I lose?
    Thanks in advance,
    Akash
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    Mr. M,

    Unfortunately on question 3i) I expanded the first binomial incorrectly by using 5C1 and 5C2, as opposed to 6C1 and 6C2. Thus my answer was 64+800x+4000x^2, which I then used in ii) and so got c=-7.25.

    How many marks do you think I would lose for this?

    Thanks a lot
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    (Original post by metaltron)
    I'm not sure whether you're interseted but I suspect the formula works out the length of BC. You know for triangle BAC, and another triangle BDC with D on the circumference of the circle, angle BDC will be half angle BAC, or half theta. From the extended sine rule, the circumcircle of BCD will be the circle with centre A and radius 16 so:

    BC/ sin(half theta) = 2R, so BC = 2Rsin(half theta).

    However, germankid has made the mistake that D is not on the circumference of the circle.
    Yes that'll be it.
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    (Original post by Akash B)
    Hi Mr M,

    Q1 - I worked out the wrong value value of h (stupid I know) but apart from that my method going forward was fine
    Q7ii - I differentiated the equation but made a silly mistake and worked out m as 6 instead of 3. From then on however, my method was fine.
    How many marks would I lose?
    Thanks in advance,
    Akash

    drop 1 and drop 2 I expect
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    (Original post by dee1)
    Mr. M,

    Unfortunately on question 3i) I expanded the first binomial incorrectly by using 5C1 and 5C2, as opposed to 6C1 and 6C2. Thus my answer was 64+800x+4000x^2, which I then used in ii) and so got c=-7.25.

    How many marks do you think I would lose for this?

    Thanks a lot
    drop 2
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    Mr M how many marks would i get for 7ii if i worked out wrong tangent by using wrong x-value, but differentiated correctly, but showed you had to integrate by doing curve- tangent. Therefore, overall wrong area under curve

    Also do you get penalised by putting to more decimal places
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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.
    RE; 8i), I put y=1 and y=4 as my answers instead of (0,1) and (0,4). Will I still get the marks?
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    (Original post by chuckynoriss)
    Mr M how many marks would i get for 7ii if i worked out wrong tangent by using wrong x-value, but differentiated correctly, but showed you had to integrate by doing curve- tangent. Therefore, overall wrong area under curve
    curve - tangent will not necessarily produce a sensible answer

    You will probably just get 1 mark for differentiating
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    (Original post by martin_brundle)
    RE; 8i), I put y=1 and y=4 as my answers instead of (0,1) and (0,4). Will I still get the marks?
    No I don't think so. Read the questions more carefully in future.
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    (Original post by Mr M)
    drop 2
    Hi Mr M,
    Do you have a rough idea how many UMS I would get for 68/72?
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    (Original post by missypink<3)
    Hi Mr M,
    Do you have a rough idea how many UMS I would get for 68/72?
    No sorry.
 
 
 
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