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    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.
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    Roughly what do you think the grade boundaries will be for an A?
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    For the area under curve question would I get some marks for workings as i worked out equation of tangent and area of small triangle but got complete wrong answer?
    thanks
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    (Original post by kiraeil)
    For the area under curve question would I get some marks for workings as i worked out equation of tangent and area of small triangle but got complete wrong answer?
    thanks
    Definitely.
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    For the area under the curve how many would I drop for, writing down it was the integral- the area of the triangle. I messed up calculating the x coordinate so in the end just guessed it as 2, in a last ditch effort :P

    also for question 8 didn't b have to be a negative number?
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    What was the working for 5ii) the question on finding the perimeter? Thanks.
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    Thanks Mr M! Looks like I made two stupid mistakes :/

    Q3) I put 64+960x +3000x ^2
    Q 4) ii a =1/2 (didn't notice the ^- so didn't flip)

    How many marks lost?
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    For question 9)ii) I inspected the cubic correctly to get a linear and a quadratic factor but somehow managed to get the + and - signs the wrong way round when I factorised by quadratic, getting (2x+3)(x-1). How many marks do you think this will cost me? Also, I then carried this error forwards into 9)iii) but used the correct method and got 2 of the 3 answers. Is there an allowance for error carried forwards or will I lose more marks here?
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    Do you think that grade boundary for A might be 54 points as well?
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    (Original post by Music99)
    For the area under the curve how many would I drop for, writing down it was the integral- the area of the triangle. I messed up calculating the x coordinate so in the end just guessed it as 2, in a last ditch effort :P

    also for question 8 didn't b have to be a negative number?
    Nope. Check your revision guide.
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    (Original post by kiraeil)
    What was the working for 5ii) the question on finding the perimeter? Thanks.
    You had to work out the missing side using the cosine rule:
    A^2=7^2+16^2-(2x7x16xcos0.8)
    A=12.2
    then you worked out arc length which is r x theta = 16x0.8 = 12.8
    then you worked out the last side which was just 16-7=9
    therefore, perimeter = 12.8+12.2+9 = 34

    hope this helps
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    Hey does anyone remember the questions for 5i), 5ii), 6i) 6ii) and 9i) please
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    I got the value for C as 24 and dont understand how it is -11 can you explain the answer in more detail please?
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    (Original post by eggfriedrice)
    Thanks Mr M! Looks like I made two stupid mistakes :/

    Q3) I put 64+960x +3000x ^2
    Q 4) ii a =1/2 (didn't notice the ^- so didn't flip)

    How many marks lost?
    1 for q3 and probably 2 for 4ii
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    (Original post by swaggy)
    Hey does anyone remember the questions for 5i), 5ii), 6i) 6ii) and 9i) please
    5i) was the area of that portion of the sector with angle 0.8 radians, radius 16 and side length 7, you had to find the area of the whole sector and subtract the area of the triangle
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    Thanks- If I worked out the missing side and arc length correctly but made a silly mistake for the length of the last side how many marks would i get. Thanks
    (Original post by Majeue)
    You had to work out the missing side using the cosine rule:
    A^2=7^2+16^2-(2x7x16xcos0.8)
    A=12.2
    then you worked out arc length which is r x theta = 16x0.8 = 12.8
    then you worked out the last side which was just 16-7=9
    therefore, perimeter = 12.8+12.2+9 = 34

    hope this helps
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    Hey does anyone remember the questions for 5i), 5ii), 6i) 6ii) and 9i) please
    (Original post by milesw97)
    5i) was the area of that portion of the sector with angle 0.8 radians, radius 16 and side length 7, you had to find the area of the whole sector and subtract the area of the triangle
    Oh thats right thanks do you know the questions for 6, 7 and 9?
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    How many marks will I get for the area under the curve if I got the equation of the tangent wrong


    Posted from TSR Mobile
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    (Original post by kiraeil)
    Thanks- If I worked out the missing side and arc length correctly but made a silly mistake for the length of the last side how many marks would i get. Thanks
    I'd imagine you would get 2/4 marks.
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    (Original post by 5foot)
    How many marks will I get for the area under the curve if I got the equation of the tangent wrong


    Posted from TSR Mobile
    Maybe 1 or 2, depends if you did any part of it right or not because you only had to work out the area under tangent and were given the area under the other graph.
 
 
 
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