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    Can someone please explain how to do 4ii, I know ive made a mistake somewhere but im not sure where
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    (Original post by Mr M)
    The trouble with squaring is that it can introduce extra incorrect solutions.

    2 marks probably.
    Ok thanks very much at least I should get a couple of marks
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    (Original post by Ozil5)
    Thank you, would you mind explaining how you get 12 too? I just tried it again and i got (x+3)(x-4). Am i just being really stupid here?
    Using previous part:
    x^2/(x+4) = 3^2 = 9
    x^2=9(x+4)
    x^2=9x+36
    x^2-9x-36=0
    (x-12)(x+3)=0
    x=12 or -3

    But log base 3 (-3) is undefined.
    Therefore x = 12
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    (Original post by Mr M)
    Mr M's OCR (not OCR MEI) Core 2 Answers May 2016


    1. (i) 10 cm (2 marks)

    (ii) 5.04 cm (2 marks)


    2. (i) \displaystyle \frac{3 \pi}{10} (2 marks)

    (ii) r=20.4 (3 marks)


    3. (i) \displaystyle 27 +27kx+9k^2x^2 +k^3x^3 (4 marks)

    (ii) k=\pm \sqrt{3} (2 marks)


    4. (i) \displaystyle \log_3 \frac{x^2}{x+4} (2 marks)

    (ii) x = 12 (4 marks)


    5. (a) \displaystyle \frac{x^4}{2} -x^3 +2x^2-6x+k (3 marks)

    (b) (i) \displaystyle \frac{2}{a^2} - \frac{6}{a} + 4 (4 marks)

    (ii) 4 (1 mark)


    6. (i) k=91 (3 marks)

    (ii) \displaystyle S_{16} = 978 (2 marks)

    (iii) N=38 (6 marks)


    7. (i) Quotient x^2-4x+3 and remainder 0 (3 marks)

    (ii) x=1 or x=-1 or x=3 (3 marks)

    (iii) Show (2 marks)

    (iv) \displaystyle \frac{512}{15} (4 marks)


    8. (i) Translation by 2 units in the positive x direction (2 marks)

    (ii) Stretch parallel to the y axis scale factor \displaystyle \frac{1}{9} (2 marks)

    (iii) Sketch showing (0, \frac{1}{9}) as the only point of intersection (2 marks)

    (iv) x=6.73 (3 marks)

    (v) 9.60 (3 marks)


    9. (i) \displaystyle \frac{2 \pi}{a} (1 mark)

    (ii) a=5 and k=0 (3 marks)

    (iii) \displaystyle x=\frac{\pi}{3a} and \displaystyle x=\frac{4 \pi}{3a} (4 marks)
    Question 9ii - the question said that k must be a positive constant and 0 isn't a positive constant?
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    (Original post by Mr M)
    Mr M's OCR (not OCR MEI) Core 2 Answers May 2016


    1. (i) 10 cm (2 marks)

    (ii) 5.04 cm (2 marks)


    2. (i) \displaystyle \frac{3 \pi}{10} (2 marks)

    (ii) r=20.4 (3 marks)


    3. (i) \displaystyle 27 +27kx+9k^2x^2 +k^3x^3 (4 marks)

    (ii) k=\pm \sqrt{3} (2 marks)


    4. (i) \displaystyle \log_3 \frac{x^2}{x+4} (2 marks)

    (ii) x = 12 (4 marks)


    5. (a) \displaystyle \frac{x^4}{2} -x^3 +2x^2-6x+k (3 marks)

    (b) (i) \displaystyle \frac{2}{a^2} - \frac{6}{a} + 4 (4 marks)

    (ii) 4 (1 mark)


    6. (i) k=91 (3 marks)

    (ii) \displaystyle S_{16} = 978 (2 marks)

    (iii) N=38 (6 marks)


    7. (i) Quotient x^2-4x+3 and remainder 0 (3 marks)

    (ii) x=1 or x=-1 or x=3 (3 marks)

    (iii) Show (2 marks)

    (iv) \displaystyle \frac{512}{15} (4 marks)


    8. (i) Translation by 2 units in the positive x direction (2 marks)

    (ii) Stretch parallel to the y axis scale factor \displaystyle \frac{1}{9} (2 marks)

    (iii) Sketch showing (0, \frac{1}{9}) as the only point of intersection (2 marks)

    (iv) x=6.73 (3 marks)

    (v) 9.60 (3 marks)


    9. (i) \displaystyle \frac{2 \pi}{a} (1 mark)

    (ii) a=5 and k=0 (3 marks)

    (iii) \displaystyle x=\frac{\pi}{3a} and \displaystyle x=\frac{4 \pi}{3a} (4 marks)
    Do you know what question 9 i and ii were?
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    (Original post by Mr M)
    Mr M's OCR (not OCR MEI) Core 2 Answers May 2016


    1. (i) 10 cm (2 marks)

    (ii) 5.04 cm (2 marks)


    2. (i) \displaystyle \frac{3 \pi}{10} (2 marks)

    (ii) r=20.4 (3 marks)


    3. (i) \displaystyle 27 +27kx+9k^2x^2 +k^3x^3 (4 marks)

    (ii) k=\pm \sqrt{3} (2 marks)


    4. (i) \displaystyle \log_3 \frac{x^2}{x+4} (2 marks)

    (ii) x = 12 (4 marks)


    5. (a) \displaystyle \frac{x^4}{2} -x^3 +2x^2-6x+k (3 marks)

    (b) (i) \displaystyle \frac{2}{a^2} - \frac{6}{a} + 4 (4 marks)

    (ii) 4 (1 mark)


    6. (i) k=91 (3 marks)

    (ii) \displaystyle S_{16} = 978 (2 marks)

    (iii) N=38 (6 marks)


    7. (i) Quotient x^2-4x+3 and remainder 0 (3 marks)

    (ii) x=1 or x=-1 or x=3 (3 marks)

    (iii) Show (2 marks)

    (iv) \displaystyle \frac{512}{15} (4 marks)


    8. (i) Translation by 2 units in the positive x direction (2 marks)

    (ii) Stretch parallel to the y axis scale factor \displaystyle \frac{1}{9} (2 marks)

    (iii) Sketch showing (0, \frac{1}{9}) as the only point of intersection (2 marks)

    (iv) x=6.73 (3 marks)

    (v) 9.60 (3 marks)


    9. (i) \displaystyle \frac{2 \pi}{a} (1 mark)

    (ii) a=5 and k=0 (3 marks)

    (iii) \displaystyle x=\frac{\pi}{3a} and \displaystyle x=\frac{4 \pi}{3a} (4 marks)
    For 7iv I integrated between -1 and 1 instead of -1 and 3 do you think I'll get any marks?
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    (Original post by Mr M)
    Mr M's OCR (not OCR MEI) Core 2 Answers May 2016


    1. (i) 10 cm (2 marks)

    (ii) 5.04 cm (2 marks)


    2. (i) \displaystyle \frac{3 \pi}{10} (2 marks)

    (ii) r=20.4 (3 marks)


    3. (i) \displaystyle 27 +27kx+9k^2x^2 +k^3x^3 (4 marks)

    (ii) k=\pm \sqrt{3} (2 marks)


    4. (i) \displaystyle \log_3 \frac{x^2}{x+4} (2 marks)

    (ii) x = 12 (4 marks)


    5. (a) \displaystyle \frac{x^4}{2} -x^3 +2x^2-6x+k (3 marks)

    (b) (i) \displaystyle \frac{2}{a^2} - \frac{6}{a} + 4 (4 marks)

    (ii) 4 (1 mark)


    6. (i) k=91 (3 marks)

    (ii) \displaystyle S_{16} = 978 (2 marks)

    (iii) N=38 (6 marks)


    7. (i) Quotient x^2-4x+3 and remainder 0 (3 marks)

    (ii) x=1 or x=-1 or x=3 (3 marks)

    (iii) Show (2 marks)

    (iv) \displaystyle \frac{512}{15} (4 marks)


    8. (i) Translation by 2 units in the positive x direction (2 marks)

    (ii) Stretch parallel to the y axis scale factor \displaystyle \frac{1}{9} (2 marks)

    (iii) Sketch showing (0, \frac{1}{9}) as the only point of intersection (2 marks)

    (iv) x=6.73 (3 marks)

    (v) 9.60 (3 marks)


    9. (i) \displaystyle \frac{2 \pi}{a} (1 mark)

    (ii) a=5 and k=0 (3 marks)

    (iii) \displaystyle x=\frac{\pi}{3a} and \displaystyle x=\frac{4 \pi}{3a} (4 marks)
    I thought q 8 iii and 9 iii were 3 marks each?
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    I'm almost certainly wrong, but how comee this doesn't work:

    Sin(api/5) =sin(a*2pi/5)

    Thus (double angle theorem)

    Sin (api/5) =2*sin(api/5)*cos(api/5)

    Thus 0.5 = cos(api/5)

    Thus pi/3 = api/5

    Thus 5/3 = a (which is smaller than 5 so is surely the first solution if it is right?
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    (Original post by Jamesk123)
    I thought q 8 iii and 9 iii were 3 marks each?
    Nah 2 and 4.
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    How come when you substitute in 4pie/3a into sin(ax) = root(3)cos(ax) they are not equal - it gives -ve value on the cos side.
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    (Original post by M99)
    I'm almost certainly wrong, but how comee this doesn't work:

    Sin(api/5) =2sin(api/5)

    Thus (double angle theorem)

    Sin (api/5) =2*sin(api/5)*cos(api/5)

    Thus 0.5 = cos(api/5)

    Thus pi/3 = api/5

    Thus 5/3 = a (which is smaller than 5 so is surely the first solution if it is right?
    This is correct.
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    (Original post by ComputeiT)
    Using previous part:
    x^2/(x+4) = 3^2 = 9
    x^2=9(x+4)
    x^2=9x+36
    x^2-9x-36=0
    (x-12)(x+3)=0
    x=12 or -3
    Thank you, I think i got that down in the exam but just realised I made a mistake whilst doing it again just now
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    (Original post by ChrisWeatherilt)
    How did you get the answer to 9ii?
    Wrongly as it happens. It's fixed now.
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    (Original post by bantersaur0usrex)
    Attachment 538249
    You were asked to solve something else though weren't you?
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    (Original post by anon326589)
    How is that the answer to 9ii
    Sorry my mistake. Corrected now.
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    (Original post by M99)
    I'm almost certainly wrong, but how comee this doesn't work:

    Sin(api/5) =sin(a*2pi/5)

    Thus (double angle theorem)

    Sin (api/5) =2*sin(api/5)*cos(api/5)

    Thus 0.5 = cos(api/5)

    Thus pi/3 = api/5

    Thus 5/3 = a (which is smaller than 5 so is surely the first solution if it is right?
    I got a = 5/3 and used a completely different method? Are we certain that a = 5/3 is incorrect?
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    (Original post by Gogregg)
    How did we get a= 5 and k=0 for 9ii ? I think i got a = 5/3 (i think) and k = (√3)/2 or something similar?
    That's correct. I didn't read the question properly. Sorry.
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    (Original post by Willdabeast666)
    I made a quadratic out of question 5 bi) and then factorised it, but earlier in my working out had written down the correct answer, do i lose any marks?
    They might ignore subsequent working.
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    Thanks for this!
    For question 9i if I've written 2pi x 1/a do I still get the mark or did I need it to be simplified?
    How many marks do you think I would get for 9iii. if I rearranged to get tan (ax) = root3 and then did tan^-1 of root3 to get ax equals 60?
    And do you think I would only lose 1 mark for 6iii. if I left N as 37.3?
    Finally, how many marks do you think I would lose for writing kx^2 and kx^3 in 3i. instead of writing k^2x^2 and k^3x^3?
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    Sir for question 3ii would i get any marks for saying that k=1/9 and 0. As well as that for 9i would i get the mark for putting the period as 0<ax<2pia
 
 
 
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