Mr M's OCR (not OCR MEI) Core 2 Answers May 2016 Watch

Mr M
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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) \displaystyle a=\frac{5}{3} and \displaystyle k=\frac{\sqrt{3}}{2} (3 marks)

(iii) \displaystyle x=\frac{\pi}{3a} and \displaystyle x=\frac{4 \pi}{3a} (4 marks)


Sorry guys I made a mess of 9(ii) at the start - I didn't notice k had to be positive. It's right now.
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bantersaur0usrex
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for 4ii) why was x=-3 not a valid answer?
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ChrisWeatherilt
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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)
How did you get the answer to 9ii?
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Mr M
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(Original post by bantersaur0usrex)
for 4ii) why was x=-3 not a valid answer?
Try finding \log_3 (-3) on your calculator.
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S.G.
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How many marks would be lost if I didn't simplify at the end. I left it as (1/3)/a.

Thanks. How do you think the paper was?
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Bobby21231
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For the last question how many marks do you think I would get, I squared everything to get sin^2ax=3cos^2axThen used the identies to get 1-cos^2ax=3cos^2axThen rearranged to get 4cos^2ax=1 ,then cos^2ax=1/4 then square rooted to get cosax=+ or - 1/2Then somehow I ended up with pi/3a and 2pi/3a as my final answer.
Thanks very much
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Mr M
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(Original post by SGHD26716)
How many marks would be lost if I didn't simplify at the end. I left it as (1/3)/a.

Thanks. How do you think the paper was?
So long as you remembered your pi, you might not lose anything.

I thought it was of standard difficulty.
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bantersaur0usrex
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(Original post by Mr M)
Try finding \log_3 (-3) on your calculator.
Name:  Screen Shot 2016-05-25 at 17.09.27.png
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Willdabeast666
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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?
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ComputeiT
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For the last question 9 iii I left my two answers as (1/3)pi/a and (4/3)pi/a - will I still get the full 4 marks?
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Ozil5
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Do you know what the question was for 4ii? Thanks
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Mr M
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(Original post by Bobby21231)
For the last question how many marks do you think I would get, I squared everything to get sin^2ax=3cos^2axThen used the identies to get 1-cos^2ax=3cos^2axThen rearranged to get 4cos^2ax=1 ,then cos^2ax=1/4 then square rooted to get cosax=+ or - 1/2Then somehow I ended up with pi/3a and 2pi/3a as my final answer.
Thanks very much
The trouble with squaring is that it can introduce extra incorrect solutions.

2 marks probably.
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anon326589
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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)
How is that the answer to 9ii)?
How did you do 3ii)? (i did 9k^2=1 how many marks would i get?)
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S.G.
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(Original post by Ozil5)
Do you know what the question was for 4ii? Thanks
2logbase3x - logbase3 (x+4) = 2
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groceryboi
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what was question 6 part i and question 1 exactly. Can't remember the answers that I put. Thanks
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czj1997
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6iii)
I did the sum of Un and the sum to infinity to Wmn.. and I ended up with 3n^2+17n-1200>0 or 3n^2+17n-1200<0 (i forgotten)... getting n=17.4
Did I do the entire question wrong or do I get any method marks?
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Gogregg
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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?
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DaKoolguy
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Thanks sir, for part 5bi. I got the correct answer and wrote it on my exam paper, however for some stupid reason I carried on and made it equal to 0 and tried to solve for a. What will happen here? And for 6iii. I worked it out and got 37.2, but rounded down to 37. How many marks will I get?
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anon326589
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How is that the answer to 9ii)? I got 5/3=a and k=(root 3)/2

How many marks would i get for 3ii) if i did 9k^2=1 and worked that out to get k=+-1/3
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Gogregg
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(Original post by Mr M)
Here's the graph:

Wasnt the question to find the minimum points on the graph, not the x-intersects?
(Sorry for double posting and sorry if I'm wrong :lol:)
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