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

Mr M
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#41
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#41
(Original post by ComputeiT)
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?
I expect so.
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Saching99
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#42
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If i did the last question not in terms of A but rather just 4Pi/5 and Pi/5, how many marks would i lose?
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Mr M
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#43
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#43
(Original post by Ozil5)
Do you know what the question was for 4ii? Thanks
Solve 2 \log_3 x - \log_3 (x+4)=2
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DaKoolguy
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#44
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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) \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.
Thank sir. For 5ib I wrote the correct answer down in the exam booklet however for some stupid reason I made it equal 0 and tried to solve for a. How many marks would that be? And for 6iii. I worked it out and got 37.2 but rounded down to 37. What will I get?
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Willdabeast666
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#45
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#45
(Original post by Mr M)
They might ignore subsequent working.
So might i lose marks
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Bobrey
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#46
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(Original post by Mr M)
Solve 2 \log_3 x - \log_3 (x+4)=2
For the last part of the last queston, why when subbing 4pi/5a into the equation do the values not equate - the cos part is negative
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Mr M
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#47
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#47
(Original post by czj1997)
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?
N needs to be an integer so you round up and your quadratic is wrong.

You should have obtained 3N^2+17N-4800.

You would get 3 marks I expect.
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Mr M
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#48
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#48
(Original post by Gogregg)
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:)
It wasn't either of those. I've corrected my solution.
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Mr M
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#49
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#49
(Original post by Ozil5)
Can someone please explain how to do 4ii, I know ive made a mistake somewhere but im not sure where
\frac{x^2}{x+4}=9

x^2-9x-36=0

x=12 or x=-3 but the second solution has to be rejected as it does not satisfy the original equation.
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Mr M
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#50
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#50
(Original post by Jamesk123)
I thought q 8 iii and 9 iii were 3 marks each?
No.
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swagmister
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#51
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#51
(Original post by Mr M)
N needs to be an integer so you round up and your quadratic is wrong.

You should have obtained 3N^2+17N-4800.

You would get 3 marks I expect.
I had that answer but wrote 17 instead of -17 when using the quadratic formula so i had 38 as negative and 43 as positive so i wrote N=43 how many marks would I lose?
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Mr M
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#52
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#52
(Original post by ellaatherz)
For 7iv I integrated between -1 and 1 instead of -1 and 3 do you think I'll get any marks?
Yes 2 marks.
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swagmister
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#53
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#53
(Original post by Mr M)
\frac{x^2}{x+4}=9

x^2-9x-36=0

x=12 or x=-3 but the second solution has to be rejected as it does not satisfy the original equation.
Would I lose a mark for including -3?
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Mr M
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#54
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#54
(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?
Interesting method. I did it wrong first time. It's corrected now. I didn't notice that k had to be positive.
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Mr M
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#55
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#55
(Original post by swagmister)
Would I lose a mark for including -3?
Yes.
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Mr M
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#56
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#56
(Original post by AlfieH)
Question 9ii - the question said that k must be a positive constant and 0 isn't a positive constant?
Yes I fixed it some time ago.
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cc262626
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#57
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#57
If it did for the period question
0 is greater than or equal to ax is less than or equal to 2api is this wrong
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groceryboi
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#58
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Do you lose marks if you forget to put cm?
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AlfieH
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#59
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(Original post by Mr M)
Yes I fixed it some time ago.
And I posted some time ago!
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Mr M
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#60
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#60
(Original post by Bobrey)
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.
They are equal. Both sides are negative.
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