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

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2 years ago
#21
Can someone please explain how to do 4ii, I know ive made a mistake somewhere but im not sure where
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2 years ago
#22
(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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2 years ago
#23
(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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2 years ago
#24
Question 9ii - the question said that k must be a positive constant and 0 isn't a positive constant?
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2 years ago
#25
Do you know what question 9 i and ii were?
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2 years ago
#26
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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2 years ago
#27
I thought q 8 iii and 9 iii were 3 marks each?
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2 years ago
#28
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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2 years ago
#29
(Original post by Jamesk123)
I thought q 8 iii and 9 iii were 3 marks each?
Nah 2 and 4.
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2 years ago
#30
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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2 years ago
#31
(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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2 years ago
#32
(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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Thread starter 2 years ago
#33
(Original post by ChrisWeatherilt)
How did you get the answer to 9ii?
Wrongly as it happens. It's fixed now.
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Thread starter 2 years ago
#34
(Original post by bantersaur0usrex)
Attachment 538249
You were asked to solve something else though weren't you?
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Thread starter 2 years ago
#35
(Original post by anon326589)
How is that the answer to 9ii
Sorry my mistake. Corrected now.
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2 years ago
#36
(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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Thread starter 2 years ago
#37
(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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Thread starter 2 years ago
#38
(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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2 years ago
#39
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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2 years ago
#40
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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