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

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#1
4
5 years ago
#2
for 4ii) why was x=-3 not a valid answer?
0
5 years ago
#3
How did you get the answer to 9ii?
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#4
(Original post by bantersaur0usrex)
for 4ii) why was x=-3 not a valid answer?
Try finding on your calculator.
0
5 years ago
#5
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?
0
5 years ago
#6
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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#7
(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.
0
5 years ago
#8
1
5 years ago
#9
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?
1
5 years ago
#10
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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5 years ago
#11
Do you know what the question was for 4ii? Thanks
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#12
(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.
0
5 years ago
#13
How is that the answer to 9ii)?
How did you do 3ii)? (i did 9k^2=1 how many marks would i get?)
0
5 years ago
#14
(Original post by Ozil5)
Do you know what the question was for 4ii? Thanks
2logbase3x - logbase3 (x+4) = 2
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5 years ago
#15
what was question 6 part i and question 1 exactly. Can't remember the answers that I put. Thanks
0
5 years ago
#16
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?
0
5 years ago
#17
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?
0
5 years ago
#18
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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5 years ago
#19
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
0
5 years ago
#20
(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 )
0
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