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OCR (Not MEI) Core 3 - Post Exam Discussion

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Reply 220
On question 9ii, i got the right answer but i didn't really show much working, and was just throwing values in and seemed to be getting close to 4. I was in too much of a rush, will I get marks for that?
Reply 221
Original post by Mr M
Hi all. I have been away from school for 3 days and only got home late last night. If anyone still needs answers I can produce them this morning or have you already got it sorted? Someone has pmed me with the questions although I would prefer to work from a copy of the real paper if anyone has one.


It would be great if you could do them and also show us how much each question was worth, Thanks!
Reply 222
Original post by Mr M (jr)
All of the above are correct. :thumbsup:

There was also another part to 5(i): f1(8)=4f^{-1}(8)=4.

The Simpson's rule answer (5(iii)) was indeed 206 as others have already said.


Ah yes, I forgot to type that one in. I'll amend. Thanks.
Original post by gossip999
It would be great if you could do them and also show us how much each question was worth, Thanks!


I can't show you how much each question is worth without seeing a paper. I may hang on until one appears if that is ok with you?
Reply 224
Original post by Mr M
I can't show you how much each question is worth without seeing a paper. I may hang on until one appears if that is ok with you?


I'm just scanning mine in - I'll send it to you when it's done.
Original post by dybydx
I'm just scanning mine in - I'll send it to you when it's done.


Fine - thanks.
Reply 226
Original post by dybydx
I'm just scanning mine in - I'll send it to you when it's done.


Would you mind posting the scans up too. Could be useful to see where we went wrong?

Although I do hate knowing that I could have actually done a question, but in the actual exam I was clueless! :mad:
I wonder if I'll get all of the marks for the last question. I did it a weird way.

|k(x²+4x)| = 20 where k is a positive constant. There are 3 distinct x values that make this true. Find k and then find all 3 x values.

What I did...

kx² + 4kx = -20 or kx² + 4kx = 20
So kx² + 4kx + 20 = 0 or kx² + 4kx - 20 = 0

Then I did discriminants for both.

16k² - 80k > 0 or 16k² + 80k > 0 (yes, I put > sign at first, even though that implies 4 roots)

But then I realised I have 2 quadratics but only 3 roots. So I put "only way for that to happen is if 1 quadratic has 2 roots and the other has a repeated root / is a perfect square".

So I found what values of k give each equation with 1 root. And it turned out to be k=-5,0,5.

So I tested each and realised it was 5. Is that gonna drop me method marks?

I got the x values afterwards...
(edited 12 years ago)
Reply 228
Original post by TooEasy123
I wonder if I'll get all of the marks for the last question. I did it a weird way.

|k(x²+4x)| = 20 where k is a positive constant. There are 3 distinct x values that make this true. Find k and then find all 3 x values.

What I did...

kx² + 4kx = -20 or kx² + 4kx = 20
So kx² + 4kx + 20 = 0 or kx² + 4kx - 20 = 0

Then I did discriminants for both.

16k² - 80k > 0 or 16k² + 80k > 0 (yes, I put > sign at first, even though that implies 4 roots)

But then I realised I have 2 quadratics but only 3 roots. So I put "only way for that to happen is if 1 quadratic has 2 roots and the other has a repeated root / is a perfect square".

So I found what values of k give each equation with 1 root. And it turned out to be k=-5,0,5.

So I tested each and realised it was 5. Is that gonna drop me method marks?

I got the x values afterwards...


That seems like a perfect answer to me. You used logic to solve the problem based on the information given - this is what they are trying to encourage and reward. I did it by a sketch, which was a bit quicker but came to the same answer.

I guess we won't know till the mark scheme is out, but I think that would be full marks.
Reply 229
Original post by TooEasy123
I wonder if I'll get all of the marks for the last question. I did it a weird way.

|k(x²+4x)| = 20 where k is a positive constant. There are 3 distinct x values that make this true. Find k and then find all 3 x values.

What I did...

kx² + 4kx = -20 or kx² + 4kx = 20
So kx² + 4kx + 20 = 0 or kx² + 4kx - 20 = 0


Then I did discriminants for both.

16k² - 80k > 0 or 16k² + 80k > 0 (yes, I put > sign at first, even though that implies 4 roots)

But then I realised I have 2 quadratics but only 3 roots. So I put "only way for that to happen is if 1 quadratic has 2 roots and the other has a repeated root / is a perfect square".

So I found what values of k give each equation with 1 root. And it turned out to be k=-5,0,5.

So I tested each and realised it was 5. Is that gonna drop me method marks?

I got the x values afterwards...



I did up to the bit in bold then lost faith, would I get any marks lool?
Reply 230
For the question about what rate was it decreasing, I got the right answer of 2.788 or w/e it was. But i just put THAT value. with working obv... but would you have to say 'DECREASING at a rate of 2.788' or will the examiner be a top bloke and give me the full marks? :frown:
Reply 231
My scan of the paper exceeds the upload limit, so I'll upload as two files.
Original post by Tukazoar
I did up to the bit in bold then lost faith, would I get any marks lool?


Same :frown:
Reply 233
Second part
Reply 234
For Question 8
i) i put 1-4sin^2xcos^2x instead of 8, cos i didn't really think about it during the exam (3 marks)
ii)I got 1-cos4x value, but forgot to divide by 8 at the end (2 Marks)
iii) I did it completely the wrong way, but attempted to use earlier results and some trig idents(5 marks)

How many will i lose in total(give it in a range if you arent sure). This is the only question i screwed up on, rest is perfect.

Looking to get as close to 100 ums as i can. Is this still possible for me?
Think I messed a lot of this paper up :\ any idea what about 45-48 would be as a grade? :s-smilie: Not entirely sure exactly what I will have got as I don't remember my answers to a few questions, but im guessing around 50 at the most? :\ thanks :smile:
Reply 236
Original post by jodiedoodles
Think I messed a lot of this paper up :\ any idea what about 45-48 would be as a grade? :s-smilie: Not entirely sure exactly what I will have got as I don't remember my answers to a few questions, but im guessing around 50 at the most? :\ thanks :smile:


well 50/72 is around 70%. which would probably get a few more UMS so around 76? - B
Original post by jodiedoodles
Think I messed a lot of this paper up :\ any idea what about 45-48 would be as a grade? :s-smilie: Not entirely sure exactly what I will have got as I don't remember my answers to a few questions, but im guessing around 50 at the most? :\ thanks :smile:


With last year's grade boundaries, 47 would get you 72 UMS. Probably a B (unless the grade boundaries are unexpectedly high or low) :smile:
Did you think that this paper was harder than june 2011???
Reply 239
Original post by Smiley Face :)
Did you think that this paper was harder than june 2011???


I don't, I got 62/72 on june 11, and got in the low 50s on this. :frown:

But i have seen quite a few people said they thought it was MUCH easier.

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