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# C3 OCR (Not MEI) June 13th Tweet

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1. Re: C3 OCR (Not MEI) June 13th
which one was question 8? where you had to equate the 2 exponential functions? It was alot harder than any of the past papers, does anyone know how likely a revision of the grade boundaries is?
2. Re: C3 OCR (Not MEI) June 13th
(Original post by XST)
I thought it was the hardest C3 yet. Didn't really know how to do that exponential one where you had to show that T was a root. For question 9, I couldn't for the life of me find the area...I got the integral (2/9)(3x-5)^(3/2) but one of the limits was 0 right, so surely you would get the square root of a minus number? :/
You can equate y function when y = 0 to find x intercept (5/3) to use as bottom limit instead of 0 (then you are not square rooting a negative number, but zero).
3. Re: C3 OCR (Not MEI) June 13th
Do you think Mr M will do a markscheme?
4. Re: C3 OCR (Not MEI) June 13th
(Original post by XST)
I thought it was the hardest C3 yet. Didn't really know how to do that exponential one where you had to show that T was a root. For question 9, I couldn't for the life of me find the area...I got the integral (2/9)(3x-5)^(3/2) but one of the limits was 0 right, so surely you would get the square root of a minus number? :/
That sounds like Q5 (or 6?) to me.

the limit should be 5/3 (Let y=0 and obtain x) as the line doesn't pass through x=0....
5. Re: C3 OCR (Not MEI) June 13th
Well considering that Jan 2011 was supposed to be very hard, and an A was 52/72 I expect the grade boundaries for this one to be about the same or maybe lower? This was definitely harder than Jan 2011, but I guess it all depends on how everyone does.
6. Re: C3 OCR (Not MEI) June 13th
the one about P=10/3 you had to first differentiate the function, to give (a not straightforward) y=mx (c=0 as tangent went through origin) then equate this new differential to the original equation. I thought I was doing it all wrong untill the very last step
7. Re: C3 OCR (Not MEI) June 13th
I'm hoping for low grade boundaries! I've made some stupid mistakes =/ grr
Hope C4 is nicer, got a lot of revision to be doing now though =/
8. Re: C3 OCR (Not MEI) June 13th
52/72 for an A - how many UMS is that?

Can you speculate as the how many marks are required for 90 UMS?
9. Re: C3 OCR (Not MEI) June 13th
(Original post by XST)
I thought it was the hardest C3 yet. Didn't really know how to do that exponential one where you had to show that T was a root. For question 9, I couldn't for the life of me find the area...I got the integral (2/9)(3x-5)^(3/2) but one of the limits was 0 right, so surely you would get the square root of a minus number? :/
No, I thought that at first too but the curve doesnt actually go through zero, it goes through 5/3, so that is the other limit. Which happens to equal zero when you plug it in anyway.

Edit: Oops just seen about 4 people have already explained it, haha.

Anyway if they lower the grade boundaries will the 90% for an A* be lowered or not does anyone know?
Last edited by R0BMAN; 13-06-2011 at 11:17.
10. Re: C3 OCR (Not MEI) June 13th
I FOUND THAT AWFULLLL! compared to the other papers, that involved a lot of thinking, most of the papers you know straight away what had to be done and got to it. but i sat there most of the time, wondering what method to use.. cryyyy..if everyone found it hard, we should do what we did for the chem4 in jan, emailing ocr saying it was too hard! LOL :/
11. Re: C3 OCR (Not MEI) June 13th
(Original post by Mousebudden)
One of the hardest core 3 papers of all time

What the hell was the whole last page about?

The exponential question was also very hard

How do you go about proving P=10/3

Incredibly hard paper!!
I nearly said WTF when I saw the question, it was definitely a mind-blower.
part one was easy though, I just used the booklet and copied the double angle rule and blahblahblah.....

I just hope that grade boundary will be as low as it can be.....
12. Re: C3 OCR (Not MEI) June 13th
(Original post by Mousebudden)
52/72 for an A - how many UMS is that?

Can you speculate as the how many marks are required for 90 UMS?
An A is 80 UMS.
Last edited by millr; 13-06-2011 at 11:18.
13. Re: C3 OCR (Not MEI) June 13th
(Original post by Mousebudden)
One of the hardest core 3 papers of all time

What the hell was the whole last page about?

The exponential question was also very hard

How do you go about proving P=10/3

Incredibly hard paper!!
This sums up my feelings perfectly
14. Re: C3 OCR (Not MEI) June 13th
(Original post by Mousebudden)
One of the hardest core 3 papers of all time

What the hell was the whole last page about?

The exponential question was also very hard

How do you go about proving P=10/3

Incredibly hard paper!!

to show p=10/3....you know that gradient at P and gradient of tangent of curve are equal therefore equate the differentials of the curve to the line OP....the line OP intersects the origin (has y intercept 0) therefore takes the form y=ax

equating differentials means a= expression for differential of curve, then manipulate to find value of a at x=10/3

you then know the equation on OP and then just a case of taking integral from area of rectangle that P makes with x and y axis

I didn't like the last page, (i) I seemed to end up with 2tan(theta)
15. Re: C3 OCR (Not MEI) June 13th
(Original post by R0BMAN)
No, I thought that at first too but the curve doesnt actually go through zero, it goes through 5/3, so that is the other limit. Which happens to equal zero when you plug it in anyway.
?
Yeah I got the limit of 5/3, but wasn't it bounded by the tangent at the point P and then 5/3? The tangent went through the origin so one of them would be 0.

Or did I read the diagram wrong?
16. Re: C3 OCR (Not MEI) June 13th
Found this so so difficult had no idea what to do for the last two parts of question 8? and for question 6 could you find the area of the integral between 10/3 and 5/3 (area under curve) and minus it from the area of the triangle under the line? ie 1/2 x route5 x 10/3? or have i messed that up too..
17. Re: C3 OCR (Not MEI) June 13th
To find the area of the shaded region

Did anyone else do this:

Find area underneath tangent using [(10/3)*(5)^0.5] divide by 2.. (using same method as area of triangle

and then use integration to find area of 'non shaded region'

subtract area of non shaded region from area of triangle to get area of shaded region
18. Re: C3 OCR (Not MEI) June 13th
Oohhhh Dear... Goodbye uni!!
19. Re: C3 OCR (Not MEI) June 13th
(Original post by XST)
Yeah I got the limit of 5/3, but wasn't it bounded by the tangent at the point P and then 5/3? The tangent went through the origin so one of them would be 0.

Or did I read the diagram wrong?
Yeh the limits for the tangent were 10/3 and zero, and the limits for the curve were 10/3 and 5/3. That what you mean?

Also for the equation of the tangent did everyone get something like 300/36root5 after you put the limits in? Before it was like, y = 3x/4root5 or something like that? Struggling to remember what I put....
20. Re: C3 OCR (Not MEI) June 13th
This was quite a difficult paper. Just managed to finish question 6 while the invigilator was collecting the papers
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