You are Here: Home >< Maths

# OCR MEI C3 Maths June 2015 watch

1. (Original post by Garn197)
isn't it reflection in y=x ?
That is what I meant lol
2. (Original post by chizz1889)
Yeah that's right 😊
Think I got that. What was the radius equal to?
3. (Original post by chizz1889)
Did everyone remember to subtract their answer to the integration from 3 for the area between y=1 and x axis? ... ☺️
Completely forgot to do that, how many marks do you think that will have costed?
4. (Original post by dandav33)
Rate one i got 0.023 or something like that

Posted from TSR Mobile
Oh that, i think it was 5/100pi = 0.0159
5. I've done some sort of a mark scheme here: http://www.thestudentroom.co.uk/show...332257&page=13

I just found the question I missed, which was the chain rule one - my answer was 1/(20pi)
6. (Original post by chizz1889)
That is what I meant lol
XD i was worried there for a second haha
7. (Original post by bruce11)
isnt it reflection in y=x?? andhow about inverse
Yeah sorry I meant that, you had to make F(x)->x and the x->f-1(x) and then rearrange to give f-1(x) = ln((x+1)^0.5 +2) or something like that
8. (Original post by pierre225)
yeah I put 0.016 to 3 s.f
Thats 2 sig fig...
9. (Original post by Duskstar)
All the functions that appeared in the paper, as follows:

(I'll try and add questions as I remember them, though I have almost no idea of question numbers :P )

~ differentiate and find the coordinate of the maximum (1.1, 4.1)

~ Integrate for:

~ Integration by parts for something I can't quite remember, but someone will bother to work out or ask about...

~ verify point (1, 1) which means sub it in and prove it still equates, then differentiate implicitly to get the gradient at that point, with was 1/2.

~ prove f(f(x)) = x, which involves subbing f(x) in in place of each x, then rearranging.
~ hence write down the inverse of f(x), which is f(x) (I personally don't understand why this was a 'hence' question - I haven't seen a rule that would describe this before, although it's obvious there is one)

~ show g(x) is even - sub (-x) for x, and show that is the same as g(x). This means that g(x) has a line of symmetry in the y-axis, or x = 0.

~ you had to differentiate this twice. I found it easier to expand the top and divide through by x than use the quotient rule, but either should work and I think the answer was f'(x) = 1 - 4x^(-2), then differentiate again for f''(x) = 8x^-3 you should fin the point Q(-2, -8), then show it is a maximum because f''(x) < 0

~ I think this was the question where you had to integrate to find the area enclosed by the function and the line y = 1 (after proving the intersections were x = 1 and 4), which was easy enough - either use a rectangle, or take the integral of 1 - f(x). I think my answer was 15/2 - 4ln4

translate (-1, -1) for:

~ you had to show this, which just meant showing f(x + 1) - 1 = g(x).
~ then, it asked you to write down the integral from 0 to 3 of g(x) with no further calculation, which is the integral from the second part (I don't know if they would want you to indicate that it was this value but negative), and then describe why this was the case. I guess you just talk about the translation and stuff ~

~ I can't remember quite what the questions were here, but I think you had to differentiate it, find x-intercept at ln3, and minimum at (ln2, -1)

I got for the inverse:

~ then sketch the inverse by reflecting in y = x. The easy way to do this was plot the images of Q and P, and make sure you cross the line y=x in the same place, and continue the line without changing gradient much (it was pretty much a flat line past the intersection with y=x).

and somewhere in the paper, two parts to a question:

Domain was -1<= x and range was ln2 <= f-1(x)
10. (Original post by StudentH817)
Thats 2 sig fig...
no it isn't
11. (Original post by JamesExtra)
Think I got that. What was the radius equal to?
12. I got really flustered after the first question and reckon I dropped 20 marks at least despite getting A*s in past papers. I want to die.
14. (Original post by hukdealz)
no it isn't
.016 is three decimal places but 2 sig fig as the 0 after the decimal place is not significant. Just as the 160 would be 2 sig fig not three.

Not that it really mattera that much.
15. (Original post by Garn197)
Ok now I'm feeling very confident I did well on this. Thanks
16. (Original post by StudentH817)
.016 is three decimal places but 2 sig fig as the 0 after the decimal place is not significant. Just as the 160 would be 2 sig fig not three.

Not that it really mattera that much.
Hallelujah !!!

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 20, 2016
Today on TSR

### AQA Mechanics Unofficial Markscheme

Find out how you've done here

### 2,644

students online now

Exam discussions

### Find your exam discussion here

Poll
Useful resources

Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE