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OCR (not MEI) C1 - Wednesday 13th May 2015

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Basically
I used the actual booklet to do the majority of my working and answers and I used the continuation booklet to redo/check some of my answers to absolutely make sure I didn't make a mistake. I used the continuation booklet to actually answer q8 part i as I messed up the graph the first time round. I did put 'check' in brackets beside each question I was checking and starred question 8 to say I was actually reattempting it. Now, someone is telling me that it is 'risky' I redid/checked my answers in this booklet and that the two sets of working may not be credited as it 'doesn't match'. Is this true? I would've thought the examiner would realize I was checking my answers as I circled/underlined most of my actual answers in the original booklet...
Would appreciate it if someone could get back to me on this :smile:
Original post by MrDilanPatel
Basically
I used the actual booklet to do the majority of my working and answers and I used the continuation booklet to redo/check some of my answers to absolutely make sure I didn't make a mistake. I used the continuation booklet to actually answer q8 part i as I messed up the graph the first time round. I did put 'check' in brackets beside each question I was checking and starred question 8 to say I was actually reattempting it. Now, someone is telling me that it is 'risky' I redid/checked my answers in this booklet and that the two sets of working may not be credited as it 'doesn't match'. Is this true? I would've thought the examiner would realize I was checking my answers as I circled/underlined most of my actual answers in the original booklet...
Would appreciate it if someone could get back to me on this :smile:


It would all depend on your quality written communication. If you made it easy for the examiner to see then you could possible be okay.
Hi - quick question, I'm trying to solve this disguised quadratic (that came up in a C2 paper?!) but I'm having trouble matching the answer. So here's the quadratic that I've been given:

Find the value of k
15 = k + 2k1/2
where k is a constant in y = 4kx such that k > 1

My solution so far:

let k1/2 = x

15 = x2 + 2x
x2 + 2x - 15 = 0
(x+5)(x-3) = 0

x = -5 or 3
k = (-5)2 or 32

k= 25 or 9

However the question says give the value of k, so only one of these two solutions is correct. I know which one from looking at the MS, but can someone explain why both are not valid? They are both > 0, so I don't get it :redface:
(edited 8 years ago)
Original post by ADotCross
It would all depend on your quality written communication. If you made it easy for the examiner to see then you could possible be okay.


But do you think it will be a huge issue? I didn't put anything incorrect down tbh its just some questions may have repeated working as I was redoing it to check
Hello Mr M, this is for the edexcel c1 ecam. I would really appreciate an answer! For the differentiation question? I put 6x instead of 6x^ -1 and therefore my dy/dx was wrong. And when I had to sub -1 the method was entirely correct however I used the same 'incorrect dy/Dx' will I not get ecf marks!???
Original post by gabby07
Hi - quick question, I'm trying to solve this disguised quadratic (that came up in a C2 paper?!) but I'm having trouble matching the answer. So here's the quadratic that I've been given:

Find the value of k
15 = k + 2k1/2
where k is a constant in y = 4kx such that k > 1

My solution so far:

let k1/2 = x

15 = x2 + 2x
x2 + 2x - 15 = 0
(x+5)(x-3) = 0

x = -5 or 3
k = (-5)2 or 32

k= 25 or 9

However the question says give the value of k, so only one of these two solutions is correct. I know which one from looking at the MS, but can someone explain why both are not valid? They are both > 0, so I don't get it :redface:



15 = k + 2k^1/2

25 doesnt work for this, 9 does
Original post by gabby07
Hi - quick question, I'm trying to solve this disguised quadratic (that came up in a C2 paper?!) but I'm having trouble matching the answer. So here's the quadratic that I've been given:

Find the value of k
15 = k + 2k1/2
where k is a constant in y = 4kx such that k > 1

My solution so far:

let k1/2 = x

15 = x2 + 2x
x2 + 2x - 15 = 0
(x+5)(x-3) = 0

x = -5 or 3
k = (-5)2 or 32

k= 25 or 9

However the question says give the value of k, so only one of these two solutions is correct. I know which one from looking at the MS, but can someone explain why both are not valid? They are both > 0, so I don't get it :redface:


k^(1/2) is, by definition, positive. So -5 doesn't work. Here's a link to one of Mr M. 's threads http://www.thestudentroom.co.uk/showthread.php?t=3173187
Thank you both :smile:
Original post by nobodycarescarla
Can anyone remember 3iii? I can't remember the question or what I put as my answer so if someone could remember the question I can work out what I put XD thankyouu 😊


Is that the surds one, 5 root 5 cubed?
Original post by Rubberduckiller
thnx so does that mean ill get 2 or 3 marks


Not sure sorry.
Hey all! What would 71/72 get me UMS wise do you think? I'm slightly concerned that I wrote 1/48 instead of -1/48 for a question. I might not have done but I don't remember. Overall I thought is was a pretty simple paper though :smile:
Reply 631
Original post by ChoccyPhilly
We had 3 months to learn all of fp1, so we resorted to 1 chapter a day (last year). At first I hated it but after a while when I FINALLY started to understand it, it became so interesting and i'd go as far as to say that it's the most interesting maths i've ever done and it really did fuel my interest for it further. Kind of took me back to gcse days when maths was so easy and fun and all I wanted to do...

And then I got a D in the fp1 exam cos the exam was balls up. Everyone else got Us and an E in my school.


I wouldn't go as far as saying it's all I want to do, but it's not *too* bad anymore. Messed up the first part of the last question on the exam today and then needed my answer for that to get the full part of the second half correct but hoping I'll get method marks. Could have gone worse anyway :smile: Though I was talking to some people in the year above at my school and apparently FP3 is an absolute killer, so I'm really not looking forward to that.
Reply 632
Original post by Tobybeta
Hey all! What would 71/72 get me UMS wise do you think? I'm slightly concerned that I wrote 1/48 instead of -1/48 for a question. I might not have done but I don't remember. Overall I thought is was a pretty simple paper though :smile:


I thought it was an easy paper too but I know quite a few people in my school and around the country messed up some questions (particularly the last) so I'd imagine the mark needed for full UMS would be 71. Otherwise it'll be very close to 100 UMS so you'll be fine!
Reply 633
Original post by nurav11
Yup exactly that, it's such a nice easy calculator to use. Had it since year 5 as well so it's trusted now :tongue:. I'm really hoping for a matrices induction Q and easy roots stuff. They'll make or break how well I do really. Btw June 09 had 49 as the A boundary but it really wasn't that challenging... Very strange



100% agree, there's a satisfaction about getting a summing series question right or an induction question right, especially if it's a nice and long 7 marker.

Take it in! Honestly no one is going to say a thing, I know someone who's walking in with a huge graphic calculator that has a home screen... It's basically a computer. They are definitely allowed. The only thing that's not allowed is if it can do indefinite integration but that doesn't even apply to FP1 so you're fine :smile:

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I stuck with my trusty basic calculator. Probably quite lucky because (for the first time ever) they were checking makes of calculators as you went in. Though it was a French teacher doing this so she probably wouldn't have known any different if I'd said it was allowed... Luckily I got the matrix question right, or at least judging by everyone else's answers I did. How did you find it?
In c2, in a geometric series does a always need to be positive? Just asking as i am confused with an answer to q7b in june 07 paper, i have two values of r and yh jts confusing ah
Original post by Tobybeta
Hey all! What would 71/72 get me UMS wise do you think? I'm slightly concerned that I wrote 1/48 instead of -1/48 for a question. I might not have done but I don't remember. Overall I thought is was a pretty simple paper though :smile:


Yeah... I think I made a similar 1 mark error, but it could just be the post exam 'Oh crap did I remember to do this?' I would estimate at around 96/97 UMS, probably not 100 though because it wasn't a hard paper.
Reply 636
Your problem does not lie with a. It is because the sum to infinity is finite I.e for the sum to infinity to work r has to be restricted where -1<r<1.

If you look at the formula booklet it says so right next to the formula for the sum to infinity in a geometric series

So imagine that one of your values for r was in the range of -1<r<1 and the other one wasn't, so you must discount this one.
http://youtu.be/0EW8N60kHkU

Some subtitles are off but I hope you enjoy it.
Who can remember the question 4)?

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