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# Edexcel C3 - 12.06.06 - The 'how did it go' thread! watch

1. (Original post by Lusus Naturae)
Could have been worse but naturally my mind went blank on a couple of trivial parts, they were "show that...4k+sin4k+2=0" and . Should shoot myself for that Quite sure rest was correct though so around 90UMS I guess.
I still don't get that...
2. Yeah. Exactly.

And can anyone remember the ln question?
was it f--> ln(x+k) and
g--> |2x -k|?
3. (Original post by BovineBeast)
I don't really think so. Negative Infinity isn't a real number.
Guess that makes sense, thanks
4. Think so.
5. is anyone remember how many marks the Teperture lost in per minutes worth? thanks
6. I'm still trying to work out why fg(k/4) is 0.5k...
I got 1.5k :x

The temp one was...4 marks I think. Maybe 3.
7. (Original post by BovineBeast)
Er, that's 1.64.
Ok so -1.64 then :P MB but its def negitive...
8. (Original post by Malik)
Definetly not 56, compared to the Jan06 and Jun05 I think larger majority would have found this paper easier. At best its just going to be the typical 60/75 80% boundary.
Do you really think this paper was easier than the past 2 years?

Besides, we have some foreigners amongst us.. who are probably taking the answers from us right now UP GO THE UMS BOUNDARIES

sigh..
9. (Original post by Rabite)
I'm still trying to work out why fg(k/4) is 0.5k...
I got 1.5k :x

The temp one was...4 marks I think. Maybe 3.
Its comes down to how you treated the modulus in the question. I not sure which way was right so I cant really comment.
10. fg(k/4) IS ln(1.5k) why are people saying its not.
11. i think it is ln(1.5).

well how many marks was that one then? cause they're gone for me now.
12. Ill show you again:

f(x) = ln(x + k)
g(x) = |2x - k|

fg(x) = ln(|2x - k| + k)

Lets x = 0.25k
Ln(|-0.5k| + k) = ln(1.5k)
13. You'll still get method marks for the earler part of the question.
14. (Original post by spacko101)
so absolutely NO ONE remembers how many marks the Sin2A thing was?

5 mark question...
15. (Original post by Sollytear)
Ill show you again:

f(x) = ln(x + k)
g(x) = |2x - k|

fg(x) = ln(|2x - k| + k)

Lets x = 0.25k
Ln(|-0.5k| + k) = ln(1.5k)
Yea thats how I did it.
16. that was quite a challenging paper.. anyone else feel like they rushed through it? i think i finished it.. although i know i didnt get that Sin2A questsion right.. the proving stuff was quite difficult but i think it was managable is u let it sit for a while..
17. (Original post by ellen8899)
is anyone remember how many marks the Teperture lost in per minutes worth? thanks
3 or 2 i fink
18. There were two questions I didn't finish, which totalled 10 marks but I think I got method marks for what I did do

I'll just have to hope there won't be too many silly mistakes in there, as they can cost me dear.
19. i got up to:

f(x) = ln(x + k)
g(x) = |2x - k|

fg(x) = ln(|2x - k| + k)

then i went dumb. and cancelled the k's. lol.
20. ok paper!

does anyone remember how they showed k=0.277 after the iterative questions? i just left it out cos i cud not do it, simple as.

i hope the c4 paper is really eas, it neeeds to be so i can get an A!

and it is Ln(1.5k) definite.

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