C3 Jan 11 Edexcel - Solutions and Paper in the first post (Now On) + TIPS

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Reply 60

Bleo

Yeah gonna attend this exam. (: I generally like doing them, but i havent done any intensive practices yet. Soo... yuh.

Reply 61

TobeTheHero

Original post by anshul95

Well I was helping someone the other day on the OCR syllabus (not MEI) for the
C3 Jan 2010 paper. Although it had some C4 edexcel in it, it had some great C3 questions. They weren't particularly hard but they were worded differently to what we would get for Edexcel.

care to post the paper...? I wanna try it :smile:

Reply 62

anshul95

here they are then

ocr_52277_pp_10_jan_gce_4723.pdf64.4KB

ocr_52509_ms_10_gce_jan.pdf661.5KB

Reply 63

1Unit1

I'm not feeling good about this. Flicked through some past papers and I really have very little idea how to answer these queations.

I have a week to rectify this, are there any particular topics which make up the bulk of the exam questions? Like differentiation and integration in C2. Which chapters in the Edexcel text book should I be looking to master?

Reply 64

Ultimate1

Original post by 1Unit1

Well C3 is an exam where you have a nice mixture of all the questions from each chapter and no specific chapter dominates. However I strongly recommend 'mastering' Trig, Logs and exponentials and Differentiation.

Reply 65

TobeTheHero

Original post by anshul95

here they are then

went through it and isn't particularly hard(the stuff we need to know for edexcel)

Reply 66

mir3a

Shall have this exam raped

Reply 67

Patelios

Original post by jam277

Yeah get the hard questions infact solomon press ones if you can. I'll find some hard ones as well. Gonna aim for at least 94 so I beat my other modules.

do you know where i could get them papers i really need them :frown:

C3 i need like 90UMS

Reply 68

soutioirsim

Original post by mir3a

Shall have this exam raped

hard

Reply 69

anshul95

Original post by TobeTheHero

went through it and isn't particularly hard(the stuff we need to know for edexcel)

I never said it was a lot harder - just they asked the questions in a different way.

Reply 70

darpan_r

Original post by jam277

You're right it is actually simple just a bit of identities, it took me a long time to do this. This would be the question I'm bogged on for most of the exam.
dx/dy=sec^2(y)
dy/dx=1/sec^2(y)
dy/dx=1/tan^2(y)+1
dy/dx=1/x^2+1
I need to get better at identities I only know cos^2+sin^2=1 and I convert from that.
What year was that paper?

it was in Jan 2010 paper. i spent my last 20min in this question, but couldn't solve it.

Reply 71

doobucks

Original post by darpan_r

it was in Jan 2010 paper. i spent my last 20min in this question, but couldn't solve it.

Oh ok. They never showed me how to do a question like that. My teachers are so unreliable.

Reply 72

doobucks

Original post by Patelios

do you know where i could get them papers i really need them :frown:

C3 i need like 90UMS

I'll send them to you. They're on my school website, the school didn't even tell us about them.

Reply 73

M_I

Hey guys

Can someone help me on a few questions. All from June 2006.

Question 6. b). I don't get it at all.
Question 7. a). The graph of g(x). I got the point (k, 0), not (k/2, 0).
Question 8. a).

Thanks.

Reply 74

Freddo:)

Original post by M_I

For 6 b) look at cosec^4(x) - cot^4(x). Factorise it first..
[cosec^2(x) + cot^2(x)] multiplied by [cosec^2(x) - cot^2(x)]

rearrange identity cot^2(x) + 1 = cosec^2(x) and you get the second bracket to equal 1..
therefore cosec^4(x) - cot^4(x) is equivalent to [cosec^2(x) + cot^2(x)] which is the right hand side.

I'll do the other two in a minute? :biggrin:

Hope that helps..

Reply 75

Freddo:)

Original post by M_I

As far as 7 a) is concerned, you need to set 2x-k (all in a modulus) = 0.
from there you solve it to find x, which is x=k/2

I hope that helps,
sorry if I sound patronising,
it's difficult to explain it on a computer :biggrin:

Reply 76

Freddo:)

Original post by M_I

8 a) So like, you know what CosA is, so you can find sinA by using a triangle..
So you'd do CosA = Adj/Hyp. You know these two values, so therefore you can work out what the missing one is, the Opposite of the angle A. (you do this by rearranging pythagorus, a^2 + b^2 = c^2)
you should get the opposite to equal root7.
in the question it says the angle A is between 270 and 360.. therefore sinA MUST be negative. (CAST diagram and/or general sinx graph)
therefore sinA = opp/hyp.. so sinA = root7/4, but because it's negative it would be..
sinA = -root7/4

so sin2A = 2sinAcosA,
therefore sin2A = -2(root7/4)x(3/4), which comes to -3root7/8

Good luck on Monday and I hope this helped.
Freddie :smile:

Reply 77

doobucks

Original post by Patelios

do you know where i could get them papers i really need them :frown:

C3 i need like 90UMS

Sorry I tried emailing it to you but the file won't attach.

Reply 78

M_I

Original post by Freddo:)

Thanks so much. Mainly silly mistakes.

For question 7, I drew y=2x and then y=2x - k (i shifted it to the right, but that would be y= (2x + k) right?)

Reply 79

Freddo:)

Original post by M_I

Thanks so much. Mainly silly mistakes.

For question 7, I drew y=2x and then y=2x - k (i shifted it to the right, but that would be y= (2x + k) right?)

no wait, the 'k' part you're talking about there is a movement downwards by 'k'.
i.e. f(x) = 2x, so what you're doing there is saying f(x) -k, which is not the same as a horizontal shift, in which you'd be doing f(x+k) to get that shift in the <--- x direction.
I see what you're saying, but I think you're confusing the two..
Because there'll still be a value of x for when y=0 if you shift y= 2x downwards by k, if you see what I'm saying.
It's hard to explain!! :colondollar: