OCR C4 (not mei) 18th June 2013 revision

I thought i'd start a thread for C4 OCR.

Can anyone help me on this question:

Expand binomially: 3/(2-x) + 3(1+x) +1/((1+x)^2)

Well i know you have to convert it into (1-x) for the first one... but yeah...

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Reply 1

Mr M

Study Forum Helper

Original post by master y

(2-x)^{-1}=2^{-1}(1-\frac{x}{2})^{-1}

Reply 2

erniiee

Need 91 in this for an A*!

Reply 3

master y

Original post by erniiee

Need 91 in this for an A*!

I need 93

Reply 4

tinkerbell_xxx

I haven't even started revising :frown:

Reply 5

erniiee

Original post by master y

I need 93

So do you think its true? That C4 is "easier" than C3? So far I much prefer C3 :lol:

Reply 6

master y

Original post by erniiee

So do you think its true? That C4 is "easier" than C3? So far I much prefer C3 :lol:

I prefer C3 too, and i've heard C4 can be a right pain in the backside sometimes. Some of my friends sat the january exam, and came out with appalling results, and they are going to study maths/economics/engineering at oxbridge! I'm just a mere medic :afraid:

Reply 7

erniiee

Original post by master y

Ah not what I want to here :lol:

Same here, I actually need 1 A* to meet my offer - wasn't originally expecting it from Maths but I think it's achievable!

Reply 8

Big Bad Boss

Anyone got a good website for C4 Vectors? Any other decent resources for C4 are appreciated too :smile:

Reply 9

master y

Original post by Big Bad Boss

Anyone got a good website for C4 Vectors? Any other decent resources for C4 are appreciated too :smile:

you may have already seen this but... http://www.examsolutions.net/maths-revision/syllabuses/OCR/period-1/C4/module.php

Reply 10

master y

Does anyone know how to answer june 2008 Q4ii, and Q9iii please?

http://www.ocr.org.uk/Images/62967-question-paper-4724-core-mathematics-4.pdf

thanks!

Reply 11

ScarlettFierce

Haven't done C3 like most people have... But I'm doing this paper... Nervous as I need to maintain my A by receiving an average of high Bs in S1 (resitting), C3 & C4... Going to try and aim for the A* though!

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Reply 12

master y

Original post by erniiee

Ah not what I want to here :lol:

Same here, I actually need 1 A* to meet my offer - wasn't originally expecting it from Maths but I think it's achievable!

I feel stupid for not doing this but how do you simplify,

(20-5x)/(6x^2 - 24x)

Reply 13

Quest

Original post by master y

I feel stupid for not doing this but how do you simplify,

(20-5x)/(6x^2 - 24x)

\frac{{20 - 5x}}{{6{x^2} - 24x}} = \frac{{ - 5(x - 4)}}{{6x(x - 4)}} = - \frac{5}{{6x}}

Reply 14

erniiee

Original post by master y

I feel stupid for not doing this but how do you simplify,

(20-5x)/(6x^2 - 24x)

As above, thanks Quest!

Reply 15

master y

Original post by Quest

\frac{{20 - 5x}}{{6{x^2} - 24x}} = \frac{{ - 5(x - 4)}}{{6x(x - 4)}} = - \frac{5}{{6x}}

ahhh cheers dude!

Reply 16

master y

sorry, i seem to be the only one who has questions haha!

A curve has parametric equations x= 3

cos^2

(t), y = sin(2t)

what is dy/dx? I cant differentiate dx/dt for some reason... well i wrote it in terms of sin(2t).. which is

\frac{3}{2}

\frac{3}{2}

Unparseable latex formula:

\sin \2

2t... but it doesn't work! I'm supposed to get dy/dx =

\frac{-2}{3}

(cot(2t))

Original post by erniiee

Original post by Quest

urgh, and how do you write in this latex thing quickly! I have to keep copying and pasting...

(edited 11 years ago)

Reply 17

JASApplications

I'm doing C4 in June! It will be my last ever A-Level exam!!! (Hopefully!!)

Personally I think it's quite a bit easier than C3! I got 79/100 in C3 and I need an A overall so get into Uni so anything over 80 in C4 will be good!

So far:
C1: 79, C2: 84, C3: 79, C4: ?, D1: 62 (Re-sitting), S1: ?

Reply 18

master y

please someone help!! last question June 2009... HOW DO I DO THIS?? I found C, but do i need to find k?
IM DESPERATE.

Reply 19

master y

Does anyone know how to know the value of x for which the expansion is valid. Eg. (1+4x)^(3/2)
I got x<= -1/4 but the mark scheme says (mod x )< 1/4.... HOW DOES THIS WORK? aaaaaaaarghhhhhhhh

Thanks