If you’re still looking for some past papers to do, then you can find loads to download here (scroll down to the Edexcel ones) and there are also lots of practice papers to download from this site from this post - usual password needed!
Also, if you want to look at some Powerpoints on any C2 topics, you can download them here. Again, usual password.
If you need help, shout here or don’t hesitate to come in. We’ve cleared some time to offer a full afternoon of C2 help on Thursday, from 1:30 so come with any questions, papers to mark etc.
Hello, just a few questions:
from the Algebra and Functions Chapter (Chapter One)
Mixed Exercise 1F
Question 14:
(a) Show (x-2) is a factor of f(x) = x^3 + x^2 – 5x – 2
so I use the factor theorem and say:
f(2) = (2)^3 + (2)^2 – 5(2) – 2 = 0
To prove that (x-2) is a factor of f(x) = x^3 + x^2 – 5x – 2
But it is the second part of the Question that I do not understand:
(b) Hence, or otherwise, find the exact solutions to the equation f(x) = 0
Oh actually, I might have just got the answer.
Is it asking for the ‘roots’ – as in where, if you were to draw this equation as a line on a graph they’d cross the x-axis?
If that’s right, then it solves my other question too…
Also just about the powerpoints. Firstly, thank you, they’re so fun with the little interactey bits – Very exciting! But also, some of them (Differentiation, Sequences and Series and trig 2) aren’t downloading properly, and I was wondering whether anyone else was having this problem too? It’s not really a big issue, becasue I’m fine with the resources I have at the moment, but I thought I’d just ask to check?
Thank you,
Grace
Hi Grace,
The roots are the x values that give f(x)=0 ie where graph meets x axis (crosses or touches), so Yes is the answer to your question.
If you know that (x-2) is a factor, then you can divide the cubic equation by x-2 which will leave you with a quadratic which you can then factorise to get the other roots.
Tea Lady
Thanks
Sorry if I wasted your time a bit; as I was writing the question I kind of worked it out.
Have a nice evening!
When should I put my calculator into radians mode? When doing any question involving radians (i.e. some of the trig questions they ask) or just the ones relating to the Radians chapter…
I keep on forgetting to take it outof/put it into radians mode and am scared that I’ll do it in the exam.
Thank you.
You’ll need radians more than degrees probably – but just RTQ!!
I’d keep it in Deg mode, and just switch it into Rads for those questions, and then deliberately switch it back at the end. Or take two calculators!
Or mark up the paper quickly in the first minute or so with reminders…
lol!
hello im a bit stuck on binomial expansion.
in the book example 3 (chapter 5)
im doing mr burfords way of nCr and i got right answers up to -810 but for next answer i got -540?? and they were wrong from there.
This is how i did it:
3^5 =243
(5/1)*3^4*-2x = -810
(5*4/1*2) *3^3*-2x = -540x^2
however the book says 1080x^2????
please help
i hope you understand….
ohhh….!!!
nevermind i just realised that i forgot to square the -2x to make 4x!!
thanks anyway….
Bernice,
(-2x)^2 = 4x^2 not 4x
Sorry, RTQ?
I was reading through someone elses notes and just thought RTQ standed for ‘read the question’… obviously not…
I think I’m going to have a very frustrated examiner.. everything wrong becasue of the degrees/radians.. Good idea actually, I might have to steal my brothers calculator and take two in…
also, if they ask for the coefficient in binomial, if it’s negative, I say that, right? (so for example if the coefficient was -810, I say -810 and not just 810)..
I’ve just had a negative numbers stress, and my brain’s not understanding dividing by neative numbers… becasue it shouldn’t work… according to my brain…
Right.
Thank you
Hi Grace,
RTQ does stand for read the question lol!
When you are asked to solve a trig equation, it will give you a range for your answers – if it’s in degrees, work in degrees; if it says 0-2pi, for example, work in radians.
For arcs and sectors, always use radians.
Re negative numbers:
-7 x -8 = 56
therefore 56 / -8 = -7
so it does make sense
Tea Lady
Ooops, forgot the binomial question:
Yes, if it’s negative, say so.
Thanks
I think my problem with negative numbers is that I still think when I look at one “but nothing is nothing, you can’t have MORE nothing than another nothing… becasue then you’re starting to have somethings, which is not nothings at all, that’s somethings…”
…and then i’m not revising
oh yeah!!! thankyou
i have another question though….
in the trapezium rule the one in the book and the one on the formula sheet are different and so i don’t know what yn is !!!
is yn the y bit of the n used in the h =b-a/n??????
Grace,
Think of negative numbers as debts.
If your purse has no cash but just a piece of paper saying ‘I owe Mum £5′, you can think of your purse as containing -£5.
If your IOU note is for £20, how many £5 debts is that? 4
ie -20/-5 = 4
Does that help or confuse??
Hi Bernice,
yn is the y value you get by plugging the nth x value into the function.
oh and also when it asks like 4 strips or something, how do you know which boundaries to use or do you just guess or always use 0 to 2
??????
thanks
as in for x in the trapezium question.
Yes, thank you, my confusion is over.
Sorry about that.
I’m really worried about this exam.
Not exciting.
sorry another question for ya…..
what if there is no y0 like in a question in the book because surely its needed to fill out the t rule????????????
Grace, you’ll be fine.
Put the kettle on, have a hot bath and an early night! See you tomorrow
same grace i poo
Too much information Bernice lol!
Bernice,
You’ll know the limits for the area under the curve so they will be your first and last x values. Put the first one into the function to get y0.
If there are 4 strips, divide the gap by 4 to get your other x values for the function.
BACK LATER.
umm with the trapezium rule when your working out h = (b-a)/n
is n the number of ordinates or the number of trapeziums?
n is the number of strips Katie.
thank you mrs tibble i understand now!!
Hello,
just working through June 2005 paper and gotten stuck on 2b, log(base2)(x+1)-log(base2)x=log(base2)7. How do I go about solving this?
Thank you,
Mariam
The left hand side goes to log(base2)[(x+1)/x]
So comparing LHS and RHS you get [(x+1)/x] = 7 (because they are both log(base2)) so just solve from there.
OK?
sorry for late reply, thank you has been solved now