Official Edexcel C3 Discussion Thread

That's not the way to solve equations with the modulus function! What are you solving here, x in terms of a? Add one to both sides and divide through by x (|2x+a|=1/x). Then do a quick sketch of the graphs |2x+a| and 1/x and consider points of intersection.

You can solve modulus functions like that, treat the modulus as a bracket, and expand, and after you found your solution, put a minus infront of the bracket, and solve again so you have two solutions. For this question as it involves "a" is gonna be hard to do, your method is useful by drawing the graph, but sometimes it will be hard to pin point where exactly the lines intersect so the method I stated will be required.

thankyou!
I was wondering if you could help with domains/range.
How do you know if it is all real numbers XER or whatever.
for the range do you put in the x values of the domain or x=0?
i can only sometimes get these right xxx

Well yeah, you can do it like that (not the way I personally do it), but he didn't include the 'putting a minus infront of the bracket' bit so I thought I'd just explain how I go about solving them. And you don't need to pin point exactly where the lines intersect, you only need to see the 'regions' they'll intersect in so you can form equations with the appropriate signs.


Do you have any questions to post in particular?

Drawing a quick sketch of the graph is always a good idea if you can't see the domain/range. For the range you would use the domain yes. However if the domain of x is simply that it's an element of the reals (or something along those lines), then you need to consider what the graph looks like. If you have any particular questions you're stuck on I'm sure myself and a few others can help.

how would you do this

if f(x)= 4-x^2

solve ff(0)
so 4-(4-x^2)^2
x^4-8x^2-12=0

i dont know how i would do it.
xxx

Owl City - Fireflies

find the range of:
(2x-3)/(x-2) XER, x>2
how would you do this?
i thought you put x=2 but this would give 1, when the answer is 2?!?!? xxx

find the range of:
(2x-3)/(x-2) XER, x>2
how would you do this?
i thought you put x=2 but this would give 1, when the answer is 2?!?!? xxx

For these questions I'd try and construct a table to help you (might not).
so x=2 y=infinity
x=3 y=3
x=4 y=2.3
then try x with a large number i.e. 100000, then y is like 2.00001 or something like that. so now you know (or already do ) that the minimum value is 2 and it is tending towards infinity, so with luck, the range should be f(x)>2.

(experiencing a bit of deja vu, i think i did this exact question for someone else lol)

hope that helps.

Wow. I worked June 2008's paper and I got 60/75 O_O

Silly mistakes and I just didn't know some things!

Can someone please simplify this:

y - 8 = 16 ( x - 0.5ln(2-1) )

(this is Q1 of June 2008 by the way).

Thanks.

You've got the bracket in the wrong place. It's this...

$y-8=16(x-0.5(ln2-1))$

$y-8=16x-8ln2+8$
$y=16x-8ln2+16$

These papers are marked online and, if a pencil is used in drawing sketches of graphs, a
sufficiently soft pencil (HB) should be used and it should be noted that coloured inks do not come
up well and may be invisible. A number of candidates gave answers which went outside the area
on the pages which is designated for answers and this caused the examiners some
difficulties.

Thanks.

Just a little reminder to everyone:



Also, how the flap do you find P?

Also, how the flap do you find P?