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Discussion in 'Mathematics' started by dydx, May 7, 2008.
Post the questions here. Will be glad to help.
Here is a link to the paper:
You get y = 8/(4 + x^2). Since x^2 is positive for all x, then the domain must be -infinity < x < +infinity.
Will work on 7(d) in morning.
Or x belongs to all real numbers. XeR
I have the markscheme - if you drop me a mail...
wibble73 @ btinternet dot com
6 c) You are given x = 2 cot t, for t values from 0 to pi/2, so domain (possible x values) is x > 0 - I think...
7 d) I drew a diagram, I think OD = a + (c-b). This give OD = 6i - 2j + 2k
I have the markscheme also but need an explanation for the two anwers wibble.
Thanks Yoggle - I understand 7(d) now but please forgive my ignorance when I ask the following for the domain on 6(c):
I thought that o - pi/2 was the values for t - how does that give us x>0 for the domain?
The function was defined using t, so that limits the values of x (the domain). Have a look at the graph of y = 2 cot x for x values from 0 to pi/2 - the range of this graph is the domain of f...
Almost got it yoggle - Why is it x > 0 and not x is greater than OR EQUAL to zero?
I just looked at the graph of cot x on...
...and changed my mind domain is x >= 0 - I think!
Domain is greater or equal to 0.
The Mark Schemes are copies of the documents which are discussed at the Examiners' meeting before marking starts. Details of any changes made to the final mark scheme are not shown so errors can appear.
I don't think anybody picked up my reply in #2.
The DOMAIN is x e R. The cartesian equation is y = 8/(4 + x^2) which is a continuous function for all real x.
Check out the graph in Autograph.
dydx, I haven't seen the wording of the question, but the content of other postings is that the function is given parametrically, with a restriction on the parameter. The Cartesian equation may well be as you have stated, but not all values of x are allowed for the stated values of t.