Why did we have to restrict the domain of g(x)? I thought that we consider it a different function on it's maximal domain. I interpreted it as, if g(x)=sign(x). Then g(x) = -4 for x<0 and 4 for x>0
I think the wording of this question is very ambiguous - \(f'\left(x\right)\) is over its maximal domain, which could mean \(g\left(x\right)\) could have any domain of which \(\left(-\sqrt{2},0\right)\cup\left(0,\sqrt{2}\right)\) is a subset. Given that the graph question didn't ask for any specific points to be labelled, I'm now more inclined to think that the intention was for \(g\left(x\right)\) to have its maximal domain, \(\mathbb{R}\setminus\lbrace 0\rbrace\).
you sure about that segment q. I got 4pi/3 - root(3)
You're correct, I've made some error somewhere that gave me an angle of \(\frac{\pi}{3}\) instead of \(\frac{2\pi}{3}\)
6e) 146.51? instead of 146.52
As has been stated, I rounded up as 146.51 is the largest value for which \(H_{0}\)
is rejected, as opposed to 146.52, which is the smallest value for which \(H_{0}\) is
not rejected.
Did 4e ask for the total time (4.1) or the time interval?
4e asked for 'what period of time' which I think is ambiguous - I only included the total time as an afterthought. I realise now that this is going to be incorrect to the nearest minute as I rounded the lower end-point up to 91.8 (as the two yachts are not within 0.2 km of one another at \(t=91.7\)
Is question 9 C or D? Two answers have been provided so far.
I've made an algebraic error on this one - it's
D