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The only part I can remember is the question about how he would react after entering the room with that knowledge. I chose the one about him pretending that nothing has changed, but on second thought I think he would have pretended to be more positive. 2 is usually my good section but I think...

I am a bit surprised that this post still hasn't been moved. Usually it takes about 5 seconds for the mods to do something about it. Nonetheless. I'm assumimg the question meant that perpendicular distance AP = 6 because it would otherwise be impossible to work out an answer seeing as no...

One of the questions in Section 1 was exactly the same as Q13 from Volume 2 of ACER's UMAT Practice Questions. I was very surprised to see that. I didn't think ACER would ever reuse past questions. :eek:

Not really sure about the number - I don't think you will have one if it's, say, a letter from your school. Leave it blank and ask the supervisors when you get there, I suppose. I'm sure you don't need to wear uniform but I'd probably still need to wear uniform seeing as I'm planning to go...

That step is wrong. Remember the basic index law: (x^m)*(x^n) = x^(m+n)

Strictly speaking, this type of questions is of the Ext-2 course, but I'm not sure to what extent it is included in the Ext-1 syllabus. Taking this as an example: find the equation whose roots are twice those of P(x) = 3x³ - 2x² + 1 = 0 The concept is very simple: Let the roots of P(x) = 0...

a) Remember the double angle results for sine and cosine. LHS = -2isin(n@/2)*[cos(n@/2)+isin(n@/2)] = -2isin(n@/2)cos(n@/2) + 2[sin(n@/2)]^2 = -isin(n@) + [1 - cos(n@)] = 1 - [cos(n@) + isin(n@)] = RHS as required. b) I think you're missing some vital information - e.g. 'z is the nth...

Then I would say auxiliary circle because there is no physical relation between the angle and the conic section except through its auxiliary circle. Most students tend to forget that the syllabus actually requires us to know how to do that. There was a question in a HSC paper some years ago that...

On second look - the exact values of those angles are known. So maybe just work out the coordinates and, well, mark them on a lovely scaled diagram?

This relates to the construction of hyperbolae/ellipses from their auxiliary circles. Or is that too brief?

This is rather difficult to describe verbally. Basically, it is known that the eccentricity of a rectangular hyperbola is √2 so prior to rotation of the axes the foci are S(±√2a, 0), i.e. S(±2c, 0). This means that after rotation counter-clockwise by 45 degrees, the new foci are now located at a...

I did consider attending preparation tutorials for UMAT for a while but decided not to. I'm planning to play bit with non-verbal-pattern-deciphering quizzes online and work through the practice questions booklets under exam conditions. There seems to be only one problem - can someone please...

Both should get you the correct answer, but the different forms vary in convenience depending on the initial position of the particle. If this is at an extremum of the motion use x=a*cos(nt+@); at the centre of motion use x=a*sin(nt+@); at anywhere else use x=a*sin(nt) + b*cos(nt). Why are...

I wouldn't complain unless there's a clash - quantum superposition cannot be relied upon. I understand the complaints but I think they are unnecessary. If exam timetables are set up to have morning and afternoon components, then the prospect of having two 3-hour exams in a day is well within...

How did you integrate lnx without using IBP?

I still have three: MX2, MX1, Adv English. Funnily, I was just complaining to a friend a minute ago about still having a handful of assessments to come before the trials. Now I guess I should be feeling lucky in comparison. SS: What on earth is "maths" doing next to "presentation"?

That is a much neater method. It's nice to learn something new. Thanks a lot for you input!

Poly: Cambridge p.136 Q10(a) Show that cos(5@) = 16(cos@)^5 - 20(cos@)^3 + 5(cos@) Hence solve P(x) = 16x^5 - 20x^3 + 5x - 1 = 0 & Deduce the exact values of cos(2pi/5) and cos(4pi/5) I've done similar deductions with nice quartics where the quadratic formula would be sufficient to work...

This might get a bit confusing using plain text.. Method 1 requires no more than 2/3U skills: Remember a = e^ln(a) and ln(b^c) = c.ln(b) so: y = x^lnx = e^ln[x^(lnx)] = e^[(lnx).(lnx)] = e^[(lnx)^2] Now you have y = e^f(x) f(x) = [ln(x)]^2 f'(x) = 2ln(x).(1/x) dy/dx = f'(x).[e^f(x)] =...

Q1. ma = -mkv<SUP>3/2</SUP> a = dv/dt = -kv<SUP>3/2 </SUP>=> dt/dv = -(1/k)v<SUP>-3/2</SUP> Integrate to find t = (2/k)v<SUP>-1/2</SUP> + C Clearly, v cannot = 0 -> the particle is never brought to rest. It might make your teacher/marker happier if you use the condition (t = 0, v = u) to...