I'll quickly post my answers, no point LaTeX-ing because I really don't have the time this close to the exam
Also no working, purely just answers to discuss.
So the format of the answers will look horrible, especially with fractions or squared's or cubed's involved.
And I haven't checked these on my calculator or anything, so highly likely there are errors everywhere.
1a) (8 - ky^3) / (3kxy^2 - 6y)
1b) -2/3
2) x = 5pi/6, 11pi/6, pi/2, 3pi/2
3) (sqrt(3) - 1) / 2
4a) two force vectors acting down (hard to show on one diagram I found) with magnitudes 2g and (2v^2) / g respectively
4b) show that question5a) [1]: 2a + b + 3c = 0, [2]: a - 2b - c = 0
5b) a = b = -g (g is the parameter, gamma)
5c) x = -gi - gj + gk
6a) area = 2pi units^2 (by symmetry)
6b) volume = 3pi^2 units^3
7a) u = ( g - 8 ) / sqrt(3)g
7b) T = 4 newtons
8a/b) You turn z + 1/z = k into a quadratic, then use the quadratic formula to solve for z.
Using the discriminant, we know that if k^2 - 4 >= 0 then z will lie on the real axis.
If not, then sqrt(k^2-4) will be an imaginary number.
Let z = x + yi, and x = k/2, y = sqrt(4-k^2)/2 (you have to times k^2-4 by -1 because you are after the imaginary PART of z).
Square both x and y, add them together and you get x^2 + y^2 = 1, showing that z either lies on the real axis or on the unit circle.
Part b is easy once you have done this. 
9a) x= -4, x = 8, y = 0
9b) y = -10 / 3(x-8) + 7 / 3(x+4)
9c) a = 17/3
10a) z = 2 - sqrt(2)i, z = 2 + sqrt(2)i
10b) a = -3, b = 2
10c) z = -1
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