Ok, I thought this was a really easy exam, but that's my own opinion- don't bite me for it!
I brought my answers out of the exam on scrap paper, so I'll do my best to translate. I'm pretty confident they're all right, if I'm wrong correct me, but I went through with teachers and other students afterwards and these seem to be correct answers. If anybody did get similar, shout up!
1)i) 12 rt.5
ii) 4 rt.5
iii) 5 rt.5
2) x^3 = 1/8 and -1 ... therefore x = 1/2 and -1
3) f'(x) = -(12x^-3) +2
f''(x) = 36x^-4
4)i) 3(x+3/2)^2 + 13/4 [Because]
ii) Vertex is (-3/2, 13/4)
iii) b^2 - 4ac = 9^2 - 120 = -39
5)i) Basically the same graph as 1/x^2... Curves symmetrical and never touch any of the axes
ii) Stretch in Scale Factor 1/2 in y (vertical) direction
6)i) x^2 + (y+4)^2 = 40 Centre is (0, -4) Radius is rt.40 = 2 rt.10
ii) B co-ordinate is (-2, -10)
7)i) x < -1/8
ii) 0 </= x </= 5 ( </= means greater than or equal to! solved this one by drawing out the curve.. )
8) m of perpendicular is 1/3, so m of line is -3.. put into formula to get 3x + y -1/2 = 0 .. x2 to get into integers, 6x + 2y - 1 = 0
9)i) Positive quadratic curve (smiley face, not sad face). Intercepts y at (0,-6) and x at (-3/2,0) and (2,0)
ii) Vertex at x=1/4 so function is decreasing for x < 1/4
iii) Submit into formula for curve, use quadratic formula to solve, find that points P and Q are (-2,4) and (2.5,4) so distance is 4.5
10)i) Solve to find that k = -5
dy/dx = -3x^2 - 6x + 4 - k
0 = -3(-3)^2 - 6(-3) + 4 - k
k = -27 + 18 + 4 = -5
ii) d2y/dx2 = -6x - 6 ... submit -3 in to get 12, 12>0 therefore it's a minimum point
iii) This one was really long winded for 5 marks, basically you had to put the formulas together, solve the quadratic to find the two points on the cubic curve that satisfied a gradient of 9, you then put those two points (0 and -2) into the y=9x-9, and see which works. 0 doesn't work, but -2 does, so A = (-2, -27)
Hope this helps! Please point out any mistakes you think I've made, would be awesome if someone could scan a copy of the question sheet too