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It wasn't as easy as the last question in our Baulko trial this year though.

In my booklet, Option D in Question 9 was printed as |x|≥tan-1x which is true, not |x|>tan-1x (difference being it has a greater than or equals sign, and thus the equality at x=0 doesn't make it false).

Ahh I misinterpreted the question quite badly then :L Yep, okay, with this, I managed to solve the problem. Got stuck with rgV02cosθ = rg(V12 - V22) - (V0V2)2sinθ for a while, though. That was rather tricky.

Holy moly I just realised an almost identical question came up in my half yearlies and I lost a couple of marks over it. Thought I'd learnt from that mistake, but apparently not. (for anyone's that wondering, it's just using the basic angle subtended by arc at centre = twice the angle subtended...

Ooh yeah this sounds good. Hopefully 50%+ in 4U (also not sure what happened to old account on this forum here…)