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Latex is not working. g'(x) = 1/(1+x) f'(x) = 1 - x + x^2 = x^3/(1+x) + 1/(1+x) - from c.i. = (x^3 + 1)/(1+x) >= g'(x) since x^3 + 1 >= 1 for x>=0 This is what I put, not sure if it is sufficient though.

2 lost in 3b, 2 lost in 4c and 1 in 5b so far. I cannot remember the signs I used in 8a, hopefully the right ones. It is only a 1 mark question though, so hopefully they 'round up'.

Edited. This team-work is quite depressing. My mark is down to 115 now :(

The pipe costs $2600/km under the water, and $1000/km on land. The line PS is underwater.

9bi) cost = 1000*5 + 2600*3 = 12800 9bii) cost = 2600*root(25+9 = root(34) *2600 =15160

Yes. Edited.

They maybe wrong...

THese are my solutions - well answers. Remember these are by a student, and aren't necessarily right. Tell me if there is something wrong, as there most probaby is. Sorry for the horrible format. It won't let my upload either. 1b) x= 4/3 1c) x=4, -6 1d) m=1 1e) θ=π/3 1f) x= 7.3891...

Yes, but the question specifies that there are 2 solutions, hence you take the next solution. Therfore @ = pi - pi/3 = 2pi/3

Area of triangle = 1/2*absinC = 2sin@ Therefore root(3)/2 = sin@ @ = 2pi/3

Yes. This is when you have to think logically. 7pi/3 > 360 degrees > a whole circle. Hence it is impossible.

Thanks for the heads up.

You needed to sum the chord length with the sector length.

My thoughts.

Most probably: 1 mark for differentation, 1 mark for finding x, 1 mark for testing, 1 mark for substituting back to find the cost. So maybe 1-2 for you.

And you know that how?

Yeah, I do 3-unit. Considering the difficulty 120 will be needed in my opinion. I am out of that race now, down to 118 :(

Well the mark means nothing atar-wise if you do 4-unit i guess. Hopefully the difficulty of 3-unit parallel's 2-unit.

Re: Would this be enough for a band 5? Don't exaggerate it's difficulty. Try looking at 2003, or q10's from 2000 or 2001.

Most probably.