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if f(x) is my function then f'(x) is my derivative, so subbing c into the derivative, i.e f'(c), will give me a constant

i know it doesn't ....as i said...i proved K can't be zero so my proof contradicts the question lol....so MVT fails here Edit: well K can be zero but i can't prove it here using the MVT..so to stop any further questions MVT fails on this question

lol...well the MVT states for some function that is both differentiable and continuous on the interval a to b there lies some c in between such that: \frac{f(b)-f(a)}{b-a}=f'(c) and this is pretty easy to picture...you draw a diagram of a continuous function on a to b and work out the...

but my contradiction is that for this to work K has to be 0, but i proved it cant be 0, which contradicts f being constant

f(x)-f(y)\leq (x-y)^{2} divide through by x-y \frac{f(x)-f(y)}{x-y}\leq (x-y) now you will notice the LHS is the MVT: f'(c)= \frac{f(x)-f(y)}{x-y}, \;for \; some\; c:\; y<c<x Therefore we get: f'(c)\leq x-y now f'(c) is a constant, say K, hence this becomes: x-y \geq K...

yep i like it...but is that from a uni book or q8 lol?

lol what a nice question from a uni book...part 2 is done very easily knowing the mean value theorem

i did 3u in yr 11 and i got 48/50 which i was happy with..but i was aiming for 98+ in 4u and only got 96 lol...stuffed up a few easy marks last yr

lol i tutor 4u...so you don't have to worry about competing with me lol

for i) we have 5 lines which are not parallel, so they all intersect, and no three lines are concurrent, so no 3 lines intersect in one point. This means the first line will intersect the other lines 4 times... The second line will then intersect the remaining 3 lines... The third line will...

three should be right...what paper is this from? Just looked at the solutions...they are wrong they forgot to square v if you look, so 3 is right

yep thats right

post your solution to the last part if you dont mind

here is another good one:

this is correct

correct :)

lets get this starting again... Determine all the values of: \sum_{k=0}^{n}i^{2k} btw i made this up so enjoy :)

For the projectile question, your eqns of motion for the slow particle should be: \!\!\!\!\!\!\!\!\!y=-\frac{gt^{2}}{2}+Utsin\alpha \\ x=Utcos\alpha For the faster one: \!\!\!\!\!\!\!\!\!y=-\frac{gt^{2}}{2}+Vtsin\alpha \\ x=Vtcos\alpha Now we want to find t when y=0 for the slower...

lol well in this case i knew it had to be right, this was one of those prob questions if you lay it out right you will 100% know you are right

i know it's correct, however my aim was for you to understand it all lol, hope it helped And yet again another great question of yours lol