By the way - for the very last bit of Q5, did you simply say that the gradient of C₂ is increasing more quickly than that of C₁ between 0 and 1, and therefore the result follows?

For q5, how important do you think the last part (namely showing that k> 1/2e^-1 (0r whatever!) is? For some odd reason, I just completely left that out, I'm so annoyed about that!

For Q5, the last part follows from e^x3 ³ 1 and e².

For the last part of question 5, Olof, yes, that's how I did it. (you should also mention that they both start at the origin.)

Tom.

So for the last part of Q5, you simply use e^-x3 £ e^-x2 for 0 £ x £ 1.

You use that to say that the C₂ is increasing more quickly than C₁ right?

Yes, or from first principles, if y(x) denotes y as a function of x then define:

L(x)=y(x)-x²e^-x2/2

Then:

dL/dx=dy/dx-xe^-x2-x³e^-x2=x(1-x²)e^-x3-xe^-x2-x³e^-x2 ³ x(1-x²)e^-x2-xe^-x2-x³e^-x2=0

for 0 £ x £ 1. And so we have L(1) > L(0) as the above inequality is strict for all x other than 0, 1 so we get y(1)-e^-1/2 > y(0)-0 and since y(0)=0 the result immediately follows.

Ah, nice!

Olof