STEP 2001 Paper 2 Question 5

This thread is an excerpt from a discussion of the STEP papers immediately after they took place.

By Olof Sisask on Thursday, June 28, 2001 - 03:24 pm :

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?

By Nikhil Shah on Thursday, June 28, 2001 - 04:10 pm :

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!

By Michael Doré on Thursday, June 28, 2001 - 04:45 pm :

For Q5, the last part follows from $e^{x^{3}} \geq 1$ and $e^{2}$ .

By Tom Hardcastle on Thursday, June 28, 2001 - 05:09 pm :

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.

By Michael Doré on Thursday, June 28, 2001 - 09:05 pm :

So for the last part of Q5, you simply use $e^{- x^{3}} \leq e^{- x^{2}}$ for $0 \leq x \leq 1$ .

By Olof Sisask on Thursday, June 28, 2001 - 09:11 pm :

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

By Michael Doré on Thursday, June 28, 2001 - 09:18 pm :

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

$L (x) = y (x) - x^{2} e^{- x^{2}} / 2$

Then:

$dL / dx = dy / dx - {xe}^{- x^{2}} - x^{3} e^{- x^{2}} = x (1 - x^{2}) e^{- x^{3}} - {xe}^{- x^{2}} - x^{3} e^{- x^{2}} \geq x (1 - x^{2}) e^{- x^{2}} - {xe}^{- x^{2}} - x^{3} e^{- x^{2}} = 0$

for $0 \leq x \leq 1$ . 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.

By Olof Sisask on Thursday, June 28, 2001 - 10:14 pm :

Ah, nice!

Olof