STEP II 2011 Discussion Thread

Could anyone please remind me what the last part of question 1 was?

For arcsin x / x > x cosec x, can someone tell me the range you needed to prove this over?

I thought that was a pretty horrible paper as II goes, but I got full solutions out for 1, 6, 8 and 9 and half of 3 plus a tiny bit of 4 in the dying moments of the exam. Though I know I made a slight numerically mistake in the last part of 1 but my method would have worked had I not added up incorrectly

So yeah I completely failed it. Only question I completed was question 1 and my graph had 6 intercepts, with 4 being the same and 2 being the other. I got -3 and 2.2 for the intercepts but I forgot to check if they actually satisfied the original. I wasn't sure whether we had to consider plus or minus square roots so took both just in case.

For question 2 I did part i, I got x = 1, y = 19 and z = 7. It took me aaaages to show z cubed was less than k squared.

For question 3 I think it was, I got stuck on showing that arcsinx/x >= xcosec x

The integration question was WTF. I didn't even attempt any more questions than that FML.

x

Thats not true, x=1.

If it makes anyone feel better, in my experience the Cambridge examiners often "overcorrect" when trying to address a previous year being too easy. So I was thinking it likely this would be a very difficult paper.

How did you answer the last part of question 9?

Here's a couple that I can remember well enough. I ended up doing all the multiples of 3 completely, plus just over 2/3 of the way through Q2 and got a fragment out of Q1 (maybe 4-5 marks). I haven't bothered to post Q3 because it was long (in fact, I made a bad decision to attempt it; even though I finished it, it had taken almost an hour...) and involved a lot of graphs. Q9 was just about algebra, which makes it harder for me to remember my working unless I write it out on paper again. I feel sorry for those of you who didn't attempt it; it was by far the easiest question on the paper!


As for the paper, it was reasonably harder than the last 2 or 3 STEP II papers, although I don't think it was as scary as the hype in the examiners report.






Good luck for STEP III everyone!

0 < x < 1

6 - I fully completed this, but it was the only one and I think I did it wrong. I wasted huge amounts of time on this. First I tried integrating the expression for f''(x), but then I did IBP with u = [f'(x)]^2, v' = [f(x)]^n, and got a really BIG result! I carried on anyway, and my verification worked! Except it didn't - I lost a factor of 2, although now I think maybe they wanted you to find k (if k = 2, then my verification works) but I thought that k was any constant, and maybe it is.

Now, here's where I got SO pissed off. I KNOW to ALWAYS check for derivatives in integrals, but I wanted to wrap up the question quickly with no time wasted, so I thought secx + tanx will never derive to anything I want, so I expanded the brackets binomially and integrated each term separately - I thought this was right because I was making use of sec^2 = tan^2 + 1, and the result I found, but then I realised I couldn't integrate sec^6 or tan^6. Just for kicks I differentiated secx + tanx - it gives secx(secx + tanx) - allowing you to use the rule with n = 4. I was livid. The answer falls out immediately, meaning I wasted 20mins for nothing.

You can show that

x - x^3/6 < sin x < x

and

1-x^2/2 < cos x < 1 in that range;

I thnk that's enough to show

$\frac{d}{dx} \arcsin x - \frac{x^2}{\sin x} > 0$

[This is a somewhat ugly solution, but I'm looking for "how would I do this quickly" here].

Wish I looked at the mechanics questions -.- Well on the positive side, we've learnt from our mistakes and it should help us with STEP III/I

I think if you let tan(x) be f(x), then the second derivative is equal to 2f'(x)f(x) so k has to be 2 when you're working with tan(x).




I feel a bit silly now, I had a similar problem with not being able to see what to do with that. Do you remember what your answer looked like? Because I did it a different (and much worse way). I wrote (tan(x)+sex(x))^6 as exp(6ln(tanx+secx)), because I was so pleased with myself when i remembered that sex(x) was the derivative of ln(secx+tanx) and then managed to integrate it in a similar way to how you derive that result in the first part of the question, using parts, although I had to use parts twice, and even then the integral I was left with was a multiple of the integral I started with, so I had to equate them and solve the equation.

It was all a bit messy and I could well have ended up with the wrong answer, did your answer have a factor of 35 in it somewhere by any chance?