AEA June 2011

Far simpler: as $\cot(x) = \tan(90^{\circ}-x)$, we have $\cot(\theta-53^{\circ}) = \tan(143^{\circ}-\theta)$; hence what we were asked to solve is $\tan(\theta+35^{\circ}) = \tan~(143^{\circ}-\theta)$ from which $\theta = 54^{\circ}+90^{\circ}n$. Obviously $n = 0$, and $n = 1$ give values of $\theta$ in the desired range!

What's yr firm? Are you majoring in math in future?

My firm is Warwick and my insurance is Nottingham and yeah I'm 'majoring' in maths, how about you?

I've got to be honest, I think I could have done much better in this exam, I guess nerves got to me .

This is how I did question 1:

But that's not true that opposite angles in a kite add up to 180 deg; it's coincidental that both opposite angles add up to 180 deg, as one of the angle is 90 deg.

Assume there are 2 similar angles of 100 deg in a kite. Do the remaining opposite angles add up to 180 deg?

Anw, do u require AEA for yr offer?

I thought it was true for 4 sided shapes.
Oh well!

Yeah I require it for my insurance.
FML

How'd you find it, man?

Absolutely crap.

I too thought Q1 was just a typo.

I found the vectors question especially difficult too.

I thought it was a very difficult paper - don't remember any of the others being as difficult at all.

Same.

I really would have liked a Merit here. ffs why was this one of the hardest years.

It was like the worst year ever for the AEA and STEP II. Just *******s really.

What do you guys reckon the grade boundaries will be for a distinction? I'm seriously hoping for sub-70.

On that last question, how the hell were you meant to find p and q?

Well, the vertical asymptote had been shifted to the y-axis, i.e. 3 to the left. So p = 3.

If you look at the original graph, the y-values of the stationary points were -2 and -10 if I remember correctly. They're symmetrical about the line y=-6. If you look to the translated graph, they both lie the same distance above the x-axis, so the line of symettry y=-6 has been shifted up 6 to the x-axis. Hence q = 6.

On that last question, how the hell were you meant to find p and q?

So is anyone doing the solutions?

I got p=3, q=6 ?