MAT Prep Thread - 2nd November 2016

You're putting 60 degrees in wrong - specifically I think you're not squaring sqrt(3)/2 to 3/4.

can anyone help out with Q4 of 2011? I just dont think i'm following the question at all. I can sketch the graphs but i dont really see any relation? ANy hints pls? thanks!

fyi the 81 mark thing is 2014 and i got 93 in 2012

Congrats, those are really good marks.

For more recent exams (I believe 2010 onwards?), there are examiner reports on the website, below the test solutions. They give you a breakdown of how many applications and offers were made to each course.

In a few of those (including the 2014 paper), there's a graph showing bands of scores (e.g 50-59, 60-69), how many people got that score, and whether they got an offer, got interviewed but didn't get an offer, or didn't get an interview.

For 2014, around 55-60 people got between 80 and 89 (and 50 got between 90 and 100). For those getting between 80 and 89, all of them were interviewed, and around 40 of them were given offers (and around 40 getting 90-100 got offers as well).

In short, you've got a really good chance of getting a place with a score of 81 in 2014. But it's not guaranteed - there's quite a few people getting 90+ that didn't get a place, and even somebody who got between 30-39 got a place. Really, it depends on how low grades you're talking about, and what qualification they are. But with that sort of score it's pretty likely you'll get an interview at least.

For doing the 2015 paper, I'd maybe do it a week or two before the exam. That gives you time to evaluate where you've done well, and to really understand where you've gone wrong. 2015 was seen to be a very difficult paper, so it'll be really good preparation for whatever comes up this year.

What graphs? Do you mean the lines x+y = constant, and curves xy = constant?

You can see from them how x+y increases. The largest value x+y can take is the value taken on the last line to intersect Q.

sorry forgot to specify. i think i get the x+y graph part how it corresponds to the y-intercept hence i guess the max would be the last line of gradient -1 that touches the quarter of the circle? what i dont really get is the next few graphs and the corresponding questions.

Yes, the bold is right.

Well the next ones are hyperbolae, can you sketch those?

And if you think about it you only need those hyperbolae to answer the rest of the second part.

I did the MAT 2014 paper and on question 1, part (j), I was confused about why $\displaystyle 3x^2\int_{-1}^1 f(t) dt$ can be integrated to . In fact, the entirety of the answer confuses me, really. Can anyone explain it to me in a simpler way than the Oxford solutions do?

I did the MAT 2014 paper and on question 1, part (j), I was confused about why $\displaystyle 3x^2\int_{-1}^1 f(t) dt$ can be integrated to . In fact, the entirety of the answer confuses me, really. Can anyone explain it to me in a simpler way than the Oxford solutions do?

Actually, Oxford's solution is quite trivial except that it didn't explicitly specify that
$\displaystyle\int_{-1}^1 f(t) dt$ is a constant (but it shouldn't be hard to deduce that), hence the integral can be treated as a number when integrating.

Yes, I did think $\displaystyle\int_{-1}^1 f(t) dt$ was just a number so I replaced the last term with $3ax^2$. But this will just integrate to $ax^3$ - I don't see the connection with $\displaystyle\int_{-1}^1 f(x) dx$.