Oxford MAT papers

Haven't finished going through all of these yet however the papers have definitely been getting harder in recent years. Last year's paper was imo quite a jump from the 2014 paper.

Not sure what others would say but my recommendation would be to do the 2015 paper earlier rather than leaving it to use as a mock. I felt it was quite different to previous years, and it would give you an indication to what the standard of this year's paper might be like.

And do you mean my solutions to the 2015 paper? The full solutions are up on the website but I don't mind posting mine too if you want.

There's also a specific MAT 2016 thread on here. There are some people on there that are a lot smarter than me

Oh I Will check that out, thank you. And I was talking about the 1990s solutions .. I don't think the website has got them.
Q.4 c of '97 was inaccessible for me.. would be great if you could help with that!

Really could do with a diagram here, but...

Label the pi/2 cos(x) curve as the A-curve, and the pi/2 cos(y) curve as the B curve.

You want to find:

Area_A = the area between the A-curve, the x-axis, and the lines x=0 and x=b, and subtract

Area_B = the area between the B-curve, the x-axis and the lines x=0 and x=b.

I'm going to assume that you're supposed to know how to integrate cos x. This allows you to find Area_A.

The problem is that to find Area_B in the obvious way, you want a way of expressing the B-curve in the form y = f(x), and you're going to end up needing to integrate arccos x, which I expect you're not supposed to know how to do.

So, instead, you need to find another way of finding Area_B.

It's not too difficult to integrate arccos(x) by parts, and since the MAT at that time did require C3 and C4 techniques as well as C1 and C2, it's possible that this was the intended method...

What did you guys get for multiple choice? Here's what I got:

1996:
a) i - Very standard - 2 distinct roots therefore discriminant>0
b) iii - By testing the first few cases eliminate the remaining options
c) iv - Not sure whether this one is iii or iv but I think a 2x2 system can't have multiple solutions anyway (?)
d) iv - Using the double angle theorem and factorizing yields cosx(2sinx-1)=0
e) ii - Not much to say here
f) i - A translation of -i then a vertical reflection
g) ii - The nth root of 10000 approaches 1 as n becomes very large and positive
h) iii - Chain rule
j) iii - Fundamental theorem of calculus
k) iii - 3(4/52)(48/51)(47/50) is approximately 0.2

1997:
a) iii - Solve or sub points in
b) i - f(-2)<0 and f(-1)>0
c) iii - Not sure about this one, really bad at combinatorics
d) ii - If 1<x<2, P becomes +(-)/(-)=+ therefore Q implies P, but solving P (by considering when 1-x is positive/negative) doesn't give Q
e) iv - In the range, cosx ranges from -1 to 1, so cos(cosx) is positive throughout. This rules out ii and iii. i is wrong because for cosx=0 we require x=pi/2, which is greater than 1
f) ii - Not sure whether it's iv - as n approaches infinity, the (-1)^n becomes insignificant
g) ii - The coefficient of x^n is (10Cn)/(2^n). Compare magnitudes
h) Not sure about this one but intuitively I think it's ii. Somebody explain?
j) iii - The nth line creates n more regions. As we have 1 region in the beginning the answer is 1+(the sum of the first n positive integers)
k) iv - Reducing the system to echelon form we find that the bottom row is all zeros

1996 question j) should be f(x)-f(0), i.e. option (i). Try it with an example function to realise this.
For 1997 question c), notice that the teams must consist of 2 people, 2 people, and 1 person. Thus the number of ways would be
5C2 * 3C2 * 1, except that this would then count e.g. (1, 2), (3, 4), (5) as different from (3, 4), (1, 2), (5), so we divide by 2 to give 15, I think.
1997 question f) is indeed 1, as you thought.
For 1997 question h), note that f(3-x) gives the same points as f(x) between 1 and 2: it's a well known result that for a definite integral of f(x) with limits a and b, you can replace f(x) with f(a+b-x) without changing the value of the integral. Now just multiply by 2 since we have 2*f(3-x), to give the answer of 2 as you thought.

Hey, thanks.

Could you explain 1996 j) a bit more? Also for 1997 c) the teams could also be 311 so we add 5C3=10 to 15, giving 25.

Actually I've thought about it a bit more and so I now think you're correct about 1996 j).

What did you get for section B for the 1996 paper?

It's very weird as section B is VERY elementary (factorise x^2+x-6). It seems implausible that 150 minutes is given to complete the entire paper whereas any able candidate would probably be able to finish the paper in about an hour.

I know the earlier papers were easier, but to this extent?