MAT Prep Thread - 2nd November 2016

Original post by boombox111

yeah i lost plenty too aargh
it was the same kinda standard as 2015/13 so i'd guess around 55 and 61 for the interviewed/accepted averages. could be a tad higher tho

I really want to know the individual mark for each question from question 2 to 5 then I can estimate my marks :tongue:

Reply 1562

Original post by boombox111

yeah i lost plenty too aargh
it was the same kinda standard as 2015/13 so i'd guess around 55 and 61 for the interviewed/accepted averages. could be a tad higher tho

couldn't wait for the mark scheme

Reply 1563

DylanJ42

18

Original post by boombox111

ikr but they ALL looked like damn molars!

all of them? youve been eating too many sweets if your molars look like a parabola :rofl:

Reply 1564

In the Q4 did you have to know that arctan sqrt(3)/3=pi/6? I see a lot of people posting with that substitution, couldnt you express it using alpha?

Reply 1565

Also in question 2 last part the m_1 and n_1 were positive so it was impossible by a wide margin (not just 94 should be 124/2)

Reply 1566

SM-

6

What was the formulae for un and tn in Q5 and how did you find it?

Reply 1567

SM-

6

Also for the multiple choice with integrals and also the one with A and B being max or something can someone show how they got they're answers?

Also for the last multiple choice what was option A and option B and was it asking which one is false?

Reply 1568

theaverage

2

When are we able to access the paper online?

Not sure if im looking forward to it going up or not XD

Posted from TSR Mobile

Reply 1569

Original post by OxfordMathsDept

Not really - the numbers are much smaller for the joint degrees, but the performance on the MAT is broadly similar. Going by the Maths averages on the website should be fine.

Hi, you have mentioned that the exam paper will be online on Friday. Is that on oxford maths department website?

Reply 1570

DFranklin

18

Original post by ShatnersBassoon

The summation question I think you're talking about, to quote myself:

1J: Might have messed up some of the options in particular because I spent more time analysing some of them than others. But this was roughly it:

Let ∏(n) denote the number of unique prime factors in n; for example, ∏(8) = 1 and ∏(6) = 2. Let x(n) equal the last digit of n. Which of the following statements is false?

(a) If ∏(n) = 1, there are some values of x(n) for which n must not be prime.
(b) If ∏(n) = 1, there are some values of x(n) for which n must be prime.
(c) If ∏(n) = 1, there are some values of x(n) which are not possible to obtain.
(d) If ∏(n) + x(n) = 2, it is not possible to determine whether n is prime.
(e) If ∏(n) = 2, there are some values of x(n) which are not possible to obtain.Haven't seen the paper, but I'm havnig trouble seeing how both (b) and (e) aren't false.

Specifically, for (b), it's not possible for x(n) to be 0, for all other digits d=1,...,9 we can find composite n with ∏(n) = 1, x(n) = d. The smallest solutions for each d are:

81, 32, 243, 4, 25, 16, 27, 8, 9,

Similarly for (e), the following numbers with ∏(n) =2 show all values of x(n) are obtainable.

10, 21, 12, 33, 14, 15, 6, 57, 18, 39.

What am I missing here?

Reply 1571

ShatnersBassoon

4

Original post by DFranklin

Haven't seen the paper, but I'm havnig trouble seeing how both (b) and (e) aren't false.

Specifically, for (b), it's not possible for x(n) to be 0, for all other digits d=1,...,9 we can find composite n with ∏(n) = 1, x(n) = d. The smallest solutions for each d are:

81, 32, 243, 4, 25, 16, 27, 8, 9,

Similarly for (e), the following numbers with ∏(n) =2 show all values of x(n) are obtainable.

10, 21, 12, 33, 14, 15, 6, 57, 18, 39.

What am I missing here?

Since I can't fault your logic, it probably indicates I messed up my re-writing the question. In fact, thinking about it, I definitely remember finding examples for (e) covering every value of x(n), and then deducing that option was true, so (e) was probably the opposite of what I said - "If ∏(n) = 2, every value of x(n) is possible to obtain." or something similar.

I guess when Oxford put the paper up we'll see for certain.

Reply 1572

Original post by DFranklin

Haven't seen the paper, but I'm havnig trouble seeing how both (b) and (e) aren't false.

Specifically, for (b), it's not possible for x(n) to be 0, for all other digits d=1,...,9 we can find composite n with ∏(n) = 1, x(n) = d. The smallest solutions for each d are:

81, 32, 243, 4, 25, 16, 27, 8, 9,

Similarly for (e), the following numbers with ∏(n) =2 show all values of x(n) are obtainable.

10, 21, 12, 33, 14, 15, 6, 57, 18, 39.

What am I missing here?

I am pretty sure that the e) was not as you quoted.

Reply 1573

Original post by goofyygoober

Anybody remembers the answer to the MCQ about the graph of （x^2-1）+（cospix）

I'm quite sure that the question was (x-1)^2 - cos(πx)

Reply 1574

Original post by 0x3bfc9a1

I'm quite sure that the question was (x-1)^2 - cos(πx)

okay, the answer is still (a) though

Reply 1575

Original post by goofyygoober

okay, the answer is still (a) though

I can't remember the choices. I think the correct answer was the only one with two global minima.

Reply 1576

I did all of the questions, and I think some of the difficult problems that appeared in past papers did not show up in the 2016 paper.

1. Logarithms did not appear in any of the problems.

2. Multiple choices did not include 'finding the greatest number'.

3. I was not asked to graph in any of the questions (except for multiple choice where I was asked to choose the correct graph of (x-1)^2-cos(πx).

4. There was not a question about finding the number of roots to polynomials.

Subjectively I think the 2016 paper is easier than ones in the past few years.

Q2 was about two commutable functions A and B, and i think it is impossible to have 214x+96 since 214>2*96

Q3 mainly dealt with functions that are symmetric against x=\alpha

Q4 was an easy one, the angle was π/6 and area was 4√3-(11/6)π i think.

Q5 was about the summation of n2^n and included ∑i<n∑k<i k2^k as the last part, i think the last part required us to use f(n) again to find the expression. It is probably
8 - 2^(n + 3) + 2n + n2^(n + 2).

(edited 7 years ago)

Reply 1577

Original post by 0x3bfc9a1

I did all of the questions, and I think some of the difficult problems that appeared in past papers did not show up in the 2016 paper.

1. Logarithms did not appear in any of the problems.

2. Multiple choices did not include 'finding the greatest number'.

3. I was not asked to graph in any of the questions (except for multiple choice where I was asked to choose the correct graph of (x-1)^2-cos(πx).

4. There was not a question about finding the number of roots to polynomials.

Subjectively I think the 2016 paper is easier than ones in the past few years.

Q2 was about two commutable functions A and B, and i think it is impossible to have 214x+96 since 214>2*96

Q3 mainly dealt with functions that are symmetric against x=\alpha

Q4 was an easy one, the angle was π/6 and area was 4√3-(11/6)π i think.

Q5 was about the summation of n2^n and included ∑i<n∑k<i k2^k as the last part, i think the last part required us to use f(n) again to find the expression. It is probably
8 - 2^(n + 3) + 2n + n2^(n + 2).

How many points do you think I would lose in q4 for putting alpha=arctan sqrt(3)/3 (I worked with tangents and thought it would be a random angle). And in question 2 m_i and n_i were positive so the margin is wider (214-96)*6>214

(edited 7 years ago)

Reply 1578