MAT 2020 | All questions | MAT Livestream Bonus
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- เผยแพร่เมื่อ 1 ส.ค. 2024
- All of the questions for MAT 2020. Do not watch all of this video, you've got better things to do! Click "show more" for links to jump to a particular question.
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0:00 MAT 2020 Q1A
2:57 MAT 2020 Q1B
4:42 MAT 2020 Q1C
7:31 MAT 2020 Q1D
10:06 MAT 2020 Q1E
13:26 MAT 2020 Q1F
15:58 MAT 2020 Q1G
18:25 MAT 2020 Q1H
22:00 MAT 2020 Q1I
24:37 MAT 2020 Q1J
29:50 MAT 2020 Q2
47:11 MAT 2020 Q3
59:46 MAT 2020 Q4
1:14:07 MAT 2020 Q5
1:24:22 MAT 2020 Q6
1:33:05 MAT 2020 Q7
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Last day before MAT 2021.Good luck everyone and wish we all good luck in the exam!
best of luck too!!
@@Chloe-qe6xp Best of luck too!
@@jimboli9400 What r u guys doing at the minute for last minute revision. Idk what to do.
Good luck g
@@kingofblox2976 Read through past papers, specifically how to answer the long questions. Hope this helps and good luck!
I really like question 2’s relation to binary!
James, you're the best!!
Taking MAT this afternoon! All the best~
tomorrow is the 2021 MAT test. Good Luck everyone
Hi, I was wondering if saying that A(theta) has a point of inflection at (0,1/4) as it is a translation and stretch of y=tan (theta) is a sufficient explanation for 4e. Also does the MAT give partial marks within parts of a question. Eg. on 3vi, I found the equation of the cricle and equated it to the equation of C, but didn't find the actual x coordinated. The solutions say this part is worth 5 marks, so would I get none because I didn't get it fully, or would I get like 3 marks for doing parts. Thanks
Yes we do give partial marks within parts of a question (we call that "method marks" or "marks for working out"). We would like to give you e.g. 3 out of 5 if you've done some but not all the work! I think that we did give marks for the claim that tan(theta) has an inflection point, which was perhaps a bit generous. ^James
Hi, James. Any chance you could do solution tutorials on the MAT papers before 2016? Been struggling with those and could not find any tutorials on TH-cam and mark scheme does not explain very well. Would be very helpful. Thanks.
Glad you like these - it's on my to-do list to work backwards at some point (they each take a long time to make!). I think other TH-cam channels have done solution videos for some of the past questions. I made these massive videos because I think it's handy to have it all in one place ^James
Hi James I am a little unsure about the marking of this paper. Is the percentage you get equal to = marks you get / questions needed to attempt or marks / Total paper marks.
As a computer scientist, I won't be doing some of the other longer questions obviously so does that mean I automatically miss out on some % of marks if it's the entire paper being taken into account?
No matter which course you're applying to, you answer questions worth 100 marks; you do Q1 (40 marks) and four long questions (15 marks each). Applicants for Computer Science at Oxford do Q1,2,5,6,7 (applicants for M&CS at Oxford do Q1,2,3,5,6 and applicants for Maths at Oxford do Q1,2,3,4,5). Everyone gets a score out of a 100, which you can convert to a percentage if you like. ^James
Hi, James, for Q4, e, could I just say by inspecting the A(theta) graph, I know that either side of theta=0, the gradient have same signs, so that at theta = 0, it is a point of inflection?
No that's not enough to identify a point of inflection. Note for example that x^3-3x^2+4x+1 has the same gradient at either side of x=0, but it does not have a point of inflection at x=0. ^James
Hi, James, I am very confused on how Q5 iv is worded. Not sure what they are asking me to do. Could you explain please?
Q5(iv) wants us to actually work out the number of ways that Santa can put 7 balls into 5 stockings. So far in the question we've just found identities involving g, but now we're trying to actually evaluate that function. ^James
Hi Dr James, for question 2 (vi), do we consider the number 1 an element of S?
Otherwise S2 is not equal to (S1 + S0 + 1). Thanks :)
For queation 6v, I tried the same reasoning as you showed, but I wrote down h(n,k)=g(n-k,k), so h(7,5)=g(2,5), which did happen to be 15 - is this method right? Thanks massively for these videos, James, they are a huge help!
yes it is right, i think mark scheme does it too
@@DotALotTroller Thx
hi! i did not understand q.2 (v) and (vi). can someone explain them again?
Hi! u_k is the number of elements between 2^k and 2^(k+1). We'd like to work out how many elements there are. Why are there any elements in S? Oh, because they come from f or g applied to smaller elements. How many "smaller" elements are there, and how much smaller are they supposed to be? Making that precise gives part (v). Then part (vi) just wants the sum of those things to deal with this new definition of the range for s_k ^James
Hi James,
Is the following argument convincing enough for some marks in 5.v?
Swapping the nth day (say its rainy) and the kth day (say its sunny), causes Adam to eat n sweets where he would have eaten k, so his total changes (increase OR decrease) by n-k sweets. This swap also causes Myriam to either eat or not eat 1 extra sweet on each of the days between the nth day and the kth day (excluding one end day, so n-k days), so she either eats or doesn't eat n-k extra sweets thus changing her total by n-k. Hence, any swap causes no change in their totals (and we have our base case where they have the same total from earlier).
I appreciate that this is clumsier than the intended proof but does it seem sufficient? Sorry for the long question. Thanks for all the MAT content!
Hi - this sounds like a sort of higher-powered version of the thing we proved in 5(iii). You've done all swaps rather than just adjacent swaps. But all the marks for this bit of the question are for that bit you've put in brackets (and we have our base case where they have the same total) - and the fact that you can start from the base case and make swaps to get to any permutation. So I think this is kind of the same as the proof in the video, but with a very helpful midway step and less detail on the last bit. I'd probably give it some marks. ^James
@@OxfordMathematicsPlus Thanks James!
Hi Dr James, your videos are ever so helpful so thankyou ever so much. Can you please do the answers for all the past papers when and if you can, seeing your train of thought as you tackle these questions is priceless.
:) I'm working backwards - this channel has full video solutions for past questions back to 2016 (give me time and I'll do more) ^James
@@OxfordMathematicsPlus Thank you ever so much, they are ever so helpcul, we will be eternally grateful :)
I am so lost
Hey there, I have a question on Q1D, when substituting u=sinx and calculating the turning point of that new quadratic, how do you know that they will have the same turning points. Surely having the sinx would change it right? Like for example, if you want to calculate the turning point of y = sinx, which happens at x = 90, if you sub in u = sinx, and calculate the turning point of y = u, wouldnt that give you no turning points?
He calculated the turning points of sin(x), not x itself
It's true that the turning points of the function aren't just where the turning point of the quadratic are. By analysing the quadratic of u, I'm finding a theoretical maximum - no matter what sin(x) is doing, we can't end up with a number bigger than the maximum of that quadratic. Then we've got to actually check that the original function attains that max value. If the question asked us for the minimum value of the function... that actually happens when sin(x)=-1, I think. The chain rule is secretly behind all this, but you don't need to know what that is for MAT. ^James
YOU ARE SINCERE,DILIGENT AND YOUR HEART IS BRIMMING WITH LOVE ,BIIZNILLAH .
For Qno.2 ii, Can I write gf(12)?
You would also need to prove that 12 is in S. ^James
2022 MAT gang
ayyyyyy
WHILE OMNIPOTENT,OMNISCIENT,ALWISE GOD HAS CREATED THE ENTIRE MULTIVERSE, ACCORDING TO HIS DIVINE RATIO, BE AWARE THAT GOD IS THE GREATEST MATHEMATICIAN AND HE HAS INSPIRED YOU TO BE A VERY SMART GUY.