MAT 2018 | All questions | MAT Livestream Bonus
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- เผยแพร่เมื่อ 1 ส.ค. 2024
- All of the questions for MAT 2018. 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 2018 Q1A
2:47 MAT 2018 Q1B
6:37 MAT 2018 Q1C
11:55 MAT 2018 Q1D
14:56 MAT 2018 Q1E
19:18 MAT 2018 Q1F
26:45 MAT 2018 Q1G
33:13 MAT 2018 Q1H
40:46 MAT 2018 Q1I
43:27 MAT 2018 Q1J
53:39 MAT 2018 Q2
1:14:01 MAT 2018 Q3
1:29:14 MAT 2018 Q4
1:53:31 MAT 2018 Q5
2:10:11 MAT 2018 Q6
2:28:31 MAT 2018 Q7
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This is so hard 💀💀
James I love you these are so helpful
For Q1C, alternatively we could use the properties of quadratic function when it has two distinct real intersection points at x-axis which is D > 0.
These were very helpful, thanks James.
1:12:58 smh 🤦♂️^James
LOL this moment is so funny
For 3v is it correct if I drew a cubic graph annotating where p is max and q is max and then as seen from the diagram from p to q the graph is a decreasing function therefore h'(X)
For question 2i is the checking step necessary? Or is it sufficient enough to say that (-y,x+1) =/= (-y+1,x)?
Q1J could be solved much simpler by realizing that the graph must be symmetrical by the line y=x due to y and x being irreplacatble in the equation, automatically ruling out A.
For Q1B; did you have to find the 3 values? Do you loose marks if you don't? Because I got k**3 = 2k and concluded they'd obviously intersect 3 times. Would that be good enough on the actual test?
For the multiple choice questions (Q1), you get four marks for the correct answer and no marks for the wrong answer. So it doesn't matter if you actually get the right values for k or not! ^James
Just a quick question , can I do Q3 (v) using Cauchy's Mean Value Theorem? We know that h(q)-h(p) is negative and the result follows immediately from the MVT. But I'm not sure because MVT only says that there exists some point x in (p,q) but not every x in the range. Thank you James!
You've answered your own question :) Cauchy's MVT does not directly prove this result for every x in the range. (note for anyone else reading this; we do not expect MAT candidates to know about Cauchy's MVT) ^James
Thank you, James. I found your videos helpful especially when It comes to how to approach and untie a problem that I have never seen before, and I think this skill will also be very useful when it comes to solving problems outside MAT.
For ques 3 ii, can't we find m(x)/slope using differentiation?
No, sorry, m(x) is the gradient of the chord between two points, not the gradient of the curve at (x,g(x)). Here's a desmos thing to illustrate the difference (green for chord, blue for tangent) www.desmos.com/calculator/rkry1erzak ^James
for question 4(iii) could i find the area by doing 1/12 of the area of s1 - 1/3 of the area of s2? your videos are really helpful thank you! :)
I'm worried that this area will include the triangles ABC and ABD, I'm afraid ^James
In Q3(ii), how can m(x) be defined for x=0 when the chord has no length? The workings for part (iv) seem to contradict this as well, because instead of taking the limits we could just use m(0) as defined in (ii), right? Thank you :)
You're right, m(x) isn't defined for x=0 (reading the line above part (i)), so it doesn't make sense to ask for m(x) in the case x ≥ 0 in part (ii). That should probably say x > 0. My working-out is even more sloppy because I wrote ≥ and ≤ when I should have written > and
Hi, for question 5 (III) as it says give an expression is my answer allowed its not simpified like yours is but it essentially states the same thing: 2(n) + (summation from i= 2 to n of (n-i)) + 1, could you please validate if this correct and if, due to the wording of the question (an expression), give any points?
I think that you want to start that sum from i=1, and I think you maybe want to subtract the 1 at the end rather than adding it. ^James
For Q1G can I let t=ks where k is the ration between them so the function only remain a variable k and then differentiate them.
I got the same ans but the method is different
For Q1H? Yes. At 38:49 I wrote u=s/t but you could instead write k=t/s if you want to. ^James
Hi James. Just wanted to ask if we would get penalised for using radians as I as much more confident with them. If I used radians for working and then converted to degrees for the final answer would that be fine?
There's no penalty for using radians! We just don't use them in the questions. You're allowed to use radians and you're allowed to give your answers in terms of radians if you like. ^James
@@OxfordMathematicsPlus yay!
for Q1C the geometric reasoning part. how would having bx^2+d grow faster then ax^2+c make them intersect? aren't they in opposite directions?
It's not a very helpful picture by that point because it's showing a case with no solutions. You're right - we don't want them to go in opposite directions. In fact, even if they're going in the same direction, if b
@@OxfordMathematicsPlus Ah I see thank you.
also for Q5, can I just use the fact the recursion is the( n-1)th Catalan number, so all parts of the question are trivial? Do I need to derive the formula of the n-th Catalan number, or can I simply use the result?
"can I just use the fact that the recursion is the (n-1)th Catalan number" - what recursion? We're trying to give you marks in part (v) for writing down a recursion relation for t_n. If you recognise that recursion relation once you've written it down then that's nice, but you already did all the work? I don't consider part (vi) to be obvious for the Catalan numbers, so I don't really see how this observation helps you, to be honest! ^James
@@OxfordMathematicsPlus Thank you Mr.James for the reply! Also, I already have sent my UCAS application but unfortunately, my referee can only send the reference letter one week later than the deadline due to her current workload. Will my application still be considered?
@@advaithkumar5966 When I went to Bath, they said they don't garner that much information from the reference letter anyway, so hopefully it's the same for Oxford. It would suck to be rejected for reasons outside of your control.
for question 1H, just use AM-GM inequality.
could you explain?
For 1E I used the area of the circle and the area of a grid (which is 1 unit sqr) to get the number of grids for radius=2 then got the 2n+5 as 13 but I couldn't use this method to get the increase in radius needed for the 13 grids. Is there something I missed?
I can't see an obvious link between the area of the circle and the number of grid points inside the circle - it's probably about pi R^2 grid points for a circle with large radius R, but it's not clear how to make that more precise. So I think this idea might be a bit of a dead-end :( ^James
@@OxfordMathematicsPlus Thanks a lot. I'll try my best to understand the method used in the video then.
@Oxford Mathematics Plus
But just to clarify my method, I thought of each grid point as a box with area 1 unit square. The area of the circle was 12.568 I think so I rounded up to 13 since the picture had the circumference overlapping with some boxes and the question considered grid points on the circle as well. So n=13. You have made me understand that this won't be very helpful for finding the radius increase but assuming we were required to find just n are there any other faults with my reasoning?
Hi, for 4iv, I interpreted it as the area covered by both S1 AND S2 rather than the area covered by either S1 or S2 or both. Would this be credited as correct, since the wording in the question isn't very clear?
Sorry, but I think that the word "total" here means that we're talking about the area covered either by S1 or by S2. ^James
@@OxfordMathematicsPlus Oh ok, thanks
In 4iv), since the question said the total area is split into two equal areas, i assumed that S1CE would be equal to CEDS2 and not equal to half of this. can anyone please explain where i went wrong?
I think you've interpreted the question correctly - my handwriting is bad at 1:51:17 but my equation is supposed to be S1CE = half of the total. That total is 6 pi plus the thing we found in part (iii) ^James
@@OxfordMathematicsPlus oh yeah okay i understand. i decided to set the halfs as equal to each other instead, which made it unnecessarily complicated
Q5(vi) You showed how t2n is odd if and only if tn is odd, how does that automatically mean tn is odd if and only if t2n is odd? (The question asks the opposite way around to how you answered)
Yeah, "if and only if" is symmetric because it means " "if A then B" and also "if B then A" ". I've proved those two things in perhaps the opposite order to what you were expecting, but I've done both. ^James
Hi James … this is a bit ridiculous but occasionally on the multiple choice questions I tick the wrong box on the first page when I have the correct working and have circled the correct answer on the actual question (I think it’s probably because I’m thinking oh I’m doing question d then end up ticking (d,D) instead of say (e,D)) … I’m going to try really hard to avoid doing it during the real exam but if I do would I still lose the marks? I will always circle the answer I think it is on the actual question .. ?
If your score is borderline (not good enough to make the choice to give you an interview easy but not bad either) then they look at your workings carefully for places where you ALMOST got marks. In this case they would recognise that you did all of the maths right but failed to tick the corresponding box, and this would give you an advantage over applicants with similar MAT scores who did not provide evidence that they actually know the maths. But Ideally make sure you tick the correct box to make your score as high as possible; every time I find the correct answer to a question I think to myself "H is d", for example, until I have turned the page and ticked the box. This really helps me personally to remember which box to tick.
For Q4（iv）I think you only proved that there were some points in between that is the solutions for the function. But the question seems to ask you for the proof for only one solution. How can we prove that ?
Good point. I think that the area on the left is a decreasing function, but I should have said that. ^James
Are there multiple answersfor q6i
Yes! I've got two expressions for 2/5 in the video. There are multiple ways to write the others too. ^James