Q1J may be solved using the 1/3, 2/3 rule, and similar shapes. For any medial line from any vertex to the opposite midpoint, 2/3 of the line lies between the vertex and centre, and the other 1/3 lies between centre and midpoint. Use that fact for a line of the vertical height, and use similar shapes (draw 2 right triangles of full height and of 2/3 of full height) and some trig!
thanks james, for q1F, is there a reason why you didn't draw the graphs of sin^3(x) and -cos^2(x) and see how many times they intersect in the range? would that method also be fine or is it harder. for me i think that method was easier
That's a reasonable strategy, perhaps even easier! I guess I didn't try it because I was SO EXCITED to use the cos^2+sin^2 fact. Hmm. Perhaps I've learned something here. ^James
Surely for Q1H, after you have done pythagoras and found that it contains sqrt(5), you can immediately rule out B and E (because they dont contain sqrt 5) and also rule out A and D because they are either complex or negative, leaving you with C?
What does "contains sqrt(5)" mean? Does sqrt(sqrt(5)^2-2) contain sqrt(5)? Does it contain sqrt(3)? It's equal to sqrt(3) so maybe it does! Oh no! Joking aside, it's OK to guess on the multiple-choice questions, but I am going to work them out properly in these videos :) ^James
In 1J., why isn’t the line pq horizontal, wouldn’t that minimise its length and it would still go through the centre. Then the answer would be 1/2 as the centre 0 point would be 1/2, root 3 by 4
Question 4 uses abit of analysis . I have to ask that on the 2020 mat test there is binary expansion but there is no mention of that in the current syllabus
Thank you very much! There is one thing I don't understand. It's about question 5(v). when you keep substituting f(2n-2, n-1) by putting in the function of f(n+1, k) how do you get to x5 x3 xf(2,1) at the end? I would think you will just get f(2n - a, n - a) where a gets bigger and bigger?
He's used the fact that f(2n-1, n) = 0 (this was part (i)) As f(2n,n) = (2n-1)f(2n-2,n-1) + nf(2n-1,n) From part (i), f(2n-1,n) = 0 So we just get f(2n,n) = (2n-1)f(2n-2,n-1) and continuing this we get the product of all odd numbers less than 2n
Just to be super clear, the things you get are f(2n-2a,n-a) where a gets bigger and bigger. From the first step, I noticed that the first argument decreases by 2 (!) each time. ^James
I have a question with Q5 (v), wouldn't it also be possible to argue combinatorially, saying that f(2n,n) = (2*n!)/(2^n), knowing that it's guaranteed that every set is only going to contain 2 elements? Or would this answer not be accepted, since it doesn't follow the previously established pattern? Thanks a lot!
Hey there Dr. James, for 1H I used a very slightly different method. Firstly I could immediately eliminate A since 1-phi=-phi and we're not dealing with complex numbers. Secondly, I split the right angle ar thing with two cases, where ar^2 is the hypotenuse like you did and one where a is the hypotenuse is r
That works, but you've got slightly lucky with your method (writing ar^2 for the hypotenuse instead of a isn't actually a different case, but it coincidentally switches the roles that the two short sides are playing, so you do end up with both solutions). I guess what I'm saying is that you've considered both cases, but could explain better why those are two different cases. Anyway, you get the marks for getting the right answer since this is the multiple-choice section we're talking about! ^James
@@alikarimp.chumsworth5674 Yes, that's what I mean. It's a very subtle difference between the two arguments! When you draw a diagram, you get to choose which corner is B, and if you're not careful you'll choose the same corner in both diagrams... ^James
For question D, I dont understand why you drew x^2+2ax+3 above the y-axis, because the roots of a-x^2 = +/- sqrt(a) , but the intersection at A, technically they should intersect at the left of the negative root, meaning some of the area which you have integrated was negative?.
I drew the graph y=x^2+2ax+a before I knew that a=3 :) For this calculation, it doesn't matter whether the two curves are above or below the x-axis, because I'm actually integrating the difference between the two functions. So I just need to make sure that I've got the curves the right way around to make sure that I'm integrating an expression that's positive. Try it out by finding the area between y=1-x^2 and y=-1+x^2; check that taking the difference and integrating gives you the correct area in between (check that it's twice what you'd get if you found the area between y=1-x^2 and the x-axis). ^James
@Oxford Mathematics Plus For Q1J, if you make PQ a horizontal line, I think you get the distance root 3/3 (option c) - why is this not the actual answer? Thanks in advance
@@rohan_142 Is your centre at the origin? I think you've assumed that the midpoint of the triangle is halfway along the line from a corner to the midpoint of the opposite side, which is sadly not the case. I try to do the relevant calculation at 48:02 ^James
Sir, can you plz upload a video based on MAT 2014? The long answer questions (particularly the 6th and 7th ) are a bit tricky. Can you plz clarify those? It would be very helpful if you respond to my request.
@@nq2c Yep, I just uploaded a maths-only edit of MAT 2014 Q7 :) . I don't think we covered many of the MAT 2014 long questions, but we did do a lot of the short questions at some point during the livestream project. ^James
Why get into IIT/other college with JEE when you can get into Oxford easily? You can even avoid the competition of JEE and get a better score than other people looking to get into Oxford.
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Thanks a lot, This was very useful as I take the 2020 test tomorrow. My god Oxford isn't kidding when they say it's hard.
Agreed
STEP is far harder
How did it go ? I hope you did well.
@@ym276 But at least we will have support programme for STEP
All the best in life and everything
thank you very much :) can't say i'm looking forward to tomorrow but these livestreams have helped loads :)
Q1J may be solved using the 1/3, 2/3 rule, and similar shapes. For any medial line from any vertex to the opposite midpoint, 2/3 of the line lies between the vertex and centre, and the other 1/3 lies between centre and midpoint. Use that fact for a line of the vertical height, and use similar shapes (draw 2 right triangles of full height and of 2/3 of full height) and some trig!
I have a maths degree and I find these questions hard, so many traps
Thank you soo much James, I just did this paper this morning.
Thank you so much for uploading this...I feel like 2019 was one of the toughest past paper!😅
and 2013!
thanks james,
for q1F, is there a reason why you didn't draw the graphs of sin^3(x) and -cos^2(x) and see how many times they intersect in the range? would that method also be fine or is it harder. for me i think that method was easier
That's a reasonable strategy, perhaps even easier! I guess I didn't try it because I was SO EXCITED to use the cos^2+sin^2 fact. Hmm. Perhaps I've learned something here. ^James
Surely for Q1H, after you have done pythagoras and found that it contains sqrt(5), you can immediately rule out B and E (because they dont contain sqrt 5) and also rule out A and D because they are either complex or negative, leaving you with C?
What does "contains sqrt(5)" mean? Does sqrt(sqrt(5)^2-2) contain sqrt(5)? Does it contain sqrt(3)? It's equal to sqrt(3) so maybe it does! Oh no! Joking aside, it's OK to guess on the multiple-choice questions, but I am going to work them out properly in these videos :) ^James
In 1J., why isn’t the line pq horizontal, wouldn’t that minimise its length and it would still go through the centre. Then the answer would be 1/2 as the centre 0 point would be 1/2, root 3 by 4
good question, i was also thinking the same
Roughly the 48:03 mark is where I realised that I could draw the line PQ horizontal. That's definitely got length 2/3 though (not 1/2). ^James
Q1J. The easiest way is to use cosine law and we can easily prove PQ has the minimal length when PQ is parallel to one side of the triangle.
"Easily": please show us?
Question 4 uses abit of analysis . I have to ask that on the 2020 mat test there is binary expansion but there is no mention of that in the current syllabus
Good luck 🤞 get a good nights sleep n don’t worry yaself 😁 we’ll be fine👍
we'll be fine 😂😂😁😁 it would be better if we were in a horror movie and the monster is looking for us rather than this real life situation 😂😂😂
Thank you very much! There is one thing I don't understand. It's about question 5(v). when you keep substituting f(2n-2, n-1) by putting in the function of f(n+1, k) how do you get to x5 x3 xf(2,1) at the end? I would think you will just get f(2n - a, n - a) where a gets bigger and bigger?
He's used the fact that f(2n-1, n) = 0 (this was part (i))
As f(2n,n) = (2n-1)f(2n-2,n-1) + nf(2n-1,n)
From part (i), f(2n-1,n) = 0
So we just get f(2n,n) = (2n-1)f(2n-2,n-1) and continuing this we get the product of all odd numbers less than 2n
Just to be super clear, the things you get are f(2n-2a,n-a) where a gets bigger and bigger. From the first step, I noticed that the first argument decreases by 2 (!) each time. ^James
Good luck everyone else doing this today :)
I have a question with Q5 (v), wouldn't it also be possible to argue combinatorially, saying that f(2n,n) = (2*n!)/(2^n), knowing that it's guaranteed that every set is only going to contain 2 elements? Or would this answer not be accepted, since it doesn't follow the previously established pattern? Thanks a lot!
That's fine, and should get the marks. The question doesn't say that we have to use the same logic/pattern/method as the previous parts. ^James
At 1:40:10, f(5,2) was already given at the start. Anyways very helpful video thanks.
Hey there Dr. James, for 1H I used a very slightly different method.
Firstly I could immediately eliminate A since 1-phi=-phi and we're not dealing with complex numbers.
Secondly, I split the right angle ar thing with two cases, where ar^2 is the hypotenuse like you did and one where a is the hypotenuse is r
That works, but you've got slightly lucky with your method (writing ar^2 for the hypotenuse instead of a isn't actually a different case, but it coincidentally switches the roles that the two short sides are playing, so you do end up with both solutions). I guess what I'm saying is that you've considered both cases, but could explain better why those are two different cases. Anyway, you get the marks for getting the right answer since this is the multiple-choice section we're talking about! ^James
@@OxfordMathematicsPlus By roles you mean e.g. opposite or adjacent to find tan right ?
@@alikarimp.chumsworth5674 Yes, that's what I mean. It's a very subtle difference between the two arguments! When you draw a diagram, you get to choose which corner is B, and if you're not careful you'll choose the same corner in both diagrams... ^James
For question D, I dont understand why you drew x^2+2ax+3 above the y-axis, because the roots of a-x^2 = +/- sqrt(a) , but the intersection at A, technically they should intersect at the left of the negative root, meaning some of the area which you have integrated was negative?.
I drew the graph y=x^2+2ax+a before I knew that a=3 :) For this calculation, it doesn't matter whether the two curves are above or below the x-axis, because I'm actually integrating the difference between the two functions. So I just need to make sure that I've got the curves the right way around to make sure that I'm integrating an expression that's positive. Try it out by finding the area between y=1-x^2 and y=-1+x^2; check that taking the difference and integrating gives you the correct area in between (check that it's twice what you'd get if you found the area between y=1-x^2 and the x-axis). ^James
Oh, and I should have said that you are correct that if a=3 then the intersection is to the left of the root :) ^James
@@OxfordMathematicsPlus Thanks for making that clearer for me :)
@Oxford Mathematics Plus For Q1J, if you make PQ a horizontal line, I think you get the distance root 3/3 (option c) - why is this not the actual answer? Thanks in advance
To work that out, I said that the top vertex is at (0, root 3/4) - is this wrong?
@@rohan_142 Is your centre at the origin? I think you've assumed that the midpoint of the triangle is halfway along the line from a corner to the midpoint of the opposite side, which is sadly not the case. I try to do the relevant calculation at 48:02 ^James
That one dislike is from someone that didn’t pass maths
Sir, can you plz upload a video based on MAT 2014? The long answer questions (particularly the 6th and 7th ) are a bit tricky. Can you plz clarify those? It would be very helpful if you respond to my request.
He has done them already probably at some point if you go through each livestream on this chanel
@@nq2c Yep, I just uploaded a maths-only edit of MAT 2014 Q7 :) . I don't think we covered many of the MAT 2014 long questions, but we did do a lot of the short questions at some point during the livestream project. ^James
You are evil for creating question H.
1J was the evil one. 1H was calm.
For jee advance aspirants this is a cake 🎂 walk. I am able to solve some of these mentally
Why get into IIT/other college with JEE when you can get into Oxford easily? You can even avoid the competition of JEE and get a better score than other people looking to get into Oxford.
You are very sweet.