Hi James, For Q1J we didnt need a lot of work to prove (e), as soon as you get x of 22 and pi of 22 you can tell the statement will be true for all numbers from 2 to 9 other than 6 since six is the only one with more than one prime factor ( we can do pi of 77 for example and the it has 2 prime factors 11 and 7 and ends with seven) so just prove 6 separately after 2. I hope it helps save time for others if a similar thing comes out again in the exams.
Oh, I see - you're thinking of the numbers n=22, 33, 44, 55, 77, 88, 99 as good examples of Pi(n)=2 with various values of x(n). That would have save time if I'd thought of it - I think I was blinkered into only considering small numbers, and only working one case at a time. Good stuff. ^James
Hello Sir, In Question 5, What if you're asked to find the simplified expression for Sn, without the (An+B)2^n+C being given? How would you go about it, as it's neither Geometric or Arithmetic? Also Tn and Un?
That sounds a bit too hard for a MAT question, and it's not something I'd expect you to guess. You would learn the general method for this on a Maths course at university though! ^James
Which bit of 5(iv)? The geometric progression n+2n+4n+8n+...+2^(n-1)n ? I think that's [2^(n-1)-1]n. Maybe you mean a different bit of the calculation. ^James
No, it's fine to use radians (we've just started asking all the questions in degrees, just in case you haven't seen radians). If the question doesn't specify a method, anything is fine. ^James
Hi James, for question 5iv, the mark scheme says that t(n) is 2^(n+1)-1 ..., but you have 2^(n)-1, so I am confused which one of these is actually right. Thanks :)
My final answer agrees with the mark scheme, and we're both correct! Note that the mark scheme writes t_n as a geometric sum minus s_n whereas I write it as a different geometric sum minus s_{n-1}. More precisely, the mark scheme includes "+n2^n" in the first bracket and includes "-n2^n" in the second bracket, and I don't have those terms (this is fine because those things sum to zero). ^James
@Oxford Mathematics Plus Hi James, for Q3 vi, the questions asks about all values of t, however does your proof not just concern values where t is greater than or equal to a (as per the description in the first statement of part v)? If so, could you please explain how you would prove for when t is less than a? Also, could you also describe briefly how to answer q4 v of the November 2009 paper and Q1G from November 2011 and Q3 iv from November 2013. Thanks so much
2016 Q3 vi Fair question, but in fact if A=-B then B=-A so we're done. Here's a substitution to show how it works here; for ta so G(u)=-G(2a-u) by (iv). Sub back in; G(2a-t)=-G(t). Multiply both sides by -1. I should probably have spotted the different domain here, and addressed this. This is straightforward because 2a-t is a reflection, and is it's own inverse.
Less briefly; 2009 Q4 is in livestream #005 here ruclips.net/video/03PQhlMH9uQ/видео.html , and 2013 Q3 is in livestream #002 here ruclips.net/video/hZv7xpEyOSE/видео.html ^James
Hi James, Can you still get marks on an extended question even if you have crossed out your working with one line? Would you be awarded marks if the correct solution was found but crossed out?
That's the plan- 2017, 2018, and 2019 should release every Thursday for the next three weeks, in parallel with the MAT livestream covering those past papers at www.maths.ox.ac.uk/r/matlive . I'm already sort of regretting this because it's an astonishing amount of work to record and edit 2.5 hours of MAT, but here we are! ^James
Hi. If suppose Q1 I was part of the subjective questions( the proof ones) , then can i just use the cauchy-schwartz inequality and directly get the answer, will i get full marks? or do i need to stick to the syllabus
For Question 3 part iv, is it not sufficient to say the integral of some f(x) is some F(x) and just use the same logic as before? Namely, F(b) - F(a) = -(F(a) - F(b)).
That maths makes sense to me, but the word "hence" in the question is important - the instruction to "hence" do something means that we're supposed to use the thing we just proved previously. I've just noticed that the version of the question that I've put on the screen has a and b the wrong way around on the RHS, which is very confusing! ^James
I would strongly encourage you to write some words to explain what your diagrams mean. Note that in part (ii) or (v) or (vi) drawing a few diagrams of ways that work is not enough, because we're asked to explain how many there are, so we need to make sure we've found them all. ^James
@@OxfordMathematicsPlus I think you are only required to explain your solution for part (i), (ii), and (iii). Is it enough to just state the answer for part (iv), (v), and (vi)?
Hi James,
For Q1J we didnt need a lot of work to prove (e), as soon as you get x of 22 and pi of 22 you can tell the statement will be true for all numbers from 2 to 9 other than 6 since six is the only one with more than one prime factor ( we can do pi of 77 for example and the it has 2 prime factors 11 and 7 and ends with seven) so just prove 6 separately after 2. I hope it helps save time for others if a similar thing comes out again in the exams.
Oh, I see - you're thinking of the numbers n=22, 33, 44, 55, 77, 88, 99 as good examples of Pi(n)=2 with various values of x(n). That would have save time if I'd thought of it - I think I was blinkered into only considering small numbers, and only working one case at a time. Good stuff. ^James
Hi James,
Thank you so much for making these solutions.
In the real MAT exam do we just write on the question sheet or we will get extra papers?
应该是国人吧。。MAT考试卷是一整本sheet 选择题答案写在选择题前面的答题表里面，后面大题直接写在题目下面的
@@LTY_CK_TS 感谢🙏 嗯我知道选择题就把字母填在答题卡上， 我就担心大题那一页半写不完呀😂我做大题都得2-3面 可能是我字太大了
趙佶 那也写太多了吧哈哈哈，一页半应该是够的
@@LTY_CK_TS 🤝你也下个月考吗
@@趙佶-b1q 嗯，哪还有一个月啊，11.4🤣
Hello Sir, In Question 5, What if you're asked to find the simplified expression for Sn, without the (An+B)2^n+C being given? How would you go about it, as it's neither Geometric or Arithmetic? Also Tn and Un?
That sounds a bit too hard for a MAT question, and it's not something I'd expect you to guess. You would learn the general method for this on a Maths course at university though! ^James
@@OxfordMathematicsPlus Okay, many thanks for that clarification. I look forward to learning more about it.
If you are here for the treasure hunt its not this video. I can't figure out where to go
Hi James, for question 5 part iv , for the geometric progression isn't it [2^(n-1)]n, I am a bit confused there.
Thanks
Which bit of 5(iv)? The geometric progression n+2n+4n+8n+...+2^(n-1)n ? I think that's [2^(n-1)-1]n. Maybe you mean a different bit of the calculation. ^James
also Q5 is just an arithmetic-co geometric progression, so the answer is s_n = (n-1)*2^n+1 + 2
Hi James, would you be penalised for using Radians or any content not on the MAT syllabus if the question doesn’t specify what method to use.
No, it's fine to use radians (we've just started asking all the questions in degrees, just in case you haven't seen radians). If the question doesn't specify a method, anything is fine. ^James
Hi James, for question 5iv, the mark scheme says that t(n) is 2^(n+1)-1 ..., but you have 2^(n)-1, so I am confused which one of these is actually right. Thanks :)
My final answer agrees with the mark scheme, and we're both correct! Note that the mark scheme writes t_n as a geometric sum minus s_n whereas I write it as a different geometric sum minus s_{n-1}. More precisely, the mark scheme includes "+n2^n" in the first bracket and includes "-n2^n" in the second bracket, and I don't have those terms (this is fine because those things sum to zero). ^James
@@OxfordMathematicsPlus Thanks, I understand now.
@Oxford Mathematics Plus Hi James, for Q3 vi, the questions asks about all values of t, however does your proof not just concern values where t is greater than or equal to a (as per the description in the first statement of part v)? If so, could you please explain how you would prove for when t is less than a? Also, could you also describe briefly how to answer q4 v of the November 2009 paper and Q1G from November 2011 and Q3 iv from November 2013. Thanks so much
2016 Q3 vi Fair question, but in fact if A=-B then B=-A so we're done. Here's a substitution to show how it works here; for ta so G(u)=-G(2a-u) by (iv). Sub back in; G(2a-t)=-G(t). Multiply both sides by -1. I should probably have spotted the different domain here, and addressed this. This is straightforward because 2a-t is a reflection, and is it's own inverse.
2009 Q4 v (briefly) Look at a point outside the parabola, like (0,-1).
2011 Q1G (briefly) x^2-1 is between -1 and 0, where f is x+1, so the integrand is x^2
2013 Q3 iv (briefly) reflect twice; once in the line x=1 to move the root from k to 2-k, and then once in the x-axis to fix the coefficient of x^3
Less briefly; 2009 Q4 is in livestream #005 here ruclips.net/video/03PQhlMH9uQ/видео.html , and 2013 Q3 is in livestream #002 here ruclips.net/video/hZv7xpEyOSE/видео.html ^James
Hi, thanks tons for this explanation!
Hi James,
Can you still get marks on an extended question even if you have crossed out your working with one line? Would you be awarded marks if the correct solution was found but crossed out?
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Hi James, are you going to do any more full MAT papers? This was great, thanks.
That's the plan- 2017, 2018, and 2019 should release every Thursday for the next three weeks, in parallel with the MAT livestream covering those past papers at www.maths.ox.ac.uk/r/matlive . I'm already sort of regretting this because it's an astonishing amount of work to record and edit 2.5 hours of MAT, but here we are! ^James
Hi. If suppose Q1 I was part of the subjective questions( the proof ones) , then can i just use the cauchy-schwartz inequality and directly get the answer, will i get full marks? or do i need to stick to the syllabus
Hi - on Q1 you get the marks for right answers only, and it doesn't matter what method you use at all. ^James
For Question 3 part iv, is it not sufficient to say the integral of some f(x) is some F(x) and just use the same logic as before? Namely, F(b) - F(a) = -(F(a) - F(b)).
That maths makes sense to me, but the word "hence" in the question is important - the instruction to "hence" do something means that we're supposed to use the thing we just proved previously. I've just noticed that the version of the question that I've put on the screen has a and b the wrong way around on the RHS, which is very confusing! ^James
Hi James,
Should we do the MAT exam using a pen or pencil, or both are pen and pencil are allowed? And are we allowed to use correction tape?
Pen, please! No correction tape, please. We'd like to see your working out :) ^James
@@OxfordMathematicsPlus Hi James, can the graphing question be an exceptional case for using a pencil? Many thanks.
In the MAT exam, if I explain my answers to the dancers questions using diagrams rather than words, does anyone know whether it will be accepted?
I would strongly encourage you to write some words to explain what your diagrams mean. Note that in part (ii) or (v) or (vi) drawing a few diagrams of ways that work is not enough, because we're asked to explain how many there are, so we need to make sure we've found them all. ^James
@@OxfordMathematicsPlus Got it, thanks!
@@OxfordMathematicsPlus I think you are only required to explain your solution for part (i), (ii), and (iii). Is it enough to just state the answer for part (iv), (v), and (vi)?
For question 1f I used the factor theorem with i, it was bit messy but it worked
I liked your way much better
@@jacobhubbard617 If you like complex numbers, go for it! ^James
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