You are correct - I made a mistake here by not including that logic, so wouldn't even have got full marks myself on this question 😅😂 I'm going to pin this comment!
I've already done my A levels in 2023. And this bring backs a lot of memories. Your videos helped me a lot. Especially a 100 days of Maths. I ended up with a B in the end where I ended up in 1st choice.
I have my Year 12 mocks next week and unfortunately, they will be predicting University grades, so alongside exam questions on the online textbook, I have also attempted some AS papers. This is a really helpful video to remind me of topics I have forgotten to do, than you very much!
Hi Bicen, I commented before about A-Level Statistics (9ST0). Would love to see a walkthrough of papers from 2022 if you might have the time. Thank you!
Thank you so much, I got a C on my a level maths. May the Lord Jesus bless and keep everyone safe and strong as they study this interesting course in Jesus name Amen
@@seanbaxter44 I am focusing on my housekeeping job and family time. My studies took so much time off my daughter. I am not too sure what to do with my a levels at the moment if I am honest. Are you doing a level maths by any chance?
@@seanbaxter44 that's lovely, I found this guy days before exams and he helped me tremendously. I will be praying for your growth and understanding throughout your studies and beyond!!!
The problem with the modelling questions ( e.g. ferris wheel) is that the model is completely and utterly wrong. A ferris wheel going round at constant angular speed would have a sinusoidal fluctuation of vertical displacement (not the modulus of some sinusoid). In this example, the velocity abruptly changes from being negative to positive at the bottom, causing an infinite acceleration that would kill the passenger and destroy the mechanic components of the ferris wheel. And why does only my carriage have the death acceleration - why me??? Surely each carriage would experience a symmetrical behaviour, so if it really was behaving like this, then there would be as many discontinuities in the gradient as there are carriages within the full cycle. It's a jerky death ride. A ferris wheel doesn't behave like a bouncing ball, that has a mechanism to absorb energy temporarily at the bottom of the bounce due to conversion to elastic potential energy. The question setters are literally telling students to forget common sense, forget what we are supposed to know about how trigonometric functions relate to the unit circle. I genuinely believe that Edexcel exam writers are not very good mathematicians and haven't really understood why we model things as trigonometric functions (mostly because they fail to understand the relationship between circular motion and SHM)...they don't reward actual mathematical insight, and an actual ability to reflect on the situation and give a sensible model. Please, do not subject humans to infinite G force.
hey seb i passed into my 2nd year of uni doing engineering and even tho i didnt get the best grade i still passed. still remember the GOAT teacher and all your amazing videos. keep it going!
For part b on the question about inflection points at 18:21, if i simply subbed in x = 4root of 27, into the equation for the second derivative and showed that it = 0 .Would that be enough to show that there is an inflection?
Hello Mr Bicen, I think I have a method for question number 2, but it may be patchy so sorry if it's wrong. But basically, here's the proof: If it's even, you can express it as 2k. (2k)^2 is 4k^2, and 1 less than this is 4k^2 -1. So 4k^2 itself obviously has a factor of 2k, and 4k^2 - 1 = (2k+1)(2k-1). So, for the square of an even number, OR the square of an even number minus one (which the question allows), you have 3 consecutive factors of (2k-1), 2k, and (2k+1). Every 3rd number must be a multiple of 3, so this is true for all even numbers. Odd numbers can be expressed as (2k+1)^2. This is 4k^2 + 4k + 1, and 1 less than this is 4k^2 + 4k. Now 4k^2 + 4k + 1 obviously has a factor of (2k+1), but 4k^2 + 4k can be expressed as 2k(2k + 2). So, we've found factors of (2k), (2k+1), and (2k+2). So, by the same chain of reasoning for the even numbers, the statement must also be true for odd numbers.
Hi, regarding number 7's point of inflection section, I was taught that I could use the first derivative and input a value slightly smaller and slightly above the stationary point to show that the gradient is constant on both sides to prove it is a point of inflection. Would this way give you marks for this question. For example I mean subbing in root 28 and root 26 and showing there is no change in sign. Thanks
When you simultaneously solve a pair of equations, and the result of that is a quadratic equation, the number of solutions it has corresponds to the number of solutions to the simultaneous equations. If there are no solutions to that quadratic, there are no solutions to the sim eqs and so they don't cross. If there are 2 solutions, they cross twice. And if there's only one solution, they only cross once, and in this context, can only mean that there is a tangent, as a tangent has one point of intersection with a curve (usually).
I have a different proof for the last one and i wonder if it would also be sufficient first write the equation as (2p)^2 - q^2 = 5^2 then rearrange to get (2p)^2 = 5^2 + q^2 we now by definition have a pythagorean triple 2p, 5, q but we also know that because it is a pythagorean triple the only combination that works with 5 at that side is 13, 5, 12 however this means that 2p = 13 and thus p = 6,5 which is not an integer, hence we've reached a contradiction I'm not sure tough if I can just say that we know only the comination 13, 12, 5 works
For question 9 part b, I used long division with g(x) to get g(x) = 3 - 1/(ln(x)-2) and I found that easy to differentiate. Which way is better in your opinion?
hey mister bicen would you please sir make a video like this one for s1? i would really appreciate it since im performing it in this oct session (oct 12) . thank you in advance mister
Dear Mr Bicen, in question no. 1 why have you only used integer factor pairs for 25, why didn't you use like 12.5 and 2 which also multiply to give 25?
Not necessarily - it would be a weird curve, but the property for inflection is that it goes from concave to convex, so we have to use that, not the gradient property.
hello! I'm asking if the video includes all the pure papers or just P1 and P2?Also is this for the WMA12 and WMA11 paper reference, thanks in advance, I'm asking before watching the video to know before getting started
My best guess is that the grade boundaries and difficulty will be in line with 2023 - but we'll only really know until we come out that first exam paper!
Hi Sir, for the last proof where 2p + q =5 and 2p - q =5, could you say that 2p + q = 2p - q which implies that q = -q which is a contradiction that q is a positive integer and end the proof there?
Yes, you could. You'd just need to put it in the standard form first, which is dy/dx + P(x)y = Q(x), so in this case would be dh/dt + 5/1200 h = 24/1200. But the normal maths method is just quicker!
For number 7 part b, can you not just find points around x and show that the gradient of these points are both negative so it is a point of inflection?
We need to show that it is going from convex to concave or vice versa for an inflection point, whereas I don't think just checking the gradient is sufficient for this!
Can't you also prove a point of inflation by showing that second derivative equals zero AND that first derivative is same sign either side of the point of inflection????
No I don't think the logic of this quite holds up, as we are trying to prove it goes from convex to concave, not just same gradient either side. I could draw some very odd looking functions that do not have a point of inflection but do have the same gradient either side (i.e. could have straight line on one side, so not convex/concave.)
Hi Mr Bicen, I was just wondering about question 9 (the differentiation question, part c specifically). What happens if 0
You are correct - I made a mistake here by not including that logic, so wouldn't even have got full marks myself on this question 😅😂 I'm going to pin this comment!
thank you for this video!! paper 1 is fast approaching now and this is really helpful
I found it very useful how you made the notes/workings out before the video and only focused on the talking/explaining them during the video :)
Thank you! I changed style for this video, and think I preferred it to writing them out live.
@@BicenMaths Probably makes the video shorter so easier for the viewer.
I've already done my A levels in 2023. And this bring backs a lot of memories. Your videos helped me a lot. Especially a 100 days of Maths. I ended up with a B in the end where I ended up in 1st choice.
That's amazing! Congrats - getting into your 1st choice uni is an incredible achievement, well done. I hope you're enjoying it!
You are the equivalent of Pep Guardiola tearing up the Premier League
hes more an ange
I have my Year 12 mocks next week and unfortunately, they will be predicting University grades, so alongside exam questions on the online textbook, I have also attempted some AS papers. This is a really helpful video to remind me of topics I have forgotten to do, than you very much!
Coming back to this video to brush up my maths skills after my gap year, very useful
its hard when bicen puts skull emoji in title lol
I wanna go back to gcse's
Same
fr man
I wanna go back to first year uni 😂 real analysis was so easy to learn compared to later stuff
@@TheBlueboyRuhan damn😓cant imagine uni
don't we all 😮💨😮💨
Can u do these “hardest” type of vids for fm too thanks
I'm ready to go with Core Pure if people want it!
@@BicenMathsplease!
@@BicenMaths I would like that too please!
@@BicenMathsYESSS PLEEASEE
@@BicenMathsanticipating the matrices one on the chimpanzees from an AS paper lol
This video saved me for paper 1 maths today, absolute legend🔥
Hi Bicen, I commented before about A-Level Statistics (9ST0). Would love to see a walkthrough of papers from 2022 if you might have the time. Thank you!
9:21 For the second q part c, can’t you also say that a < e^2 ? (by making both the denominator and numerator negative)
Yes, that is correct! Someone else pointed this out too, I made a mistake here. I've just pinned their comment so hopefully people will see it :)
Thank you so much, I got a C on my a level maths. May the Lord Jesus bless and keep everyone safe and strong as they study this interesting course in Jesus name
Amen
what are u doing now
@@seanbaxter44 I am focusing on my housekeeping job and family time. My studies took so much time off my daughter. I am not too sure what to do with my a levels at the moment if I am honest. Are you doing a level maths by any chance?
@COSTIETVbehappy oh okay fair enough and yh im doing a level maths I'm in year 12
@@seanbaxter44 that's lovely, I found this guy days before exams and he helped me tremendously. I will be praying for your growth and understanding throughout your studies and beyond!!!
@@Masowe.Your daughter?
The problem with the modelling questions ( e.g. ferris wheel) is that the model is completely and utterly wrong. A ferris wheel going round at constant angular speed would have a sinusoidal fluctuation of vertical displacement (not the modulus of some sinusoid). In this example, the velocity abruptly changes from being negative to positive at the bottom, causing an infinite acceleration that would kill the passenger and destroy the mechanic components of the ferris wheel. And why does only my carriage have the death acceleration - why me??? Surely each carriage would experience a symmetrical behaviour, so if it really was behaving like this, then there would be as many discontinuities in the gradient as there are carriages within the full cycle. It's a jerky death ride. A ferris wheel doesn't behave like a bouncing ball, that has a mechanism to absorb energy temporarily at the bottom of the bounce due to conversion to elastic potential energy. The question setters are literally telling students to forget common sense, forget what we are supposed to know about how trigonometric functions relate to the unit circle. I genuinely believe that Edexcel exam writers are not very good mathematicians and haven't really understood why we model things as trigonometric functions (mostly because they fail to understand the relationship between circular motion and SHM)...they don't reward actual mathematical insight, and an actual ability to reflect on the situation and give a sensible model. Please, do not subject humans to infinite G force.
🤓
hey seb i passed into my 2nd year of uni doing engineering and even tho i didnt get the best grade i still passed. still remember the GOAT teacher and all your amazing videos. keep it going!
I’m so pleased to hear this! Thank you for commenting and getting back in touch, it’s lovely to hear from you. Hope you’re well!
@BicenMaths im doing well! Lots of my modules could use someone like you explaining them with such a clear and easy to understand format such as yours
@kuba2466 that’s such a nice thing to say - thank you. I hope it continues to go well for you, I know you’ll put in the work and do great!
Hi, love your content, been watching you for all of my year 1 maths, was wondering if you plan on doing content on MAT/TMUA exams
I am hoping to do some in the future - possibly in time for next year’s exams - but currently working on finishing all the many modules for FM!
@@BicenMaths sounds good! thank you
Dreading the day I do that
@@chaska8144 good luck bro
For part b on the question about inflection points at 18:21, if i simply subbed in x = 4root of 27, into the equation for the second derivative and showed that it = 0 .Would that be enough to show that there is an inflection?
No, it wouldn't - you do need this full method for all the marks.
Hi Mr Bicen, can you please also do one of these hardest question for statistics and mechanics? Thank you
I am aiming to do this before the exams this summer!
Hello Mr Bicen, I think I have a method for question number 2, but it may be patchy so sorry if it's wrong. But basically, here's the proof:
If it's even, you can express it as 2k. (2k)^2 is 4k^2, and 1 less than this is 4k^2 -1. So 4k^2 itself obviously has a factor of 2k, and 4k^2 - 1 = (2k+1)(2k-1). So, for the square of an even number, OR the square of an even number minus one (which the question allows), you have 3 consecutive factors of (2k-1), 2k, and (2k+1). Every 3rd number must be a multiple of 3, so this is true for all even numbers.
Odd numbers can be expressed as (2k+1)^2. This is 4k^2 + 4k + 1, and 1 less than this is 4k^2 + 4k. Now 4k^2 + 4k + 1 obviously has a factor of (2k+1), but 4k^2 + 4k can be expressed as 2k(2k + 2). So, we've found factors of (2k), (2k+1), and (2k+2). So, by the same chain of reasoning for the even numbers, the statement must also be true for odd numbers.
Hi mr Bicen,
The videos for further pure 1 videos says I need to get a membership. Is it a bug or do I really need to pay to get help now?
Hi, regarding number 7's point of inflection section, I was taught that I could use the first derivative and input a value slightly smaller and slightly above the stationary point to show that the gradient is constant on both sides to prove it is a point of inflection. Would this way give you marks for this question. For example I mean subbing in root 28 and root 26 and showing there is no change in sign. Thanks
I think I've addressed this is another comment here - check it out! :)
i was just wondering, sorry if this is a stupid question, but why if its a tangent, is the discriminant equal to 0
When you simultaneously solve a pair of equations, and the result of that is a quadratic equation, the number of solutions it has corresponds to the number of solutions to the simultaneous equations. If there are no solutions to that quadratic, there are no solutions to the sim eqs and so they don't cross. If there are 2 solutions, they cross twice. And if there's only one solution, they only cross once, and in this context, can only mean that there is a tangent, as a tangent has one point of intersection with a curve (usually).
I have a different proof for the last one and i wonder if it would also be sufficient
first write the equation as (2p)^2 - q^2 = 5^2
then rearrange to get (2p)^2 = 5^2 + q^2
we now by definition have a pythagorean triple 2p, 5, q
but we also know that because it is a pythagorean triple the only combination that works with 5 at that side is 13, 5, 12
however this means that 2p = 13 and thus p = 6,5 which is not an integer, hence we've reached a contradiction
I'm not sure tough if I can just say that we know only the comination 13, 12, 5 works
at 37:37 the mark scheme doesn't say to do that method would the examiner still mark it correct?
Yes, it's probably in the full details of the mark scheme for all the approaches that can be taken!
For question 9 part b, I used long division with g(x) to get g(x) = 3 - 1/(ln(x)-2) and I found that easy to differentiate. Which way is better in your opinion?
Hmmm I like that way too! Totally up to personal preference.
hey mister bicen would you please sir make a video like this one for s1? i would really appreciate it since im performing it in this oct session (oct 12) . thank you in advance mister
hi - would you ever do hardest further maths questions?
Yes, it's on my list and should be something coming soon!
Dear Mr Bicen, in question no. 1 why have you only used integer factor pairs for 25, why didn't you use like 12.5 and 2 which also multiply to give 25?
Because we are only investigating p and q as integers, there's no way that 2p+q could be equal to a non-integer, hence only testing integer pairs!
@@BicenMaths I see, thanks
For #7 solving the point of inflection, can you instead use the first derivative? As the sign would be constant either side of the stationary point.
Not necessarily - it would be a weird curve, but the property for inflection is that it goes from concave to convex, so we have to use that, not the gradient property.
with q15, if I know the standard result for arctan can I just use that to solve it? or is that not "showing that"?
I don't think that would be 'showing that', but good question!
@@BicenMaths ah ok thank you!
hello! I'm asking if the video includes all the pure papers or just P1 and P2?Also is this for the WMA12 and WMA11 paper reference, thanks in advance, I'm asking before watching the video to know before getting started
These are for the A-Level sat in the UK, not the International A-Level!
do you think this year the grade boundaries will stay the same as last years and wb the difficulty of the paper? thanks
My best guess is that the grade boundaries and difficulty will be in line with 2023 - but we'll only really know until we come out that first exam paper!
Defo using this after I finish all the questions in the textbook LOLL
can you please do a video on stats and mech 🙏thank you sm
Good idea!
Quick question, are even and odd function in the Edexcel exam specification?
No, they aren't!
Hi Sir, for the last proof where 2p + q =5 and 2p - q =5, could you say that 2p + q = 2p - q which implies that q = -q which is a contradiction that q is a positive integer and end the proof there?
Yes, that's also a good way of finding a contradiction!
For the second differential equation can we use first order from further ?
Potentially, yes, but I don't think it would be easier necessarily. Needs to be in the dy/dx + Py = Q form first of course.
Actually, the hardest question is so hard that no one can even come close to a solution.
Hi,
Will you be uploading core pure 2 chapter 4.
Thanks.
Yes, I will do at some point soon - it's just that there's nothing new in that chapter, so I've always forgotten to get round to doing it!
Hello bicen, are any of these questions AS level or all year 2
All year 2
13:49 50sin(1/4 t
hi mister, is this A2 maths only? (like p3 and p4 only) or does it include AS maths? (p1 and p2)?
The questions are primarily P3 and P4, but there's one or two from P1 and P2!
could you please inform me which ones are from p1 and p2 ? im so sorry for the nuisance and i really appreciate your help
@@bonekichi607 I think it's just Number 5 and Number 2!
okie thank you sir i very much appreciate your help
could you answer the third question using methods from further maths, such as finding the integrating factor and using that?
Yes, you could. You'd just need to put it in the standard form first, which is dy/dx + P(x)y = Q(x), so in this case would be dh/dt + 5/1200 h = 24/1200. But the normal maths method is just quicker!
@@BicenMaths thank you!
Love your mullet 👏 ❤
It’s even longer now! Video out on new years featuring it 💁🏻♂️
what tablet do you use?
iPad Air - think it's the new gen, not sure though!
can we use integrating factor in number 3
I would avoid it as it doesn't usually make things that much nicer - give it a go and see, I've not tried it!
I'm kinda confused with number 8 why one full revolution is 180 degrees? Shouldn't it be 360?
because it's the modulus of a sin graph so if you draw it out what's beneath the x-axis will now be above so the graph now repeats every 180.
For number 7 part b, can you not just find points around x and show that the gradient of these points are both negative so it is a point of inflection?
We need to show that it is going from convex to concave or vice versa for an inflection point, whereas I don't think just checking the gradient is sufficient for this!
Do
Ah ok yep, makes sense. Thank you.
Will this channel still help me if I do AQA
Yes! Content is so similar
Coming from Russian/Soviet mathematical education, the last 2 questions were the the easiest ones for me.
Interesting! It's always fascinating to hear how different education systems prepare students in different ways!
Can't you also prove a point of inflation by showing that second derivative equals zero AND that first derivative is same sign either side of the point of inflection????
No I don't think the logic of this quite holds up, as we are trying to prove it goes from convex to concave, not just same gradient either side. I could draw some very odd looking functions that do not have a point of inflection but do have the same gradient either side (i.e. could have straight line on one side, so not convex/concave.)
@@BicenMaths OHHH ok that's always confused me but now I understand - Thank you!!!
If x cube - y cube = xy + 61 find ( x , y )
Hi there Mr Bicen,
I believe you got #9 c) wrong/incomplete. You forgot to consider the situation where the denominator is negative.
Agreed!
proofs are just hard its such a stupid concept
the hardest but is knowing where to start... ig maybe im stupid idk
'Do something' is my mantra for proof - just see if you can factorise, expand, rearrange, etc. Then something might become clearer of what to do!
definitely because the proofs are at the end
literally slay
🐐