This video is a masterclass in exam pragmatism, distilling the dense Further Maths syllabus into a high-efficiency survival guide. It is an indispensable final check for students looking to convert complex theory into immediate marks under pressure.
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A Level Further Maths | Core Pure | Paper 1 - Last Minute Revision (WATCH THIS BEFORE PAPER 1)Añadido:
Hello and welcome back to another Alevel Further Mass revision video. Now, first things first, a quick apology here. Um, because it has been a while since I've done one of these types of videos. As you can see by kind of looking behind me here, um, we've had an office upgrade, which is all well and good. I'm still not fully settled. Um, however, I kind of keep neglecting these videos, right?
We've got exams coming up. Um, and I want you guys to do the best that you possibly can. Right? So, that is the whole reason I put this video here together um, for today. Right? So, as you can probably tell from the title, this will be some last minute revision here for Corpure Paper 1 coming up this week. Now, Corpure Paper One, there isn't a set list of topics here. It just features everything um, or a mix of things from Scorpio 1 and Corporio 2.
So, as you can see from the video here, I have put together the topics from Corpo 1. So that's complex numbers, argon diagrams, series, roots of polinomials, volumes of revolution, matrices, linear transformations, proof by induction and then finally vectors.
Right? So that's the year one material.
Corp 2 covers the second year material here for the full A level further math.
Right? So corpor 2 material features the following. So complex numbers, it's complex numbers in more detail such as the mos theorem, um roots of unity, geometric problems, stuff like that.
Right? series. Again, it's just kind of like a continuation from the year one material such as Mclloren series, method of differences, stuff like that. So again, just a little bit more challenging methods in calculus, volumes of revolution. So again, that is a continuation from the year one material just slightly more challenging. Polar coordinates, hyperbolic functions, methods in differential equations, and modeling with differential equations.
Please don't neglect any topics that are on both of these lists here. Anything that's on either of these two lists here could feature in corporate now. Clearly they can put absolutely everything in corpor paper one. Um so they're going to pick kind of like a handful from both of these lists here and put it into paper one. So it's up to you how you go about kind of revising here. Um you should know your own strengths and your own weaknesses. That's what you need to prioritize in terms of revision. These are the topics here you need to be familiar with and confident with before paper one this coming week. Right now, quick disclaimer here. This video features just eight questions. I know that doesn't sound like a whole lot, but um I think the actual total length here of the questions that I've answered. So, I've already done the questions. This is just my introduction, right? But the questions here, I think it's about 80 minutes already. Um just for eight questions roughly around there. So, 80 to 90 minutes. Um so, as you can see straight away, it does take quite a bit of time. Now, I could have done more questions and the video be longer, but I want this just be a last minute revision, right? I have covered everything on the channel here in full detail. So, you know, for example, if you're struggling with matrices, then I've covered that in more detail on the channel. Um, and another quick disclaimer here, the questions that I featured, they're all from the old spec.
Now, the old spec was a little bit easier um in a way in that the question didn't really feature modeling as such.
So everything that you see here in this video is still relevant to what you're saying, but the questions are kind of written a little bit easier, right? So what I would say here for this video is if you're aiming for an A star, you're probably not going to find much use from this video. I would say go back to all of our other videos. So go through all the exam revision videos if you're aiming for an A star. Um but if you're aiming for like a B, for example, um you're around there. So a C, a B or maybe even an A, you should still find this quite useful. Right? So what I tried to do for uh this video here today is I picked out questions where it's the key concepts from that topic. So for example, the very last question in this video here, question eight, is based on matrices. So what have I chosen for matrices? Well, it's finding the determinant and the inverse of a 3x3 matrix. Right? Those are key things that comes up pretty much every single year without fail. So you do need to be confident with that. If you can't find the determinant and the inverse of a 3x3 matrix, you're going to lose easy marks, right? It comes up every single year.
So, just make sure then you are confident and you do know how to find the inverse and the determinant of a 3x3 matrix. Same again, the question that I've chosen for proof by induction is a divisibility result. Now, there's a few different things that they can ask for proof by induction. Um, summation results, divisibility results, and matrices. So you do need to be confident with all three, but they usually like to feature, you know, at least two of those three, right? So make sure again that you are confident with all three of the individual topics. So to finish with, I just want to make a couple of quick points, right? So for the first point here, once paper one has been sat, I'm going to review that paper, right? I'm going to make a list of the topics that did appear within paper one. And from there, we should have a better idea as to what to expect for paper two. Because for example, then I'm just picking this out at random. this isn't going to be true or necessarily true, right? But let's just say that polar coordinates doesn't appear on paper one. There's no guarantee of it. So let's just say that polar coordinates doesn't appear on paper one. And what we would expect here is that polar coordinates does appear on paper two. That is generally how it works. So I'm going to review the topics that did appear within paper one once it's been sat and then from there what I will do is make a second video last minute revision for paper two right again in a similar fashion just working through um exam style questions from the old spec based on the missing topics right so please do keep an eye out for that if you're new here if you've never seen me before then welcome to the channel please do consider subscribing we have covered all of Alevel maths and A level fever maths in quite a bit detail right So that's the first kind of key point there. The second point because I'm getting a lot of questions about it. Um I am not going to do predictive papers for Corpure this year.
So we did those last year. So I've got Corpure um paper one and paper two on the channel from last year. So if you've never done those papers before, please do have a go at them. They are brilliant for revision and you can download the papers over on our website ajmass.co.uk and I'll leave a link in the description down below somewhere in there. Okay. So the link will be down below. Um but I will be doing predictive papers for um further mechanics one, further stats one and FP1 this year. So if you are doing any of those as your optional modules, please do consider subscribing again just so you don't miss those. They will be perfect for your revision. And again, the pers will be available to download over on our website agentmask.co.uk.
We do have to plug it right. Um other than that, there's not really too much more to add. Over on our website, we've got the worksheets if you're looking for any additional revision in the run-up to Copio one, Copio 2, and the optional modules. Again, the worksheets are available on our website. Um, and if you're looking for additional revision videos, then they are also available on our YouTube channel by becoming a member. Again, I'll leave a link in the description down below where you can click the join button down below as well, giving you access to all of our additional revision videos for Alevel Maths and Alevel Further Mass. Other than that, I don't think I've really got anything else to say other than best of luck for Corporio one this week and Corporio 2 is exactly a week after that.
So again, I will be doing a second video for Copio 2 on the, you know, in the run-up to that next week. And again, just generally speaking, there will be plenty of additional revision videos here on the channel over the coming weeks. So again, please do subscribe and keep an eye out for that. Anyway, I'm kind of just waffling here and I don't really know what I'm saying. Um, so let's cut the waffling and let's get started then with the maths here. So this is last minute revision for paper one again best of luck with corpor.
So to get us started here then question one as you would expect is a pretty straightforward question right what we've got here is this cubic polinomial which is made up of following two factors one linear and one quadratic factor where a and b are real constants.
So for the first part of this question then a let's just do this at the very top here all we want to do then is just find the value of a and the value of b.
Now there's two ways to do this, right?
The first way is the long way which I wouldn't advise which is to use long division. Um so you would take this cubic polinomial here and divide it by 2 zus 3 and that would just give you this quadratic factor here. That's absolutely fine. If you do like to do long division who like to do long division, I don't know. Um so I'm just going to do this by inspection. But like I said, if you do want to do long division for whatever reason, then feel free to do that. Um EA's fine, but like I said, inspection is a much quicker way to do this. So what I can see straight away here is for B then I know that -3 * B must be equal to -6. So as we just said then - 3 B is equal to -6. In that case then just divide both sides by minus3 and as we can see then nice and easy B is equal to two. So that's the value of B we need the value of A here.
So to obtain the value of a here, what I'll consider then is the z^ squ term here. So - 5 z^ 2. How do we obtain that then um from this product here? Well, we take the minus 3 and we times it by z^ squ. That is one part of it. The other part then comes from 2 zed * a z. Okay.
So we've got - 3 z^ 2 like so. And also then 2 a z^ 2 so + 2 a z squared. There's no more ways here to obtain a z^ squ term. So we know that this here this sum must be equal to - 5 z^ 2 right? So - 5 zed^ 2. So um I can take the minus 3 z^ squ over to the other side to so to do that I will just add 3 z^ squ to both sides hope that's obvious what I get here then is 2 a z^ 2 is equal to minus 2 z^ 2 if we just equate coefficients here then what I can see is 2 a is equal to minus2 and in that case then a here nice and easy I'm also losing my voice A here is equal to minus one. Perfect. Okay, so that gives us the value of A and the value of B giving us a solution to par A. And now for the final part of this question then part B hopefully I will just make it before my voice gives way.
Um it just says given that zed is a complex number find the three exact roots of this equation here. So this equation is just a cubic polinomial from part A. Right? Now we know then the factorized form here. So this is the product of factors here that gives us the original um cubic polinomial. Right?
So what I can do here is write this as 2 zed minus 3 times then this here this quadratic factor which we now um know is z^ 2 minus z + b where b is 2. So let's just write that down in full. So zed^ squ um minus zed + b which is 2.
Perfect. Okay. And this here is equal to zero. Now straight away then we get one um root here from the linear factor. So set that equal to zero. I get 2 zed is equal to 3. In that case then I get here that zed is equal to 3 / 2. Nice and easy. Right?
So that's the first of the three roots here for this equation. The next two roots here come from this one. Now clearly these will be um in this case here complex roots and they will obviously occur in a conjugate pair.
Okay. So all I need to do here is take this quadratic factor here, set that equal to zero and then just solve for zed. Now clearly I need to do this either using the quadratic formula or by completing the square. I'll just use the quadratic formula here. But if you you know if you want to complete the square that's also fine as well. Okay.
So um if I complete the square sorry if I use the quadratic formula here what I would get then is I would get um 1 plus or minus here the square root then of b ^ 2 that is - 1^ 2.
So min - 1^ 2 minus them.
I'll just put all of this here in a bracket so we avoid any sill mistakes.
So it's 4 * a which is 1 * c which is 2 and that's all over then 2 a. So that's 2 * 1 giving us 2. Right? Let's just simplify this here. And this is for zed by the way. So zed is equal to all of this. Um so I get 1 plus or minus then um this here. So that's 1 - 8. So that gives us the<unk> of - 7. So the square root of - 7 then I can write that as I um * the<unk> of<unk> 7. Okay.
So that is I * the<unk> of 7 just like that. And this here then is all over two. Perfect. Okay. So what I've now got here is the three exact roots of this equation here. What I'll do then to finish with here is denote these as zed 1, zed 2 and zed3. So zed 1 then I'll just take to be 3 over two.
So zed 1 here as we just said is 3 over two.
Zed 2 I will take that to be the positive of this route here of this solution here. So it's going to be 1 + then i<unk> 7 that's all over two. And then finally here Z3 will be the negative of this solution here. So 1 minus then I<unk> 7 all over two. Perfect. Okay, we just about made it before my voice gave way.
So there we go then. That gives the solution here to question one.
So for the next question here, question two, we have everybody's favorite topic, right? Proof by induction. Now, even if you don't enjoy proof by induction, it comes up every single year without fail and it is usually worth a good amount of marks, right? So, just make sure that you're confident with it, even if you do hate it. Um, anyway, what does this question want us to do? So, for this question then, what we want to do here is prove by induction that for n belonging to the set of positive integers here, f of n, which is equal to the following. So, 8 n - 2 the n is divisible by 6. Right? So to begin with here as is always the case with any proof by induction right always follows the same four steps here we have the base case or the basis step the assumption or the assumption step the inductive step and then finally the conclusion which isn't really a step but you do need to give the conclusion for all proof induction questions. Okay so let's just start then with the base case here or the basis step whatever you want to call it. We'll just call it the base case here. So for the base case then what are we dealing with here?
So we just need to prove that this holds then for a given value of n. So the easiest case here then it's just let n equal to 1. Okay. If I let n be equal to one here what do we get? So f of one then is equal to a ^ 1 minus then 2 ^ 1. So that's just 8 minus 2 which gives us six. And clearly then six is divisible by itself. Right? So six is divisible by six and therefore it's true.
So therefore true or n being equal to one. Perfect. So that's the first step there in question two. So for the next step here we have the assumption. Right?
So for the assumption here, let's just do that underneath then.
So for the assumption here, what do we assume then? What I'm assuming here is that this is going to be true for n equals k. Okay, so all of this here will be true when n is equal to k. Let's make a note here. So assume true.
So assume true.
R and being equal to K. So what does that give us then? Well, in that case what I get here is f of K is equal to 8 ^ K.
So a K - 2 K. What we're saying then is that if this is true for n= K, this here is divisible by six, right? So this here is divisible divisible by six. Okay. So now this is where we get on to the more challenging part of the question. Here we have the inductive step, right? Or the induction step.
So using induction here.
So for induction then we now consider f of k + one. Okay. So let m equal to k + 1. So in that case then what I get here is f of k + 1 and f of k + 1 here would give us 8 the k + 1.
So we get 8 the k + 1 minus then 2 to the k + one. Now if you ever have if you ever have I can't speak. If you ever have a question such as this proof induction here where you want to prove a divisibility result such as this one here, what you want to do is take f of k + one and subtract f of k from that.
Okay. So we now consider f of k + 1.
So f of k + 1 minus f of k. Okay.
What does that give us then? Well, this here is f of k + 1. So that is 8 to the k + 1 minus then 2 to the k + one. Okay, so that's the first part there. That's f of k + one minus f of k and this here is f of k. So that is minus then. What I'll also do here is just put this in a bracket.
You'll see why in just a second. So minus then all of this here. So a the k minus 2 to the k.
Okay. The reason I put this in a bracket here is to avoid any silly potential errors with the sign here because it's minus of a minus. It's very easy to make that mistake. So just be very very careful for that. What I'll now do here as well is group these terms together.
So a to the k + one and the minus a to the k here. and then this minus2 to the k + one and this positive2 to the k.
Okay, so doing that underneath then what I get here is and I'll put these in a bracket just to make it look a little bit neater.
I get a the k + 1 minus a k minus a k there like so. What I'll do here is I'll subtract from this. So minus and just be very very careful your signs here. Right? So if we're minus in here, what I get then is pos2 the k + 1.
So 2 k + 1 and then this would be minus 2 to the k. Okay. Because that is positive like so. Okay. Why do I do this here?
Well, what I want to do then is just kind of use a little bit of clever manipulation here. get this into a form which will then prove that this here f of k + one um is eventually divisible by six. Okay. So what I can do here is write this as so just using the basic rule of indices here um 8 the k + 1 is the same as 8 the k * 8. So what I've got here is eight lots of 8 to the k minus then a k like so I'll do the same here as well.
So minus them and what I'll do is I'll just put this in a bracket so again I don't make any silly any errors here. So this is the same there's two lots of 2 to the k two lots of two to the k minus then 2 to the k like so obviously just be careful as you do work through these questions as well is very very easy to make just a tiny little mistake which if you do make will unfortunately throw everything as you continue through the question so just take your time just be very very careful so if I just simplify here this is 8 um * 8 the k minus 8 to the k. So I've got seven lots of a to the k there. Right?
If I subtract one lot of a to the k away from this here, I've got seven lots of a to the k. I hope that makes sense.
Seven lots of a to the k like so. And doing the same here. Then this is two lots of 2 to the k. So if I subtract 2 to the k from that, I just simply get 2 to the k. Right? So minus then 2 to the k. Okay.
Perfect.
Again, what I want to do here is just use a little bit of manipulation.
What we can see here is we're kind of getting close to f of k here, right? A to the k minus 2 to the k. I've got minus 2 to the k here. But I've got seven lots of a to the k here. So is there anything I can do then with this here would actually give us f of k here.
Okay. Well, what I could do here is split this up as the following.
So I could write this here as 6 lots of a k plus then a to the k minus 2 the k. Okay, again just a little bit of kind of clever manipulation here giving us the following. And as we've just said then what we've now obtained here and I'll just highlight this in a different color is this here. And this here is just f of k. Right?
Okay. So this here then is f of k.
Okay.
So what I've now got here is six lots of a to the k.
So six lots of a to the k plus then f of k.
Okay, perfect. Now what we actually wanted here is f of k plus one. That's what we actually want to consider. What I've got here is f of k + 1 minus f of k. So we just want f of k + one here.
Let's just add f of k to both sides. So therefore then um I'm running out of room.
So therefore then f of k + one here.
f of k + one is equal to so it's six lots of a to the k here.
So six lots of a to the k plus then two lots of f of k. Right? If I add f of k to f of k, I've got two lots of f of k, right?
So plus then two lots of f of k.
And this here is perfect, right? I already know then that f of k here is divisible by six given that we assumed it was true for n equals k. So we just kind of work on that assumption. So this here we already know then that it's divisible by six and this here so this 8 to the k is multiplied by six. So it's a multiple of six. So therefore then it must also be divisible by six. So what that means then is the full expression here for f of k + one is actually divisible by six. Perfect. That is exactly what we want here. So to finish with then let's just give a quick conclusion a quick summary what we've actually just shown. So um you know it doesn't need to be an absolute um novel.
Just quickly summarize what we found here. So therefore then it's true for n equals 1.
So therefore true.
So true for n= 1 and it's true for n= k + 1 as we've just shown here.
So it's true for n= k + 1 if and we kind of work on this assumption here if it's true for n= k.
So it's true for n= k + 1. If true for n equals k right I'll just say here by mathematical induction. Oops my pen is not working. by mathematical mathematical induction. You could just say induction here. I'm being a little bit pedantic but never mind. So by induction and therefore then it must be true all n belong to the set of positive integers.
So therefore then therefore true for n belonging to the set positive integers. I can never do a neat integer symbol, but that will do, right?
So there we go then. As you can see, these proof induction questions are usually kind of structured in a pretty similar fashion. We always start with the base case. We then have the assumption, the inductive step or the induction step, and then finally a quick conclusion here. Um, so as I said before, proof induction comes up every single year. Please do make sure that you are confident with this. We do have plenty of videos on proof induction. So if you are still struggling with proof induction, go and check out those videos. Again, I'll leave a link in the description and below. Okay, so there we go then. That gives the solution here to question two.
So for the next question here, question three. As we can see, then we have a question on the method of differences.
And these types of questions here are usually pretty fun, right? The layout is very very similar between question to question. It usually just starts with a partial fraction decomposition as this question here demonstrates. The second part is normally just the method of differences. again as this demonstrates here and then for the third and final part here just a quick follow on based on series and summation. So let's just get started then with part A here. So for part A then it is just one mark and hopefully this doesn't cause any issues.
It's just a partial fraction decomposition. So just basic A level maths here, right? So I won't go into a massive amount of detail here as to how to do this. This should be very very trivial as corpor um or as a level for math student sitting in Corpio 1 in a few days. This really shouldn't be a challenge, but let's just quickly run through it anyway. So what I get here is so I'm going to get a over r + 3. So it's a over r + 3 plus then b over r + 4. Okay, that is identical to this here. So 1 all over r + 3 * r + 4.
Perfect. So getting this over a common denominator here I get a * r + 4 plus then b * r + 3 and this here is equal to 1. So if I let r be equal to -4 for example.
So let r be equal to -4. What do I get here? So that just disappears. I get -4 + 3. So that's minus uh 1. So minus b there is equal to 1. Therefore b is minus one. And in a similar fashion here if I let r equ= to minus3. This just disappears. I get um just positive one there. So a * 1 is a.
We get a is equal to 1. Perfect. So therefore a is one. Okay. So if we express that then in partial fractions, what we've got here is one all over this. So r + 3 * r + 4 is identical to um so what we get here then is a over r + 3. So that's 1 / r + 3.
1 / r + 3. So it's um plus b then. But b here is - 1. -1 all over R plus for B posh. Good stuff. So that gives the solution to part A as I said then hopefully nice and easy just for the first mark there. So moving on to the next part of this question here we have part B. So for part B then it just says hence using the method of differences show that this summation here is equal to N all over A * N + A where A is just a constant to be found. Right? So for the method of differences here, what we want to do is just go through the first few terms here and the last few terms because what you would expect here with the method of differences is basically pretty much everything will cancel down.
Okay, that is the whole point of doing the method of differences here. It's the beauty of it, right? Everything just cancels down. We just get left with one or two terms, sometimes three terms um and we can work from there, right? So let's just make a note here. So if we let r equal to 1, what do we get here?
So we're using this right the partial fraction decomposition. That's the whole point of expressing this here in partial fraction form. Right? So what I get here then is 1 / 4 minus m1 over 5. Right? If r is 2, what do we get here? I would get 1 over 5 - 1 / 6. And you can clearly see then the pattern that will form here. If r is 3, what I get here is 1 / 6 - 1 / 7.
Right? That's all you know uh good.
Happy days. Uh but what about the last couple of terms here? So for the last couple of terms then this would be r is equal to n minus one. And the final term then would be r is equal to n. So what do these give us here?
So for n minus one then that's going to be 1 all over n minus 1 + 3. So that is 1 / n + 2.
So 1 / n +2 and then this would just be 1 / n + 3 right? Just increases by one.
So -1 / n + 3 just like that. And then for the very last term here this one r is equal to n. I would get um what would I get here? So that's going to be 1 / n +3 1 / n +3.
Oops, there we go. My pen isn't working.
And then - 1 / n +4.
Perfect. Okay, now hopefully at this point here, it's pretty obvious what's going to happen.
If not, just think about this logically.
Right, I've got 1 over 4 - 1 over 5.
I've then got 1 over 5 - 1 over 6. I've got 1 over 6 - 1 over 7. So on and so on. And hopefully what you can see here is this - 1 over 5 cancels with this 1 over 5 here. The - 1 over 6 cancels with this 1 over 6 here. The - 1 over 7 would cancel with the 1 over 7 when r is equal to 4. That would cancel there like that.
The 1 / n +2 would cancel with the previous lines. That would be when r is equal to n minus2. That would cancel there. - 1 / n + 3 that cancels with this 1 / n +3 here. And these two terms here, there's no way to cancel those, right? They just get left over. So what we're saying here is this summation here. So therefore then the summation here is from r = 1 to n of this here that's 1 all over r + 3 * r + 4 like so that is equal to then so it's 1 / 4 - 1 / n + 4.
Now you might be thinking, well that doesn't look right. You are correct because they want it in this form here.
Basically all they've done there is just got it over a common denominator. That's all they've done there, right? So if we just get this over a common denominator here, I would times this by n + 4.
That's going to be n plus 4. I would then times this fraction by 4. So n + 4 minus 4. That just gives us n, right?
Which is what we would expect here for the numerator. Perfect. That is what we want. So we get n there. And then for the denominator here, it's going to be 4 * n + 4. Right? So 4 * n + 4. And that is exactly what we want here. Right?
I've got a * n plus a where a here.
So therefore a is equal to four.
Perfect. Okay. So that's always promising, right? And you get the desired form. If you clear don't get something like this, just go back and double check everything. Clearly something has gone wrong. If you do not get something of this form here, right?
If you don't get just n, for example, in the numerator, just go back, check what you've canceled, check your substitutions are actually correct, it's a very easy mistake to make, but it can happen on the day, right? Um, you know, these tiny little things do happen. So, there we go. Then that gives the solution to part B. And then finally, for the very last part here, let's just box this off ever so slightly. We'll do part C here in the bottom right corner. So for part C then it says find the exact value of this summation here. Now clearly then we can just use our answer from part B to help us answer part C. That's the whole point of this question. Right? So I've got this summation here from R= 15 to 30 of this here. So 1. So 1 all over r + 3 * r + 4.
Right? So what do we do here? Well, just be careful then with the subscript here um on the summation. So from i= 15 to 30. That is the same then as doing the summation from 1 to 30 minus the summation then from 1 to 14. Okay. As we just said then summation from r = 1 to 30. I won't write down this part here of the summation the actual thing that we're summing. I'm hoping it's just implied minus then this summation here from r equ= um 1 to 14. Okay like that.
What this does here basically is it gets rid of the first 14 terms and gives us the sum from the 15th term to the 30th term. Okay. So all you simply to do here is just use the result that we obtained in part B or part C. Right now in your exam what they will normally do is give you the desired form here rather than you know saying A is a constant to be found. The reason why is because if you can't actually derive this here then obviously your answer to part C will definitely be wrong. Um they'll still give you a method mark normally but generally like I said it will just give you this anyway just to avoid that as a you know a problem on the day. Okay, but anyway, I digress. What would I get here then? So from 1 to 30, all I would simply do here is just substitute 30 into this here, right? So that's going to be 30 all over. It's 4 * 30 + 4. So 4 * 34. I would then subtract here. This is from 1 to 14. So all you need to do here is just substitute 14 into this here. So - 14. So 14 all over. So it's 4 * then 14 + 4 which is 18.
So just put this into your calculator here. And what do we get then? So if you do this correctly then what you should get here is 4 all over 153. So 4 all over 153. Perfect. And there we go then. So, as you can see, you know, these questions, I quite enjoy them. I don't know about you, but you know, I think for the most part, these questions are quite nice questions. Um, you know, you can certainly get more challenging types of these questions here. And we did cover that in last year's predicted paper. So, go and check that out. Again, I'll leave a link in the description down below. If you haven't already attempted our, you know, predicted paper, mock exam paper from last year, please do go have a uh an attempt at that, have a go at it before paper one um this week. Right. Well, there we go then. So, that gives us the solution here to the third question in this video.
Now, for the next question here, question four, I've chosen a favorite for everyone. Vectors. Everyone loves vectors. Um, you know, I never hear any complaints about vectors. Everybody loves them. The favorite topic um for most students, right? If only that was the case. Anyway, um this question here, to be honest, isn't too bad. Um, as I said kind of at the start of the video, the questions I've primarily chosen here are just testing the key points for these topics. Right? So for this question here is testing points of intersection. And then also for the final part of the question here, finding the acute angle between two intersecting lines. Right? So again, these are kind of like the key things for vectors here.
On the day your question will probably be more challenging. It would be nice if it was quite this straightforward. Um but you know the key kind of concepts here are still relevant even to your questions. You just kind of usually address it with a little bit more context um on the day. But either way you should hopefully find this question to still be some decent revision. So before we do anything else then let's just actually quickly read through the question. So what have we got here? So it says relative to a fixed origin O the lines L1 and L2 by the following equations. I won't bother reading out each equation. You can just quickly read them here. lambda and mu are scalar parameters and p is a constant. We're given that L1 and L2 intersect. So for the first part here, I mean part A and part B kind of go hand in hand here.
Want to find the value of P. So for part A then we're told here that they intersect, right? So we know then if they intersect here, what we can do is form a system of equations here. So what I will do here just to make my life a little bit easier is I will represent L1 and L2 in column vector form. You don't have to do this but I just find it kind of easier to see what's actually going on. So L1 here this is for R here right that is equal to so it's 3 I PJ 7 K as a column vector and that is 3 P and 7. So 3 P and 7 like so plus lambda then so plus lambda times by this effect here. So 2 - 5 4. So 2 - 5 and 4. And then for L2 here, this is again for R here is equal to I've got a - 25.
So A - 2 5 and then mu * by this common vector here. So mu * 4 1 and two. Perfect. Now again, you don't have to do this here. If you want to just kind of crack on with the question, that's absolutely fine. But I just find this easier here to see what we're actually doing. Right? So to form the system of equations here, what I know then is 3 + 2 lambda must be equal to a + 4 mu. Right? That is if they intersect.
Again, we can do the same here with the next line. So p um - 5 lambda must be equal to -2 + mu. And same again here for the final line. Right? So 7 + 4 lambda must be equal to 5 + 2 mu. Right?
So that gives us the system of equations here. So let's just quickly denote system of equations. So 3 + 2 lambda is equal to a + 4 mu.
So a + 4 mu there. Um p - 5 lambda is equal to - 2 + mu - 2 + mu. And then finally here 7 + 4 lambda 7 + 4 lambda is equal to 5 plus then 2 mu perfect right what I'll also do here is denote the top um equation as equation one the middle equation here as equation two and the bottom equation here as equation three right good stuff so what I will now do here is take equation equ one and equation three.
Right? Don't forget what I want to do here is just find the value of p. So clearly what I need to do here is solve this system of equations here to obtain the value of lambda and mu. And once I've done that, I can just substitute those back into equation two to obtain a value for p. Right? But as we said, I do need the value of lambda and I do also need the value of mu before we can find the value of p.
Right? So um as we just said then let's just take equation one and equation three here. What we will do then is just solve these simultaneously.
Um so what I'll do here is I will times equation 1 by two. So that's going to give me 6 + 4 lambda is equal to 16 + 8 mu like so. And then equation three here that stays the same right? So 7 + lambda is equal to um 5 + 2 mu. Perfect. Okay.
So let's just solve these simultaneously here. Um obviously there's a couple of different ways that you can do this. You could just use substitution if you really want to. You could just use your calculator or we can just use um elimination here. Right? These are pretty, you know, easy simultaneous equations. So we'll just use elimination here. We've matched the lambda, so we can just go straight into um subtracting here. The sign is the same. Um so 6 - 7 is -1.
They just cancel. 60 - 5 is 11.
And then 8 mu - 2 u gives us 6 u. Sorry, mu. Can't speak. So they are mus not u's that is embarrassing. So there we go. - 1 is equal to 11 + 6 mu. There we go.
Too many Greek letters. Um, right. So again, let's just keep solving here.
Um, we want to solve for mu, right? So subtract 11 from both sides. In that case, then what I've got here is 6 mu is equal to -1 - 11 giving us -12. Then we just want mu here. Just divide both sides by 6. So -12 / 6 gives us mu is equal to -2. Perfect.
Now, what I can do here is just substitute this into um either equation one or equation three. It's completely up to you. It's probably just easier if I do it into um equation one, right? So, the reason why I say equation one is because once I substitute that into equation one here, four lots of -2 is - 8. 8 + - 8, they just cancel. So, the right hand side here is just zero. So I've got 3 + 2 lambda is equal to 0. That case then 2 lambda is equal to minus 3. And then finally here lambda is equal to minus3 over 2.
Perfect. Everything up to now looks good. So we're pretty much there now. So for the value of p here, um I'm not going to have a ton of room. I'm just going to quickly finish that up here. So for P then what we can do here is rearrange equation two making P the subject that is minus 2 um plus mu plus mu + 5 lambda 5 lambda. Um so what does that give us then? So it's -2 plus mu which is also -2. So -2 + -2 is just -2. Then five lots of this here.
So five lots of um minus 3 over two.
Apologies I kind of run out of room here. There we go. So just throw this into your calculator here. If you do all of this correctly, then what you should get here is - 23 - 23 over 2, right? And that is for the value of P. Okay, so there we go. Then that gives us the solution to part A. That would be the first four marks there for this question. For part B then it just says find the position vector of point of intersection. So as I said then part A and part B pretty much go hand in hand because once you've done all of the work for part A you can just get the solution to part B really with just an extra line of working. So where would this intersect here?
So for part B then this would intersect.
So intersects that.
So all we need to do here is take either the value of um lambda or the value of mu here and then substitute that back into um one of these equations here.
Right? So L1 or L2. Given that lambda is - 3 over2, it's probably just easier to use um L2 here where mu is minus2. It's an integer. Obviously you can just do it with lambda as well. Obviously, I'm probably going to need a calculator to do that. So, it's just easier to use um the equation for L2 here. So, again, if I use column vectors here, what do I get? I've got a - 25.
So, a - 2 5. All I'm doing here is just using this equation, right? But mu now is minus 2 as we've just solved to find here. So minus two lots then of minus two lots here of um 41 2 4 1 2 like so let's just quickly evaluate this here then what do we get so 8 - 2 lots of four that's 8 - 8 which is zero um - 2 - 2 lots of 1 so - 2 - 2 is -4 and then 5 - 2 lots of two so that's 5 - 4 giving us 1 and we get the following here. So I get 0 - 4 1 which if we represent then in um as a unit vector here that is - 4 J - 4 J and plus K right so it looks a little bit neater to give it like that is the final solution. Okay so as you can see then that gives us the solution to part B right giving us the position vector of the point of intersection. So, good stuff there.
We're pretty much done now with the question, right? So, for part C, then hopefully we should have just right enough room here to finish um this question off question four. Right. So, for the final part then all we want to do here is just find the acute angle between L1 and L2 giving your answer in degrees to one decimal place. Right? So for the acute angle here, so for the acute angle here, acute angle between two intersecting lines here.
So it would be between L1 and L2 here.
Um let me just quickly give the formula here that you would need to use. So the formula here is cos theta cos theta is equal to then the absolute value here the modulus of would be a dotted with b all over then the magnitude of a time the magnitude of b right so obviously if you want the angle you would take cos inverse of both sides or out co of both sides right but we'll cross that bridge when we get to it what we taking for a and b here. So for a then I will take this to be this vector here. So 2 i - 5 j + 4 k and for b that would be this vector here 4 i + j + 2 k. Okay. So let's just quickly dot these together. Um so a dotted with b.
Um my notation is a little bit inconsistent. I did squiggly lines here.
I'm doing kind of just you know um a straight bar here. But you know what I I'm just representing these as vectors.
Um, apologies for being inconsistent, but never mind. Anyway, let's just continue here. So, a then, as we said before, is 2 - 5 4 2 - 5 4. We dot that with this vector here. So, 4 I + J + 2 K. That's 4 1 and 2. Dot these together. Then, what do we get here? So, 2 * 4 is 8. Um, - 5 * 1 is - 5. And then 4 * 2 is 8. So 16 - 5 gives us 11. What I also now need here is the magnitude of a and the magnitude of b. So the magnitude of a here.
What is that? So for the magnitude of a here, right, that would be the square root then of 2^ 2 + - 5 2 + 4 squ. So that's going to be the square root. Then just got to do that here. That is the square root here of um 2 square which is 4 um minus 5 squ which is 25 right? So + 25 and then um 4^ 2 which is 16.
Add these together here I get 41 + 4 which is 45. So that is roo<unk> 45 which obviously we can simplify here as well if we want to. That is root 9 roo<unk> 5 which is 3<unk> 5. We'll do it anyway. It might not be relevant, but we'll just quickly do it here anyway. And then also we need the magnitude of the vector B here.
So for the magnitude of the vector B here, um again B is this vector here. So that would be the square root then.
So the square root here of 4 square which is 16 um 1 square which is just one. And then finally here 2 which is 4 like so. And that gives us <unk>e2 which we can simplify into something neat like this. Right? So we just leave it as it is. So putting that all together then what we know here is cos theta is equal to this here. So what I get then is the absolute value of a dotted with b which is 11 as we evaluated here. It's 11 all over then magnitude of the vector A which is 3<unk> 5. I will give it in this form here. It's a little bit neater anyway. So 3<unk> 5.
So 3<unk> 5 um times by the magnitude of the vector B which is roo<unk> 21.
So roo<unk> 21 there. Okay. Just extend that a little bit. Perfect. Um so if we evaluate this here then um it's something a little bit neater. What I get here is 11 all over then. So 3<unk> 5 *<unk> 21 that would give us 3 * the<unk> of 105.
Perfect. Okay. Now this is cos theta.
Just be careful. We want the acute angle between L1 and L2. So we want theta, right? Not cos theta. So that is cos theta. So therefore then beta here is equal to r cos or cos inverse of this here cos inverse of this here. So 11 all over 3 *<unk> of 105.
Quickly throw this into your calculator here. If you do that correctly then and again just be careful here. It does need to be in degrees to one decimal place.
What we get here then is 69.0°.
Beautiful. Okay, there we go. Um, so as you can see, it's just working through the kind of key things here for vectors, right? So, finding the points of intersection here, um, as we've done for part A, um, and part B, really, they kind of just go hand in hand. And then part C there, just finding the acute angle between two intersecting, um, lines, right? So, there we go then. So, hopefully not too bad there. And that gives us the solution to question four.
So moving on to the next question here question five as we can see then the curve C has this equation here. Now for the first part of this question then part A all we simply want to do here is just find the integral of y is this fraction here with respect to x right.
So for part a then let's just do this at the very top here. So part a how do we integrate this here then? Well, if you see something of this form here, straight away what you should be thinking of is completing the square on the quadratic expression here. Why is that the case then? Well, basically if I can complete the square on the quadratic um expression here, then the integrant once I actually denote it as an integral, then the integrant here becomes a standard result. Okay. So, let's just complete the square here then on this um quadratic expression here.
So, x squ + 2x - 3. Well, that is identical to. So, this should hopefully be very easy. I'm just completing the square on this, right? Just basic GCSE mass. I get x + 1 all squared. I would then get -1 - 3 giving us -4.
Okay. So, this is the completed square form this quadratic here. So, let's just replace this here then for um this fraction here. Just replacing this quadratic here with it completed square form. Right? So we'll just do this underneath. Then the integral here of y with respect to x here is the integral then. So it's the integral here of so it's one all over the square root of this here.
So the square root then of so it's x + 1 all^ 2 - 4 like so. Okay. And don't forget here we are integrating with respect to x. Okay.
Now once I've got it into this form here as we said then this is just a standard result right so just refer to your formula book if you're not too sure. But once it's of this form here, this is just simply then r. So it is r kosh here.
So r kosh then of so we take the square root of this. So if it's x + 1 square the square root of that is just x + one, right? So it's r kosh of x + one.
So it's r of x + one in a bracket here.
So x + 1 all over then. So if you take this as positive then n that would be four again take the square root of that and we get two.
So we get r kosh of x + 1 / 2. Now just be very very careful and we don't have any limits here on this integral. Right?
So do not forget then the constant of integration. It would be a shame to forget that and drop a silly mark.
Right? So there we go then. Hopefully nice and easy there just to get started with part A. So moving on to the next part of the question here. Part B.
We've got a little bit of waffle to quickly read through. So the region R is bounded by the curve C, the Xaxis, and the lines of equations X= 2 and X= 3.
The region R is rotate through 2 pi radians about the X-axis. What we want to do then is find the volume of the solid generated given your answer in this form here. So P pi L Q where P and Q are rational numbers to be found.
Right? So we're looking at volumes of revolution, right? Hopefully that is obvious. um if not the big clue is that you're looking for the volume of the slide generated, right? So that should hopefully be obvious.
So we're looking at volume here. So V is equal to now we're looking at here um a region R that is rotated through 2 pi radians about the x ais. So we're rotating about the x-axis here. So the formula then v is equal to pi * the integral here. So it's the integral then I'll come back to limits in just a moment. So it's pi * the integral then of y^2 here because we're rotating about the x-axis. Okay. So the integral of y^2 with respect to x. Now for the limits here then that is dependent on the context of the question because we're rotating here um through 2 pi radians about the x-axis and the region r is bounded by the curve c the x-axis and the lines then with equations x = 2 and x = 3. What that tells me here then is my lower limit is 2 and my upper limit is three. Just like that. Okay. So let's now go ahead with evaluating this integral here. So that is pi * then the integral here from 2 to 3 of um it would be this here right so this here just squared. So that is 1 all over the square root then of x + 1^ 2 x + 1^ 2 - 4 just like that again we square full expression there because it's y^2 right this is y here so we square that we're integrating here with respect to x okay right so if you square this expression here. Well, we would just square the numerator and the denominator separately. So the numerator squared that is 1 square which is just simply one again. So let me just do this on the line underneath then. So I get pi time the integral here then from 2 to 3 I get one all over them. So that's one square which is 1. If you square the denominator here if you square square root they just cancel right they're the inverse of each other. Just get left with the argument here. So I get x + 1 all^ 2 - 4 here in screen then with respect to x. Now once I get my inrand into this form here what you need to recognize then is the integrant.
So the integr here is of the form. So integr is of the form. Then so what do I have here? So it's going to be 1 all over u ^2 - a^ 2 right it's 1 all over as we just said then u ^ 2 - a 2. Okay.
So again, this is a standard result here and again just refer to your formula book if you're not too sure. So the result here will be 1 over 2 a. Um learn again I'm just referring to my formula book here because this is something I do forget myself. So learn of u minus a all over u plus a. Okay, that is the result here.
once we integrate something of this form. Okay. So, putting that all together then what do I get here? So, a in this case here then um will be two, right? So, 1 over two lots of two would be 1 over 4. Let's just start putting this all together. I've got pi times by here. Um so, it's pi * by here. It's going to be 1 over 4 as we just said 1 over 4. I then get l. So learn here of oops why is my pen I'm doing that I don't know never mind. So it's u minus a. So u here what is u? Um well u would be x + 1 right? This is u 2. So u is just x + 1. Okay. So um that's going to be x + 1 - a. So that's um x + 1 - 2. So that is x -1.
x - one there for the numerator. And then for the denominator here that is u + a. So u again is x + 1 and then a is 2. So x + 1 + 2 gives us x + 3 stuff. Okay. We'll close that. And don't forget here the limits are from 2 to 3.
Why my square brackets are so off? I'm not really too sure. Um I'll just quickly redo them because that's kind of annoying.
There we go. It's a little bit better.
It'll do. Right. So, we're pretty much there now, right? Let's just continue here. And all we all we now need to do here is just substitute in the limits.
We start with the upper limit and subtract the lower limit. So, what I'll also do here, just to make my life a little bit easier, let's take the 1 over4 outside here. Um, because obviously it has no material effect here inside once we start substituting in the limits. So, what I've got here is pi over4. So, all I've done there is just took out the 1 over 4. So it's p<unk> / 4. Um so this is for v by the way the volume. So v is equal to p<unk> / 4 inside here. Now what do we get? So I get l of 3 - 1 over 3 + 3. So 3 - 1 is 2. 3 + 3 is 6. That is l of 1 over 3.
So l of 1 over 3 minus them. So we end substituting the lower limit here. So 2 - 1 is 1. 2 + 3 is five. So I get learn the fifth one over five there. Okay, we'll close that. Now what they've got here is just a single natural log, right? So what they've done here is they've just expressed that as a single natural log. So clearly I said the next step here with the inside of the square bracket, right? Let's just express this as a single logarithm. In doing so, what I get here is pi / 4 times then um so it would be learn of is going to be um 1 over3 and because we're subtracting here this would be a quotient. So 1 over 3 over 1 over 5 just like that. And obviously 1 over 3 / 1 over 5 that just gives us 5 over 3. Right? So in that case then the volume V here is equal to so we get p<unk> / 4.
So I get pi over 4 times then the natural log here as we just said of 5 over 3. Perfect. Right. What I've now got here is my answer in this form here.
Right. P um in case it's not obvious is one over four.
Let's just make a quick note here. So therefore then P is 1 over 4 and what else do we want here? Q. So Q is 5 over 3. Perfect. And there we go then. So hopefully that kind of demonstration there as I to kind of work through that question did make sense.
You could probably do it a little bit quicker than I'm doing it. Obviously I'm kind of explaining it as I go. Does just take a little bit more time. Um but there we go then. So that gives us the solution there to question five.
So let's just continue here then. As we can see now we have question six here.
And to be honest, question six is a nice little break um because we're already approaching about an hour here just for the first five questions of this video.
As I did mention at the beginning as well, these questions do generally take quite a bit of time. Um obviously the level of math that we're kind of looking at here, you know, it's a little bit more involved than just Alevel math for example. So um nearly an hour just for the first five questions here. But anyway, let's just continue with this question here. As I said, a bit of a nice break. Then there is just three marks for this question here. And it is quite a nice easy question all things considered. So let's quickly read through the question here. So what we're given then is y here. So y is 3x ark sin 2x for x between 0 and 1 /2. So for the first part of this question here, then part a, all we want to do here is just determine an expression for dy by dx.
Just two marks for this, right? So shouldn't really be expecting too much work. Now if we just look at y then I'm hoping it's obvious here that to evaluate dy by dx here we do need to make we do need to make use I can't speak anymore we do need to make use of the product rule here we have a product of functions right so um I'll do it underneath then just so I've got a little bit more room so for part a then let's just do that here as we said we do need to make use of the product rule here take u and v here as my two functions going from left to right then I'll take u as 3x here so u is 3x and v here will be ar sin 2x ark sine 2x there okay what we also need here is u prime and v prime so urime here nice and easy and that's just simply free and v prime here well just be careful with this Right? It's r sin 2x. But we can just simply use our formula book here for r sin 2x. V prime then would just simply be 2 all over here. The square root then of it's going to be 1us then. So it's 2x which is 4 x^2 there.
So that's going to be 2x all squ giving us 4x^2 there like so. Okay. Once we've got this here then we can just simply now use the product rule here. So for dy by dx then so for dy by dx here what do we get then? So we do u * v prime going to be u * v prime and we'll then also do u prime * v. Okay like so we add those together.
So um u * v prime so that's 3x * this fraction here. 2 all over the<unk> of 1 - 4x^ 2. That would give me then um 6x here. So I get 6 x all over the square root here of 1 - 4 x^2 just like so. Okay. So that's u * v prime. We now just need u prime * v. So that's 3 * r sin 2x. So that is just simply plus 3 sin 2x.
Okay, like so. And there we go. Then there's nothing else that we can actually do with that. So once you get it into, you know, this form here, that's it. Job done. Giving us the first two marks for this question. As I said, it is quite a nice easy question just to break up the um pace of things, right?
So that's a solution to part A.
Hopefully nothing too challenging there.
So for the next part then of this question here we have part B. So for part B then it says hence determine the exact value of dy by dx when x is equal to a quarter or 1 over4 here given your answer in the form a pi + b where a and b are fully simplified constants to be found.
Right? So let's just make a quick note here.
This is when x is equal to 1 / 4. Right? So this is pretty straightforward, right? All I simply need to do here is just substitute x= 1 over 4 into dy by dx here. So let's just do that underneath then. So when x = 1 4 d y by dx here is equal to so I get 6 * 1 over 4. So that is 6 over 4 which is 3 over2 I get 3 over2 there that's all over then so it is the square root here but it's the square root here of 1 minus them so it's 4 x^2 so x here is 1 over 4 so that is 1 4 which is 1 over 16 * 4 so I get 4 16 which is 1 over 4. So I've got 1 - 1 / 4 just like that. Hope that does make sense what I've done there.
And then 3 r sine of 2x. So if x is 1 over 4, two lots of one over four is uh 1 over two. Right? So um what I get here is 3 ark sine of 2x that is as we just said then arine of 1 /2.
Okay, just like that. So, um where do we go from here? So, let's just simplify this fraction here. Um what we can do here is deal with the denominator, right? So, the square root of 1 - 1 4.
So, 1 - 1 4 is 3 over 4. So, if you take the square root then of four here, that's going to be <unk>3 over two, right?
So what I've got here um I'm kind of running out of room so I'll continue up here. So dy dx here is equal to so it's 3 over2 all over <unk>3 / 2 like so and then we've got three lots of ark sin 1 /2.
So ar sin 1 /2 here that's just simply p<unk> / 6. So ar sine of a half is<unk> / 6. So 3 * p<unk> / 6 is 3 p<unk> / 6 which is p<unk> / 2. So I've got this here plus then as we just said um p<unk> / 2.
So obviously I can't simplify p<unk> / 2 any further. We've just already done that here all in one step with the free ar sign of a half. That's fine. So this part here is dealt with. But let's just simplify this here. And again this is pretty straightforward, right? If we've got 3 over2 all over roo<unk>3 over two, then these twos here just actually cancel, right? I get 3 over<unk>3.
So again, what I want here, so I've got 3 over roo<unk>3, but what I want here is a pi where a and b are just fully simplified constants. So let's just rationalize here. I get 3<unk>3 over 3. Three just cancel. We simply get roo<unk>3 there. Okay. So al together then here.
So when x is equal to a quarter here and x equal to 1 over 4 in that case then d y p i dx here is equal to what we just got here right so roo<unk>3 so roo<unk>3 + pi over 2.
Okay and there we go then. So hopefully that wasn't too bad there. I can see that this question took me just just under eight minutes even just for the free mark. So maybe I'm just slow. Um but there we go. Right. So hope that wasn't too bad. Um again if there's any issues at all just leave a comment. Right? I will try my best to help. But there we go then. So that gives the solution here to question six.
So we're pretty much there now guys.
Right? This is the penultimate question in this video. This is question seven here which is based on roots of polinomials. Now these types of questions here are really really common.
So, please do make sure then that you are confident and comfortable with these types of questions here. I'm not saying it's going to definitely come up, but Ed Excel really do like these types of questions here and they often do make their way into one of the two papers.
Okay, so before we do anything else then let's quickly read through the question.
So, what we're dealing with here. So, what we're told then is this cubic equation here has roots alpha, beta, and gamma. Now solving the equation, what we want to do here is find the cubic equation whose roots are alpha + 3, beta + 3 and gamma + 3, giving your answer in this form here where pqr and s are all integers to be found. Like I said, very standard stuff here. Now, there's a few different ways that you can kind of approach a question such as this here, but the method that I will be using here is substitution. It's just the way that I like to do these types of questions here, but any of the other methods here are absolutely fine, right? doesn't specify um an exact method to use. So the substitution then what we look at here is this form here right as we can see this is a cubic equation as well in terms of w it's pw cub + qw ^2 + r + s is equal to zero. So we start with w here being equal to what it's equal to here is based on these roots here right for the new cubic equation. So it's alpha + 3, beta + 3, and gamma + 3. So what we do here is we set w equal to x + 3. Okay. And now what we want to do here is rearrange this and make x the subject. Right? So x here will be equal to w - 3. Okay. Why is it x then? Well, the original um cubic equation here is in terms of x, right?
is 2xqub + 6 x^2 - 3x + 12 is equal to zero. So if we make x subject here, we can just substitute this here into this cubic equation. So let's just do that.
And what I'll do here is I'll do it underneath then just because we do need quite a bit of room here for the full expansion.
So we start with 2x cubed here. So that's two lots then of x cub. So x here is w - 3. So that's two lots of w - 3 cubed plus 6 lots then of x^2. So x here is w - 3. So that's 6 lots of w - 3 squared. If you want to you can do the expansion as you go but generally I just like to kind of write down the full expression first and then deal with the expansions as I go. So um that is 6x^2 - 3x then is - 3 lots of w - 3 w - 3 + 12 and all of this here is equal to zero.
Right now, this is probably the most challenging part of this question here is dealing with the expansions, which all things considered is just basically a level mass, right? And GCSE mass for expanding triple brackets here, double brackets there, and a single bracket there. It really shouldn't be too bad.
So, if you want to just use the bin expansion, that's absolutely fine. If you want to expand the free brackets by hand, that's also fine. But, I'm assuming this is pretty trivial for most students. I'll just kind of do it um as I go here. So this would be two lots then of well in the interest of saving time here right because nobody wants to watch me expand triple brackets or use the binary expansion right I have already expanded these triple brackets beforehand um same for these double brackets here as well because that is pretty trivial nobody really needs to watch me expand triple brackets here so hopefully nice and easy then what do we get here so I get w cubed I get - 9 w^2 I get + 27 W and then finally here min - 27 that is the triple brackets there um expanded I'll times it by two a little bit later on I then got six lots here of wus 3 squared so again hopefully nice and easy right as a level for math students I don't think I need to explain how to expand double brackets I hope anyway um so I get w^ 2 here so w ^2 get - 6 w - 6 w uh + 9 + 9 there. And again, I'll multiply by the 6 a little bit later on. -3 * w is -3 w.
And then -3 * -3 is + 9. To be honest, the only thing that you can really kind of mess up with here is the signs more than anything. Um, it is, you know, quite easy to make a silly sign mistake here. So, just be very very careful for that. Right. So, -3 * -3 is + 9 + 12 here.
And all of this is equal to zero. Right?
Let's multiply through now by these um two terms here. So two with this um bracket here and then six with this bracket here. So al together then what we get here is 2 w cubed - 18 - 18 w^ 2 + 54 w - 54.
I've then got 6 w ^2.
So 6 w^ 2 - 36 w and then um + 54 - 3 w and then + 9 + 12 is + 21. And again all of this here is equal to zero. And we're pretty much there, right? To finish with here, let's just collect like terms. And at that point there, we should have something in this form here where hopefully PQR and S are all integers, right? Which, you know, by quick inspection here, that will definitely be the case. So, um, what have we got here?
I've got 2 w cub. There's no more um w cub terms. So, it's just simply two w cubed. I've got -8 uh w ^2. I've got + 6 w ^2 there. So, that is -2.
So -2 w^2 that's simplified there. We can't do anything uh further with that. I've got 54 w here and uh - 36 w here and also - 3 w here. So that is uh 54 - 36 is 18 - 3 is 15. Right? So 15 w 15 ws quick math there. Look at that. Um so what else have I got? I've got minus 54. That cancels with the 54.
Um is it just plus 21? Yeah, plus 21, right? So plus 21 here and there we go then. So that is all equal to zero. And just know it's here then it is of this form here. So to finish with then let's just quickly state PQ uh and S. So P here is equal to um so P is a coefficient for W cubed here. So that is 2. Q then is a coefficient for W ^ 2. that is -12.
R here is just simply 15.
And finally then S here and S here is just simply 21. And there we go. Perfect. Okay. So again these types of questions here are really really common. Please do be confident with these types of questions here because all things considered these are you know quite a nice easy five marker to have um at your disposal. Right. So there we go then. I guess the solution here to the penultimate question. I can't speak him up, but we're pretty much there. The penultimate question for this video here, question seven. So we make our way to the very last question now to finish with.
And as we just said then to finish with here, we have the very last question for this video, question eight. Now it wouldn't be a corpure revision without a question on matrices and more specifically then finding the determinant and the inverse of a 3x3 matrix. Again, similar to the previous question here, this type of question is bread and butter stuff. Please do be confident with finding the determinant and the inverse of a 3x3 matrix. I appreciate it is long and tedious. Um, but they are quite easy marks. Um, they're just painful to work through, but that is the nature of matrices.
Anyway, um, I digress. Let's take a look at this question here, then question.
What are we dealing with here? So the matrix M is given by the following 3x3 matrix here. So for the first part of this question here then part A I will definitely start at the top here just because um space will be needed for part B right for the inverse. So for part A then all we want to do here is just show that dem the determinant of matrix M here is the following 5k minus 10 just two marks for this right. So what I'm going to do here is just expand along the top row. You can kind of just go along any row here, but it's easiest to go along the top row here because the middle element is zero. So that will just cancel, right? So just saves a little bit of kind of working in between here. So how do we do this then? So the way to do this here then is imagine. So we go along the top row here, right? So I start with the top left element here and imagine then we cross that row that it lies on and also the column. What 2x2 matrix do we get left with? Well, I get left with this 2x2 matrix here. 3 2 1 and k. So I want the determinant of that 2x2 matrix. But what I also want here is this element here. So we times by this element here. And when you expand here it goes plus minus plus like that. Okay.
So the first one here is positive2. I get two lots then of this 2x two um determinant here. So 3 2 1 k. So 3 2 1 K like so middle one here is zero. So that would be 0 times by the 2x2 u matrix left over or the determinant of that 2x2 matrix. But because it's 0 times by something result is just clearly zero.
So we can just skip that one. And then for the last one here it's plus minus one. So obviously that would be negative we get minus one lot then of again if you're struggling to see what the 2x2 matrix is just cross it off. So we cross off that row that it lies upon and also this column here and we get left with this 2x2 matrix here which is oops I want the original pen color here. Sorry.
So my 2x two determinant here is k 3 - 2 and 1. So k 3 - 2 and 1. Perfect. Okay. So let's just evaluate these 2x two determinants here. So I've got two lots then of two lots of 3 * k which is 3k and then minus here 1 * 2 which is -2.
Perfect. What I've then got here is -1 law of so k * 1 is k. Just be careful here now. So we subtract then - 2 * 3.
So that's - 6 which is + 6. Right? Just be very very careful for your signs here. It is very easy to make a mistake with these types of questions and the signs. To finish with them, let's just simplify this here. So 2 * 3k is 6k.
Um 2 * - 2 is -4 - 1 * k is - k and then -1 * 6 is -6.
So if we collect like terms here, what do we get? Well, 6k - k is 5k and then uh - 4 uh - 6 here that is - 10 exactly as we wanted here. So as required there good stuff.
So as required happy days because if you get the same thing you know you pretty much got the marks there unless you've somehow flooked it which is pretty rare. Um so there we go then 5k minus 10 as required there giving us a solution to part A.
Now for part B then it says given that K is not equal to two find the inverse of the matrix M in terms of K. Again this is very standard stuff here right so for part B then let's just do that here underneath. Now this will take quite a bit of room. So finding the inverse of a 3x3 matrix here is a long and tedious process. There isn't really a quick way to do this um unless you can use your calculator which clearly we can't do here because the original matrix I'm going to do that sorry the original matrix here as you can see then contains elements that are not numerical. So I've got K here and K again. So we can't just use our calculator right so we do need to do this by hand. So do make sure you can find the inverse of a 3x3 matrix by hand. You will get caught out if you can't do this. I promise you um this kind of question comes up every single year without fail. So how do we find the inverse? What I'm going to do here is I start by finding the minus. Okay, the matrix of minors. What is that I hear you ask? Well, we start with a 3x3 matrix of the individual 2x2 determinants. So that might not help if you've never really seen that before. Um so what I'll do is I'll just kind of do it and explain it as I go. It's probably just easier to do it like that, right?
So I'm going to do a very large matrix here because it does take up quite a bit of space. Um hopefully it should be just about enough room. So what we do here is we imagine then that we go along each individual element. So let me just get rid of the plus minus plus here at the top just so I don't cause any confusion.
Imagine then that we go along each and every element here of this 3x3 matrix.
Right? And what I want then is the respective determinant of the 2x2 matrix that is left over once you delete the respective row and column. So for example then right the first one here I delete this row and this column and I get left with 3 2 1 K. Okay the top left element here it's going to be um as we just said then 3 2 1 K because I delete this row and this column.
So three 2 one k like so for the next element here the 2 x2 determinant then I delete this row and this column I get k 2 - 2 and k. So k 2 uh minus 2 and k.
I'll do the next two here and then I'll just kind of give the answers um or the two 2x2 determinants rather than explaining each and every one otherwise we will be here for quite a while. Um so minus one then so delete this row and this column I get k 3 minus 2 and one.
So k 3 minus 2 and 1.
So close the matrix here like so. So underneath then that corresponds to this element here. So we delete this row and this column. So again I'll just do in a different color just so you can see how this actually works. Delete this row and this column. So what I get left with then is 0 - one 1 and k. So 0 - one 1 and k. And now in the interest of saving time here because I don't want to take all day just doing the matrix of minus I will just give the respective 2x2 determinants. Hopefully you've got the exact same as me here for each individual element. You shouldn't get something different at this point here just for the matrix of minus. Right? So for the next one here I get two minus one minus 2 and k. For the next one here you should get two 0 - 2 and one. The next one here you should get zero minus one 3 and two. The next one here you should get two minus one mk and two there. And then finally for the very last uh 2x2 determinant here we get two zero k and three. Let me just quickly double check then I continue here because if I continue I don't know it's a mistake then this will be painful. So up to now this all looks good. So what we do here then is we evaluate each individual 2x2 determine yes it is painful but that is just the nature of finding the inverse of a 3x3 matrix. Right? So let's just continue here. Again we get a pretty large matrix right so it's a 3x3 matrix for the result here just evaluate then each individual 2x2 determinant so for this one here we get 3k minus 2 right I've just noticed so that this will run into the mark so I'll just get rid of that I'll just do it underneath here just so we don't run out of room so as we just said then the top left element here I get 3k minus two like so the Next one here I get k^ 2.
Again, just be careful with your signs.
This is going to be k^ 2 minus minus 4.
So that's k^2 + 4. Right?
So k^ 2 + 4. I'll do the next one here and then I'll just give the answers again just in the interest of saving time. So I've got k * 1. And again, just be careful with your signs here. So k minus - 6 here. So that's k + 6.
Okay. Like so. Again, just save time here. to give the next um six determinants here as I go. So this one here that gives us one. I get one there.
The next one here is 2k - 2. The next one here is two. Next one here is three. Next one here is four + k. And then finally here six. Perfect.
Okay. So this here then is what we call the matrix of minors. Okay, what I now need here is the matrix of co-actors.
Okay, so this here is the matrix is the matrix of minors as we just said. Then what we now need here is the matrix of co-actors. So where does that come from? Um so let's make a quick note here. The matrix of co-actors.
the matrix of co-actors.
Well, I've got a 3x3 matrix again. Okay.
Now, what we did here along the top row when we were finding the determinant is we went plus minus plus. That continues for the full 3x3 matrix. Right? So, it is plus - plus - plus - plus uh minus plus. There we go. Okay. That is a matrix of co-actors. So we go back to the matrix of minus and we now apply this here to it. So wherever it's a plus that stays as it is and wherever there's a minus you just put minus in front of the full expression. So minus bracket k 2 + 4 becomes minus k ^ 2 minus 4 right I'm hoping that is pretty obvious. So the matrix of co-actors here we just usually denote that as c. So c here again will be a 3x3 matrix. What I get then is 3k minus 2. that doesn't change.
Um, I probably should have done a little bit larger here.
Um, let me just quickly redo that. I apologize. My art skills. Um, apparently I can't even draw a bracket. So, let's try again. So, something like this here should hopefully do. I get 3kus 2 as we said.
Then this one here becomes - k^ 2 - 4 - k 2 - 4. This one here stays the same.
So k + 6. This one here becomes negative. So that's min -1.
Like so. This one here stays the same.
So that's 2k minus 2. This one here becomes negative.
So min - 2. Um that one is positive. So that stays the same. Positive 3. This one becomes negative. So -4 minus k.
Then finally this one here is positive as well. Perfect.
Okay, we're pretty much there. Now, what I now need here is the transpose of this matrix here, the matrix of co-actors, right? So, what I now want here is the transpose of the matrix C. Again, this will also be a 3x3 matrix. Um, so what will this be? So, the kind of the easiest way to do this really is just to take each row then and flip it into respective columns. So, the first row becomes the first column. The second row becomes the second column and then the third row here becomes the third column.
Right? So this row here becomes this column. So I get 3k - 2 I get - k^ 2 - 4 and then k + 6. Perfect. The second row becomes the second column. So it goes -1 2 k minus 2 and then the minus two here at the bottom. And then finally the bottom row here becomes the third column. So I get three I get minus 4 minus k and then finally six in the bottom right corner.
As you notice here hopefully you do notice anyway the leading diagonal will not change. Right? So you should get the same leading diagonal here as the original matrix of co-actors. Perfect.
Let me just quickly double check this then just make sure that I haven't made any silly mistakes here for matrix of co-actors and the transpose. Um, so 3K - 2 - 1 3. Yep. - K square - 4. Yep. 2K - 2. Uh, yeah. - 4 - K. Yep. K + 6 - 2 and six. Yes. Perfect. This is all good. And we're pretty much there now for the final bit of work in here. This is for the inverse of the matrix M here. Right, we're done now. So for the inverse of the matrix M here, what we now need then is one over the determinant of M. So that's 5k - 10 as they gave us in part a.
It's 1 all over 5k minus 10. Like so. We then times this here by this matrix here C transpose. Okay. So it's the matrix of co-actors transpose which is this matrix here. So again for the final time let's just write this down. So 3k - 2 -1 3 uh - k^ 2 - 4 2 k - 2 -4 - k near there guys k + 6 - 2 and six there we go right pain is over and there we go that is the inverse of the matrix M in terms of K right if you really want two, you can give each term here in this matrix over 5k minus 10, but I wouldn't recommend it, right? Just leave it like this. This is a much nicer way to give the inverse of the matrix M in terms of K, right? And there we go. This question here alone took 15 minutes. These questions are never quick to be fair because there's a lot of kind of like little things going on as you can see here the matrix of minors um which we get here and then the matrix of co-actors transposing it and then finally times it by one over the determinant. It is a tedious process but for four marks it's not too bad right but again unfortunately is something that you do need to do by hand unless the full matrix is given numerically which in this case here it's not and that is often the case to ensure you can't just use a calculator right but there we go then. So there we have it, right? That is eight questions covered here for some last minute revision here for paper one. So um if you've made it all the way to the end of the video, thank you so much for watching. I do appreciate each and every one of you that does watch these videos here. Um we have covered pretty much all of Scorpio on our channel. So if there is anything that you're still struggling with, right, please do go and check out um the videos on our channel. We've got the individual videos covering the topics and also exam revision videos for the full topics. So, for example, if you're struggling with matrices, then we've done a full exam revision video for matrices. We do also have members videos on our channel here. So, if you do want to watch those extra videos, just become a member or alternatively here, go visit our website, ajmass.co.uk, become a member over there. That will give you access to our worksheets, our mock exam papers. So, we have also done um predicted papers for corpure. We did those last year. So, there's two papers there for corpure. And we will also be releasing in the coming weeks here um predicted papers for uh further mechanics one further stats one and further P1 maybe decision one um I'll consider that in the near future but yeah possibly decision one as well um but definitely FP1 FS1 and FM1 as well. So there we go.
I am waffling quite a bit but the only thing I've got I've kind of got to say here before we finish then is best of luck for sitting Corp 1 um this week. um if you are still watching at this point here. Um and what I also said at the beginning is I will do a second last minute revision video once paper one has been sat that will be useful because I will know what topics are going to come up then for paper two roughly based on the topics that appeared on paper one.
Right. So there we go then. As I said I'm waffling quite a bit. But as I said best of luck and thank you for watching.
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