Edexcel A Level Maths 2026 | Paper 1 - Last Minute Revision (WATCH THIS BEFORE PAPER 1)

Añadido: 2026-06-03

[00:00:09]Hello and welcome back to another Alevel mass revision video. Now, as we can see then in the bottom left corner here, this video is going to cover some last minute revision for paper one, which is later on this week. And we did do something similar last year, but for papers two and three. But I do think students find those videos quite helpful, especially with their revision writing. You know, I'm hoping that's going to be the case again this year with this video and the last minute revision videos for papers two and three. We've also got our predicted papers. So, paper one's already been released, right? If you haven't seen that already, go and check that out.

[00:00:39]I'll leave a link in the description down below somewhere down there. Right?

[00:00:42]So, go and check that out. Papers two and three, they'll be released shortly.

[00:00:47]Obviously, I'll release paper one after, sorry, I'll release paper two after paper one has been sat and then paper three once paper two has been sat. Now, paper two is somewhat dependent on paper one because they don't generally reassess topics again. So, once we've seen paper one, you know, we kind of have an idea of what to expect there within paper two. Paper three is different because it's the applied paper. So, stats and mechanics and obviously whatever appears in papers one and two doesn't really affect paper three, right? So, um paper 3 will just be released once paper 2 has been set anyway. Um just because that's how most students kind of revise it. Um so, yeah, keep an eye out for that. If you're new here, please do consider subscribing um just so you don't miss those videos as well. Now, I don't want to yap for too long here, just a few kind of quick points to run through. All of the questions featured in this video here from the old EdXL Alevel math specification. So, the mark allocation might be slightly off, right? Um and I do highlight that a little bit later on with one of the trade questions. Um you know, you're getting six marks for something which isn't too challenging, which you won't get anymore, but they did used to kind of go out like that. Um so different times different era but yeah that's the way things work now. Um so that's the first kind of key point there. The second point here is there's only 10 questions in this video here. So the questions featured in this video here they are somewhat basic. There's no real kind of like modeling questions here. If you are looking for modeling type questions then do go and check out all of our predictive papers where there's more emphasis on the modeling type questions. As I said then this video here it's just covering the basic concepts behind a few of the topics here within a level math. up um you know differentiation an integration question a couple of trick questions um you know stuff like that there's a proof by contradiction question in here as you can see question one is based on cartisian equations so you know I won't go into too much detail about what topics are actually featured here I will leave kind of the list in the description down below as well and I think that's pretty much it in terms of an introduction here as I said right I don't want to yap for too long um the final thing to kind of mention here then is basically if you're looking for any additional A-level mass revision then go and check out our website ajmass.co.uk.

[00:02:51]At this point here, we have 10 predicted papers live on the website. We've got additional worksheets, additional revision videos, challenge papers, or sorry, challenge worksheets um for Alevel math. So, go and check that out if you are looking for any additional Alevel math revision in the run-up to your Alevel math exams. Right. And I think that is pretty much everything. As I said, um this isn't a predicted paper.

[00:03:12]Please don't comment and go, "This is a really easy paper. I'm not bothered."

[00:03:16]Um, it's not a predicted paper. If you are looking for a challenge, like I said, go and check out our predicted papers. Um, and yeah, that's pretty much it. Like I said, the video here for last minute vision for paper two. That will definitely be more kind of adapted based on what we've seen for paper one. And it will definitely feature more questions than what we've got here. I've only got 10 questions for this video. It does run over an hour already. Um, so I've already recorded the question. I think it's about an hour and 10 hour and 15 in total. So either way, should give you some good last minute vision here for paper one. If you're aiming for an A star, this video is probably not for you. Um, I would definitely go and check out the predicted papers before this video here. But if you're aiming for like a B maybe, um, you know, this video is probably all right for you and you should get some decent revision either way. Okay, so I think that covers everything. That's enough of the app in.

[00:04:02]Let's just get started then with question one.

[00:04:06]So to get started here then, we have question one, right? So as you would expect here with question one, quite a nice easy question just to ease us into things, right? So for question one here, what we've got then is a curve C which has the following parametric equations here. So for this question then what we want to do here is just show that all points on C lie on the curve with this cartesian equation here where A and B are just constants to be found right marks for this here. So we shouldn't really be expecting too much work. So to begin with here then as we can see right the cartesian equation here is y equals the following. So we've got our two parametric equations here. X equals this and Y equals this here. So it seems logical then to rearrange this parametric equation here and make T the subject and then substitute that into this parametric equation here. Right? So we have X is equal to so it's t all over TUS 3 like so. And as we just said then all we want to do here is just make T the subject right. So I get x bracket tus 3 is equal to t. Okay. Expand this single bracket here. Then what do we get? So I get xt - 3x is equal to t. And what I'm going to do here then is just subtract t from both sides and add 3x to both sides as well.

[00:05:32]given us then xt uh minus t is equal to 3x. Okay. Now on the left hand side here we can play a t as a factor. So I get t b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b.

[00:05:47]It's going to be x - one there is equal to 3x. Okay. And again all we want to do here is just make t the subject. So just divide both sides by x - 1. And therefore then t here is equal to 3x all over x - 1. Perfect. And now what we can do here is substitute that into this parametric equation here. So therefore then y is equal to 1 all over t which is this here is 3x over x - 1 + 2.

[00:06:26]Like so. If you've got one over something like this here, so one over a fraction, then the result here is just the reciprocal of this bottom fraction here. So what I'm going to get here then is x -1 over 3x + 2 like so. Now clearly for this result here, they've got that over a common denominator. So let's just do the same here. I just need to times 2 by 3x, right? So I get x - 1. I've then got 2 * 3x that's 6x there.

[00:07:01]So plus 6x and this here is all over 3x.

[00:07:06]Okay, let's just simplify here. What do we get then? So 6x + x is 7 x.

[00:07:13]We got 7 x there - 1 and that is all over 3x. And just notice here then what we've now got here is of the required form. Right? I've got ax here which is 7 x - 1 over bx where bx is 3x. So perfect then. So therefore what we've got here is y is equal to 7 x - 1 over 3x should make it look a little bit neater there.

[00:07:41]As we can see then a here is 7 and b here is three. And there we go. Nice and easy. Right? That gives us the solution here to question one.

[00:07:57]Now question two here as we can see on the screen then is on proof by contradiction. And as proof by contradiction questions go, I don't think this one is too bad. It's just four marks. Right? So let's just work through this question together and let's just see how we get on. So for this question here, we've got two curves, right? I've got the curve C1 and the curve C2. So curve C1 has this equation here and the curve C2 has this equation here. So for this question then what we want to do here is use algebra to prove by contradiction that C1 and C2 do not intersect here. Okay. So it's proof by contradiction, right? And as always with proof by contradiction, we start with an opening assumption here. So what am I going to assume here? Well, I'll just assume then that C1 and C2 do intersect, right? Let's just do that at the very top here.

[00:08:49]So assume then so assume that C1 and C2 the two curves here intersect.

[00:09:03]Okay. Now if C1 and C2 intersect here then X 4 + 10 X2 + 8 must be equal to 2x^2 - 7. Right?

[00:09:12]So in that case then I get x 4 + 10 x^2 + 8 is equal to 2x^2 - 7. Okay let's just make the right hand side here equal to zero. So just subtract 2x^2 and add 7 to both sides here. So I get x 4 there.

[00:09:36]I've got a 10 x squ here but we are subtracting 2x 2 from that. So 10 - 2 is 8. So I get 8x^2 there and 8 + 7 is 15 and this here then is all equal to zero.

[00:09:52]Now where do we go from here? Well clearly at some point we do need a contradiction. Right? Now what I would hope here is at some point I can show either the discriminant or um if I attempt to solve this here then the equation has no real solutions. Right?

[00:10:09]So you might notice here that we can actually factoriize this here, right? So I get x^2 + 5.

[00:10:18]So x2 + 5 and x^2 + 3. And this here then is equal to zero. And you might start to see here then how the contradiction is going to form here, right? Because I get two solutions here.

[00:10:34]So my first solution then is x^2 is = -5.

[00:10:42]I've also got here x^2 is equal to minus3. Right? And this here is a big problem because as a level math students we can't solve these two equations. Okay. So this gives us a contradiction here. So therefore then what I'm going to say here is um this equation here so x 4 + 8 x2 + 15 being equal to 0 has no real roots okay or no real solutions whichever you prefer. So x 4 plus then 8x^2 + 15 being equal to zero has no real solutions.

[00:11:29]So this has no real solutions which isn't possible right because we assumed here that C1 and C2 intersected but for C1 and C2 to intersect then well it's not possible with these two given equations here right so this here has no real solutions which isn't possible so therefore then my pen works C1 and C2 why is my pen not working there we go so C1 and C2 here do not intersect do not intersect.

[00:12:07]So this here forms the basis then of our contradiction here right because um this isn't possible for exponent set of real numbers and same again here for cur2 right so there we go then just give it some kind of conclusion to finish with and you can probably summarize that a lot better than me but there we go right so that gives the solution there to question two now the next question here question three is really quite straightforward it's just a simple application of the trapezium rule right so for the first part of this question here just one mark it just wants us to find the single value of y here in other words complete the table and then for part B just actually using the trapezium rule right so again pretty straightforward here so let's just start then at the very top here with part a so for part a then it just says complete the table above given the missing value of y to six decimal places so that is something that we do need to be careful of here so to find this missing value of y here we simply need to then just substitute x=<unk> / 3 into this equation here So y then is equal to so it's sec here of so it's 1 / 2x with x is p<unk> / 3 that's 1 2 *<unk> / 3 which is p<unk> / 6 right so sec of<unk> / 6 just throw this here then into your calculator if you do this all correctly then and again just be careful here this is to six decimal places what you should get here then is the following so we get 1.154 71 one. Nice and easy, right? So that's the solution there to part A. Also included in the table here like so. So 15 or 7 01. Okay. So there we go then. Nice and easy. That gives us the solution to part A. And now for part B then let's just do this underneath here.

[00:13:56]We're just going to use the trapezium rule here. Now don't forget the result for the trapezium rule is given in your formula book but I don't think the result is particularly challenging to kind of memorize right. So we start with the area here.

[00:14:09]So the area here is approximately then so what we have here is a half h or h over two. So what is h here? Well h is the gap then between the x values here. So we can easily see then that the gap here is pi / 6. Right? Just take this x value here and subtract the first one. So p<unk> / 6 - 0 is just p<unk> over 6. You can verify that as well doing p<unk> over 3 -<unk> / 6 and also p<unk> / 2 - p<unk> over 3. Again, you'll get p 6 here for each individual difference. Right? So what I've got here then is 1 / 2 *<unk> / 6.

[00:14:47]So h here is the pi over 6, right? And then what I do here is I times all of this then or I times this product here then by all of this I should say English is hard. So what do I get here? So for the trapezium rule here we do the first y value. So that would be one here and also the last y value that's 1.414 214.

[00:15:12]So 1 plus then 1.414 414 2 1 4 I should have done this a little bit further down with a little bit more room because I am definitely going to run out of room here. Um but never mind.

[00:15:25]And then what we also do here is two lots then of the remaining y terms here. So the middle terms essentially. So it's these two terms here, right? We add these together, times those by two and add on the last term. That's this one here. And also the first term, right? So I've got 1.03 03 5276 plus then what else did we have here?

[00:15:53]This missing Y value that we just found here in part A, right? So 1.1 1.1 547 547 01. Okay, perfect. So we'll close that bracket there. Apologies. Um oops, I'm in room. There we go. Actually made it look a little bit neater, but you can follow along. Right. So we get this here. Now at this point here, just throw this into your calculator. You don't need to do this by hand. You don't need to kind of show um you know the exact answer here as such. Just throw it into your calculator because we're going to give this to four decimal places here. So again, the only thing to really be careful of is the required accuracy.

[00:16:33]So in this case here, it's to four decimal places, right? So in that case then an approximation for the area of R here is going to be um well it's just going to be this here right so the area then so the area of R here is approximately so if you do all of this correctly then what you should get here um to four decimal places is 1.7787 so the actual kind of um preliminary kind of work in here before we round it is 1.77870902 23 or two four decimal places and what we get here is 1.7787.

[00:17:13]So 77 7 87 right and that is the beauty then of the trapezium rule here it gives us an approximation for you know these kind of integrals here which aren't the nicest to evaluate by hand right so there we go then four marks for the question there as I said nothing particularly challenging with it is just a simple application of the trapezium rule and again don't forget the relevant formula here for the trapezium rule is given in your phone book so not something you do need to worry about memorizing but again I don't particularly challenging to kind of, you know, remember that or memorize it. But there we go. I'm just yapping. So that gives us the solution to question three.

[00:17:51]Now, moving on to question four here.

[00:17:53]This is where things start to get a little bit more difficult, right? Again, nothing particularly crazy here. But as we can see, this question does involve rates of change. So, you know, straight away there will be a little bit on differentiation. So, definitely more interesting than the previous few questions, right? But anyway, let's just begin here with question four. So what I can see straight away is figure one of this hemispherical bowl. Amazing. So water is flowing into the bowl at a constant rate of 180 cm cubed per second. When the height of the water is h cm as we can see on the diagram here the volume of water v cm cubed is given as the following here for h between 0 and 30. So for this question here, what we want to do then is just find the rate of change of the height of the water in centimeters/s when h is equal to 15 given our answer to two significant figures.

[00:18:43]Right? Okay. So as already kind of noted here, this question is based on rates of change. If that wasn't obvious, then then the big clue is this here. Okay. So find the rate of change. And again, these kind of questions are always pretty similar. It will usually be something to do with a bowl or a tank or some kind of like cone and water is kind of, you know, going into the bowl or the cone or whatever it is, right? Or some kind of fluid. Okay. So again, they're always pretty kind of similar in a way.

[00:19:13]So what I've got here is V, right? V here is 1 over 3 pi h 2 * 90 - h. What some students might do here is they might see this as a product of functions, which you can do. So obviously if I want to differentiate this here, you can use the product rule.

[00:19:30]But that is really kind of over complicating things because what I can just actually do here is just expand this out. Right? Now we will need to differentiate V here with respect to H at some point. So before I do that, let's just expand this out here. Let's just make my life a little bit easier.

[00:19:45]Right? Math is already pretty challenging. So let's make it a little bit easier. And let's just expand this out here. So in that case then v is equal to so it's 1 3 p<unk> h ^ 2 * 90.

[00:19:57]So what I'll do is I'll just write down the original expression. It's 1 3<unk>i h 2 * 90 minus h like so. So now let's just expand it here. Um so 90 uh if we 3 that's 30. So what I get here is 30 pi h 2. So that's the first term there.

[00:20:21]And then we've got minus h * this here.

[00:20:23]So 1 over 3<unk> h ^ 2 * - h. Well, nice and straightforward. Then um what I get here is - 1 over 3. So - 1 3 pi h cubed. Okay. So all we've done there to begin with then is just kind of expanded this out, right? And just simplified it. And it just makes the following differentiation here much easier to work with. Okay, as I said before again you can use the product rule but you're making life more challenging than it needs to be. So as we said before then what I will need here is the rate of change then for the volume here with respect to h the height and what I'll also need as well is the rate of change then for the volume here with respect to time t. So let's just start with DV by DH here. So we can easily obtain that then by differentiating V here respect to H. So that's this expression here. So therefore then dv by DH what is that? Well just differentiate term by term here nice and straightforward then. So 30 times by the two here that gives me 60 and then we reduce the power by one here. Right? Pi is just a constant. So it's 30 pi. So we get 60 pi and then I get h on its own right. Um because this power reduces by one. So we get 60 pi h there 60 pi h. Do the same here then with the second term. So - 1 over 3 * 3 is just -1. And again just reduce the power by one here. So nice and straightforward.

[00:21:55]Then we simply get minus pi h 2 just like that. Okay. So that's dv by dh. And as we just said then we also need dv by dt. So that's the rate of change here for the volume respect to time t. So where does that come from? So we just said then dv by dt.

[00:22:18]So that might not be quite as obvious here. But just read through the question carefully because what we're told here is water is flowing into the bowl at a constant rate of 180 cm cubed per second. So I know then that dv by dt here is just simply 180 right.

[00:22:35]So that is just simply 180. Now for this question here what we want then is the rate of change of the height of the water in centimeters/s when h is equal to 15. So what we actually require here is dh by dt.

[00:22:52]Right? So make a quick note here.

[00:22:56]So we need So we need dh by dt, right? That's what we want here. So how do we get dh by dt? Well, nice and straightforward then. Let's just do it over here on the left hand side. So dh by dt then.

[00:23:16]So dh by dt. That is just simply then.

[00:23:20]So it's going to be um dv by dt.

[00:23:24]So, dv by dt. And what I can do here is times that by dh over dv, right? dh over dv.

[00:23:37]In this case here, obviously the dvs would just cancel. We just get left with dh over dt, right? So, nice and straightforward here. What do we get then? So, dv by dt is 180.

[00:23:49]So we get 180 times then um DH um so what have I got there? I've got DV by DT. So that's 180.

[00:23:57]Perfect. That's that part dealt with.

[00:23:59]I've also got DH by DV. Now what I've got here is DV by DH. So I need the reciprocal of that. So just write that over one and then flip it. Right. So it's 180 * then 1 all over 60 pih -<unk> h^ 2 just like that. Okay, let me just simplify that here. Then what I get is 180 all over 60 pih - 10 pi h 2. Okay, just like that. So we're nearly there now, right? What we want here is the rate of change of the height of the water centimeters/s when h is equal to 15. So what we don't need to do here to finish with then is just substitute h= 15 into this here. So this is dh by dt.

[00:24:57]So to finish with them just um h= 15 into this here.

[00:25:03]I'm just going to kind of do that over here.

[00:25:07]So when h is equal to 15 what do we get here? So when h is equal to 15 dh by dt and that is equal to so it's 180 all over then we get 60<unk> * 15.

[00:25:28]So 60 pi * 15 minus then pi * 15 squared. Okay, let's just extend that a little bit. As you can see, my working is never the prettiest, but it'll do. Right, so um just throw this into your calculator here, right? And then we're pretty much there. So this actually does simplify quite nicely to 4 over 15 pi. So 4 over 15 pi. But obviously I can't give the solution like this because it wants it to two significant figures. So if you want it in exact form, this here would be the solution. But what I want here is the answer to two significant figures. So again, just throw into your calculator.

[00:26:10]And in that case then dh by dt here is 0.085 there. So 0.0 0 85 and the units here then would be centimeters/s.

[00:26:25]So 0.085 cm per second. Okay.

[00:26:31]Just like that. Okay. And there we go then. So as you can see like I said a little bit now of an increase in difficulty. Um you know it's a little bit more involved than the previous three questions here. Not quite as straightforward but it is just five marks for this. So again just take your time with it. Just be very very careful as you do work through these questions here. I don't think there should be anything too tricky within this question. Right? But there we go then.

[00:26:53]So that gives us the solution here to question four. Now for question five here we have a question on series and summation. Right? So for part one a and b here is based on this given recurrence relation here. Three marks that is pretty straightforward. I don't think anyone should really be struggling with that too much. And then for the second part here we have this summation which is straight less than zero. This could potentially cause some problems. So we'll cross that bridge when we get to it. As we can see three marks for that, right? So let's just start then with part one A and B. And let's just do that at the very top here. So part one then and part A, right? So what we're given here is U1 is equal to four. We've got this given recurrence relation here. And all we want to do then for part A here is just find U2. Just one mark for this.

[00:27:39]We're not really expecting too much work, right? So u2 then how do we find that here? Well as we just said then we just use this given recurrence relation here. So I want u2.

[00:27:51]So for u2 here that's when n is 1. So I get u1 over u1 - 3. So that's four all over them. U1. So u1 is 4 again. So 4 - 3. Well 4 - 3 is one. I get four over 1 giving us four. And well, we've just got the exact same as U1 here, right? So, as you can see, then every term here for this given recurrence relation will just simply be four because U3 would be the exact same working giving us four. Same for U4, U5, U6, so on and so on. Right?

[00:28:26]So, that's the first part complete there. Again, that's the first mark for this question here. And now for part B, then it just wants the summation here of UN from n= 1 to 100. So, don't over complicate this, right? Let me just write down the summation here. So from n = 1 to 100 of u n well as we've just said right every term for this given recurrence relation here is just four. So this is four plus four plus four plus four so on and so on 100 times.

[00:29:04]Right? So as we just said then it's just four * 100. Nice and easy. um is just simply 400 there. Okay. So as I said right hopefully for the first three marks here that doesn't cause too many problems. So for the final part then I will just do this at the very bottom here. And as I said before then this could potentially cause some problems.

[00:29:27]Um so what we're given here is this summation then. So the summation of 100 minus 3 r is strictly less than zero and this is from r = 1 to n. All we want to do here then is find the least value of the positive integer n the for free marks here. Right now if you've done or are aware of any of the alle f mass kind of uh material you cover series information in that and you can employ a method from that to answer this part here. But I'm kind of doing this video here as if you've only studied a level mass right? Um some of you will be aware of that. I'm sure some of you are studying aloof firm mass alongside this.

[00:30:02]But like I said, I'm doing this video here for Alevel math students only. So I won't show you that method here. I will show you the method that you're expected to use here just as an A level math student. Okay? So if you get something like this here and you're not really too sure what's going on, just start with the first few terms of the series, right? So when R is one here, what do we get then for this given summation here?

[00:30:23]So when R is one, what I get here then is 100 minus three lots of one. That's just 97. So the first term then in this series here is 97.

[00:30:35]I do the same here then for when r is two.

[00:30:38]When r is two here, what do we get then?

[00:30:41]So we get 100 - 3 lots of two. So that's six. 100 - 6 is 94. Perfect. And let's just do one more here. And then hopefully the penny should drop, right?

[00:30:50]So for r equals 3. Then what do we get here? So I've got 100 - three lots of three. So that's 9. 100 - 9 gives us 91. Perfect. And normally what you can see here then is we actually have an arithmetic series, right? It's a decreasing arithmetic series. So 97 94 91. What we can see here then is a common difference is minus three, right? It's decreasing by three every time. So what I've got here then is an arithmetic series.

[00:31:23]So arithmetic series. I can see then that the first term here is 97. So a is 97 and as we've already noted here the common difference then d here is just simply minus3. Now how does that help then? Well what I've got here is a summation right? So this is a sum here um well basically from r= 1 to n this needs to be straight less than zero.

[00:31:48]Okay so what I can do here is use the sm formula here for an arithmetic series.

[00:31:53]Don't forget this is also given in your formula book. So not something that you need to worry about memorizing. So let's just go a little bit further up here. So Sn then what is the formula here? So again it is given in your formula but I'll just write it down anyway just so we can see on the screen. So it's n /2.

[00:32:10]So n / 2 * then so it's 2 a + n minus one * d. There is a modified form of this here. if you know the last term in the series, but I can't bother finding the last term. So, I'm just going to use this here, right? So, what do I do here then? Well, just substitute a and d into this here. But this needs to be straight less than zero, right?

[00:32:36]As we're told here. So, I don't know what n is. So, I leave that as m, right?

[00:32:42]So, I've got n / two times then. So, two lots of a. So, a is 97. So two lots of 97.

[00:32:51]I've then got n minus one. Again, we don't know what n is, so I just keep it as n. So n minus one times by d, which is -3. So times that by minus 3 there.

[00:33:02]Okay, close that. And again, this here needs to be straightly less than zero.

[00:33:08]I'm not really too sure why I'm kind of writing on an incline, but never mind.

[00:33:11]Let's just continue. So what do I get here? So I've got n /2 again.

[00:33:17]2 lots of 97 is 194. Yeah. So I get 194.

[00:33:24]I've then got n * -3. So that's minus 3 n.

[00:33:29]So - 3 n there. And just be very very careful with the signs here. -1 * -3 that's + three. Right.

[00:33:36]Plus three there. Close that. And again this here is strictly less than zero. So let's just simplify here inside the bracket. 194 + 3 is 197. Right? So what I've got here is n /2 * 197 - 3 n just like that. And this here is less than zero.

[00:34:00]Now what we need to do at some point here is just basically solve this inequality for n. Right? Um so to make life easier here, let's just times through by two, which doesn't really do anything other than get rid of this fraction here, right? because the right hand side is 0. 0 * 2 is just zero again. So um once I do that what I get here is n * 197 - 3m is straight less than zero. Okay, so you've got two solutions to consider here. You've got n being straight less than zero which doesn't help us in anyway. So we'll kind of omit that solution. And then the second solution here comes from 197 - 3N being strictly less than zero. I've kind of run out of room. Um, which is something I always do. So, let's just continue at the very top here. I'm hoping that you can kind of follow along. Apologies, by the way, my handwriting and my just general presentation is shocking, but never mind. So, I've got 197 minus 3N is strictly less than zero. So, I would add 3N to both sides here. So 197 is strictly less than 3 n which is just the same then as saying 3 n is strictly greater than 197 right divide both sides by three and what I get here then is n so n here is strictly greater than 65.6.

[00:35:28]Now just be very very careful here. What we're looking for then as we're told here, right, is the least value of the positive integer n. Now, what I've got here is a decimal. So, clearly it's not an integer. I'm hoping that is obvious.

[00:35:42]Um, and with that being the case, then what I need to do here is round this up.

[00:35:48]So, the least value of the positive integer n here is going to be 66.

[00:35:53]Therefore, then n here is equal to 66, right? And there we go. So you can see that the final part of this question here is a little bit more challenging, you know, than the first two parts. I can't speak anymore, but it's certainly more challenging than part A and part B here. You know, definitely more work involved. Um, but again, it's just an application of summation formula here for an arithmetic series. But that is a key thing here, just recognizing that it is an arithmetic series. Um, and then we can just easily go from there. As I also said, if you're aware of the um series in summation chapter from the first year of Alevel Fast, then you can also use um results from that to solve this here.

[00:36:34]But again, we're Alevel math students, not a level mass student. So, I won't show that method here. I'll just show you the kind of um not the required method, but the method that you're expected to use. Okay. So, there we go.

[00:36:45]Then I'm just yapping. But that gives us the solution to question five. Now, for question six, then as we can see here, two parts this question. in the first part wouldn't look out of place in a GCSE math paper. So please don't struggle with that. Right? Um as I said it is just basic GCSE math for part A.

[00:37:02]Part B is definitely Alevel math. So um as we can see then natural logs involved. So let's just start with part A then and let's just see what we're dealing with for the full question.

[00:37:12]Right? So for part A then it just says simplify fully this quotient here. Now as I said then this wouldn't look out of place in a GCSE mass paper. Why is that the case? Well, all we simply need to do here is factoriize the numerator, factoriize the denominator, cancel the common factors, and we have the solutions part A. Right? So, let's just start then by writing down the original quotient here. I've got 2x^2 + 9x - 5. That's all over them. So, it's all over x^2 + 2x - 15.

[00:37:52]Now again, no one should really be struggling here with these two factorizations, but you just want to make life a little bit easier, just start with the denominator here. So once I factoriize the denominator here, what I get then is x - 3.

[00:38:07]So we get x - 3 * x + 5. Now I know then that the numerator here requires one of these two linear factors here, right? So the numerator will either have an x - 3 or it will have an x + 5. Okay. So clear then given that the last term here is minus 5. It must have an x + 5 as well.

[00:38:27]So it's x + 5 there like so. So the second term here then this is x + 5.

[00:38:34]This must be 2x - one. Right?

[00:38:38]So again that shouldn't be challenging but you just want to make life a little bit easier just start with the easier of the two um quadratic expressions. Right?

[00:38:47]As we said, then cancel the common factors which is the following and that gives us 2x -1 all over x - 3. Nice and easy, right? So that's the solution to part A. Let's continue underneath then um because we will need a little bit of room here for part B.

[00:39:07]So for part B then we're given the natural log of 2x2 + 9 x - 5 is equal to 1 plus this natural log here. So x^2 + 2x - 15 x not being equal to -5. What we want to do here then is find x in terms of e. Now what you should hopefully recognize here is the argument then of this natural log here is numerator that we had in part a. And then the argument of this natural log here is the denominator of this quotient here. So clearly it's going to be relevant then or the answer to part a will be relevant here for part b. Okay. So what I'm going to do here then is start by getting um the two natural logarithms on one side.

[00:39:50]Let's just subtract this here from both sides. Right?

[00:39:54]So what I get here then is learn of 2x^2 + 9 x - 5us then l of x^2 + 2x - 15. Right? And this here is equal to one. Okay, I've got a difference of logarithms here. They clearly have the same base um base, right? They're natural log. So we can write this here then as a single logarithm, right? So if we do that then what do we get here?

[00:40:28]So we get learn of. So we get a quotient here, right? Because we're subtracting logs here. So once we simplify that and we write as a single log room here, it becomes a quotient. So this here would be the numerator.

[00:40:41]we get 2x^2 + 9 x - 5 and then for the denominator here we get x^2 so x^2 + 2x - 15 and this here is equal to 1. Okay.

[00:41:01]And would you look at that right? It's almost a coincidence that we get this quotient here which is the exact same quotient here or the exact same fraction. Right? Well, that is useful because we factorized in part a, right?

[00:41:14]We got 2x - 1 over x - 3. So, we're clearly going to use that. Okay. So, we get l of, so as we just said then 2x -1.

[00:41:26]So, 2x -1 all over x - 3. Okay. And this here again is equal to 1. Now clearly at some point here we do need to get rid of the natural logarithm because I want to express um x in terms of e. So how do we get rid of the natural logarithm here?

[00:41:46]Well what I do then is I write both sides here of this equation um with e as the base. So I get e to the power then of learn of this here. So e to the power then of learn well they're the inverse each and they just cancel. So once we do that then what I get here is just the argument. So I get 2x -1 / x - 3.

[00:42:07]So my tablet's not looking. There we go.

[00:42:09]2x -1 all over x - 3 like so. And that is equal to then so again we take e as the base here for both sides. That's e to the 1 which is just simply e. Right?

[00:42:22]Like so. Okay.

[00:42:25]Now if we want to find x in terms of e here, what I need to do at this point here then is get all of the x's on one side. So let's just times both sides by x - 3 here. So I'll just go towards the top here. So we get 2x -1 is equal to e bracket x - 3. Now don't get me wrong. You can kind of skip this step here and just expand it straight away. But I'm just I'm just trying to show all my steps are working here as I go. Right? So I get 2x -1 is equal to so x * e is x e or ex whichever you prefer minus 3 e okay like so. So let's subtract x from both sides here and add one to both sides as well.

[00:43:13]What this will do here then is it will guarantee all the x's are on one side and then everything else over on the other. Perfect. Right. So I get 2x we get 2x there min - x e is equal to 1 - 3 like so. At this point here we've now got all of the terms involving x on one side. So I've got 2x - x e. Perfect. So what I can do here then is factor out an x. Right? So I get x bracket 2 - e is equal to 1 - 3. Okay, just like so. And then finally, we want x in terms of e here. We're pretty much there now, right? All I need to do here to finish with then is just divide both sides by 2 minus e. So once we do that, what I get here is x is equal to So it's 1 - 3 e.

[00:44:08]So 1 - 3 e. Apologies, by the way, I'm losing my voice here. Um, hay fever is getting me. So then divide by 2 minus e like so. And there we go. That is the solution. Bosch, let's highlight that quite enjoy that question. Um there we go. Right. So again, just using the laws of logs here as we did in part B, expressing that as a single logarithm and then from there just using properties of um you know logs and the exponential function to simplify and then express it as x in terms of e.

[00:44:42]Right? And there we go then. So four marks for that. Not too bad, right? And there we go then. So that gives us the solution here to question six. So for the next question here, question seven, what we're given then is the curve C which has this equation here. Now for the first part of this question in part A, we just want to show that dy by dx is equal to the following here for the first four marks. Right? So this question is clearly on differentiation.

[00:45:06]If you can't figure that out, then you're going to struggle. Um but basically if I've got a fraction such as this one here, right? And I want dy by dx. And hopefully it's obvious that we do need the quotient rule here. I've got a quotient of functions, right? 3 + sin 2x over 2 + cos 2x. So yes, we do require the quotient rule. Now the quotient rule is given in your formula book. So you don't need to worry about memorizing it, which is always nice, right? And um the only thing to kind of mention in that aspect is the way it's presented in the formula book is slightly different to how I will present it in just a moment. Um they do it as functions. I will use u and v notation.

[00:45:43]Either is fine. Right? So as we just said then what I need here to begin with is U and V. Now with the quotient rule is always the same. U is the numerator and V is always the denominator. Right?

[00:45:53]That never changes. So U here and this is part A, right? So for part A then U here we just said is the numerator. So that's 3 + sin 2x 3 + sin 2x. And then for V here that is 2 plus cos 2x.

[00:46:15]So 2 + cos 2x. Okay. Now what I also need here is u prime and v prime. So u prime here.

[00:46:26]So just differentiate u with respect to x. Then the three is a constant. So that's just zero once we differentiate.

[00:46:31]Sin 2x becomes 2 cos 2x. Okay. And we also need v prime. And so in a similar fashion here, you're going to get zero.

[00:46:40]That's a constant. And then the cos 2x here that differentiates to give us minus 2 sin 2x like so. Okay. So now for dy by dx then as I just said here I will present that using u and v. Um but again in the format that's given in terms of functions but e is fine here whichever you're more comfortable with. So you by dx here then what is that equal to? So in terms of u um u prime v and v prime what we get here then is it's u prime * v. So u prime * v minus then would now be v prime * u. So v prime * u and this here then is all over v^ 2. Okay. So yes, the order does matter here with the quotient rule um unlike the product rule. So just be very very careful when you're working with the quotient rule, right? So just going to use this here and hopefully at some point we should obtain this here.

[00:47:42]If we don't then it's not looking good.

[00:47:44]So let's give it a go. Um so u prime then is this here times by v. So um I might need a little bit more room here.

[00:47:54]I'll just continue down here. So dy by dx then what do I get here as we just said then it's u prime which is 2 cos 2x is 2 cos 2x times then um v which is 2 + cos 2x so 2 + cos 2x you then subtract here v prime * u. So v prime here is - 2 sin 2x. So if you're doing minus or minus here is just the same as plus. But I'll put it in a bracket here for now just to avoid any potential mistakes. Right? So what I get here then is - 2 sin 2x times then u which is 3 + sin 2x. So 3 + sin 2x there just like that. Okay. And this here then is all over v ^ 2 which is 2 + cos 2x all squ right. So 2 + cos 2x all 2 like so. Okay. So at this point here it looks a little bit grim. Um let's just start expanding here and let's just see what happens. I'll continue underneath then just so I've got a little bit more room here.

[00:49:18]So 2 cos 2x * 2 that's 4 cos 2x.

[00:49:22]So 4 cos 2x there. I've then got 2 cos 2x * cos 2x. So what I've got there is 2 cos^ 2x.

[00:49:34]Right. Like so. That's the first bracket dealt with there. I've then got minus of - 2 sin 2x. That's the same as + 2 sin 2x. As I was saying before, just be careful with the signs there. Right? It is easy to trip up with that. So 2 sin 2x * 3 that's 6 sin 2x + 6 sin 2x and then um 2 sin 2x * sin 2x that is um 2 sin^ 2 so 2 sin^ 2x like that. Okay, that's the numerator.

[00:50:12]And then again, this here is all over 2 + cos 2x all^ 2 like so. Okay.

[00:50:25]Right. Well, it's starting to shape, you know, shape up here. Um, I've got 6 sin 2x. We've got the same here. I've got 4 cos 2x. We've got the same here.

[00:50:34]Perfect. But I've got this here, right?

[00:50:36]I've got 2 sin^ 2 2x and 2 cos^ 2x.

[00:50:40]They've just got plus two. So what happens here then? Well, basically what you need to recognize here is I can express then and I'll do this in a different color. Let's highlight this in green. I can take this term here and this term here. And if you put those together then what I get is 2 sin^ 2x + 2 cos^ 2x. And hopefully what you recognize there then is that the same as two lots of sin^ square 2x + cos 2 2x which that's just two lots of one right sin 2x + cos 2x is just one that's just two lots of one. So make a note here.

[00:51:16]So that's two lots then of as we just said sin^ 2 2x plus cos^ 2x okay which is just the same then there two lots of one giving us two. So to finish with then underneath here what I've got is 6 sin 2x. So 6 sin 2x + 4 cos 2x + 4 cos 2x um plus two. So 2 + 1 is just two, right? So plus two and this here then is all over this here, right? So two plus cos 2x all squared like so. So that is for dy by dx. I forgot to kind of denote it again here.

[00:52:08]UY by DX. But there we go. Just wait for it on. So as required there and we'll give that a smiley face. It's always nice to get the correct solution.

[00:52:18]Want to show that. So as required happy days. So that is part A complete. Now as we can see we kind of took up the whole screen here just for part A. So what I'll quickly do here is just get rid of all of this. Feel free to pause if you want to kind of make any notes as we go, but I'm going to get rid of it now. So adios right part B then let's have a go at that again at the very top here. So for part B here it just says find an equation of the tangent to C at the point on C where X is equal to P<unk> /2 write your answer in this form here. So Y= A + B where A and B are exact constants. Right? So when X is equal to<unk> / 2.

[00:52:59]So I need a few things here. What I need then is the respective y-coordinate here. And I also need then the value of the gradient at this point here. x= p<unk> /2. Well, for the value of the gradient then that comes from dy by dx.

[00:53:15]Now even if you couldn't derive this result here in part a, you can still go on to use this here because they give you the full result. Sometimes it might be something like a sin 2x plus b cos 2x plus c. And you've kind of got to figure out a, b, and c. If that's the case, then you're cooked. But for this one here, because they give you, you know, the full expression here, there's no problems, right? We can just use that without proof, basically. So, um, let's just continue then with part B here. So, as we said, we need we need y here. I really can't speak anymore. Um, I'm losing the ability to function. Let's continue. So, y here is what is that?

[00:53:51]So, y then is 3 + sin 2x. That's three plus then. So, sine of two lots of pi / 2. So 2 lots of pi / 2 is just pi. So that's 3 + sin pi like so. And that is all over then 2 plus cos 2x that's um cos pi again. Right? So cos pi there like so. Well sine of pi is 0 cos of pi is minus one right? So what I get here is three for the numerator and then for the denominator here that's 2 minus one which is 1. So I get 3 over 1 giving us three. So that's the y-coordinate, the respective y-coordinate here. As we also said as well, I need the value of the gradient at this point here when x=<unk> / 2. So to obtain that, as we just said, then we just use dy by dx here. So this is dy by dx when x is equal to pi over 2. Okay. So for that then just substitute x =<unk> / 2 into dx. So I get 6 sin. So if it's sin 2x again it's just s of pi. So I get 6 sin pi.

[00:55:09]I've then got 4 cos 2 pi. Sorry 2 pi over 2. So that's pi, right? So 4 cos pi mass is hard. 4 cos pi. There we go. You know what I mean? 4 cos pi + 2. I need to go to sleep. And that's all over them.

[00:55:26]2 + cos pi all squared. So 2 + cos pi all squared. Okay. Well, as we said before then sin pi is 0. So that's 6 * 0 which is just zero.

[00:55:42]I get 0 4 cos pi. So that's - 4 + 2. And that then is all over um so 2 - 1 giving us 1. 1^ 2 is 1. So get -4 + 2 that's minus 2 - 2 over 1 is just simply minus2. Right? So we're looking for the equation of the tangent here not the equation of the normal. So you don't need to take the negative reciprocal of this here. We do just keep the value of the gradient as minus2. Right? So just be careful for that as well. So the equation of the tangent then we will just use here it's going to be y - y1. It's just the equation of a straight line, right? So y - y1 is equal to m * x - x1.

[00:56:30]So what is x1 and y1 here? Well, x1 is pi / 2 and y1 is just three, right? So in that case then what I get here is y - 3 is equal to m where m is -2.

[00:56:45]So - 2 * then x - x1 where x1 is p<unk> / 2 p<unk> / 2 there. Okay. Now we want it in this form here. So um let's just start by expanding on the right hand side here. I get y - 3 on the left hand side.

[00:57:04]Now on the right hand side here I get -2 * x that's minus 2x.

[00:57:09]So minus 2x there. And then minus two.

[00:57:11]So minus two here times -<unk> / 2.

[00:57:14]That's just simply plus pi, right?

[00:57:18]So plus pi there. Okay. Now what I want here is y is equal to ax + b where a and b are exact constants. So to finish with then just add three to both sides. And therefore then y is equal to so it's - 2x + 3 + pi. And there we go then. So this here is the equation of the tangent to C at the point on C where X equ= P<unk> /2.

[00:57:45]As we said, you need two things. You need the Y1 respective Y1 and also the value of the gradient at this point here. X= P<unk> /2 and again just use dy by dx from the previous part um to find the value of that. Right? Put it all together then and we get this here and there we go. So as you can see a little bit more work involved with that question there, especially part A. Um, so yeah, just be careful as you work through questions such as that one.

[00:58:10]Well, there we go. Then that gives the solution here to question seven. Now for question eight here, what I'm given then is tan of 40° is equal to P. What we want to do here then is find in terms of P the following three things here. So C of 40°, sec of 40°, and tan of 85°.

[00:58:27]Right now, if you've seen a question like this before, you might find this quite straightforward. If you haven't seen a question such as this one here before, it might look a little bit different and you might find one or two parts of this a little bit more challenging. But once you've seen kind of one example of this here, it's really not too bad, right? Let's just start then with part A. We can probably start here on the top right corner. I don't think we need tons of room.

[00:58:50]So for part A then we have C of 40°. So I'll just start here by writing down tan of 40°.

[00:58:57]So tan of 40° here.

[00:59:01]is equal to p. So for part a then we have cot of 40°.

[00:59:09]I'm hoping for part a then this is nice and easy right just one mark for this here. So you should just recognize then c of 40° that is the reciprocal of tan of 40°. Right? So this here is 1 / tan 4° and we know then because we're told it at the beginning of the question here tan of 40° as we denote it here as well is just simply P. Right? So for this one here then for part A is just simply 1 over P Bosch nice and easy right so that's part A done. Now for part B then what we've got here is sec of 40°. So sec of 40°.

[00:59:51]Now you might try things like 1 / cos of 40° so on and so on. But you probably won't find much luck with that. What you're kind of looking at here is basically is there an identity that you can use here that does involve tan to obtain some kind of expression here for sec. Right? In this case I want se of 40°. So what you should recognize here then is se 2 x = 1 + tan^ 2 x right we're going to use that identity here um to obtain then an expression in terms of p for sec of 40° right so um let's go a little bit further down here so as we just said then right if I do this as se x here um or in this case here se 2 of 40° so se 2 of 40° here that will be equal to then 1 + tan of 40° just like that, right? And we know here that tan of 40° is equal to p. So if I just square both sides here, I get tan square 40° that would just simply be p ^ 2, right? So se 2 of 40° that is equal to 1 + p ^2. But we don't want se square of 40°. What we do want here is sec of 40°. Right? So therefore then we just want sec of 40° here. Just take the square root of both sides. Right? And this here then gives us the square root here of 1 + p^ squ.

[01:01:21]Okay. And there we go. Then that gives us the solution to part B. And then finally here we have part C. And I'll just do that over here on the left hand side. Right. So for part C then we have tan of 85°. Okay. Now as you can see then for part C here that's definitely different because the argument here for part B and part A was 40°. That's the given angle. For part C then it's tan of 85°. So the argument is different here.

[01:01:50]So it might require something slightly different to part A and part B to give us some expression here for part C.

[01:01:56]Right? So what you do need to recognize here is that we can rewrite the argument then as a sum. Okay. So I can write this here then as tan bracket. So I can write this here then as 45° or 45 + 40 we'll take the angle of the full thing like that. Okay.

[01:02:17]I'm hoping that's kind of obvious because we do know that tan of 40° is equal to p. So what do you add to 40 to get 85? It's just 45, right? and tan of 45 is just one. So that's going to become quite useful here. Okay. So from here this is just the compound angle result then for tan right. Um so just refer to your formula but you don't need to memorize this result here as it's given then in your formula book here. This is tan of a + b. So tan of a + b. If you just refer to your formula book then it's given as tan a.

[01:02:54]So it's tan a plus tan b and that is all over then 1 minus tan a tan b just like that. Okay. So if we've got tan 45 + 40 let's just do that underneath here. So tan of 45 + 40.

[01:03:23]I'm just going to use this result here.

[01:03:24]Right. So, what I've got then is tan of 45° plus tan of 40°.

[01:03:30]So, tan of 45° plus tan of 40° like so. And this here then is all over 1 - tan a so that's tan of 45° tan of 45°* tan of 40° like so now tan of 45° here as we just mentioned before then is just one right so what I've got here is 1 plus tan of 40° which is p so I've got 1 plus p there as a numerator and now for the denominator here I've 1 minus m. So tan of 45° again is 1 * that by tan of 4° which is p. So 1 * p is just p, right?

[01:04:17]So 1 minus p there. Okay. So I've got 1 + p all over 1 minus p there for tan of 85°. So just finish with here quickly denote it again. So tan of 85° is equal to 1 + p all over 1 minus p. And there we go then. So again, if you've seen a question like this before, you might find this quite easy. If you haven't really seen a question in this layout before, then you might have found, you know, maybe part B and part C a little bit more kind of difficult potentially.

[01:04:52]But hopefully once you've seen an example such as this here, then it's not too bad, right? So there we go then.

[01:04:56]That gives us the solution here to question A. Now moving on to the next question here, question nine. As we can see here, right, we do have a question on interlash on by substitution. Now, as integration by substitution questions goes, this question here is pretty straightforward. I don't think there's really anything that's going to kind of catch us out here. Um, but nonetheless, it's still good practice, right?

[01:05:17]Especially as last minute revision. So, where do we begin here then? Well, let's just start with the given substitution here. So, u is equal to 2 to the power of x. Now, why am I starting with u= 2 to the x here? Well, at some point, right, we need to rewrite the integrant here, the thing that we're looking to integrate. And as part of that, we have dx. Okay, so I need something for dx here. So to start with, then let's just differentiate u here with respect to x.

[01:05:45]So I get du by dxm.

[01:05:50]So d by dx here. Now just be careful once you differentiate u here with respect to x. This is a special result.

[01:05:56]Um once you differentiate here so if you differentiate a to the x then we get a to the x learn a okay so in that case here I get 2 to the x learn 2 so 2 to x learn 2 like so okay now we want dx on its own so let's just start by flipping um both sides of the equation here we get dx over du is equal to and 1 all over 2 to the x learn 2. Okay. And then finally here if we just want dx, right? Let's just times both sides by du. So on the right hand side here, then what we get is du all over it would be 2 to the x here.

[01:06:46]But rather than writing this as 2 to the x here, I'm going to replace that with it substitution of u. And you'll see why in just a moment. I get du over then u learn 2 just like that right and we're nearly there now in terms of the foundations of this question here what I also need then is the new limits here so for the new limits here let's just do that underneath as well so for the new limits here where do they come from they come from the original lower limit and the original upper limit okay what we do then is just substitute these into the original um or not the original substitution but the given substitution I should say. So when x is zero here what do we get then? So when x is zero we get 2 to the 0 which is one that's a new lower limit here and when x is 1 we get 2 to the 1 which is two. So I think at this point here now we do have everything that we need then to evaluate this integral here. So let's just go from this point here.

[01:07:54]So we have then the integral here. We're going from one to two. So let's just start with the the limits here like so.

[01:08:02]So 2 to the x then that's just simply u.

[01:08:05]I've got u here all over them. So this is u + 1^ 2.

[01:08:12]So u + 1 squared like so. And we also have dx. Now dx here is du over u learn 2. So I've got du all over then u learn 2.

[01:08:30]You might be able to see at this point here then why I've expressed this as you learn two rather than 2 to the xarn 2 because this u here will cancel with that u there. Right? So what I can also do here as well is um write this as one over l two. Now the one over and two that's just a constant, right? So what I'm going to do here is say that outside the integral just makes life a little bit easier. You don't have to, but I just think it makes it much easier to work with. So what we've then got here is 1 over learn 2 outside the integral.

[01:09:04]We've now got the integral here from 1 to 2 of so it's 1 all over u + 1^ 2 like so integrating here with respect to u right now the integrant here you can write that in index notation so that's u + 1 to the minus2 and I think at that point there becomes trivial to integrate right so if you integrate corre correctly here I can't speak but if you integrate correctly here um what we get then is get 1 / learn 2 inside the square brackets here because we do need to account for the limits, right? I get then minus 1 / u + one like that going from 1 to 2. Okay, I'm hoping the limits here are nice and easy to deal with. Right, don't forget we start with the upper limit and we subtract the lower limit. So I get one over two outside one over two outside here. Now inside then what do we get here? Well when u is 2 I get -1 / 2 + 1. So that's - 1 over 3.

[01:10:16]So - 1 over 3 there. Like so we now subtract here the lower limit. So for the lower limit here that would be - 1 / 1 + 1. So that's - 1 / 2. So we do minus min - 1 /2 like so. So just be very very careful here with your signs.

[01:10:37]We've got 1 over then two times then. So - 1 over 3 + 1 /2. Um so - 1 3 + 1 over2 that gives us 1 over 6 there like so. And then finally we want the exact value here right of this. So don't give this a decimal. Make sure you give it in exact form. Just be careful for that. Right? Would be a silly place to drop the marks. Um or the final mark here. So to finish with then what I get here is 1 * 1 which is one and then 6 * learn 2 giving us six learn 2.

[01:11:14]And there we go. So this here then is our solution. Nicely done. Okay. So, as I said, right, all things considered for an integration by substitution question, it was pretty straightforward, but more than anything, we're just demonstrating the key concepts here of integration by substitution. Right, so there we go then. That gives us the solution here to question nine. And then finally to finish with here, right, we have the very last question for this video, question 10. So, we're at just over an hour now, right? And hopefully you find the previous nine questions to be quite useful. As I said at the beginning here, there's very little kind of modeling context to the questions here. It is more the basics behind um the topics that are featured. But either way, hopefully some last minute revision, you still find this useful. So for the final question here, then question 10, as we can see on the screen, then it's based on trigonometry, right? So part A and B is trig identities. And then part C here, just looking to solve this trig equation. So let's just start then with part A, right? Hopefully we should have enough room in the top right corner here. So it just says then for part a here using sin^ square theta plus cos square theta is identical to 1. Show then that cosec 2 theta minus co theta is identical to 1. Very standard stuff here. You should recognize what to do straight away for something such as this here.

[01:12:34]So I'll just start then by writing down the original identity. So sin^ square theta plus cos² theta is identical to 1. Now if we want to obtain something like this here, we need to divide then either by sin^ square theta or cos^ 2 theta. Okay? Now if you're not too sure, just take a guess and divide at random and you'll quickly figure out then whether you've got the right one because you'll either get this here or you won't. Um if not just take a look at what you need here. So what I need then is 1 / sin^ square theta cosec theta right minus then cos^ 2 theta is identical to one. So the one here comes from sin^ square theta / sin^ square theta. The cose theta here is 1 / sin^ square theta. And then for the minus co theta here that is from cos square theta / sin square theta. So you can kind of just figure it out by inspection. But if not, just do it at random if you really can't figure it out on the spot. Right?

[01:13:37]So either way, um, as we just said, right, we're just dividing by sin square theta here on both sides of the equation.

[01:13:46]Divide both sides by sin^ square theta.

[01:13:49]What do we get then? So as we just said then sin square theta over sin square theta is 1 cos square theta over sin square theta that is cos square theta plus cos² theta and then 1 / sin^ square theta is cosec theta that is identical to cosec^ squ theta perfect now what I want here is cosec theta c square theta is identical equal to one. So clearly at this point here all we now need to do then is just subtract co square theta from both sides. Right? And in that case then what we get here is cose 2 theta minus then cos^ square theta is identical to 1 as required. Perfect.

[01:14:42]Okay. So as required there.

[01:14:44]Happy days.

[01:14:47]Good stuff. So that's the solution to part A. Part B um is another trig identity part of the question here. So let's just work through that as well. So for part B then it says hence or otherwise prove that cosec 4 theta minus c of 4 theta is identical to cosec 2 theta plus c^ 2 theta. Right? So how do we answer or how do we prove this result here.

[01:15:12]So what you hopefully recognize straight away here then is the left hand side of this trig identity. That's actually the difference of two squares here. It might not be immediately obvious, but it is something you do need to be kind of recognizing with these types of questions here. This isn't uncommon per se. So again, keep an eye out for this.

[01:15:31]So what I can do here then is, you know, as we just said, right, and express this as the difference of two squares. So what I've got here um is I've got I'll do it in brackets. Sorry.

[01:15:43]So I've got cosec square theta.

[01:15:46]So cosec^ squ theta minus then cos² theta like so. And then in the second bracket we've got cosec theta cosec theta plus c² theta. Right?

[01:16:03]C^ 2 theta. As we said then it's just simply the difference of two squares.

[01:16:08]And this here then is um going to be useful because as we obtained in part a then I've got cosec theta minus co theta is identical to 1. Well I've got this bracket here which is cosec theta minus co theta exactly as we obtained here in part a. Right? So this bracket here is just simply one. So I've got 1* then this bracket here. So that's cose 2 theta plus co square theta like so which is just simply then cosec theta cose 2 theta plus c square theta like so. So so as required there. Bosch.

[01:16:57]Good stuff. So that's part B complete.

[01:16:59]Happy days.

[01:17:01]And then finally to finish with here we have part C. Right. So for part C then let's just finish here underneath. So it says solve 4 ft between 98 and 108° here. Cosec of 4 theta minus C^ 4 theta is equal to 2 - C theta. Now this question here and as I mentioned at the beginning of this video right all of these questions here are taken from the old Alevel math at XL specification and as you can kind of see right the mark allocation here is a little bit criminal six marks for part C here is crazy work because as a trig equation it's really not too bad right and we've kind of done all of the preliminary work in part A and part B so it's really not six marks worth of work it would be nice if you got that on your exam but Yeah, you definitely won't get six mark of something like this on your exam. So unlucky. Um, but let's just kind of work through it anyway. Let's just see what we're doing here. So for part C then let me start by writing this down here.

[01:18:03]Cosec or theta like thatus c 4 theta is equal to so it's 2 - c theta. Now the left hand side here is what we had in part B is the trig identity and that is just simply cos^ 2 theta co^ square theta right so we can just replace that there so that is cos x squ theta cos x 2 theta plus square theta and that is equal to 2us theta.

[01:18:44]Now where do we go from here? Well, what we had then in part a was this identity, right? cosec theta minus co square theta is identical to 1. So what I can do here is replace the cosec theta here with 1 plus c square theta. Right? So what we've got here then is cosec^ 2 theta as identical to 1. plus c^ square theta.

[01:19:15]Okay. And I can just substitute that then into this here.

[01:19:20]So if we do that then what do we get here? So I get 1 + c square theta another square theta there like so is equal to this 2 minus co theta. Perfect. And hope you can kind of see what's going to happen here. Then we're going to form a quadratic equation in terms of co theta. Right? So let's just take everything over to one side here. If I just simplify the left hand side first, what I get then is 2 c^ 2 theta + 1 is equal to 2 minus c theta like so. Okay. So let's subtract two from both sides and add c theta to both sides here. It'll ensure then that the right hand side is equal to zero.

[01:20:09]So on the left hand side here then I get two cos^ square theta um plus c theta so plus c theta there - one is equal to zero and at this point here then everything looks good right I can't spill the mistake so that's always a positive so let's just continue from here we'll go back towards the top then so we now want to solve this quadratic equation here for c theta and the hope here then is that we just factoriize this right and the good news is we can indeed just factoriize. So we get double brackets here like so this is equal to zero. So in the first bracket here then what we get is 2 c theta.

[01:20:54]So 2 c theta minus one like so and then in the second bracket here we get cot theta.

[01:21:01]So cot theta + one. Okay. And from here then we get two solutions. So for the first bracket here then we get two cot theta minus one is equal to zero. Just solve for copia here and we get cot theta is equal to a half right 1 / two. Okay I'll come back to this in just a moment.

[01:21:22]Let's just go to this bracket here for now. And then for this bracket here again um just imagine the other bracket is not here. You've just got c theta + 1 is equal to zero. In that case here coffee the is equal to minus1. Now kia on its own here isn't particularly useful but this is the reciprocal then of tania right.

[01:21:45]So we can write this here as tania. So therefore then tania here is equal to two right the reciprocal here of the right hand side as well. And then we can do the same here with this solution. So cot theta is equal to minus one. But in that case then tan theta is also just minus one. Okay. So tan theta here.

[01:22:09]So tan theta is equal to minus one.

[01:22:13]Perfect. Okay. Now at this point here then just grab your calculator. Right now for this question here we're working in degrees not radians. So just ensure then the calculator is in degrees not radians. Um that would be a silly mistake to make. And um obviously at this point here right just do at tan for both of these solutions here. So for this one then vo is equal to tan of 2 which gives us 63.4° there. Now there's a problem here because this here is the domain for theta. So anything below 90 anything greater than 180 here is a problem. So the 63.4 is a problem. Now for tan don't forget the next solution here would just be um this angle. So 63.4 plus 180. So that would clearly fall outside of this domain here. So we would reject that solution as well. So b being equal to 63.4 is no use to us. So I'll keep it there but can like lightly cross over it like so just to indicate then that we won't use this as a solution. Do the same here then. So arc tan of minus one here. And what we get then is theta is equal to so our tan of minus1 that gives us - 45° so - 45° okay now again on its own here this - 45° isn't useful because it's less than 90° right so we would reject that solution however again the next solution here would be this angle plus 180° right so - 45 + 180 that gives us 135° °.

[01:23:50]So we get 135° there. The next solution here would be this angle. So 135° plus 180. So clearly that would now fall outside of this given domain here. So we're going to reject that as well.

[01:24:02]Okay.

[01:24:03]So we reject this one here because it's less than 90. So it's outside the given interval. Same for this solution here.

[01:24:10]And what that means then is we just have one single solution here for this equation. Right? So therefore then theta here is equal to 135°.

[01:24:21]So theta is equal to 135° as um as we kind of highlighted here is between 90° and 180° as the is between 90 and 180° there.

[01:24:38]Okay. Like so. And there we go then. So as you can see sometimes it might involve one or two parts on a trig identity. It is pretty standard to be kind of laid out laid out in that way and then the final part is usually just solving an equation based on the previous trig identities as we can see then with this question here. But there we go then. So I'm hoping you know this video here has been helpful in some last minute revision here for paper one.

[01:25:04]There was just the 10 questions for this video here. We're going to produce something similar for paper two very very shortly once paper one's been sat and we've kind of got an idea as to what topics will come up in paper two. Right?

[01:25:15]Anything that kind of didn't appear in paper one that I've already featured here. I'll just kind of pull those questions over into the same video. Um so there might be a little bit of overlap between the two videos, but I'll make a note of that again in the beginning of the next video here. So last minute revision for paper two, right? Um and that would definitely be a little bit more detailed than this video here again because we know what topics are going to pretty much come up um based on paper one. So yeah um anyway that's enough of the yapping pretty much. Um as always if you are looking for any last minute revision here for Alevel maths including 10 and it'll be 11 predicted papers very very shortly.

[01:25:49]Then head over to our website ajmass.co.uk. We have plenty of extra revision there for your Alevel math needs. um including worksheets, additional revision videos, challenge worksheets, all that good stuff, right?

[01:26:01]So, we're here to help you boost your grade um and get the grades that you do deserve. Okay, but other than that, that's pretty much everything I wanted to say here. As always, best of luck with paper one. Um I'm hoping, fingers crossed, you'll get a good paper, right?

[01:26:15]But anyway, that's enough of the app.

[01:26:17]Best of luck and thank you for watching.