Hi Yeah how you doing an today? Excellent. So what I wanted to do today is kind of tie up everything we've looked at so far, because I really want to focus on you having a clear idea of how all the ideas are separated and when you need to use what, and also on having a really deep understanding rather than just glossing over lots of content. So I'm going to spwith a bit of a roundup of differentiation work we did last time. And then what I thought I might do is pick out some questions from a tmua paper. So tmua is an admissions test used by the very top universities for maths related courses. So it's not quite as hard as the tests that the top universities use for matths admissions. But what it does do is it tests. Do you understand the aim of our ideas deeper than the average a level students? Can you think for yourself a bit more and not just do things by roso? I'm going to pick into a few of those. But for starters, I'm going to send you a worksheet. And I'm going to pick up a few questions on the differentiation stuff we did last time. So if you have a look at that in the chat. And look, start. Yeah, let's start with question one conventional place to start. There's the permission recovery through see how far you can get. So tying together a lot of the differentiation work we've done in the last two weeks. Sorry, question two, question one, start at the beginning this time. You know what we need we haven't looked at have for that question too. Sorry, miss tutwo. Yeah sorry, let's do question two. So sorry, let's do a question to we actually need another rule that we haven't covered yet. I haven't looked at chain rule with you because I think it's actually a lot conceptually tricky than the rest of the subso. There's some other things I'd like to do first. I don't know if you know what chain rule is, but there's five questions too. And my usual policy applies that if you can tell me there's some formula you need, I'm happy to call. I remember. Oh, good. I just Yeah, I just mean that I think at this point, it's way more important to be aware what the formulas are than to have precisely memorized. So nearly memorized. It's a traction on the top, but good stuff. We've very nearly memorized that. Just pick that step we've done there, yebut, that's quite right. I think you can correct yourself then, really. Nice. This one with the gradient b for this. So you found the derivative spot on. The question is to find the equation for the tangent at the point with x coordinate minus one. So how are we going to use the already? I could just plug it in there. Absolutely. But where are we going to plug it in? What are we plug ging in super size? Excellent. So what about part b? Okay. Where I need that equation back. Oh yes, well, we can just just the little simplications how to scribble that down. It was just three over. We didn't really need to. And the brackets stantly. Very good. Yeah, nice. Excellent. So then what about part scene? I'm. Chnext questions, have a look at turning. We should be there. Yeah having it four a. Can you scroll down the witness? Oh, maybe we need sorry, scratch that, winneed chain roll for that. Keep forgetting we haven't done chain one. Instead, we need to me, teach to me, we could do. I think it's one. Is it going to take too much time? It's, I think the trickiest bit of a level differentiation and there are easier bits that I'd like to teach you first. I think that it doesn't make a very sensible order to do that now when you haven't done any trick. So I'm going to hold fire on that one if that's okay. Yeah, I think most of the more interesting questions here, one chain rule. So let's leave that there. A little bit of practice, the differentiation stuff. And instead, let's have a look at this tmua paper. I will send to you. So I'm just going to pick out the questions that we've covered the material for. So let's get that link. Keep bear in mind that these questions often involve a little bit more sort of problem solving, which tests whether you've really deeply understood the ideas. Just to say that straight ightly. So the first one that we're in a position to have a go at is question two. Question to yeon, that paper I've sent a chat, all of them or just so, Oh, if I not sent you a spit's multiple choice. It's multiple choice. Yeah, okay, okay. I thought it was like five, six of them. There's multiple choice. And just to say, this is bringing together all of the work we've done of stuff beyond gcse. So this could cover any topic that we've looked at together. So how we're going to do with that unknown, that's where our standard longer vision method slightly leaves us. How are we going to deal with that? A im. There it is. Not to fit. So. I twelve exactly. Can you explain why? Because it says it's a factor. So this must be able to be divided. So it must be equal to twelve, twelve or to think about the sign. But you got you got six. Sorry, six. No, no, no. Look like, Oh, you've got rid of it. Now mine is twelve is minus twelve. Just the sigthing. I think you got the just there. Excellent. Plus x plus D X so it's F E Yeah. I think you wrote it down, right? Yeah, Yeah, excellent. Now that was a perfectly good method just to check whether you remember the idea. If the question was just to find out the value of a, how would you have done that? So if you if the question wasn't to factorize it completely e, it was just to find out the value of a, what would the quickest way of doing that be comparing? But it needs to be an equation though comparing, did you say not like I meant it should be an equation though if I were to find out the value of a not quite sure I see what mean. There's a technique we've looked at that if the question was just definfind A, I think would be by long way the quickest way of doing that. Any ideas what I might be getting at? Yeah the one where you look at the coefficients, not quite because you're quite right, but we don't really have coefficients to compare. We don't have an equation. There's another technique that I want to check. You remember that we've looked at. I forgot the name with this one equals something something those along those lines. Yeah the factor theories. Remember how that goes? Something Yeah Yeah that okay, could you show me how the factor theorem would apply here? You're kind of quoting bits of it. Yeah. Can you string that all together into a full statement of the theorem f of minus a negative a constant equals zero than that negative constant plus x would be a factor of f of x almost. You've written it down correctly. So if f of minus c is zero, then x plus c is a factor of f. Yeah, nice. So you could also have used that here potentially, but your method was perfectly good. Just wanted to check you remember the fact theorem? Nice. Next up, let's have a look at question three. Negative four, right? That's the three. Nice. Click that eight, you've drothat your calculation that's for contract check where you've gone wrong with that next? Like a bit of a slip to me. Is f of one is just two right? Yeah absolutely. Yeah. So so when you sub that in. Ten two times worse. Yeah, not quite. Just don't don't rush. Good. Back to basis. So two equals a quarter plus c. You're quite that scrub the south because that's going we're already just back to basics two equals a quarter plus c. Don't don't overcomplicate. Yeah, and we're just doing two minus a quarter, right? 你想去吧。Yes. Is this formula right? This yes. So this is very nearly right, this one. But just tell me what you mean by this notation. Yeah so we're doing so what I would love to have seen was to write the individual points. So we're doing the distance between these two points. So you've Yeah you've worked out the the vector between them and we're looking for the magnitude. Another way of looking at it is the distance between two points is the difference between the x is squared and the difference between the y two y squared. All in all, basically correct. Apart from one missing set of brackets. Can you spot where I want brackets? Well, tell me, tell me where this what you've written down. Tell me where is it coming from? Do you have an intuition behind it? Or is it something which part, this part or this? No, thtell me where they're coming from. I want all these symbols to really make sense to I want to mean something. It's the vector. So it's a so you've worked out the vector and now you're looking for the magnitude. Okay. Could you quote for me the general formula for the magnitude the magnitude of X, Y to what. The thing you said, the. How do you say? So I'm looking for the formula for the magnitude of a vector. That's where you told me you got the idea from. So I want to check your understanding there. What's the magnitude of the vector xy? This is from the oriyeah. And what's the formula? You've used the formula, but not quite right. So I want to check your understanding of the formula. 嗯。Since it just that's the axis and Yeah, so what's the so here, you've used a formula but you've not used it. Quite right. So I want to check where the misunderstanding is. So when you quote the formula you've used, it's just plus b. Yeah if we've got if we want the magnitude of the vector ab, no, no, no. If we want the magnitude of the vector ab, then you're quite right. It's the square root of a squared plus b squared. So you understand the formula, right? Can you see what's wrong with how you've applied it? It's subtle, but it will mean you get a totally wrong answer. Actually, you won't even get an answer with this mistake, even though it's a subtle one. Yeah can you see that without the brackets you're going to get 13 quarters squared. So bit under four -49 which is a negative number. So you just get an error. So I know it's this way you get negative seven squared. It does 49. This does not equal 49. No This equals -49 this equals 49. Without the bracket of seven times negative seven equals 49. Yeah but we need the brackets. Might this notation here does not mean minus seven times minus seven because you attach the square just to the seven and not to the negative. I think brackets might be a bit of a thing to focus on actually, and it might seem like a fiddly thing, but it makes a world of difference. If we don't put the brackets there, then we're not squaring the minus sign, just the seven. Does that make sense? Are you happy with that because it that button, don? No, because I just learned I just learned it like that. So I just okay. Well, that's really important that we correct that because that's wrong and we'll lead to issues and we'll lead to errors. Yeah. And the final piece of the puzzle is I would love you to evaluate that without a calculator. See how neat form you can get it in without a calculator. Because this is actually an non calculated paper. Yeah, I think something was wrong, the little thing. That final step. So this is correct. You pointed out, so when you square root it, we want to square root each part. So this is the same as roseventeen roforty nine as as so that 17 gets rooted. Excellent. All makes sense. Any questions on any of that? No. Okay. Moving on to question four and to give you a kind of heads up. What we're looking at here is one, I want to see that you understand sigma notation, which we've had a look at, but also this is a bit of a problem solving E1. So there's a little bit of thinking outside the box. So have a look at question four, see how I get. So it goes on like minus and then plus and then minus and then plus. What exactly you mean by that? What you know is, I'm sorry, it's just okay, not. Could you give me the rule? There's a rule of some summation. We don't need one here. We don't need so the two types of formulas we looked at is we looked at some formula for arithmetic series and we looked at some formula for geometric. This is actually neither of the above. No, this is neither arithmetic nor geometric. But we still cancel that. And my hint would be that this is a pattern spotting situation. So why don't you try evaluating the first few terms and then we'll see if we can spot the pattern. Absolutely. Always jumps from negative three to view and then back looks like it. That seems to be the pattern. So my next challenge view is, can you prove that? Can you prove that for all the odd ones equal to three and for the even ones, it's my other answer. Could you prove that? So do I prove by contradiction? I wouldn't say so. I wouldn't think that was a very, because what would the contradiction look like? What makes you think contradiction? But did a three equals three, then would be contradiction. So I would not find the value of inso. That's not how proved by contradiction work. We have. We've covered proof by contradiction, haven't we? So proof by contradiction is if we're trying to prove a statement, the way we do that is we assume it's not and then work towards some contradiction. Now I wouldn't say that will be a good choice here because if this is our statement that we're trying to prove, the negation of that is just that there exists is quite complicated. It's not a natural starting point. It would be that there is some odd number such that is not equal to minus three, and there's some even number such that it's not equal to three. And that's quite a strange starting point. I can't really see a wave following that along. So I would say contradiction wouldn't be very useful here. I would just proceed directly. Say that if I have any odd number, then a subscript, that number is minus three, I would proceed directly. See? Yep, let's see. Two c minus one Yeah p even better absolutely or plus one Yeah that represents any old odd number. This has to be have to be there. And yet we want it to be bigger than one. We're already dealing with numbers bigger than one. Absolutely. Now can we evaluate that? So this will be always negative one because it's thought let's and this will be. Because this is always even this is also a negative one. And because. This is also audits, also naked one. Excellent. How can you do the same for the even scenario? And this one's always going to be one. This is also going to be and one. Excellent. So we've proved that pattern that we suspect it was going on. So we know it's just flipping back -33, -33 and so on. So using that pattern, can you evaluate the sum. Three, did you say? Yeah, pull me through. Why? Because when the first 38 cancel out, Yeah, so negatively, absolutely. We can greet them in two and then have a minus three left over. Excellent. Nicely done. What's. Can ever go at five if you want. We haven't covered integration, but this is quite basic stuff, which you might have already seen. How much do you know about integration? You know, both definite and indefinite. Yeyou know what they are? You know, houses. Okay. Do you want to have a good five? I don't mind. We haven't looked at it a yet. If youlike, okay, let's have a go then. Could you scroll a bit down then? So there's a couple of tleties, Gary. Wait, sorry, no, no, no. This is integration. Sorry, no, no. I got. So. Just miss. So that's the correct integral absolutely Yeah. Yeah so this is absolutely what the average a level student would do. And it shows me that you get the basics to understand why this isn't quite right. And this is a sneaky tmua question. And it be a little harsh for a level. Can you draw the curve for me? Can you draw y equals x squared minus one and the lines -22? Can you show me what that looks like? Excellent. X equals to one and zero. So. So one a negative one. Yeah so we could think about that in terms of the root. Another way we could think about it is we take y equals x squared and we just move it down one. It's a graph transformation. Yeah. Yeah. Okay. So a couple of ways to think it. But either way, we got one minus one. So let's pop two minus two there. So what we're after is these areas then so this one, this one and this one. So what? Okay, so these probably there. Okay. So I thought it was the whole thing. Well, can you explain why? What's slightly wrong with what you did first? It's a subtle point that most a level students would miss. Can you spot what we need to tweak? Why is it not just equal to the integral from minus two to two? It's quite subtle. 嗯。So you did this, which isn't quite right. You tell me why? Because it would cancel out sort of tell me exactly what you mean by that. I mean not not wouldn't, but it would subtract this area pretty much. So this integral here is going to be negative. If we wanted to just know about the area negative gvaridon't exist, we'll flip it and take the positive one. So if we just integrate the whole arrange minus two up to two, we'll be subtracting off the area. Absolutely. So the takeaway from that is when we're doing areas between curves and the axes, before we just plow in with an integral, we need to know what it looks like to know whether we've got any negative areas. So now that you've seen that, how would you tweak your method? What would you do now? So I would find I would still use this and plus two times that. Yeah, that would be one way of doing it. Can you think of a more direct way. I would find that though. Could do. Can you think the more direct way or take less time. Flip the graph. No, that wouldn't. I don't know, can you spot a symmetry in this graph? Yeah. Symmetrical only divix equals zero. Yeah so it's reflective about the y axis. So what we basically need to do is we need to consider these three plus one and then take the absolute value. Not quite. So on the basic level, what we need to do is calculate these three areas separately. So we need to do minus two up to minus one. We need to do minus one up to one and remember to flip the sign. And we need to do one up to two. On the basic level, that's what we need to do. We can save a little bit of time by spotting that in a new color. This area is the same as this area. So we could just double this one to save a little bit of time. But that's in general what we need to do when we realize that part of our graph is under the axis. We have to do the areas separately, the ones that we know positive and the ones that we are negative, and then flip the sign of the negative ones makes sense. So do you want to have a go at that? You scroll a bit down. Thank. What two, I think about that. Yes, indeed. Just a little poll at the last hurdle or semetic, the very penultimate thing you've written is right. That was a minus two thirds. Yeah, just double check your rhymetip there. Nice. Excellent. Nicely done. So I know we haven't really done integration, but Yeah, an important lesson there about the signs. Excellent stuff. Nice. So we'll probably run out of time, but I think we should make a start on question six and then that could maybe be quite a nice bit of homework. What nice. Quite an interesting bit of homework, if you like. Question six was really great because it doesn't involve any a level MaaS. It's really hard. So it really gets you thinking without needing advanced knowledge. So let's make a start and then maybe you could try and finish it off the homeway. This what is 30% of twelve again, four or zero point 4:12 percent of twelve. So no. So then it be 36 over ten 23.6, so 33.3233% of perbefore. Is it possible to achieve that result? As in the question general, Yeah, it's possible. Yeah, Yeah. So it's not f, no, it is possible. So tell me about what you've written down so far. So from the proportion of red paint in these. So that it would be 4.6 plus x. 20. So this video was fully very present, which are not quite actually maybe we could do it that way, possibly. But I think it's important we know exactly what we're labeling. So 3.6 and one. So that's for each of our for each twelve, five, three we have for each kind of unit like that, we have 3.6 red from the P1 read from q and how much red from R are. We calling that x. All of so let's get clear on exactly what x stands for. Can you define x for me? The amount, the percentage of volume of red painint and R? Yeah, I think that's a good way to define it. So this is percentage red in R. So can we do a similar thing to what we did here and here? If we had 30% red and p, that forms twelve parts, then we did 30% of twelve together, 3.6 similarly with q. So could we do the same thing for R then how much red paint is there in those three units? This, let's do one step at a time. Let's do it one step at a time. So if x is the percentage red in R and how much? Not quite three. Think about the fact that x is the percentage. Yeah. So it's a three times x quite because if it's say 30%, then we don't want 30x, do we? What do we want slightly different. 0.03 times. It's three times I X over 100. Absolutely. And then tell me about the rest of this equation. Tell me about why we're doing it over 20, for example. To get the percentage of volume that is red for the whole mixture. Yeah, absolutely. Because we're doing it for that's just one, twelve, 53. Absolutely. Yeah. And then 0.25 is then we're told it three, two. That's looking good. Why we're out of time. Why don't you follow that through, see if you can get to an answer. But that's looking pretty good, I reckon. So just take a snap of the screen copthat down quickly, I might as well tell you. So you know, if you've got it right, it should be c so see if you can get to c just for row. A little bit of homework, but good stuff today. Lots of tricky questions. I will see you. Actually, I think we haven't got the next one in the diary, but I think I'm doing that with the agency as we speak. So great. So try and finish off that question and I'll see you very soon. Bye. Thank you, miss. Bye.