Even My A/A Students Make These A-Level Maths Mistakes

If you're thinking that A level maths is really really easy and you've covered chapters one to four and maybe seven as well then you want to watch this video because I'm going to show you the really common exam question mistakes that students make and I'm basing this off of first of all I gave this test to around 35 year 12s in school and I also gave it to my tuition students as well and here's what the common mistakes now remember some of these students I know for a fact are going to go on to get A stars but they all made the same errors in the questions. So this is much more about exam technique and deeper understanding than it is about just knowing the overall method. So let's just jump straight into it so you don't get too bored. So the first one is not spotting factor theorems. What do I mean by that? Well, here's the question. So just very quickly, okay, you have these two equations. This is verify that when x equals one, the curves intersect. Perfect. That was fairly easy. Sub in x equals one, you get the same y-coordinate. But this was the problem here, this five marker here. The curves also intersect Except when x equals k, given that k is less than zero, use algebra. Now why do I say not spotting factor theorem? Well if you notice this x equals 1 is a solution to an equation. Now the question is which equation is it? Well it's when these two curves intersect, it's when these two curves are equal to each other. So when they say here the curves also intersect what they want you to do is to equate them right so you have x squared minus 12x plus 14 equals and then you know minus x cubed plus 6x squared plus 10x sorry minus 10x And then plus 8. And you move everything to one side, let's move it all over to the left, so we get x cubed, and then we get minus 5x squared, and then you get minus 2x, and then you get plus 6 equals 0. And x equals k is the solution to this. But the problem is, how do you actually work it out? Well, here's the thing, this here is a solution to that expression, which means I can divide it by x minus 1. And that's the part that a lot of students didn't spot. And so they spent a lot of time trying to figure out another way to do it. And to be fair, some of them actually managed it. So some of them actually managed to guess the other solutions to this, which is fairly difficult because k is actually a third. It's something like 2 minus root 5 or whatever. But again, not spotting the factor theorem is a really, really common mistake. So they're not going to make it obvious when you have to use factor theorem all of the time. That's part of getting an A star. It's knowing what the math is telling you. The next is a discriminant mistake, and bear with me for a second, but here was the question. So prove that x minus 4 is a factor here, hence using algebra show that the equation only has two distinct roots. And what a lot of students did is they just subbed in for the discriminant and got bigger than zero. The problem is that doesn't show it has two roots, but let me show you. So first of all, x minus 4 is a factor, so I'm just going to assume it's a factor. So what we're going to do is we're going to very quickly, we're going to do this division, and what I'll do is I'll come back once I've finished it. Okay so if you look here some students did the discriminant incorrectly and got that this quadratic has two solutions is bigger than zero and they said cool that means it has two solutions but here's the problem is if this quadratic has two solutions then this is also a solution so overall this equation will have three solutions which means you haven't actually shown that there's only two solutions because you've actually shown there's three. So say it one more time. This is one solution. If this quadratic has two more solutions, that's three solutions in total. So personally, what I'll probably do is maybe factorise it yourself. The reason, and this is so, so tricky, right, because the discriminant does give you two solutions. The problem is one of the solutions is x minus 4. So there's only two distinct ones overall. So, for example, if we factorise this, we get, so we're timesing to negative 24. We're adding to minus 5, so it'd be, you know, 2x, you get the idea, so it'd be x minus 4, and then you get 2x plus 3. So, if you notice, and this is what they intended you to do, this is why there is only two solutions, because one is 4 and one is minus 3 over 2. If I use the discriminant on this, so b squared minus 4ac, minus 5 squared, minus 4a, and then because this is, you know, all negative, I can clearly see it's bigger than 0. That is not showing there's only two solutions. That's showing the quadratic has two and then x minus four is another one. So there could be three solutions. So that was the big mistake that they made there. So you have to be really careful when using the discriminant. It doesn't just give you the number of solutions. It gives you the number of solutions for a quadratic, not a cubic. So you need to check that third solution. So in this case, that's how you show there's two solutions, not by using the discriminant. Cool. Transformation sadness. So this was the original expression here and part C of this question gave you this. So they drew this graph over here. By the way, you can't just use the graph to say how many distinct solutions there are. You have to show it. So I'm just going to write that again one more time. And OK, this is a big problem. Firstly, the number of real solutions to this This equation is not two. So they give you this graph. It crosses twice. But that's this graph, not this one. So you need to actually have a look and think, okay, what's the difference here? Well, if I call this f of x, then what have they done here? Well, the only difference, the 2x cubed is the same, 13x squared is the same, the 8x is the same, but 48 becomes 46. So what they've done is they've done f of x minus 2. They've subtracted 2 from the expression. What does that do? Well, it moves the whole thing down by 2. So if I were to draw this, it would kind of go like this, roughly like that. So now how many solutions are there? Well now there's one, two, three. There's three solutions because the graph has been moved down twice it's going to intersect three separate times and that was the really again another problem that students had. It's not good. Remember they're not going to ask you the same question twice. When they say sure it has two solutions they're not going to say state why it has two solutions because you've already done that. So they're never going to ask you the same thing twice. I'm also going to say part b as well. There's actually another question in the test that used the same technique. I haven't put it in here because it's the same thing. It says here the curve has equation f of x plus k, so that means it's minus in k from x, passes through the origin, find the two values. So once more, this is not the graph we're talking about, it's this one here. So the first thing you're supposed to do is work out these two intercepts which are going to be Positive 4 and minus 3 over 2. And that's positive 4. So, if we think about it, we know that it passes through the origin, which means one of those two roots is now 0, 0. So, either minus 3 over 2 has become 0, or 4 has become 0. And you just think to yourself, well, how do you do that? Well, for this one, you add 3 over 2. But remember, we're minusing k. So, we have to do it as a minus, which means k could be negative 3 over 2. Because remember, x minus 3 over 2 would add 3 over 2, because it does the opposite. Or you minus four. Which means k could equal four. There you go. Inequality. This is probably the funniest one because part A actually links to part B. So part A, when you sketch this out, and please do forgive me for this slightly scuffed drawing, you get something along those lines. It says hence. If you see the word hence in a maths exam from this point onwards, it always means use the previous part. So they're telling you to use the graph and they've even given you the graph because that's the answer to part A. They say solve 16 over x is less than equal to 2. Here's what 99% of students do. Cool. That's your answer. There you are. That only gives you two of the three marks because there's actually another value that doesn't work. In fact, it's more than one value. So here's how I do inequalities. I do them graphically. I'm going to show you how to do this. Well, I mean, I mean, basically, you can't do it algebraically. Well, can't being a strong word. So here, it says 16 over x is less than 2. So if I draw the line of y equals to 2, what we've worked out is we've worked out that that value there is 8, and everything bigger than 8 is below 2, which is true. But isn't it also true that everything over here is below 2? So isn't it also true that x is less than 0? And it's not equal to 0, because remember, we can't have x equals 0 in this graph, because if I do 16 divided by 0, it's undefined. There we go. So that's the full solution for the third mark. Now the other way to do this is by just looking at the actual thing here. So 16 over x is less than 2. Well okay 16 over 8 is 2. So 16 over 9 is less than 2. 16 over 10 is less than 2. Okay what else? Well what if it was a negative number? If I sub in 16 over minus 1 that's also less than 2. So both of those ways you can use to do it, but again use the graph especially because of this. Because there's a discontinuity here, it's not a continuous graph, there are more than one solution as compared to that. So that's another place where some students would again make some common mistakes. Yeah, modelling sucks in general, but this question in particular was fairly tricky. These first two parts were really, really easy. So for this one, so it says here they started mining in 2019, so 2020 you sub in n equals one. Yeah, that makes sense. Deduce the maximum total mass of tin that could be mined. Okay, well this is incomplete in the squared form, so it's just 1,200 and that happens when n equals 20. But calculate the mass of tin mined in 2023. Now, Now, unfortunately, this part of maths involves a lot of reading and comprehension as well. And this is going to be true in some of your entry exams and in all of A-level maths and in A-level applied maths as well. So it says the amount that's mined in 2023. So students just went, OK, well, N equals four. I'm going to sub that in. I get T equals twelve hundred minus three. 4 minus 20 all squared to give me, well let's quickly have a look. Now there is a problem with this, which is, haven't they already asked me to sub in a value into the equation? You should bear in mind, and by the way this is their answer, and they said yeah that's cool, and I think if I remember rightly this part was three marks, and they're like yeah that's worth three marks, of course right? The problem is they've already asked you to just sub in a value. They will never ask I've asked you to do the same thing twice. You don't repeat yourself. So what does this mean? Well look what this says, up to, because t is the total mass in the n years after the start of mining. So what I've worked out here is I've worked out all of the mass mined in 2019, 2020, 2021, 2022 and 2023 added up, but I only want 2023. So what do I need to do? Well I need to sub in n equals three to get rid of all of these values over here. So up until that point, so you sub in n equals 3 and then you subtract the two answers because again n equals 4 gives me everything up to 2023. I want to get rid of everything up to 2022. I do that by subbing in n equals 3 and then subtracting the answers. The other problem here was the limitation on the values of n. Now a lot of people said quite nicely that n has to be bigger than 0 because you can't have negative time, perfectly fine, but they neglected to draw the graph. And they neglect also to think about what the variables mean. And that's a big thing in modelling is you have to know what the modelling actually means. There's another question after this on modelling where it had the profit of a company, right? And its profit equals, I'm just going to say 2x. I can't remember the exact question, right? This P, though, was in thousands of pounds. So one of the questions said, find the value of X. Well, it says, I think X was the selling point of the toy or whatever. So find the value of X when the profit is 80,000. And they subbed in P equals 80,000, not 80. But P is in thousands of pounds. You're meant to sub in 80. So here, as you can tell, it's a negative quadratic. So it's going to come at some weird arc like this, where this point here is the turning point, 20, 1200. This here is the total mass mined up to and including that point. So as you can see, it's levelling off. That means in that year, they didn't actually mine any tin. Now it starts to go down. The only way for that to be true is if they mine a negative amount of tin in that year, which is impossible. So n was supposed to be less than equal to 20 as well. What students did instead is they worked out the x-intercept, thinking that t can't be less than 0. You're right, t can't be less than 0, but it also can't ever go down. It can't decrease, right? Because think about it. Think about it in terms of number of steps. If I added up the number of steps you've ever taken in your life, that number can never decrease because you can't take a negative number of steps. The same logic here. You can't mine a negative amount of tin. Anyways, those are some really common chapter one to four, and I guess chapter seven mistakes that I see all of the time. Hopefully you found that useful. I've linked in the description a set of really hard chapter one to four questions that I made myself, just to give yourself a test as well. But anyways, see you in the next one.