Mathematics Practice with Mr. Barrow

Hello and welcome to lesson 51 of Additional Maths with Mr Barrow. In this lesson I'm going to go through questions 1 to 10 of the sample paper on OCR's website. Okay so the link to that paper is here so it's on their website it's in the assessment area and it might be called specimen paper or sample paper. On the paper it says sample question paper. Make sure you have tried that paper and those questions before watching this. It's important for you to have had a go and have your own thoughts about each question before running through the solutions with me, okay? So let's have a go at question one. A sequence is defined by the rule un plus one is equal to two lots of un minus one. So this is what's called a recurrence relation. So this formula allows us to get the next term in a sequence, okay, based on the previous term. The next term is the n plus 1th term. And to get it, we do two lots of the nth term, so double the nth term, and then take away 1. So to get the next term in the sequence, we double the previous term, and then take away 1. In this question, we want to find the sixth term in the sequence. sequence, all right, in order to get the sixth term, we would need the fifth term, because we need the previous term to get the next term. We are given the third term in the sequence, which is 12. Well, with the third term, we can get the fourth term. Then with the fourth term, we can get the fifth term, and with the fifth term, we can get the sixth term. Okay, so let's do that. So to get the fourth term, we double the third term. and then take away 1. So we double 12 and then take away 1 which is 24 take away 1 which is 23. So that's the fourth term. So then the fifth term will be double the fourth term so 2 lots of 23 take away 1 which is 46 take away 1 so that's 45. And then finally the sixth term is double the fifth term so double 45 and then take away one so that's 90 take away one so the sixth term is 89 and there's your answer so the mark scheme you get a method mark for trying to find out the fourth and the fifth terms and then an accuracy mark for that 89 for the sixth term if you want to do more practice on recurrence relations my video lesson eight is on that okay let's have a look at question two Okay, so in question two, find the coefficient of x cubed in the expansion of 2 plus 3x to the power of 5, giving your answer as simply as possible. So this is a binomial expansion because there are two terms in this bracket. So that's what a binomial is, two terms. And we're expanding a certain number of these brackets, so five of these brackets in Aronian. in effect because we're putting it to the power of five. So in effect we have 2 plus 3x multiplied by 2 plus 3x multiplied by 2 plus 3x and then another two brackets like that. So there are five brackets in a row each of them with 2 plus 3x. If it was just two brackets in a row I would be very confident in your ability to expand that. You've practiced I imagine different methods of expanding two brackets. But when it's five brackets in a row, you need a more systematic way of doing this, because there are many different combinations. When you multiply out these brackets, you have to choose one thing from each bracket to multiply. So I could choose the 2 times the 2 times the 2 times the 2 times the 2. That will give me one of the different possible results of this expansion. And that's 2 to the power of 5. What I want, though, is I want the x cubed term. In order to get an x cubed term, I have to multiply the x term by another x term by another x term. And then, if I have chosen to do that, if I've done 3x multiplied by 3x multiplied by 3x, in order to keep it as x cubed and not then turn it into x to the power of 4 or x to the power of 5, I need to then choose the 2 from the other two brackets. And this is one of the different possible ways of doing this. by choosing the 3x from each of the first three brackets and then the 2 from the last two brackets. But I could have done that in many different ways. I could have done the 3x times the 3x times the 2 times the 3x times the 2. There are, from 5 brackets, choose 3 different ways of doing this calculation, of getting this x cubed term. 5 choose 3 is the number of different possible ways of choosing the 3x from 3 of the 5 brackets. So from five brackets choose three things, the 3x three times. So that's how many different ways there are of doing this calculation. And the calculation is 3x times 3x times 3x, so 3x to the power of three, and then times by two from the other brackets. So there are two brackets left, so two to the power of two. So that is the calculation that would give me all the different possible ways of getting the x cubed term. So this is one of the ways of getting the x cubed term. And this is the number of different ways I could get the x cubed term. And so that would give me all the different possible ways which we could combine together. So 5 choose 3 is 10. There are 10 different possible options of getting the x cubed term. 3x to the power of 3 is 27x cubed. 3x times 3x times 3x. And 2 squared is 4. And then when I combine those together, that's 1080x to the power of 3. But the question was, find the coefficient of x cubed. If you left your answer like this, I imagine they would give you the full marks. But I should be more exact in my answer. If I want the coefficient of x cubed, the coefficient is 1080. It is the constant in front of the x cubed. It is the... what x cubed is being multiplied by. So that is my answer. So let's have a look at the mark scheme. So method mark for trying to do binomial expansion. So they've got it in a sort of slightly weird order here. And accuracy mark for getting, I imagine, all three parts of that expansion correct. And another accuracy mark for simplifying the coefficients of those ignoring the x cubed they have here and then 1080 is the final mark. If you want more practice on binomial expansion lesson 24 is on this specific skill. Lesson 23 will build you up to that as well. Okay let's have a look at question three. You're given that y is equal to x cubed plus 2x minus 7. Part a is find dy by the x. dy by dx is the derivative of the function. It is the gradient function. So we have to differentiate y with respect to x to get dy by dx. So the way we differentiate is we take each term and we bring the old power in front and then we drop the power by one. So x cubed, the old power is three, That comes in front, it becomes three lots of x, and then the power drops by one from three to two. So x cubed differentiates to three x squared. 2x, 2x in effect is 2x to the power of one, so the old power is one. It multiplies in front, so one lot of 2x becomes two. x to the power of one drops to x to the power of zero, and x to the power of zero is just the number one. So this differentiates 2x to the power of 1 just differentiates to 2, the number 2. Then minus 7, minus 7 in effect is the same as minus 7x to the power of 0. So the old power is 0. If I bring that in front and multiply by it, it turns everything into 0. 0 multiplied by anything is 0. And therefore that just becomes 0. So when you have a constant, that differentiates to nothing. So that's our answer. 3x squared plus 2. Part b. Use your result part a to show that the graph of y equals x cubed plus 2x minus 7 has no turning points. Okay a turning point is a point on the graph where the curve has a gradient of zero. So if I drew a function, so here's a maybe a cubic function, that has two turning points. That has a turning point there. and another turning point there. Okay, the gradient of those points is equal to zero. So at a turning point, so at those points. the gradient equals zero. Okay so at those specific points there and there this would be true. So it's asking us to show using what we found out in part a that there are no turning points. Okay well if I wanted to find turning points I would try and solve when the gradient is zero. So 3x squared plus 2 is our gradient function and I want to solve when that equals zero. And because it's asking me to show that there are no turning points, this should lead me to, I should not be able to find a solution to this equation. So 3x squared plus 2 is 0, so 3x squared would equal negative 2, and therefore x squared would equal negative 2 thirds. Now there are no real solutions to that equation because if I square a real number, it'll be 0 or above. and therefore x squared cannot be minus two thirds and have a real solution and so there are no turning points okay let's have a look at the mark scheme part a just method mark for attempting to differentiate and accuracy mark for both those terms um part b is just saying show that this equation does not equal zero so that that's what that means equals with a line through it does not equal so there are no turning points. If you want more practice on differentiation and stationary points see lesson 38 in my sequence. Okay let's have a look at question four. In this question you must show detailed reasoning. The reason they ask this the reason they say this is because on some of your calculators you might be able to just type that into your calculator press equals and it'll give you an answer. They don't want you to do that, they want you to show that you can do this skill, this technique algebraically without the use of a calculator. So we need to integrate the function x squared plus 3 with respect to x between the limits of 1 and 2. What if effectively that means... although you don't need to draw this. y equals x squared plus three looks like this. That's the x squared graph but three higher, so it goes through at three there. And it means finding the area below the curve between the values one and two. So finding the area of this region is effectively what it's asking you to do. But you don't need to show that drawing, it's just I'm showing you what it's asking you to do. So the skill involved in doing definite integration is integrate first. So the integral between 1 and 2 of x squared plus 3 is equal to x squared. When we integrate it, we raise the power by 1 to x cubed and we divide by the new power. It is the opposite of differentiation. So x squared becomes x cubed divided by 3. Then 3, which effectively is... 3x to the power of 0. We raise the power by 1 to 3x to the power of 1 and we divide by 1 so it becomes 3x to the power of 1 or just 3x so plus 3x and that is between the limits 1 and 2. At this point we now have to substitute the two limits in and then subtract. We substitute the upper limit first so we substitute 2 in first so 2 cubed divided by 3 which is 8 thirds you plus three lots of two which is six. We have substituted two in and we've got those two values. Then we subtract the result we get when we substitute one in. One cubed divided by three is one third. Three times one is three and if we do that calculation we get five and one third or better than that sixteen thirds. We could put if we wanted to units squared because it's an area but it hasn't really talked about the fact that this is talking about the area below a curve so we just leave our answer like that. Let's have a look at the mark scheme. So it gives you a method mark for attempting to integrate, a method mark for attempting to substitute, and accuracy for accurately putting both parts of the substitution with the subtraction between accuracy. accuracy mark for five and a third. If you want more practice on definite integration, see lesson 41, which practices that skill. Okay, let's have a look at question five. Now, this question is the one that most of my pupils struggled with in this paper. Okay, I believe it's probably the most difficult question in this paper because conceptually it's difficult to understand what it's asking you to do and it's difficult to find answers to it. The London eye can be considered to be a circular frame of radius 67.5 meters, so the distance from the center of that circle to the edge, anywhere on the circumference, is 67.5 meters, which tells me that the height of that circle is 135 meters because it's two radii, on the circumference of which are capsules carrying a number of people around the circle. Take a coordinate system where O, the origin, is the base of the circle and O to Y is the diameter. At any time after starting off around the frame the capsule will be at height h meters when it has rotated theta degrees. So when it is rotated zero degrees we are at the origin. When it is rotated 90 degrees we are over here on the on the right side of the circle. When it's rotated 180 degrees we are at the very top of the circle etc. Okay so sketch a graph of h against theta. Okay so usually we sketch a graph of y against x so the y-axis and the x-axis. Here we're doing a graph of h against theta so h is our vertical axis and theta is our horizontal axis. Well theta will go between zero all the way around to 360 degrees. Okay so I'm going to be doing my x-axis in effect as being my theta axis between 0 and 360. And h is going to be the height of the capsule above the ground, so the minimum height will be 0 when it is at the very bottom. The maximum height will be as I said 135. So I'm going to draw an h-axis between 0 and 135. It's only a sketch so we don't need a graph paper for it and I'll put 67 and a half somewhere in the middle, hopefully exactly in the middle, and then my horizontal axis will go to 360 degrees and that will be my theta axis. Okay now as I increase the at the very start when theta is zero the height of the capsule is at zero. So I know I start off on the x-axis. I'm going to do this in green so it comes up a little bit more. As theta increases, my height starts to increase slowly. So as I get to, let's say, 45 degrees, so about there, the height of the capsule is pretty small, but it's getting there. As I reach 90 degrees, the height of my capsule is 67.5 meters. And the height is changing more quickly. The height is increasing more quickly at this point than it was over here because when I'm at the start most of my movement is horizontal not vertical. When I'm at 90 degrees most of my movement of my capsule is vertical not horizontal. So it's like I'm slowly gaining height and then I'm more quickly gaining height at 90 degrees. And at 90 degrees I'm at 67 and a half meters high. And then it's quite quickly gaining height, but then that height is slowing down up to 180 degrees where it's at 135 meters off the ground. So it's going quite quickly and then slowing down. And then at 180 degrees, I am at 135 meters. And then I, in effect, have the opposite. It's now, the height is decreasing, but slowly. And then it's gathering more speed at 270 degrees. And then It's going quite quickly, decreasing the height, and then that's slowing down until 360 degrees. I'm back at zero. So that, in effect, will look something like this. Pretty poorly drawn, but you'll get the full marks for something like that. So 67 and a half meters, that's 270 degrees. It doesn't have to be very accurate drawings, but that is, in effect, what we've got. Okay, so that's my graph of h against theta. The next thing I want to do is find an expression. for h in terms of theta. So find a basically an equation which links h and theta. Okay and this is slightly more difficult to do but what I'm going to do is I'm going to focus on this circle and theta and h and try and link them. I'm going to draw like just a right angle triangle here. h is the value that takes me is the height of this part there to there. I know that this part from there to there is what's left over to make 67.5. The radius is 67.5 in total. So this part here, let's call it 67.5 minus h. I know also that this hypotenuse is 67.5. Okay, so have a think. How could you link theta, this side of a right angle triangle, and then this side of the right angle triangle? What ratio would you use to link those three things? So think about labelling with opposite or adjacent or hypotenuse, and think about which ratio then would you use. Okay, hopefully you've thought, well... This side is the adjacent and this side is the hypotenuse. And therefore I'm using the cosine ratio. So cos of theta is equal to the adjacent over the hypotenuse. So 67.5 minus h over 67.5. And now I've got an equation with theta and with h. But they asked me for an expression for h in terms of theta. So that means h needs to be the subject of my formula. So I'm just going to rearrange this until I've got h as the subject. So multiply by 67.5, I get 67.5 cos theta equals 67.5 minus h. Add h to both sides and then subtract 67.5 cos theta from both sides, I get h is equal to 67.5 minus 67.5 cos theta. And there we have it. We have an equation for h in terms of theta. Okay, so let's write that down. So h equals 67.5 minus 67.5 cos theta. So that is my equation for h in terms of theta. Part c, find values of theta when h is 100. Well, I've got an equation with h and theta in. I'm told the value of h, which is 100, I need to then find the values of theta. So that's solving an equation using substitution. So I'm substituting instead of h, I'm going to substitute 100. So 100 is equal to 67.5 minus 67.5 cos theta. So all I need to do is solve this equation and find the values, so multiple values of theta between 0 and 360, which work in this equation. Okay, so first thing we're going to do is I'm going to add 67.5 cos theta to both sides and subtract 100 and then divide by 67.5 and I will get cos theta is equal to negative 13 27ths. So that's the value of cos theta. So if I want to find all the values of theta, good idea if you draw a sketch of the graph cos theta between 0 and 360. So it goes between 1 and minus 1 and I want to find the values of theta which give me a height of negative 1327 which is around there just just above minus 0.5 so there will be two there'll be one here and another one and they will be symmetrical. Once I've found A, to find B all I need to do is know that that distance between 0 and A is the same distance between 360 and B. So all I need to do is take away A from 360 and I've got B. To find A I just need to use my calculator. I need to do the inverse cos of negative 13 27ths which is 118.8 degrees. So that's that angle. So to find that angle, I just do 360 minus 118.8. So the other solution, both to one decimal place, I'm going to give them 241.2 degrees. Okay, so those are the two solutions. Let's have a look at the mark scheme. So there is their diagram, which is a lot nicer than mine, but I feel mine has an elegance. And they have a... factorized version of my equation and then they have the two angles but to three significant figures rather than one decimal place. If it doesn't tell you what to round it to in the question three significant figures is a good good rule of thumb or for angles one decimal place both should be fine. So if you want more practice on this have a look at lesson 14 on trigonometric equations which is helpful for finding different answers when you have a trigonometric equation. Parts A and B, they're more about just knowing trigonometry in general. Okay, so let's have a look at question six. Aha, algebraic fractions. So simplify x over x plus 2 minus 6 over x minus 1. So if I subtract two fractions, in order to do so, I need a common denominator. If I have them in algebraic form, unless one is an easy multiple of another, it's often best if I just multiply the two denominators together. That will definitely get me a common denominator. So for the left fraction, x over x plus 2, I'm going to multiply the top and the bottom, so the numerator and the denominator, by x minus 1. So I'll get x multiplied by x minus 1 as the new numerator, over x plus 2 multiplied by x minus 1. And I could expand those out, and I will do so in a second. With the second fraction, 6 over x minus 1, I'm going to multiply the numerator and the denominator by x plus 2. So 6 lots of x plus 2 over x minus 1 times x plus 2. I'll write it the other way around, though, because I want them looking the same on both, even though just reversing them will keep it the same. Okay. So let's expand the numerators so that we can then subtract. So the numerator of the first fraction is x squared minus x. And the numerator on the second one is 6x plus 12 over x plus 2 x minus 1. And I could write that in expanded form, but I'm going to keep it in factorized form. So now I need to be very careful because I'm... subtracting. I'm subtracting 6x and I'm subtracting positive 12. A common error would be to subtract 6x and then to add 12 but no I am subtracting the entirety of this numerator from the other numerator. So x squared minus x if I take away 6x that negative x becomes negative 7x so I get x squared minus 7x and then if I take away 12 I have no I've got zero there so if I take away 12 I get minus 12. over x plus 2 x minus 1. So that's my answer. If you want to you can also write it as x squared minus 7x minus 12 over and then expand the brackets x squared 2x take away 1x is 1x and 2 times negative 1 is negative 2. So either one of those is fine. You might think, oh, could I simplify it further? Could I factorize the numerator and then find a common factor and divide? But you find out that you cannot factorize x squared minus 7x minus 12. So that is as simple as it gets. Okay, part B. Solve the equation where that thing that we have simplified is equal to 4. So clearly they want you to use the result from part A in part B. So that's what we're going to do. We're going to write down that fraction from part a, x squared minus 7x minus 12, over x squared plus x minus 2 is equal to 4, and solve that. Okay, so I'm going to multiply by the denominator, both sides. So I will get x squared minus 7x minus 12 equals 4 lots of x squared plus x minus 2. So that becomes 4x squared plus 4x minus 8. Next, this is a quadratic equation it's easier if I have the quadratic equation equaling zero. So I'm going to subtract x squared from both sides, add 7x to both sides, and add 12 to both sides, and I will get zero is equal to 3x squared plus 11x plus 4. And I will think, okay, firstly, if I have a quadratic equation like this, can I factorise? And I'll try and do so. The way I like factorising is I like thinking in terms of trying to deduce what two binomial expressions will multiply together to make this quadratic. So I know I'm going to get 3x squared by multiplying this term by this term. So I know it's going to be 3x multiplied by 1x. And I know I'm going to get 4 as the constant by multiplying these two constants together. So then I have to think, are there two constants which I can put there and there, which would then combine in these two terms to make 11x? My choices are that it's definitely going to be both positive because I'm going to get a positive amount of x and two positives multiply to make a positive answer. Also two negatives would multiply to make a positive answer but that would not give me my positive 11x. So my choice is either 4 and 1 or 2 and 2 and if I try any of those So I'll just start putting things in 2 and 2 there that will give me 6x that will give me 2x, that gives me 8x, which is not the result I wanted. So 2 and 2 does not work. And then 4 and 1, if I put 4 here, I would only get 4x, and then 1 there, I would get 3x, that would make 7x, not 11x, so that doesn't work. And finally, if I try 4 there and 1 there, I get 12x, which looks promising, and 1x, that looks promising, but I'm adding them together, I'm combining them, that makes 13x. So... This is not factorizable, so I'm not going to use that as a method to solve this. So what I'm going to do instead is I'm going to use the quadratic formula. I could try and complete the square to solve it, but this is quite a difficult quadratic to complete the square for. So I'm just going to use the quadratic formula. So my a is 3, my b is 11, my c is 4. So x will equal minus b, so that's minus 11, plus or minus the square root of... b squared so that's 121 minus 4 times a times c so minus 4 times 3 times 4 all over 2a all over 6 and when I simplify that down I get minus 11 plus or minus the square root of 73 over 6. So my two solutions are x is equal to minus 11 plus root 73 over 6 or minus 11 minus root 73 over 6. And I probably should have noticed that this would probably be non-factorizable because it says solve this giving your answer in exact form. Okay so it's looking for a surd form in effect okay rather than some rounder decimal which is not exact. This is beautiful and exact any rounder decimal that you get on your calculator would not be so. Okay let's have a look at the mark scheme. There it is in all its glory. If you want more practice on this, have a look at lesson one in my sequence of lessons on algebraic fractions. Okay, let's have a look at question seven. Aha, completing the square, lovely. So in this question you must show detailed reasoning. Express 2x squared plus 8x minus 12 in the form a lots of x plus p all squared plus q. Okay so that's the completed square form. So when I have a quadratic expression which isn't 1x squared it's slightly more difficult to complete the square form but with practice you should get used to doing so. What I do is I focus on the first two parts the x squared part and the x part and I ensure that I have an expression which is just 1x squared by taking a factor of however many x squareds I have. So 2x squared plus 8x minus 12. I'm going to focus on these two parts. That is the same. 2x squared plus 8x is the same as two lots of 1x squared and 4x. So I've written exactly the same as I had above. So those two lines are equal to each other. But what I have now is I have something which is very easy. to complete the square for. So I'm now going to complete the square for that part. So two lots of. Now for that part, this part here, I'm going to just do this in a slightly different colour. So I'm going to complete the square for that purple part there. x squared and 4x, complete the square is x plus 2 all squared because that would give me my x squared. It would give me 2x and 2x which would give me 4x but it would also give me 4. I don't want 4. I have nothing else in that bracket so I have to subtract 4 from that. So that is what that part would turn into and I've still got the negative 12 at the end. Okay now I want to get rid of these outer brackets by multiplying by 2. So I get two lots of this x plus 2 all squared. I also get two lots of negative 4, so that's negative 8. And I've still got the minus 12 at the end, which wasn't being multiplied by the 2. Now I just simply combine the minus 8 and the minus 12. So two lots of x plus 2 all squared take away 20. There we go. We have it in that form. Part B, hence find the minimum value of this quadratic expression. Often that might be in the form in questions, you have this function, this graph of y equals 2x squared plus 8x minus 12. Find the vertex of this function. OK, so that's a positive quadratic. So it'll look like that. OK, or something like that. And you'll find the minimum point. If it's in completed square form, which we have here. it is very easy to find the coordinates of that minimum point. Because if I have y is equal to 2 lots of x plus 2 all squared minus 20, I know that when I square something, so x plus 2 all squared, that orange part cannot be below 0. If I square any value, it'll be 0 or above. And then if I double it, still 0 or above. So this part here is at minimum 0. So the smallest I can make y equal to is negative 20. When this part is 0, I get y equals minus 20. So the lowest part of that graph will have a y coordinate of minus 20, which is in effect what they're asking for. The minimum value of this, since I've... made y equal to that I'm finding the smallest value of y. So minus 20 is our answer. The smallest value that expression can take is negative 20. The x value that would give that is negative 2 but that is not what they asked for. They asked for simply the outcome of that expression not the input that would make that outcome. So negative 20 is the answer. Let's have a look at the mark scheme. There it is. They do it in a slightly different way to the way I did it. Still effective. I prefer my way because in my way, it doesn't matter if the final term, so minus 12 in this case, was divisible by 2. It did happen to be divisible by 2 in this case, but it didn't matter if I use my method in future. So if you want more practice on this, lesson five on completing the square. So let's have a go at the next question, question eight. triangle ABC is such that AB is 5 centimetres, BC is 8 centimetres and CA is 7 centimetres. Show that one angle is 60 degrees. Okay well let's let's visualise it properly. So I'm going to draw the longest line to 8 centimetres and then the other two lines so 7 centimetres, let's say it's something like that and so 7 centimetres there and 5 centimetres there. So approximately to scale not great because the 7 and the 8 look about the same length in that case it looks almost like an isosceles triangle but ignore that pretend that they are different lengths now in a triangle if you have three different lengths you're gonna have three different angles inside the largest angle inside that triangle will be opposite the largest side okay so this angle there will be the biggest angle The smallest angle will be opposite the smallest side because a smaller angle will open out onto a smaller side and so therefore what we're left with is opposite this 7 is the median angle. Okay, so there are three angles. If we put them in order of size, the one opposite the 7 will be the median. Okay, if we want to find the 60 degree angle, well, if there are three different angles, 60 degrees cannot be the biggest, because if the other two are smaller than 60, then you wouldn't get a total of 180. Similarly, it cannot be the smallest, because if you have a 60 degree angle and then two which are bigger than 60 that would be over 180 degrees in total and hence the median is the one we're looking for okay so the 60 degrees must be the median angle so we're looking for the angle opposite the seven centimeters so this is now a cosine rule question so if i call that angle theta then i know that cos theta if i know my cosine rule and the rearrangement of it so that cosine rule a squared equals b squared plus c squared minus 2bc cos a you should it is helpful if you know the rearrangement of this into cos a is equal to b squared minus plus c squared minus a squared over 2bc it makes things simpler than having to rearrange it every time yourself if you're trying to find an angle so here in this case since i'm Angle a is opposite little a. I know that angle theta is opposite 7. Therefore, I'm going to have 5 squared plus 8 squared minus 7 squared over 2 times 5 times 8. And that, if I simplify that, I get cos theta is equal to 40 over 80, which is the same as a half. And then inverse cos of a half is 60 degrees. There we go. QED. I've shown what I wanted to show. Okay, so let's have a look at the mark scheme for this. Yeah, middle angled, one required, etc. So there's no specific baromaths video on this because this is GCC content, but I hope you enjoyed it anyway. Right, question nine. The factor theorem. Brilliant. So in this question, you must show detailed reasoning as normal. Show that x minus 3 is a factor of this cubic, x cubed minus 5x squared plus x plus 15. Okay, the fact-tip theorem states that if you have a factor, if x minus a is a factor, then f of a will equal 0. Because if I replace x with a, this bracket will be 0. And if one factor is made 0, you are multiplying by zero and then the rest of the expression would equal the result of the entire expression would be equal to zero. So if I replace x with a I've got zero in that case. So here you have to think well if that's a factor if x minus three is a factor what value of x would make that factor zero and therefore make the entire expression zero. If x minus three is a factor then f of three will be equal to zero. So let's find f of three. So that's 3 cubed minus 5 times 3 squared plus 3 plus 15. So 27 minus 5 lots of 9, so minus 45, plus 3 plus 15. And that does indeed equal 0. Hooray! QED, quad error at demonstrandum, which was what was wanted. I am done. Okay, part B, hence solve the cubic equation. Solving quadratics, you know. you factorize solving cubics is a similar method we factorize but it's more difficult to find that initial factor however they have given us that initial factor so let's use that to factorize the cubic into a linear part and a quadratic part because x minus 3 will multiply by a quadratic x minus 3 is linear it would multiply by a quadratic to end up with a cubic answer okay so i'm gonna factorise it like this. x minus 3 is one of my factors. What do I multiply it by to get a result of x cubed minus 5x squared plus x plus 15? So I know so far this much. I know I'm going to get a result of x cubed there. I know I'm going to get a result of 15 there. The rest is going to be deduced. So let's deduce that. x multiplied by what gives me x cubed? Well that's going to be x squared. So now I know what this part is. Minus 3 times x squared is minus 3x squared. Since these two will combine to make the x squareds, and I know I want minus 5x squared in total, I have minus 3x squared already. I need another minus 2x squared. And hence, I now know this part. x multiplied by what gives me minus 2x squared? Well, that would be minus 2x. So now I can fill in this part. Minus 2x times minus 3 is positive 6x. These two parts will combine to make positive 1x. I have 6x already. I have to then have a negative 5x to combine to make 1x. And hence, the final part would be minus 5. Minus 5 times x makes minus 5x. And I just check using this 15. Does minus 5 times minus 3 give you 15? Yes, it does. So I've now managed to factorize it into a linear and a quadratic. So I had x cubed. minus 5x squared plus x plus 15 equals 0. I've now factorized that into x minus 3 multiplied by x squared minus 2x minus 5 equals 0. So I know the solution to this will be either when x minus 3 equals 0, so either when this bracket is 0 and that's when x would equal 3, or when this quadratic is zero. So can I factorize that? No I can't. There's no way of factorizing x squared minus 2x minus 5 into simple rational values inside the brackets. So I'm going to use the quadratic formula. Okay so from this I'm going to get that x is equal to minus b. So that's positive 2 plus or minus the square root of b squared which is 4 minus 4 times 1 times minus 5 all over 2a which is just 2 and that would simplify to 1 plus or minus root 6. So my solutions are x is either 3 or 1 plus root 6 or 1 minus root 6. So those are my three solutions to this cubic equation. So we'll look at the mark scheme. There we go. They've given the solutions in decimal form. How horrible. And they haven't even written what they've rounded it to. Despicable. But if you've given it three significant figures like they have, great, well done. You get the full marks. So if you want more practice on the factor theorem, see lesson four on the factor theorem. Okay. One more question to go in this video. Let's have a look at question 10. So here we have a security keypad which uses three letters, A, B and C, and four digits, 1, 2, 3 or 4. A passcode is created using four inputs. So I have four things to input into these four spaces. in order to get into the keypad system. If there are no restrictions how many different passcodes are possible? Well if there are no restrictions except for these restrictions that they've already given us, because that's what all the things I can press on the keypad, well I have four digits and three letters so I have seven characters I could put in into these inputs. So in the first input I have seven choices. that I could put in. In the second input, well I have no restrictions, I can still put seven things. I could have a a a a as my password. Wouldn't be advisable but I could. And then for the third input, seven choices. And for the fourth input, seven choices. So that's seven to the power of four. Okay so seven to the power of four is the different the how many pass words how many passcodes that are possible in the first one. Part B, if there must be exactly two letters and two digits with no repeats, how many different passcodes are possible? Okay, this takes a little bit more thought. So I have letters I could choose. Okay, so what I could... choose. If I must pick exactly two letters, I could choose A with B. Or I could choose A with C. Or I could choose B with C. I'm not talking about what order they're going in yet. I'll come back to that. But those are the three possible choices of two letters with no repeats. Okay? So I have three possible choices of which two letters I am going to choose. And for... the digits, again no repeats and choosing two digits, I could choose one with two, I could choose one with three, I could choose one with four, I could choose two with three, two with four, and I could choose three with four. So there are six different possible options of which two digits I choose. I'm not saying which order I'm choosing them in and where they're going, but I'm saying these are the six possible options. So I could choose this AB with the 1, 2, or the AB with the 1, 3, or the AB with the 1, 4, etc. Okay, so for each of these three choices, I have six possible choices there. So there are three times six, so there are 18. choices of character. Okay, so there are 18 possible choices of the types of character that I'm going to choose. Now, for each of those, so let's say I chose A, B, 1, 2. I can arrange these four characters in four factorial ways. So each of the 18 choices can be arranged in four factorial ways. So, it's like I've got 18 lots of four factorial different possible options, which gives me 432 options. Okay, and that's how you think about this. With these types of questions, they require a lot of common sense and they are very difficult. Maths teachers struggle. immensely with permutations, combinations, these types of problems. And it's about experience with these types of problems that will allow you to solve them more easily. Okay. So the answers were, oh, I forgot to say what 7 to the power of 4 was. That was 2,401. And here's the way they did it. Okay. Which is kind of similar. So 432. If you want practice on this, see my lessons 20... for factorials for ordering and maybe even 21 and 22 on permutations and combinations as well. Okay so that is questions 1 to 10 of this paper. Okay the next video I'm going to do, I'm going to do questions 11 to 16 from this sample paper. Okay so make sure again before you watch the video you have done or attempted those questions. and then watch the video to get a sense of if you're struggling with certain questions, how I might approach them. Okay, enjoy.

Make sure you have tried that paper and those questions before watching this. It's important for you to have had a go and have your own thoughts about each question before running through the solutions with me, okay? So let's have a go at question one.

A sequence is defined by the rule un plus one is equal to two lots of un minus one. So this is what's called a recurrence relation. So this formula allows us to get the next term in a sequence, okay, based on the previous term.

The next term is the n plus 1th term. And to get it, we do two lots of the nth term, so double the nth term, and then take away 1. So to get the next term in the sequence, we double the previous term, and then take away 1. In this question, we want to find the sixth term in the sequence. sequence, all right, in order to get the sixth term, we would need the fifth term, because we need the previous term to get the next term.

We are given the third term in the sequence, which is 12. Well, with the third term, we can get the fourth term. Then with the fourth term, we can get the fifth term, and with the fifth term, we can get the sixth term. Okay, so let's do that. So to get the fourth term, we double the third term.

and then take away 1. So we double 12 and then take away 1 which is 24 take away 1 which is 23. So that's the fourth term. So then the fifth term will be double the fourth term so 2 lots of 23 take away 1 which is 46 take away 1 so that's 45. And then finally the sixth term is double the fifth term so double 45 and then take away one so that's 90 take away one so the sixth term is 89 and there's your answer so the mark scheme you get a method mark for trying to find out the fourth and the fifth terms and then an accuracy mark for that 89 for the sixth term if you want to do more practice on recurrence relations my video lesson eight is on that okay let's have a look at question two Okay, so in question two, find the coefficient of x cubed in the expansion of 2 plus 3x to the power of 5, giving your answer as simply as possible. So this is a binomial expansion because there are two terms in this bracket.

So that's what a binomial is, two terms. And we're expanding a certain number of these brackets, so five of these brackets in Aronian. in effect because we're putting it to the power of five.

So in effect we have 2 plus 3x multiplied by 2 plus 3x multiplied by 2 plus 3x and then another two brackets like that. So there are five brackets in a row each of them with 2 plus 3x. If it was just two brackets in a row I would be very confident in your ability to expand that. You've practiced I imagine different methods of expanding two brackets.

But when it's five brackets in a row, you need a more systematic way of doing this, because there are many different combinations. When you multiply out these brackets, you have to choose one thing from each bracket to multiply. So I could choose the 2 times the 2 times the 2 times the 2 times the 2. That will give me one of the different possible results of this expansion. And that's 2 to the power of 5. What I want, though, is I want the x cubed term. In order to get an x cubed term, I have to multiply the x term by another x term by another x term.

And then, if I have chosen to do that, if I've done 3x multiplied by 3x multiplied by 3x, in order to keep it as x cubed and not then turn it into x to the power of 4 or x to the power of 5, I need to then choose the 2 from the other two brackets. And this is one of the different possible ways of doing this. by choosing the 3x from each of the first three brackets and then the 2 from the last two brackets.

But I could have done that in many different ways. I could have done the 3x times the 3x times the 2 times the 3x times the 2. There are, from 5 brackets, choose 3 different ways of doing this calculation, of getting this x cubed term. 5 choose 3 is the number of different possible ways of choosing the 3x from 3 of the 5 brackets. So from five brackets choose three things, the 3x three times.

So that's how many different ways there are of doing this calculation. And the calculation is 3x times 3x times 3x, so 3x to the power of three, and then times by two from the other brackets. So there are two brackets left, so two to the power of two.

So that is the calculation that would give me all the different possible ways of getting the x cubed term. So this is one of the ways of getting the x cubed term. And this is the number of different ways I could get the x cubed term.

And so that would give me all the different possible ways which we could combine together. So 5 choose 3 is 10. There are 10 different possible options of getting the x cubed term. 3x to the power of 3 is 27x cubed.

3x times 3x times 3x. And 2 squared is 4. And then when I combine those together, that's 1080x to the power of 3. But the question was, find the coefficient of x cubed. If you left your answer like this, I imagine they would give you the full marks. But I should be more exact in my answer. If I want the coefficient of x cubed, the coefficient is 1080. It is the constant in front of the x cubed.

It is the... what x cubed is being multiplied by. So that is my answer. So let's have a look at the mark scheme.

So method mark for trying to do binomial expansion. So they've got it in a sort of slightly weird order here. And accuracy mark for getting, I imagine, all three parts of that expansion correct.

And another accuracy mark for simplifying the coefficients of those ignoring the x cubed they have here and then 1080 is the final mark. If you want more practice on binomial expansion lesson 24 is on this specific skill. Lesson 23 will build you up to that as well. Okay let's have a look at question three. You're given that y is equal to x cubed plus 2x minus 7. Part a is find dy by the x.

dy by dx is the derivative of the function. It is the gradient function. So we have to differentiate y with respect to x to get dy by dx. So the way we differentiate is we take each term and we bring the old power in front and then we drop the power by one.

So x cubed, the old power is three, That comes in front, it becomes three lots of x, and then the power drops by one from three to two. So x cubed differentiates to three x squared. 2x, 2x in effect is 2x to the power of one, so the old power is one.

It multiplies in front, so one lot of 2x becomes two. x to the power of one drops to x to the power of zero, and x to the power of zero is just the number one. So this differentiates 2x to the power of 1 just differentiates to 2, the number 2. Then minus 7, minus 7 in effect is the same as minus 7x to the power of 0. So the old power is 0. If I bring that in front and multiply by it, it turns everything into 0. 0 multiplied by anything is 0. And therefore that just becomes 0. So when you have a constant, that differentiates to nothing. So that's our answer. 3x squared plus 2. Part b.

Use your result part a to show that the graph of y equals x cubed plus 2x minus 7 has no turning points. Okay a turning point is a point on the graph where the curve has a gradient of zero. So if I drew a function, so here's a maybe a cubic function, that has two turning points.

That has a turning point there. and another turning point there. Okay, the gradient of those points is equal to zero. So at a turning point, so at those points.

the gradient equals zero. Okay so at those specific points there and there this would be true. So it's asking us to show using what we found out in part a that there are no turning points.

Okay well if I wanted to find turning points I would try and solve when the gradient is zero. So 3x squared plus 2 is our gradient function and I want to solve when that equals zero. And because it's asking me to show that there are no turning points, this should lead me to, I should not be able to find a solution to this equation.

So 3x squared plus 2 is 0, so 3x squared would equal negative 2, and therefore x squared would equal negative 2 thirds. Now there are no real solutions to that equation because if I square a real number, it'll be 0 or above. and therefore x squared cannot be minus two thirds and have a real solution and so there are no turning points okay let's have a look at the mark scheme part a just method mark for attempting to differentiate and accuracy mark for both those terms um part b is just saying show that this equation does not equal zero so that that's what that means equals with a line through it does not equal so there are no turning points.

If you want more practice on differentiation and stationary points see lesson 38 in my sequence. Okay let's have a look at question four. In this question you must show detailed reasoning. The reason they ask this the reason they say this is because on some of your calculators you might be able to just type that into your calculator press equals and it'll give you an answer.

They don't want you to do that, they want you to show that you can do this skill, this technique algebraically without the use of a calculator. So we need to integrate the function x squared plus 3 with respect to x between the limits of 1 and 2. What if effectively that means... although you don't need to draw this.

y equals x squared plus three looks like this. That's the x squared graph but three higher, so it goes through at three there. And it means finding the area below the curve between the values one and two.

So finding the area of this region is effectively what it's asking you to do. But you don't need to show that drawing, it's just I'm showing you what it's asking you to do. So the skill involved in doing definite integration is integrate first. So the integral between 1 and 2 of x squared plus 3 is equal to x squared.

When we integrate it, we raise the power by 1 to x cubed and we divide by the new power. It is the opposite of differentiation. So x squared becomes x cubed divided by 3. Then 3, which effectively is...

3x to the power of 0. We raise the power by 1 to 3x to the power of 1 and we divide by 1 so it becomes 3x to the power of 1 or just 3x so plus 3x and that is between the limits 1 and 2. At this point we now have to substitute the two limits in and then subtract. We substitute the upper limit first so we substitute 2 in first so 2 cubed divided by 3 which is 8 thirds you plus three lots of two which is six. We have substituted two in and we've got those two values. Then we subtract the result we get when we substitute one in. One cubed divided by three is one third.

Three times one is three and if we do that calculation we get five and one third or better than that sixteen thirds. We could put if we wanted to units squared because it's an area but it hasn't really talked about the fact that this is talking about the area below a curve so we just leave our answer like that. Let's have a look at the mark scheme.

So it gives you a method mark for attempting to integrate, a method mark for attempting to substitute, and accuracy for accurately putting both parts of the substitution with the subtraction between accuracy. accuracy mark for five and a third. If you want more practice on definite integration, see lesson 41, which practices that skill. Okay, let's have a look at question five. Now, this question is the one that most of my pupils struggled with in this paper.

Okay, I believe it's probably the most difficult question in this paper because conceptually it's difficult to understand what it's asking you to do and it's difficult to find answers to it. The London eye can be considered to be a circular frame of radius 67.5 meters, so the distance from the center of that circle to the edge, anywhere on the circumference, is 67.5 meters, which tells me that the height of that circle is 135 meters because it's two radii, on the circumference of which are capsules carrying a number of people around the circle. Take a coordinate system where O, the origin, is the base of the circle and O to Y is the diameter. At any time after starting off around the frame the capsule will be at height h meters when it has rotated theta degrees. So when it is rotated zero degrees we are at the origin.

When it is rotated 90 degrees we are over here on the on the right side of the circle. When it's rotated 180 degrees we are at the very top of the circle etc. Okay so sketch a graph of h against theta. Okay so usually we sketch a graph of y against x so the y-axis and the x-axis. Here we're doing a graph of h against theta so h is our vertical axis and theta is our horizontal axis.

Well theta will go between zero all the way around to 360 degrees. Okay so I'm going to be doing my x-axis in effect as being my theta axis between 0 and 360. And h is going to be the height of the capsule above the ground, so the minimum height will be 0 when it is at the very bottom. The maximum height will be as I said 135. So I'm going to draw an h-axis between 0 and 135. It's only a sketch so we don't need a graph paper for it and I'll put 67 and a half somewhere in the middle, hopefully exactly in the middle, and then my horizontal axis will go to 360 degrees and that will be my theta axis. Okay now as I increase the at the very start when theta is zero the height of the capsule is at zero.

So I know I start off on the x-axis. I'm going to do this in green so it comes up a little bit more. As theta increases, my height starts to increase slowly.

So as I get to, let's say, 45 degrees, so about there, the height of the capsule is pretty small, but it's getting there. As I reach 90 degrees, the height of my capsule is 67.5 meters. And the height is changing more quickly. The height is increasing more quickly at this point than it was over here because when I'm at the start most of my movement is horizontal not vertical.

When I'm at 90 degrees most of my movement of my capsule is vertical not horizontal. So it's like I'm slowly gaining height and then I'm more quickly gaining height at 90 degrees. And at 90 degrees I'm at 67 and a half meters high. And then it's quite quickly gaining height, but then that height is slowing down up to 180 degrees where it's at 135 meters off the ground.

So it's going quite quickly and then slowing down. And then at 180 degrees, I am at 135 meters. And then I, in effect, have the opposite. It's now, the height is decreasing, but slowly.

And then it's gathering more speed at 270 degrees. And then It's going quite quickly, decreasing the height, and then that's slowing down until 360 degrees. I'm back at zero.

So that, in effect, will look something like this. Pretty poorly drawn, but you'll get the full marks for something like that. So 67 and a half meters, that's 270 degrees. It doesn't have to be very accurate drawings, but that is, in effect, what we've got. Okay, so that's my graph of h against theta.

The next thing I want to do is find an expression. for h in terms of theta. So find a basically an equation which links h and theta.

Okay and this is slightly more difficult to do but what I'm going to do is I'm going to focus on this circle and theta and h and try and link them. I'm going to draw like just a right angle triangle here. h is the value that takes me is the height of this part there to there.

I know that this part from there to there is what's left over to make 67.5. The radius is 67.5 in total. So this part here, let's call it 67.5 minus h. I know also that this hypotenuse is 67.5. Okay, so have a think.

How could you link theta, this side of a right angle triangle, and then this side of the right angle triangle? What ratio would you use to link those three things? So think about labelling with opposite or adjacent or hypotenuse, and think about which ratio then would you use.

Okay, hopefully you've thought, well... This side is the adjacent and this side is the hypotenuse. And therefore I'm using the cosine ratio.

So cos of theta is equal to the adjacent over the hypotenuse. So 67.5 minus h over 67.5. And now I've got an equation with theta and with h.

But they asked me for an expression for h in terms of theta. So that means h needs to be the subject of my formula. So I'm just going to rearrange this until I've got h as the subject. So multiply by 67.5, I get 67.5 cos theta equals 67.5 minus h.

Add h to both sides and then subtract 67.5 cos theta from both sides, I get h is equal to 67.5 minus 67.5 cos theta. And there we have it. We have an equation for h in terms of theta.

Okay, so let's write that down. So h equals 67.5 minus 67.5 cos theta. So that is my equation for h in terms of theta. Part c, find values of theta when h is 100. Well, I've got an equation with h and theta in. I'm told the value of h, which is 100, I need to then find the values of theta.

So that's solving an equation using substitution. So I'm substituting instead of h, I'm going to substitute 100. So 100 is equal to 67.5 minus 67.5 cos theta. So all I need to do is solve this equation and find the values, so multiple values of theta between 0 and 360, which work in this equation.

Okay, so first thing we're going to do is I'm going to add 67.5 cos theta to both sides and subtract 100 and then divide by 67.5 and I will get cos theta is equal to negative 13 27ths. So that's the value of cos theta. So if I want to find all the values of theta, good idea if you draw a sketch of the graph cos theta between 0 and 360. So it goes between 1 and minus 1 and I want to find the values of theta which give me a height of negative 1327 which is around there just just above minus 0.5 so there will be two there'll be one here and another one and they will be symmetrical.

Once I've found A, to find B all I need to do is know that that distance between 0 and A is the same distance between 360 and B. So all I need to do is take away A from 360 and I've got B. To find A I just need to use my calculator. I need to do the inverse cos of negative 13 27ths which is 118.8 degrees. So that's that angle.

So to find that angle, I just do 360 minus 118.8. So the other solution, both to one decimal place, I'm going to give them 241.2 degrees. Okay, so those are the two solutions. Let's have a look at the mark scheme. So there is their diagram, which is a lot nicer than mine, but I feel mine has an elegance.

And they have a... factorized version of my equation and then they have the two angles but to three significant figures rather than one decimal place. If it doesn't tell you what to round it to in the question three significant figures is a good good rule of thumb or for angles one decimal place both should be fine. So if you want more practice on this have a look at lesson 14 on trigonometric equations which is helpful for finding different answers when you have a trigonometric equation. Parts A and B, they're more about just knowing trigonometry in general.

Okay, so let's have a look at question six. Aha, algebraic fractions. So simplify x over x plus 2 minus 6 over x minus 1. So if I subtract two fractions, in order to do so, I need a common denominator. If I have them in algebraic form, unless one is an easy multiple of another, it's often best if I just multiply the two denominators together.

That will definitely get me a common denominator. So for the left fraction, x over x plus 2, I'm going to multiply the top and the bottom, so the numerator and the denominator, by x minus 1. So I'll get x multiplied by x minus 1 as the new numerator, over x plus 2 multiplied by x minus 1. And I could expand those out, and I will do so in a second. With the second fraction, 6 over x minus 1, I'm going to multiply the numerator and the denominator by x plus 2. So 6 lots of x plus 2 over x minus 1 times x plus 2. I'll write it the other way around, though, because I want them looking the same on both, even though just reversing them will keep it the same. Okay. So let's expand the numerators so that we can then subtract.

So the numerator of the first fraction is x squared minus x. And the numerator on the second one is 6x plus 12 over x plus 2 x minus 1. And I could write that in expanded form, but I'm going to keep it in factorized form. So now I need to be very careful because I'm...

subtracting. I'm subtracting 6x and I'm subtracting positive 12. A common error would be to subtract 6x and then to add 12 but no I am subtracting the entirety of this numerator from the other numerator. So x squared minus x if I take away 6x that negative x becomes negative 7x so I get x squared minus 7x and then if I take away 12 I have no I've got zero there so if I take away 12 I get minus 12. over x plus 2 x minus 1. So that's my answer. If you want to you can also write it as x squared minus 7x minus 12 over and then expand the brackets x squared 2x take away 1x is 1x and 2 times negative 1 is negative 2. So either one of those is fine.

You might think, oh, could I simplify it further? Could I factorize the numerator and then find a common factor and divide? But you find out that you cannot factorize x squared minus 7x minus 12. So that is as simple as it gets.

Okay, part B. Solve the equation where that thing that we have simplified is equal to 4. So clearly they want you to use the result from part A in part B. So that's what we're going to do. We're going to write down that fraction from part a, x squared minus 7x minus 12, over x squared plus x minus 2 is equal to 4, and solve that.

Okay, so I'm going to multiply by the denominator, both sides. So I will get x squared minus 7x minus 12 equals 4 lots of x squared plus x minus 2. So that becomes 4x squared plus 4x minus 8. Next, this is a quadratic equation it's easier if I have the quadratic equation equaling zero. So I'm going to subtract x squared from both sides, add 7x to both sides, and add 12 to both sides, and I will get zero is equal to 3x squared plus 11x plus 4. And I will think, okay, firstly, if I have a quadratic equation like this, can I factorise?

And I'll try and do so. The way I like factorising is I like thinking in terms of trying to deduce what two binomial expressions will multiply together to make this quadratic. So I know I'm going to get 3x squared by multiplying this term by this term.

So I know it's going to be 3x multiplied by 1x. And I know I'm going to get 4 as the constant by multiplying these two constants together. So then I have to think, are there two constants which I can put there and there, which would then combine in these two terms to make 11x? My choices are that it's definitely going to be both positive because I'm going to get a positive amount of x and two positives multiply to make a positive answer.

Also two negatives would multiply to make a positive answer but that would not give me my positive 11x. So my choice is either 4 and 1 or 2 and 2 and if I try any of those So I'll just start putting things in 2 and 2 there that will give me 6x that will give me 2x, that gives me 8x, which is not the result I wanted. So 2 and 2 does not work. And then 4 and 1, if I put 4 here, I would only get 4x, and then 1 there, I would get 3x, that would make 7x, not 11x, so that doesn't work.

And finally, if I try 4 there and 1 there, I get 12x, which looks promising, and 1x, that looks promising, but I'm adding them together, I'm combining them, that makes 13x. So... This is not factorizable, so I'm not going to use that as a method to solve this.

So what I'm going to do instead is I'm going to use the quadratic formula. I could try and complete the square to solve it, but this is quite a difficult quadratic to complete the square for. So I'm just going to use the quadratic formula.

So my a is 3, my b is 11, my c is 4. So x will equal minus b, so that's minus 11, plus or minus the square root of... b squared so that's 121 minus 4 times a times c so minus 4 times 3 times 4 all over 2a all over 6 and when I simplify that down I get minus 11 plus or minus the square root of 73 over 6. So my two solutions are x is equal to minus 11 plus root 73 over 6 or minus 11 minus root 73 over 6. And I probably should have noticed that this would probably be non-factorizable because it says solve this giving your answer in exact form. Okay so it's looking for a surd form in effect okay rather than some rounder decimal which is not exact. This is beautiful and exact any rounder decimal that you get on your calculator would not be so. Okay let's have a look at the mark scheme.

There it is in all its glory. If you want more practice on this, have a look at lesson one in my sequence of lessons on algebraic fractions. Okay, let's have a look at question seven.

Aha, completing the square, lovely. So in this question you must show detailed reasoning. Express 2x squared plus 8x minus 12 in the form a lots of x plus p all squared plus q. Okay so that's the completed square form. So when I have a quadratic expression which isn't 1x squared it's slightly more difficult to complete the square form but with practice you should get used to doing so.

What I do is I focus on the first two parts the x squared part and the x part and I ensure that I have an expression which is just 1x squared by taking a factor of however many x squareds I have. So 2x squared plus 8x minus 12. I'm going to focus on these two parts. That is the same. 2x squared plus 8x is the same as two lots of 1x squared and 4x.

So I've written exactly the same as I had above. So those two lines are equal to each other. But what I have now is I have something which is very easy. to complete the square for.

So I'm now going to complete the square for that part. So two lots of. Now for that part, this part here, I'm going to just do this in a slightly different colour. So I'm going to complete the square for that purple part there.

x squared and 4x, complete the square is x plus 2 all squared because that would give me my x squared. It would give me 2x and 2x which would give me 4x but it would also give me 4. I don't want 4. I have nothing else in that bracket so I have to subtract 4 from that. So that is what that part would turn into and I've still got the negative 12 at the end.

Okay now I want to get rid of these outer brackets by multiplying by 2. So I get two lots of this x plus 2 all squared. I also get two lots of negative 4, so that's negative 8. And I've still got the minus 12 at the end, which wasn't being multiplied by the 2. Now I just simply combine the minus 8 and the minus 12. So two lots of x plus 2 all squared take away 20. There we go. We have it in that form. Part B, hence find the minimum value of this quadratic expression.

Often that might be in the form in questions, you have this function, this graph of y equals 2x squared plus 8x minus 12. Find the vertex of this function. OK, so that's a positive quadratic. So it'll look like that. OK, or something like that.

And you'll find the minimum point. If it's in completed square form, which we have here. it is very easy to find the coordinates of that minimum point.

Because if I have y is equal to 2 lots of x plus 2 all squared minus 20, I know that when I square something, so x plus 2 all squared, that orange part cannot be below 0. If I square any value, it'll be 0 or above. And then if I double it, still 0 or above. So this part here is at minimum 0. So the smallest I can make y equal to is negative 20. When this part is 0, I get y equals minus 20. So the lowest part of that graph will have a y coordinate of minus 20, which is in effect what they're asking for. The minimum value of this, since I've...

made y equal to that I'm finding the smallest value of y. So minus 20 is our answer. The smallest value that expression can take is negative 20. The x value that would give that is negative 2 but that is not what they asked for.

They asked for simply the outcome of that expression not the input that would make that outcome. So negative 20 is the answer. Let's have a look at the mark scheme. There it is. They do it in a slightly different way to the way I did it.

Still effective. I prefer my way because in my way, it doesn't matter if the final term, so minus 12 in this case, was divisible by 2. It did happen to be divisible by 2 in this case, but it didn't matter if I use my method in future. So if you want more practice on this, lesson five on completing the square.

So let's have a go at the next question, question eight. triangle ABC is such that AB is 5 centimetres, BC is 8 centimetres and CA is 7 centimetres. Show that one angle is 60 degrees.

Okay well let's let's visualise it properly. So I'm going to draw the longest line to 8 centimetres and then the other two lines so 7 centimetres, let's say it's something like that and so 7 centimetres there and 5 centimetres there. So approximately to scale not great because the 7 and the 8 look about the same length in that case it looks almost like an isosceles triangle but ignore that pretend that they are different lengths now in a triangle if you have three different lengths you're gonna have three different angles inside the largest angle inside that triangle will be opposite the largest side okay so this angle there will be the biggest angle The smallest angle will be opposite the smallest side because a smaller angle will open out onto a smaller side and so therefore what we're left with is opposite this 7 is the median angle.

Okay, so there are three angles. If we put them in order of size, the one opposite the 7 will be the median. Okay, if we want to find the 60 degree angle, well, if there are three different angles, 60 degrees cannot be the biggest, because if the other two are smaller than 60, then you wouldn't get a total of 180. Similarly, it cannot be the smallest, because if you have a 60 degree angle and then two which are bigger than 60 that would be over 180 degrees in total and hence the median is the one we're looking for okay so the 60 degrees must be the median angle so we're looking for the angle opposite the seven centimeters so this is now a cosine rule question so if i call that angle theta then i know that cos theta if i know my cosine rule and the rearrangement of it so that cosine rule a squared equals b squared plus c squared minus 2bc cos a you should it is helpful if you know the rearrangement of this into cos a is equal to b squared minus plus c squared minus a squared over 2bc it makes things simpler than having to rearrange it every time yourself if you're trying to find an angle so here in this case since i'm Angle a is opposite little a. I know that angle theta is opposite 7. Therefore, I'm going to have 5 squared plus 8 squared minus 7 squared over 2 times 5 times 8. And that, if I simplify that, I get cos theta is equal to 40 over 80, which is the same as a half. And then inverse cos of a half is 60 degrees.

There we go. QED. I've shown what I wanted to show. Okay, so let's have a look at the mark scheme for this.

Yeah, middle angled, one required, etc. So there's no specific baromaths video on this because this is GCC content, but I hope you enjoyed it anyway. Right, question nine. The factor theorem. Brilliant.

So in this question, you must show detailed reasoning as normal. Show that x minus 3 is a factor of this cubic, x cubed minus 5x squared plus x plus 15. Okay, the fact-tip theorem states that if you have a factor, if x minus a is a factor, then f of a will equal 0. Because if I replace x with a, this bracket will be 0. And if one factor is made 0, you are multiplying by zero and then the rest of the expression would equal the result of the entire expression would be equal to zero. So if I replace x with a I've got zero in that case.

So here you have to think well if that's a factor if x minus three is a factor what value of x would make that factor zero and therefore make the entire expression zero. If x minus three is a factor then f of three will be equal to zero. So let's find f of three.

So that's 3 cubed minus 5 times 3 squared plus 3 plus 15. So 27 minus 5 lots of 9, so minus 45, plus 3 plus 15. And that does indeed equal 0. Hooray! QED, quad error at demonstrandum, which was what was wanted. I am done.

Okay, part B, hence solve the cubic equation. Solving quadratics, you know. you factorize solving cubics is a similar method we factorize but it's more difficult to find that initial factor however they have given us that initial factor so let's use that to factorize the cubic into a linear part and a quadratic part because x minus 3 will multiply by a quadratic x minus 3 is linear it would multiply by a quadratic to end up with a cubic answer okay so i'm gonna factorise it like this.

x minus 3 is one of my factors. What do I multiply it by to get a result of x cubed minus 5x squared plus x plus 15? So I know so far this much. I know I'm going to get a result of x cubed there.

I know I'm going to get a result of 15 there. The rest is going to be deduced. So let's deduce that.

x multiplied by what gives me x cubed? Well that's going to be x squared. So now I know what this part is. Minus 3 times x squared is minus 3x squared.

Since these two will combine to make the x squareds, and I know I want minus 5x squared in total, I have minus 3x squared already. I need another minus 2x squared. And hence, I now know this part. x multiplied by what gives me minus 2x squared? Well, that would be minus 2x.

So now I can fill in this part. Minus 2x times minus 3 is positive 6x. These two parts will combine to make positive 1x. I have 6x already.

I have to then have a negative 5x to combine to make 1x. And hence, the final part would be minus 5. Minus 5 times x makes minus 5x. And I just check using this 15. Does minus 5 times minus 3 give you 15?

Yes, it does. So I've now managed to factorize it into a linear and a quadratic. So I had x cubed. minus 5x squared plus x plus 15 equals 0. I've now factorized that into x minus 3 multiplied by x squared minus 2x minus 5 equals 0. So I know the solution to this will be either when x minus 3 equals 0, so either when this bracket is 0 and that's when x would equal 3, or when this quadratic is zero. So can I factorize that?

No I can't. There's no way of factorizing x squared minus 2x minus 5 into simple rational values inside the brackets. So I'm going to use the quadratic formula.

Okay so from this I'm going to get that x is equal to minus b. So that's positive 2 plus or minus the square root of b squared which is 4 minus 4 times 1 times minus 5 all over 2a which is just 2 and that would simplify to 1 plus or minus root 6. So my solutions are x is either 3 or 1 plus root 6 or 1 minus root 6. So those are my three solutions to this cubic equation. So we'll look at the mark scheme.

There we go. They've given the solutions in decimal form. How horrible.

And they haven't even written what they've rounded it to. Despicable. But if you've given it three significant figures like they have, great, well done.

You get the full marks. So if you want more practice on the factor theorem, see lesson four on the factor theorem. Okay.

One more question to go in this video. Let's have a look at question 10. So here we have a security keypad which uses three letters, A, B and C, and four digits, 1, 2, 3 or 4. A passcode is created using four inputs. So I have four things to input into these four spaces.

in order to get into the keypad system. If there are no restrictions how many different passcodes are possible? Well if there are no restrictions except for these restrictions that they've already given us, because that's what all the things I can press on the keypad, well I have four digits and three letters so I have seven characters I could put in into these inputs. So in the first input I have seven choices.

that I could put in. In the second input, well I have no restrictions, I can still put seven things. I could have a a a a as my password. Wouldn't be advisable but I could.

And then for the third input, seven choices. And for the fourth input, seven choices. So that's seven to the power of four. Okay so seven to the power of four is the different the how many pass words how many passcodes that are possible in the first one. Part B, if there must be exactly two letters and two digits with no repeats, how many different passcodes are possible?

Okay, this takes a little bit more thought. So I have letters I could choose. Okay, so what I could...

choose. If I must pick exactly two letters, I could choose A with B. Or I could choose A with C.

Or I could choose B with C. I'm not talking about what order they're going in yet. I'll come back to that. But those are the three possible choices of two letters with no repeats.

Okay? So I have three possible choices of which two letters I am going to choose. And for...

the digits, again no repeats and choosing two digits, I could choose one with two, I could choose one with three, I could choose one with four, I could choose two with three, two with four, and I could choose three with four. So there are six different possible options of which two digits I choose. I'm not saying which order I'm choosing them in and where they're going, but I'm saying these are the six possible options.

So I could choose this AB with the 1, 2, or the AB with the 1, 3, or the AB with the 1, 4, etc. Okay, so for each of these three choices, I have six possible choices there. So there are three times six, so there are 18. choices of character. Okay, so there are 18 possible choices of the types of character that I'm going to choose. Now, for each of those, so let's say I chose A, B, 1, 2. I can arrange these four characters in four factorial ways.

So each of the 18 choices can be arranged in four factorial ways. So, it's like I've got 18 lots of four factorial different possible options, which gives me 432 options. Okay, and that's how you think about this.

With these types of questions, they require a lot of common sense and they are very difficult. Maths teachers struggle. immensely with permutations, combinations, these types of problems.

And it's about experience with these types of problems that will allow you to solve them more easily. Okay. So the answers were, oh, I forgot to say what 7 to the power of 4 was. That was 2,401. And here's the way they did it.

Okay. Which is kind of similar. So 432. If you want practice on this, see my lessons 20...

for factorials for ordering and maybe even 21 and 22 on permutations and combinations as well. Okay so that is questions 1 to 10 of this paper. Okay the next video I'm going to do, I'm going to do questions 11 to 16 from this sample paper.

Okay so make sure again before you watch the video you have done or attempted those questions. and then watch the video to get a sense of if you're struggling with certain questions, how I might approach them. Okay, enjoy.

Transcript for:Mathematics Practice with Mr. Barrow

Transcript for:
Mathematics Practice with Mr. Barrow