Mathematics Grade 11 Paper 2 May/June 2025 Limpopo | Full Explanation

Added: 2026-05-30

[00:00:00]Welcome back guys. On this video, we're going to take a look at Libumbo grade 11 mathematics which is paper two this time around and this is June 2025 you know.

[00:00:13]Yeah. So this paper it consists of almost seven questions. It's out of 100 you are required to write it for 2 hour.

[00:00:22]So if you want discussion paper you can go to the description box and if you need some certain questions you can also check there on the description box. I'm not going to read the instruction. I'm just going to get straight to question one. So question one is this one. The first question is analytical geometry.

[00:00:38]This is easy. So this one carries 12 marks. Let me just cut off this question so I can be able to start uh with 1.1.

[00:00:47]So here it is. They say calculate the length of uh AB. So the length of A and B. So when we need length, we use the distance formula. So we're going to look for the distance of A which the distance we all know the distance formula guys. I'm going to start with -3 - 4² + -1 - 6² just like that. And if you p this in your calculator, you're going to get 7 <unk>2 units just like that. And then for 1.2, they say find the coordinate of the midpoint of AB. So they want the midpoint of AB.

[00:01:31]So we're going to have X of A. Let me start with X of A. It's -3 + 4 / 2. And then this one is -1 + 6 / 2. Yeah, good.

[00:01:47]And then the coordinate of of day. Okay.

[00:01:50]So a lot of student always ask me, do we punch all of this in a calculator? No.

[00:01:55]What you do, you punch this first, it gives you a coordinate and you punch this one secondly over there. So if you punch the first one, they going to get 1 / two and the second one is going to give you 5 / 2. So this is the coordinate of D. And then the third one is to calculate the angle of inclination of line AB. So the angle of inclination of for example this one is the there's a positive gradient obviously. So we use this formula theta is equals to tan inverse where here we put the gradient of gradient of the line that we need the angle of inclination of. So for this one we need the gradient of AB. So let's say find the gradient of AB.

[00:02:40]So let me start with the I hate negative. That's always I start with the negative. I think it makes me avoid making mistakes. So this one is going to give you one. So you're going to have 10 inverse where we put one which you here they're looking for the alpha actually not theta. So we're going to look for the alpha. Our alpha here is going to be equals to what? It's going to be equals to 45° you know. So this is how we find the angle of inclination of AB. We need a gradient and that's all. But for the negative one you can see on the next question look at the next one they say 1.4 calculate the angle of inclination of beta which is this one over there. So for beta for this one since this one is a it has a negative gradient what we do we say for beta is going to be equals to 180 minus tan inverse then we put the gradient when we put the we don't include the negative so what we need is just the gradient only in order for us to find what we need there. So we're going to find the gradient of BC this time around. So the current of BM start with the -2 -6 and then 6 - 4. Yeah, good with that. So this one is going to give you -4. So our gradient here, we're going to have 180 - tan inverse. We don't put the negative, we just put four. Some other teachers, they also include the negative. So if you understand that kind of a method, please do what you understand or what your teacher has taught you. So if you punch all this going to get 104 just like that you know perfect. So then the next question is calculate the size of ABC.

[00:04:29]So A B C. So the one that there's a cap is the angle that they're looking for over there. So they're looking for this angle over here. Look at this. We have this one which is one. Oh, this one is 104 04. 041. We round into two decimal places. 104 comma 04 you know perfect.

[00:04:50]So to find this beta, you can just simply use the exterior angle of this triangle over here, we can say beta is going to be equals to the addition of alpha plus a b c. Why? This is an exterior angle of a triangle. You know remember analytical geometry is still geometry nuclear you know. Yeah. So beta is 104 which is equals to alpha. We calculate it is 45 plus a b c. So I'm going to take this one to the other side. A b c is going to give you 59a 49. 59 04 actually not 49.

[00:05:39]Just like that you know perfect. So our next question here they say determine the length of line passing through AB.

[00:05:48]So they just want the equation of this line over here. So when we need an equation we just need remember our equation of a straight given as MX + C.

[00:05:59]So we need the gradient and we also need the value of C. So the gradient of A already has it. We calculate it on the first question. I think it's the second question. So we have y which is = x + c.

[00:06:13]To find c, we just need to substitute one point between this. Let me sub b. So I'm going to put 6 here and then put four here plus c and then 6 - 2 is going to be c. If you don't know how to find the equation guys, I have videos on geometry. Please check it out. So c here is going to be equals to 2. So my equation is going to be x + 2. This migraine is one and then the other one is two. So this is going to be the equation of a e. Let's move on to the next one. So now we're moving on to question two. So question two, it carries minimum of 12 marks. Yeah, good.

[00:06:55]Let me just quickly try to make some space and start with the first one. The first one here this Okay, the statement they say ABC is a parallelogram with the vertexes A, B and C and D. So for a on the question paper was not clear. I just errated that the first question is determine the length of AC. So what is AC? So they want the length of AC over there you know. Yeah. So to find it we just need to make use of our distance formula of AC. So here I'm not going to write the formula guys but for you so that you may not make mistakes in your exam please rewrite the formula first before you know and the reason why I'm not writing a formula is because it's only two marks. It's three marks that I need to write the formula. So I'm going to just substitute I will start with the negative side. Oh all of them they have negative. Okay, let me start with this one. Square + 6 - - 2 and then close bracket. Need to be always be careful on the other side. So if you pack this up, we are going to get 4<unk> 5 units just like that. 2.2 determine the determine E the midpoint of AC. So E is somewhere over there. I mean E is somewhere over here. So it is a midpoint of AC. So to find the midpoint of AC, we're going to make use of our formula. So X of A is -4 + X of this one is 0 / 2. And then X Y of the of A is 6 - 2 because it's going to be + 2 over there. And then E, if you punch the first one, they going to get -2. And then if you punch the second one, you're going to get two. So this is the coordinate of E which is the midpoint of AC. So the next one they say determine the coordinate of D which is this D over here and it is worth four marks. So what you must understand is that we're given a parallelogram. What do we know about the parallelogram? We know that the diagonal lines of a parallelogram bisect each other. In other words, a Even these diagonals they share a common midpoint. I mean the midpoint of AC is the same as the midpoint of BD. Why are they the same? is because of the diag of palm. They bisect each other. So you can just see the diagonal lines of palm, you know. So it means that this point E is also a midpoint of D. So if now we have the midpoint and we're looking for this, remember guys, what do we do when we have the midpoint, but we're looking for the last one? We're going to split out our midpoint formula into two halves, you know? So we're going to split it on the x value. So x of d plus x of this point over there which is b / 2. So the x value going to take the midpoint which is -2 and then the x of d is x + x of b is 3 / 2. So I'm just going to cross multiply this multiply by 1 is going to be x + 3. And then that one is going to give us -4. X is going to be = to -4 -3 which is going to be = to -7 you know. So also going to do the same thing on the y value. So y of d minus I mean plus y of b / 2. So now we take this y value which is 2 and then y of d is y + y of b is 5 / 2. we cross multiply y + 5 is going to be = to 2 * 2 is 4. So y is going to be = to 4. We take this one to the other side is -5 and y is going to be = to1.

[00:11:22]So the coordinate of d all in all is -1 and -7 over there. Yeah. So this is it.

[00:11:30]Let's move on. The last question here of analytical geometric. This one they say determine the equation of a straight line from B. So from this point but this from this point is perpendicular to AC.

[00:11:43]So AC is that one. So I don't know if this it touches D or not. But I'm just going to just extend it a little bit over there. But what do I know? I know that they say that it is perpendicular to this side over there, you know.

[00:11:58]Perfect. So if it is perpendicular what do we know about the perpendicular we know that the product of the gradient is equals to1 you know and in order for us to find the equation of the line that's passing through B we need a gradient you know so we're going to find the gradient of AC first so the gradient of AC and what will be the gradient of AC let me start with -2 -6 / by um this system with negative so -4.

[00:12:33]So if you went on PN we're going to get -2. So this is the gradient of AC. We know that multiply it with a grad of let me just name that. Let me assume it passes through point D. So I will say BD it has to give me -2 I mean1 I'm sorry guys. So the GR of this one and why do I say so is because AC is perpendicular to BD. I'm given this they have said it on the statement over them. So yeah multiply by the grand of B D which is equals to -1. So I'm going to divide by -2 and then the gradient of B D is going to be equals to 1 / like that.

[00:13:19]So now I'm looking for the equation of this one. Now I do have the gradient which is y is going to be = 1 / 2x + c.

[00:13:28]To find C, I'm going to substitute this point 5, which is = to 1 / 2 3 + C. And then 5 is going to be, if you punch this in your calculator, I'm going to get 3 / 2 + C. So I'm going to take this one right now to the other side so that I can be able to find the value of C, which is 5 - 3 / 2. C is going to be equals to 7 / 2 actually 7 / 2. So my equation is going to be y 1 / 2x uh + 7 / 2 just like that. So this is going to be the equation of a point b that is perpendic.

[00:14:13]Yeah. So now let's move on to question three. So question three caris minum of did one mark. So now we're starting trigonometry. Let me just start with 3.1.

[00:14:25]3.1 they say if tan 2 theta minus roo<unk> 3 is equals to 0 where theta is between 180 and 6 determine with an aid of a diagram the value of sin² theta plus cos square we know in grade 11 that this is equal to one when we add them this is a grade 10 question actually yeah this is a grade 10 question so this kind of a question to answer it you need to make use of your diagram you know you need to draw a diagram but the big question is Where do we draw our diagram? So you you have two information here that is going to help you to identify that. The first one is this one. The second one is that one, you know. So the first one is not simplify. We need to simplify it. We need to make one trig ratio subject of the formula. So I'm going to take this to the other side. So I'm going to have 2 tan theta which is equals to roo<unk> 3. Then I divide both side by 2 and then tan of theta is going to be equal to<unk> 3 /2. So what does this mean?

[00:15:30]This it means that we're going to draw a diagram where tan is positive because we have a positive ratio on the other side. So we have to draw our graph where tan is positive. So our diagram is something like this.

[00:15:46]Where is tan positive? turn is positive on quadrant number three and also on quadrant number one. So this is not specific guys because we cannot draw a diagram on two quadrants. We need to draw it on one. How do we know which one between these two is correct? Well, that's why you need this information you know. So you need to understand that a cardigan plane it moves in an anticlockwise direction. It moves from this point around over there you know.

[00:16:12]So remember here we have zero, we have 90, we have 1180, we have 270, we have 360 and it keeps on moving you know. So between 1080 and 36060 a lot of people make mistake they think just because we say 360 and 18 is this side and also this side. No, you need to understand it moves in an anticlockwise direction. So from 18 is this one and then it goes to 360. That's what we mean actually. So which quadrant covers 180 and 360 is this quadrant and also this quadrant. So the quadrant there's a lot of these that's where you draw your diagram. You know I have videos on this in grade 10 playlist trigonometry. If you still have trouble go and watch it.

[00:16:58]So we're going to draw our thing on quadrant number one uh two uh three. I mean why is because that's where we have a lot of the these actually. So we're going to draw it over here 90° this is theta we're going to put theta over here and then we need to understand uh the so okay let me just write the so just in case and then we need to understand that this is the same as opposite over adjacent. So our opposite is square root of three and then our adjacent this is opposite. Now our adjacent is two you know. So since this is our diagram opposite is going to be this one and then adjacent obviously is going to be that kind of a number there.

[00:17:45]But look at this here we have a positive number. How can it been ne how can it be positive on this? Because on our cardian pen here we have negative numbers. Here we have negative numbers also. So what you need to understand is that this that is here is the same as when we have tan of theta which is equals to<unk> 3 /<unk> 2 because when we divide this is going to give you the same number as that one.

[00:18:12]So this one we can put 3 over here and -2 over it does not have any effect because when we divide those two we're still going to get that one. So the reason why we put it kind of minus is because we want it to fall under that kind of diagram of ours over there. So now we are left with what with the I mean what is this hypotenuse which we need to make use of our theorem of Pythagoras.

[00:18:43]So our r is what we're looking for. So to be quick, let me just also put the square root and then x is -2² + that one is uh let me put it correctly<unk> 3 squar. So if you punch this in your calculator, you're going to get<unk> 7.

[00:19:04]So roo<unk> 7 is going to be this one.

[00:19:07]So now we can be able to answer our question. So we have sin^ 2 theta + cos² theta. So a common mistake I saw of my student what they do is that they're going to say sign since they know they're going to put like something like this they just only substitute on theta and this is not what you do on this one.

[00:19:26]What you do you're going to open brackets put square and open bracket put square and then what is the definition of sign? S is opposite over hypotenus.

[00:19:37]This is adjacent over hypotenus. So now we're going to just substitute using the diagram. What is our opposite here? Our opposite is -3. So we're going to have <unk>3 I mean so divide by our hypotenus is what we just calculated square plus our adjacent is going to be -2 and then our hypothesis is going to be this one.

[00:20:01]So what you need to do on this one guys let me just finish it over here. Punch this in your calculator one by one. You know that's how because they say you must not use a calculator. So the first one is going to give you 3 / 7 plus the second one is going to give you 4 / 7.

[00:20:17]So you can see now you can pract anything but you can see that these are the same bases. So going to just 3 + 5 we're going to just say 3 + 4 / 7 which is 7 / 7 which is equals what to 1. So this is how we solve this question and we take this form marks. Let's move on to the next. The next one gentlemen is this one. So they say if cost 20 is equals to P determine in terms of P the value of cost 340 you know. So these are those kind of type of coins whereby you need to draw a diagram on quadrant number one and then you put it way in here and then you divide this one by one and then you have um this co is adjacent. So this is one and then p and then you find this one by using the theorem of pythag to give you something like this. So when I saw this I thought maybe it needs this diagram but however it does not need that kind of a diagram because if we expand here we can be able to get D. So for here we're going to have cost 360.

[00:21:19]What number they going to subtract with 360 give us this number over there. It's going to be 20. So we're going to minus it with 20. So if we interpret this 360 is found in quadrant number four where co is positive. So going to have cos of 20. What does cos of 20 equal to? Is equals to p. So this is going to be equals to p. And we take this two marks question. The next one gentlemen is a simplify completely which is that one is worth three marks. So if you can pay to attention to what they want us to simplify here. This is a difference of squares. Let me just slow down over here. So we're going to have um okay let's just distribute one by one. 1* 1 is going to give us 1. 1 * this going to give us cos theta. cos theta* 1 is going to give us cos of theta. Negative cos theta m* cos is going to give us cos² theta. So this you can see right now let me not make this big because this is going to cancel out with that one.

[00:22:29]You're going to be left with 1 - cos² theta. But remember what is 1 cos square theta equal to? It is equals to sin square theta. So this square root is going to cancel this square and I'm going to be left with sin of theta as our answer. And we took this three marks and we go. So yeah, let's move on. So 3.3.2 is still a simplified question.

[00:22:53]But this one it is like this and then it is worth six marks. So yeah, this is an easiest question ever. Uh first we need to change this. If you didn't teach you this, this is how we do we deal with this one because this one we cannot find it in a diagram. What we do uh we add6 is if it was below 16 is to add like if it was below 108 we add 108 something like that. So for this we add if like if it was like for example if we have sine of x - 180 for this one we can add 160 but if it was below uh 18 let's say we have x - 90 you just add 180 over there and then it's going to cancel each other out and you can be able to interpret it. So for this one what we do I'm going to have sign some of the teachers they do it in different ways but for me that I always teach my student is this one at 360 over there it's going to be very easy and then this is a co function it's going to be cos actually so this is going to be cos x so this one is also a negative angle is going to be tan of xid by this is going to be a co function but this One guys this is you need to be very very careful 90 + theta is find here you know and here cos is negative so when this one change is going to be negative sin of x you need to keep that in mind that one is the one that is a little bit tricky so next here look at this we're going to have a sine of x and then this look at this this is going to cancel out with that one and I'm going with sin x and then multiply by cos x I mean cos x and then this one we can change it into sin x over cos x just like that and sin x but look at this this is going to cancel out that one but don't forget the negative so we're going to be left with cos of x * sin sin of x cos of x. So if we divide this, this one is going to cancel out with that one. So we're going to be left with what? We're going to be left with this negative over here which is multiplying sin x. So which our answer is going to be sin x. So the next one gentlemen is this one. They say hence determine the value of x for which x is between 0 and 360. If and then you are given this. Remember this is what we were doing on our previous question. So this we solve it. We end up finding that this is equals to sine x which is equals to what? To 0a 5. You know perfect. So this is just a general solution that we need to do. So we're going to look for a general solution where s is positive. So sign is positive on quadrant number one which first we need the reference angle.

[00:26:08]So a reference angle is going to be equals to sin inverse. We'll put this 0A 5 which our reference angle is going to be equal to 30° and then now we can be able to like find. So if you are looking for a general solution and quadrant number one is included it means that your reference angle is your first solution. So I'm going to have x which is equals to 80 + k 360 k element of integers or where is sign positive? It is positive on quadrant number two. So on quadrant number two we have 180 minus the reference angle which is 30 plus k 360.

[00:26:56]So x is going to be equals to 150 + k 360 k element of integers. So yeah these are the two solution. This is 150 and then this one is 30. So the x values here are going to be uh 30 and 150 just like that. So take these three marks. Let's move on. So the next one gentlemen they say prove that uh this is equal to all of this. So obviously you're going to pick one side and improve. So I'm going to pick this one because this one is the one. Always pick the one that you're going to change a lot of stuff on it, you know. Yeah. So I'm going to pick this one. And then here we have one sin theta. So I can also change this one first which is sin theta / cos theta.

[00:27:48]So from here on I need to h add this fraction. So what I do, we take this one, we multiply with that one, which is going to give us cos square theta and then we take all this, we multiply with that one. You know, let me just show you on the other side. So we have sin theta which we multiply with 1 + sin² theta over there. So okay, let me just write it here. Okay, we have sin. Let me just write it here.

[00:28:29]Just like that. And then here we have uh let me start with cos of theta + sin of theta. So now I can be able to deal with the other one. So this is going to give me plus sin of theta. s multip by s is going to give me sin square theta. So the denominator will always advise you to focus on it at last. Don't focus it when you're studying. Just focus it on at last. So look at this. This and this is going to give us one. So we're going to be left with 1 + sin of theta / cos of theta / 1 + sin of theta. And then boom, we've done.

[00:29:16]Look at this. This is going to cancel out with that one. But we're going to be left with one at the top. And then downwards are going to be left with cos of theta. So the left hand side is equals to the right hand side. And we're done proving this one. Let's move on to the next. The next one they say for which values of act of theta actually in the interval theta between 0 and 60 is the identity in 3.4 four which is this one here is undefined. So what we do when we're dealing with undefined is we're going to take in order for us to have undefined the denominator has to be equal to zero. So what is our denominator? Our denominator is cos of theta it has to be equal to 0 or 1 + sin theta it has to be equal to 0. You know for this one for it to be undecated I will explain it last. Even if I don't explain on this question, I don't think you see relevant because they're going to come up all of them. So now we're going to just solve for those kinds of values. So going to find the reference angle because this is a general solution right now. So reference angle is going to be cos inverse. Then we'll put zero which is going to give you 90° over there. and then uh our x is going to be equals to 90 + k 360 k element of inteious. So here we're looking for a general solution where cos is positive but cos is also positive on 360.

[00:30:48]So 360 - 19 + k 360.

[00:30:54]So our value of x and other one is going to be 2170 plus k 360.

[00:31:02]On the other one uh here we're going to have sin theta which is equals to -1.

[00:31:08]Then our reference angle is going to be sin inverse of 1 is 19. So we're going to look for general solution where s is negative. So s is negative on where we have oh why did I put x here? Sorry guys. This is theta. This is theta. So theta here is going to be equals to 180 + 90 because in quadrant number this is for quadrant number three and then plus k 360 and then three it's just going to be equals to 270 again plus k 360 or k element of inteious or on quadrant number four it's also positive it's also negative on quadrant number four also so it is theta 360 - 90 which is going to give us what is going to give us a I mean 270 also plus k 360 so these our solutions here theta is an element why is it undefined it is undefined on 90 and 270 because they keep on repeating each other so another thing I didn't explain this question whenever you see let's say here H you didn't find this kind of okay whenever you see tan the only way that this like it will be undefined when we have a turn remember guys if you take a calculator and just put tan 90 you're going to get undefined and turn 270 going to get undefined so whenever you see tan on that kind of any equation that are going to be given you know if they want the values of x where x will be undefined you don't You have to calculate your answers are going to be x which is = to 90 or x which is = to 270.

[00:33:00]Whenever you see tan and they need the values of x that are undefined because they already know tan if there is tan that the only value of theta or x that going to substitute that going give us undefined is 90 and 27. So whenever you see t just assume you don't have to calculate you know if you go and check on the me they already return all this but I didn't explain on the other ones.

[00:33:25]So I was explaining all the other one just in case you solve a question and you don't have this t of theta. This is what you do actually you know. Yeah. So the last one here gentlemen they said determine the general solution of this one. So here we're looking for a general solution. What we need to do is to divide by two both side. So we're going to be left with sin of theta which is equals to<unk> 3 / 2. So we're going to look for a general solution where s is negative. So first we need a reference angle which are going to theta is going to be sin inverse and we put this.

[00:34:01]Remember we don't include the negative when we are looking for this. We don't put negative there. You just like punch that but the negative don't include it.

[00:34:09]So our reference angle here is going to be equals to 60. This is our reference angle 60. And then this it means they're going to look for a general solution where s is negative. So s is negative on quadrant number three and also on quadrant number four. So we're going to say theta is going to be equals to 180.

[00:34:34]In quadrant number three we have 180 plus you know and then we add the reference angle plus k 360. So theta is going to be equals to 240 plus k 360 k element of inteious. So in quadrant number four we're going to have 360 minus the reference angle which is 60 + k 360 and our theta is going to be equals to 300 because one will subtract that one there plus k 360 and then k element of inteious. So our values here are going to be uh 2 240 and 300. So this is it. Let's move on to question four. Now let's move on to question four. So this is a trig graphs you know. So there's in a sketch below a graph of f of the functions of f ofx which is a and sin and g of x which is cos x + what what what it is between this is an interval where the graph starting and start from 180 up till 90. So that's just an interval.

[00:35:50]So something to be aware of here. You see this alphabet that they use? This is the same this they're actually telling you like if you find the value of a is still the same as the a of that other one. Hence they use the same alphabet you know. Yeah. So the first question they say determine the values of a k and p. So a okay let's start with a. What is an a? A is our amplitude guys. is the resting position of the graph in order for you to find the value of let me just give you a little bit trick what we do you're just going to go to its origin for example this is the origin and it start from zero and then it ends from zero so it means that it is it is at its origin like that or maybe let me just say it like this let me not use this the resting position of this f ofx graph is this one it rest over here onto and then also So downwards it rest on -2 you know. So the amplitude is this distance over here you know. So how long is it to the resting position you know? So the resting position there is going to be two you know. So in other words our value of a here is going to be equals to look what I'm doing.

[00:37:10]Okay the value of a is going to be equals to two. So this a is the same as the other a. So going to K.

[00:37:18]So for K you need to understand there is a formula that I actually use when whenever I'm teaching this in class we use this formula because we know in order for you to find the period of a graph what we do we say if it is a sign and cos is 360ide by the value of b the this is what we do when we have for example we have y which is equals to sin 2x or sin half x you know because in grade 10 what we normally do for a sign and cos graph the period was always equals to 360. But in grade 11 grade 12 what we do to find the period we take to find a period we take 316 divided with the value of the number that is here always you know yeah good so this is the formula we're going to use actually here in order for us to find the value of b how so is because we understand what is a period guys if you have forgotten the period is how long it takes for a graph to form one wave so this graph it start from here to form a complete wave here it start and uh it start from 0 until 180. So it takes 180° to form a one complete wave. So in other ways the period here is going to be 180.

[00:38:35]So our period is 180 which is equals to 360. So here our B on this graph they put K not B. So let me just put K. So now let me cross multiply. So this is 180 K which is equals to 360 / 180 and 180 K which is equals to 2. So my K here is going to be equals to two.

[00:39:00]Let's move on to the next one. So the next one here is the value of P. Let me just also give you a trick. This that I'm going to tell you right now. It never changes guys. It never changes. So what you must understand first is to see which graph they're talking about. So they're talking about the g of x is this one over. Look what I'm doing guys. So let me just try to write it. We're talking about this graph actually you know. Perfect. So in order for you to be always find P guys if it is one graph what you must like take a look it's to understand the steps of those kinds of graph. The value of P is the steps that the graph it goes with these steps over here. How many does it the distance between these steps you know? So it goes 30 by 30 by 30 by 30 you know. So those are your value of p. But the only problem with that is that we don't know whether it is positive or negative. So what I always tell my student to do when it is like this look at this we have g of x which is equals to the value of a we know it but the value of this one we don't know just like that. But I'm telling you that the value of p is steps that it goes. So it goes with 30° 30° they add 30° over and over and over again you know. So the value of P we already know is is 30 but we don't know if it is negative or positive. How do you know is negative or positive? What I will advise you to do is take out your calculator there. Start to sketch the graph of g of x that does not have any movement. and the original one that there's an amplitude of two like 2 cos of x and find that kind of a a graph.

[00:40:43]Let me just redraw it because I already draw it over here. So if you go and draw this graph you will see got that kind of a graph it goes where is it? Let me just try to take it. It come it comes from 90. Let me draw it.

[00:40:59]So here it is. To calculate it guys you just have to take a calculator and on the value of x put zero on the value of x also put date and find those kinds of coordinate over if you do this on your book you're going to understand this kind of a graph so when I try to draw it it's done something like that so if you pay close attention this graph it is this is the red one is the original one it has moved from this position to that position to that position to that position so it has moved toward to the right hand side. So here's something that you need to understand also. When a graph it move this is like an hyperola guys when the graph move to the right instead of us putting plus we put negative when it move to the left instead of us putting positive I mean negative we put positive. So this is what we're going to basically do here.

[00:41:52]So it means that this graph the red one it is shifted with 30 units toward to the right hand side. So our P here I mean hey our P is going to be negative.

[00:42:05]Why is because it shifted to the right position. So here our P is going to be equals to30 just like that. How do we find out that it has move with 30 H units? What I I would always advise you is just to check the the steps that this graph move on there. So once you find the steps everything is going to be settling. So let's move on to the next question. So our next question they say what is the period of f? So f I already say it guys period of f is going to be equals to 182. So how we find it? So or if you still uh don't understand the period is how long it takes for a graph to for to complete one wavelength you know. So this one it takes 180 uh degrees to form you see this is 0 to 180 to form that kind of one complete wavelength you know. So our period here is going to be equals to let me just write it as P. So period is going to be equals to 180° for this one. And then what is the range of g? So g what is the range guys? A range of this one is how far does the graph exist on the yaxis you know. So check the y-axis. So our graph of g is this graph over here you know. So we want the minimum. Okay let me just also tell if you have forgotten to find the range of a graph. what we range is equals to h the minimum and the maximum value of the graph. This is going to give you the range that you're looking for. But when you write it, I'm going to say it y is an element and then you put the minimum. The minimum here is going to be -2. So it's going to be -2 and the maximum there is going to be two as you can see and two.

[00:44:02]So this is going to be our range for this graph you know. Yeah. Good. So now let's move on to the next question. So our next question over there they say what are the values of for what values of x for which for x is an element between8 to 90 will f of x be equal to g of x? So when two graphs are where two graphs are equal to each other is the point of intersection you know. So I once teach this to my student they don't believe while I was teaching them I told them like the like when we are looking for the general solutions we are looking for the point of intersection between two graphs and I promise them that they're going to see doing trig functions. So even you if you don't know the purpose of doing general solution is that you are equating two graph like when you solve a general solution you will have two different trig ratio sun and cos you know what is basically happening over there is that you are finding the point of intersection between those two graph you know so in order for us to find this kind of point of intersection we need to like uh equate those kinds of uh f and g so remember our f ofx x here is equ= to 2 sin 2x and then our g of x is equ= to 2 cos x - 3t just like that. So what I'm going to do I just going to take this 2 sin 2x which is going to be equals to 2 cos x - 30 over there. So now we need to solve for the general solution you know.

[00:45:55]So this is what I was telling you whenever you are solving a general solution keep in mind each and every time go like you are finding the point of intersection between this graph and also that graph. So when you do you're doing this in January you'll never realize it because you're still not seeing any graph. So we can divide both side by two there. If you divide both side but you can see that this two is going to cancel out with that kind of a two there. You need to try to pause over there because I'm not going to divide.

[00:46:22]So I'm going to be left with this cause.

[00:46:25]Someone will ask himself where is the two? We divide both side by two. It's going to cancel each other out both side. No. Perfect. So our rules of solving the general solution is that we need to make one. We need to like make these kinds of equations have one single trig ratio, you know. So what we're going to do on this one, we're going to make we can make s be like cos by using co functions. Do you know that when we have sine of 90 - x this is equals to what?

[00:46:58]It's equal. Let me not show you cos. Let me just show you with cos actually.

[00:47:02]Okay. When we have cos of 90 - x this is equal to what? This is equals to sin x.

[00:47:10]So in order for us to change this one into co we just need to just uh make use of our um co functions actually you know so we can also do that over here we can just change this one into cos 90 - 2x you know so if we interpret this by using the co function it's going to bring us back where we come from over there. So this is a trade we use over here. So here we have cos x - 30t - 30t.

[00:47:40]So remember whenever we're doing this we need a reference angle. So here that is something that you need to know since our trig ratio here are the same our reference angle here is going to be this number that is over there. don't need to look for the reference angle here and each and every time when we're solving a general solution we need to solve a general solution where we need to know which um I mean which quadrants are we finding our general solution so if it is this one and we have positive on this other side we use the positive side to just interpret our things so we're going to look for general solution of course where cause is positive actually you know so the first one is going to be uh this one so we're going to take this one as our 90 - 2x which is going to be equals to this which is x - 30 and they're not going to give you marks for this screen because it is three they just want those kinds of point of intersections. So so then let me just say plus k 360 so this is for quadrat number one then I'm doing for quadrat number one. So what can we do here? Uh I'm going to take this one. Okay, I'm going to take I want to take this one to the other side and take this to the other side over there. So, I'm not going to just because I just I want more space. I'm going to do it aside. So, since here we have -2x, we take that one is going to be -x over there. And then -2 -1 is going to give me -3. So, this is going to be -3x over here. And then on the other side, I already have -30.

[00:49:15]So, I'm going to take this one to the other. is going to be 90 over there. So -3090 is going to give me -1 120 you know. Yeah. Good. So plus k 360 just like that. Then now I can divide both side by -3 -3 -3. Some other teachers don't divide there. But for me guys I always teach my student to divide. But if your teacher taught you in different way, please listen to him or her. So then I'm going to divide this by -3 which you're going to get 40° and then plus this is going to give you 120 + k 120 and then k element of integers you know. Perfect. So the other one we need to also look for the other one. Allow me guys to just erase over this. So on fourth quarter I'm going to take that one. Remember you have 90 - 2x so which is equals to we need to take 360. Remember our reference angle is what is the other side over there. So it's going to be - x - 30. This is going to be our reference angle always. So I forget to write 90. So here we're going to have 90 - 2x which is equ= to 360.

[00:50:32]And then I'm going to distribute this -x + 30 + k 30. I don't think this one is necessary, but yeah, let me just do it.

[00:50:44]And 16. And then what can I do? Okay, so we're going to have 90 - 2x, which is this is going to give us 390 - x + k 360.

[00:50:59]And then what can I do? I'm going to take this negative to the other side is going to give me -x and then uh this is going to give me 300 on the other side because I'm going to take 9 to the other side. So plus k 360.

[00:51:15]So I'm going to divide both side by so x is going to be equals to -300 minus k 360.

[00:51:25]Yeah, good. Something like that. But this one we don't need it because our interval here it ends at90 and positive 9. So this one is not going to be valid. This one you know we're not going to use it because it's not uh supporting our interval. It does not fall under our interval over there. Let me just also erase all of this actually.

[00:51:51]Yeah. So is it 40? Yeah. 40 is going to be where is going to be 40? default is going to be somewhere over here. But there's also this one and also this one.

[00:52:03]How can we find the other one? So to find the other one, we can make use of our calculator guys. You can take out your calculator then. And then do I have a calculator over here? Okay guys, so I made a mistake here. I just realized it now that I'm editing this video. So on x which is equals to 40 + k. It was supposed to be negative because there the top we divide both side by -3. So I forgot to change negative on K. I put positive. So hence here when I p in the calculator hence I first get the positive numbers and instead of getting the a negative number. So here as you can see it was supposed to be 80 and 40 and 160. That's what it was supposed to be. So I hope you can see that the mistake I've made on K plus on on 40 plus was supposed to be 40 minus K 1.

[00:52:54]The ones that falls under this interval is going to be our x intercept. So which is what which is uh this one this one and that one. So for this one our answer is going to be an x is an element between -18 14 and 160 you know. Yeah. So this is how we find this one. Let's move on to the next one. So our next one is this one. They say for which values of x let me just highlight it. And now we're coming to this one. For which values of x where f ofx multiply by j of x going to give us less than zero. So some other people are still getting trouble with this type of questions. So here we have four marks. Look at this. It's a lot of marks because they know you need to interpret this. But first you need to understand what do they mean by f ofx multiply by g of x is going to give you less than zero. Remember f and g are y values. You know when we say f of x is the same as y but y of what y of f you know when we say g of x is still y but what is the y of g of x. So what is happening basically in simple terms here they want the values of x whereby when you multiply these two graph is going to give you a negative number because less than z is going to give us negative number you know but that one is a little bit vague let me just put it in simple terms in order for us to have negative what we must have is positive and negative or positive and negative in order for us to get negative in simple terms if we can have different signs signs we can be able to achieve what the negative try to just pause and just replay that kind of a part again. So if we can be able to have different signs between these two graph then we can be able to achieve the negative. So in other words what they're looking for here they're looking for the values of x where this graph have different signs you know some the other one is positive the other one is negative that's what they need you know how can we find that so what I always advise my student to do is that draw the vertical line where on their x intercepts actually you know so starting with let me just color this one with yellow and then using that other one with pink.

[00:55:13]So yeah, so I'm going to this one the x intercept is here which is x that is equals to 90 and then the other one.

[00:55:25]Okay, let me just try to move this one to the other. So let me just write this one at the top. Okay, but even here so this is x which is equals to 90. And then the other one is this one. Draw a vertical line. And you're going to interpret this very very very easy. If you draw vertical lines, oh this is -9.

[00:55:45]This is -60.

[00:55:48]This at zero. So this is x y is at zero.

[00:55:51]Yeah. And then we end. And then also you don't forget guys that you also need to draw it on the end values because it ends you know. So don't forget to also because it does not continue it ends over there. So this is where it ends is x where is = to -1 180 and then this one is x = to 90 I mean just like that. So this is9 over there. Yeah perfect. So here we can be able to interpret them carefully. So what do we need? We need where these two graph have different signs. Let me just write the signs of each and everyone. So when we talk about the y values here they are positive positive positive positive positive. But here they are negative, negative, negative, negative, negative. But here they are positive, positive, positive, positive, positive. Going to the other one. Yes. Negative, negative, negative, negative, negative, negative, negative.

[00:56:49]Yes. Positive, positive, positive, positive. And then there, you know. So we want to where they have different signs. So you can see between -1 180 and 90 they have different signs. This is what we need. And then between 90 and9 -6 they have they're negative both of them. So this one does not qualify.

[00:57:11]And then between -60 and 0 you can see that this one is positive. This one is negative. This one it does qualify here.

[00:57:18]They're both positive. We don't want that one. It has failed. So what are our x values here? Our x values um we can put x we put these things here that this it means in between. So the find between 180 and then 90 or what is the next one?

[00:57:38]The next one is -60 and we put in between this it means in between and zero. So this our interval for this one they're going to give you two and two for the other one. These are very very easy. So yeah, moving on to the next one. Let me just erase this so that everyone can be able to see and let me just also erase that one. They say for each of x where f ofx / g of x will be undefined. Okay, look at this guys.

[00:58:09]So we have f ofx / g of x where it will give us undefined. to give us undefined that went by. But when we put this kinds of values in a calculator, we get an error or scientific maths error something like that you know. So you need to pause for a second ask yourself how can you get an undefined when you're dividing two things like what will possibly happen like what can you do in order for it to be able to have undefined. So if for example you take I don't know do you have Yeah. Yeah. If you take a calculator there and just say divide any number by zero actually you know you're going to get undefined. So in other words here what we need is the since this is y of f this is g of f I mean this is y of g. So we don't care about the number of f in order for us to get undefined is when we have uh y of g that is that is equal to zero because let's say for example we go to this point here f here is zero the y value is zero you know and then the y value of this one let's say it's two for example is two this is not going to give us any difference it's going to give us zero but if we find the value of let's say the value of y is two and then we divide by Z. That's where we're going to get undefined. So on our graph, we're just going to look where the graph of g is equal to zero on the yaxis, you know. So it's going to be this point over here.

[00:59:49]That's where it is equal to zero, which is 60 actually, you know. Perfect. So that's why it's going to be equal to zero. So this one, that's what they put one mark. So when x is equals to 60, look at this. If you come to this point, let's just assume this one was two. F it was two. f ofx / g of x. Let's say f the and since is the y value is two. And then that one it is at zero. We're going to get undefined. So the one that kind of an x value that when we substitute over there is going to give us undefined. So it's this x which is equals to uh 60. So going to the last one they say what will be the equation of h of x if h of x= to okay remember guys let me just rewrite the equation of g.

[01:00:48]So here there's equation of h of x that when g is = to 2x +. So whenever there is x on this equation of uh g, we're going to put 2x + 8 in order for us to achieve what h of x. So we're going to have 2 cos whenever there is x we're going to substitute what they've given us there which is 2x + 30 and then we have minus 30 over there and then we just need to simplify this. So this is going to be 2 I mean 2 cos and then 2x + 30 - 30 so h of x is going to be equals to 2 cos x I mean not 2 cos x 2 cos this is going to cancel this so 2x just like that so this is going to be the equation of h of x and this is it for question four which it carries 15 marks Let's move on to the next.

[01:01:52]Okay. So now we're moving on to question five. So question five is mutian geometry which this one it is worth minimum of 11 marks you know. Yeah. So yeah let me just quickly start with the first one. Let's just read the statement first. There in a diagram below a a am a am a am a am a am a am a am a am a am a am a am is a diameter. What is am?

[01:02:15]What is am? So am is this one is a diameter of a bigger circuit. A M we can see that okay they say P R S is the common tangent. So this see what is the common between those two circles and then this the other one are not important. So they say give reasons the size of P1. So what is P1? Is this one? Oh okay cool. P1 is going to be equals to 90 because we have a diameter over there. So we can see 5.1 angle P1 is going to be equals to 90° Y angle in the same second just like that and then 5.2 is show that BN is a diameter of a small second. So what is BN that they're talking about here? So BN is this one over there. They want us to show this is the diameter. So if this it was a diameter, what will we conclude? Like if like let's just pretend for now B and it was a diameter which uh like what will we conclude on this? What we will conclude it was that uh this it was supposed to be 90° if this was a D. So this is what we're going to use actually. So if you can be able to show them that this angle is 90° and then this is going to be a diameter you know. So how can B how can this be a diameter? So if you pay attention they say that MPN is a straight line which is this one is a straight line and then this one also is a straight line that you can see over there. So if this is straightened means that even this angle is equals to 90 because of the vertical opposite angles.

[01:04:13]So angle P4 is equal to 90. Why vertical opposite angle just like that and then 5.3 they say h if M1 is equals to 7 in M1 already given us what will be angle A? Angle A is this one or sum of angles in a triangle guys.

[01:04:38]So this one is very very easy. So 5.3.1 angle A is going to be equals to 20 guys. Why? Because of sum of angles on a triangle. Look here is 70 here is 90. So this one is going to be equals to 20 obviously you know. Perfect. So next is P6. What is where is P6? So P6 is this angle over there. So this is going to be 70. So 5.3.2 and angle P6 is going to be equals to 70° because of the turn cord theorem. This is turn cord just like that. So if this one is 70, what else can we prove here? So they want angle B. Oh, okay. Angle B. Where is angle B? Is this one over here? How can we finding LB?

[01:05:34]Uh we have to go in a long way.

[01:05:37]So since this one is um ah that one is not going to do.

[01:05:45]Oh, we can be able to find this angle P5 because we have 90 and 70. I didn't do this when I was rewriting this. I just like figured out right now. Yeah, I think now we can be able to go with that. Or maybe Oh, here's another way.

[01:06:05]Here's another way. Since we have this ah even all of them, they're the same.

[01:06:11]This angle is going to be equals to 70 because of the vertical opposite. So 5.3.3 angle P3 is going to be equals to 70 because of the vertical opposite angles, you know, because of a straight line over them. So if this is 70 it means that angle N is going to be equals to 70 because we have a triangle over here you know so angle N is going to be equals to 70° because of turn cord remember that's that's a common chord so if this one is 70 this one is going to be 20 angle B is going to be 20° why sum of angles on a triangle if here we have 70 90. This one is going to be 20 just like that. And for this we take this H 11 marks. Yeah.

[01:07:03]Let's move on to the next one. So now we're moving to question six. They say in a figure below O is the center and uh of the circle and P R ST is a cyclic quadrilateral. Okay cool. There's a proof that the angle R what is angle R is this one is equals plus angle T is equals 280. So this is a simple simple proof guys. So what you do here? What can I do? Okay, let me just Okay, first we need to construct now. So what we do in construction? We're going to join this. So this but when I join this this is a construction. So I have to let me just try to cut all this so that it can be able to seems as a construction. So yeah, good. So let me just which one can I name one? Let me just name this one.

[01:07:53]This one O2. Yeah, put. So, let me just also write it for the construction. So, what have I done? I joined H I joined OP and OS.

[01:08:11]That's it. And now I can start to prove.

[01:08:15]So my proof to prove is very very easy.

[01:08:18]What you need to understand is the two theorem which are very important that this angle is twice that kind of an angle over there. What do I mean? I mean angle A1 is equals to is twice angle R actually.

[01:08:35]Why? Because of angle at the center which is twice angle at the circumference just like that. And also this angle is twice that one. I mean angle O2 is twice angle T. Why? Angle at the center which is twice angle at the circumference. Just like that you know perfect. So with this will you believe when I tell you you are done also like this. Now look at this. What is the angles when we add O1 and O2? What do we get? I mean when we add O1 plus angle O2 what do we get? Remember what what what is the sum of the angles around the point? It is equals to 360.

[01:09:27]So the reason why here you have two options. You can either say revolution or some other people say angles around the point. It's up to you. But what is angle O1? O1 is equals to this which is 2 angle R1 plus what is angle O2 is 2 angle T which is not R1 is angle R yeah 360 and then if we take a common factor we're going to be left with angle R plus angle T which is equals to 360 and then we can divide both side by two and angle R + angle T is going to be equals to 80 and we are done with our proof for five marks. Yeah, let's move on to the next. So the next one is question seven. So this is the last last question guys. I'm so tired. So this one carries 14 marks. So okay, cool. So they say complete the following statement.

[01:10:23]The exterior angle of a cyclic quad literally is is equal to the interior opposite. So you can just say is equal to the interior opposite angle of a cyclic quad. So this is what you see on this one. So for 7.2 let me just try to create some space. 7.2 is this one. So this one needs me to read the statement. There a diagram. M is the center. We can see and there's a diameter. We can see that ME is drawn perpendicular. We can see that CDE is a tangent. So, oh, so this one is a tangent. Okay, cool. And then ME is a code. There's no need for me to read the code. So, they say if D4 this one is equals to X, write down reason with reasons. Two other angles are equal to X. So as you can see guys here this one is going to be equals to x because we have a triangle that is this.

[01:11:34]So turn cord. So angle a is going to be equals to x because of turn cord you know perfect. But look at this also since this one is the same it means that this is equal to this. I mean AM is equals to MD. These are the D. I don't think they will give you marks for this but I'm just explaining so you can be able to see. So if this is equal it means that even this one is equal to X angle D2 is equal to X Y angle opposite equal sides just like that. And these are the two angles that equal to X.

[01:12:19]Let's move on to the next. The next one they say prove that CM is the tangent at M. Prove that CM is a tangent at M. So CM is that one that I car with pink. So they can be able to see. So when they say at M and E and D, they're talking about that kind of a triangle that is yellow that I just illustrated over there. In other words, whenever they ask you this kind of question, you need to first bring out the triangle that is formed by these words and also that kind of they're talking about just like the way that they did here. So what you must understand in order for you to be able to put something tangent mostly this type of questions if you can be able to tell them that this is equal to this then you are there you know show them somehow somehow that M1 is equals to angle E then you are done with this.

[01:13:10]How can we show them that? So like we have to go in a long way for this one. I tried to just do it in on the and the with the other method, but yeah, it's just going to confuse some other people.

[01:13:23]So I'm just going to go in a long way to solve this one. So how am I going to solve it? So look at this guys. Remember on our previous question, we prove that this is equal to x. So this m_sub_2 this one is going to be equals to 2x. I mean angle m_sub_1 is going to be equals to 2x because of angle at the center which is twice angle at the circumference just like that. What do I mean? Look at this. We have a center here is twice this angle that is there. Remember guys that kind of a thing we we have that kind want to have a center like this and subtain the same angle there then we always conclude that. So since that now we have this one we can find m_sub_2 you know why do I want m_sub_2 is because once I find m_sub_2 I can be able to okay yeah we will see as we go uh m_sub_2 let me just see m_sub_2 + angle m3 + angle m1 is going to be equals to 1 each y this is a sum of angles on a triangle and then m2 M2 is what I'm looking for.

[01:14:45]So I'm going to take all this to the other side, guys. So M3 is 90 is going to be - 90. M1 is 2x going to be - 2x over there. And then angle M2 is going to be equals to 180 minus this going to be 90 - 2x. So I found this angle 90 - 2x. So look at this.

[01:15:11]Since here we have an a tangent and a line from the sun, it means that this one is 90. I mean m d e is equal to 90° y 10 radius just like that. So if that one is 90 and then this one is 90 - 2, we can be able to say angle E plus angle M_sub_2 plus angle M D.

[01:15:41]It has to give us 180. This is a sum of angles on a triangle.

[01:15:49]So angle E is going to be equals to 1 E.

[01:15:54]I want to be first. I want to take M2 to the other side. So m_sub_2 is okay let me just be slow so that other people can see okay plus okay m_sub_2 is 90 - 2x mde is + 90 which is equals to8 so if you can pay to attention angle e is going to be plus this one is going to be 180 - 2x which is equals to 180 so if I take this all of this with the other side. E is going to be equals to 2x. Why is because 1/8 is going to be cancel out with this one. This one's going to be positive. So this is 2x. So look at this. Boom. Angle m is equals to angle e. Y both equals to 2x. So now we need to conclude. The conclusion is very very important more than what we have done.

[01:16:55]So our conclusion we can say CM is a tangent.

[01:17:01]Why is it a tangent? The reason is because of the converse of turn code theorem turn code. This is very very very important and this is how we solve this and we take this formax.

[01:17:14]So yeah let's move the last question guys of the day is this one finally. So there's a proof f prove that the fmd fbd is exactly quad. I already highlighted it there just to save time.

[01:17:29]So this is very very very easy. Look at this. Since we are given that angle m3 is equal to 90. We are given this. This is a given information.

[01:17:42]Look at this. This is equal to 90°. I mean a dB b is 90°.

[01:17:52]Why? Because of angle in a semi just like that. And then if the is 90°, look at this. This is equal to this.

[01:18:05]Remember guys, this is going to be cyclic quad. Why wise? Because the exterior angle of a cyclic quad are equal. So this exterior angle is equal to that kind of an angle. So angle m3 is equals to a db both equals to 90°. So if they are equal therefore f m b d is a cyclic quad. Why? because of an a converse. Remember this is very important converse of an exterior angle of a cyclic quad.

[01:18:43]Remember guys when you are proving a um a liquid the property that you use when you conclude you just say converse that kind of for example if like we have used let's say we prove m this one was going to be 90 then we say that this is equal to this one or just going to say converse of opposite angles of a cyclic quad just as you see on the previous one we just say converse of t theorem yeah so this is it guys we've reached the end of this video. Thank you so much for watching. I'm Patil. If you haven't subscribed, make sure that you subscribe to the channel, you know. Perfect. My next video shop.