Let me begin with the first application which helps you to know the nature of the roots of a cubic polynomial equation. So I will be talking about one of the applications of maxima and minima which helps you to know the nature of roots of a cubic polynomial equation. So nature means how do you know what kind of roots does a polynomial equation has. That means how many real roots, how many imaginary, it also tells you how many repeated roots could be there. So all those concepts can also be figured out by the use of maxima and minima. Indirectly you can use calculus to know the nature of the roots of a cubic polynomial equation. So let us say I have a cubic polynomial equation like this. A x cube plus a x square plus b x plus c equal to 0. Now many people ask me sir why not a x cube b x square c x plus d, why did you choose your a as a 1. See it doesn't matter. Your roots position will not change even if you divide throughout with a. You anyways know in a cubic polynomial your leading coefficient will never be 0. So I am dividing throughout with that a. For the simple reason that I have to deal with lesser number of unknowns. Instead of dealing with a, b, c, d I choose to deal with a, b and c only. Just remember your nature of the roots is not going to change if you are going to divide the entire polynomial by a given non-zero constant. So let's say I want to know the nature of the root of this given equation. So let me call the polynomial on the left hand side. So in this polynomial equation this is your polynomial. So I am calling that polynomial as f of x. Now what do I do? I find its derivative. I find its derivative. And when I do its derivative I get a quadratic and there is lot of things which is dependent on this quadratic. In fact, lot of things will be dependent on its discriminant. Okay. Just a second. Yeah. Sorry. The discriminant of this quadratic equation plays a very vital role. So what I will do? I will take three cases of its discriminant. One, when your discriminant is negative. So I will start with the fact that the discriminant is negative. Now for a quadratic equation whose leading coefficient is positive, whose leading coefficient is positive, okay? So three is already positive and your discriminant is negative. What does it imply? It implies that f dash x will always be greater than zero. If your f dash x is greater than zero that means your f of x is an increasing function. So an increasing function and especially when it is a cubic function and it is increasing, it can only cut the function at one point, okay? So the function will only be able to cut at one point. So basically it will be like this, okay? It will never fall anywhere. It is always positive. So anywhere on this curve if you take the slope it is always be greater than zero. Right? So what does this imply about the, or what does this tell you about the nature of the root of this cubic polynomial equation? It tells you that it will have only one real root. One proper root. Yes. It will have only one real root. So it implies that f of x equal to zero will have only one real root. That means other two roots are imaginary because in order to create another root it must come down and cut the x axis again which cannot happen if it is always increasing. It cannot come down.  It will not come down. Okay? So if it is not coming down, no more real roots will be generated. Okay? Yeah. Sure Shashank. I'll just call up. Shashank wants to copy something. Okay? So this is the starting point Shashank. You can copy from here. Now let us take the case, let us take the case where your discriminant is greater than zero. If a discriminant is greater than zero, what does it mean? It means f dash x equal to zero has two real and distinct roots. See, I'm talking about f dash x equal to zero. Please don't get me wrong. I'm not talking about f of x equal to zero. So if the quadratic discriminant is positive, we can conclude that the quadratic equation that is this equation equated to zero will have two real and distinct roots. If it is two real and distinct roots, can I say this quadratic over here can be factorized like this. So x1 and x2 are the two roots. I'm assuming here for the sake of simplicity that x1 is lesser than x2. Okay? Let x1 is lesser than x2. So x1 and x2 are the two roots, two real and distinct roots of f dash x equal to zero, not f of x equal to zero, f dash x equal to zero, okay? Now what does it mean? It means five situations can happen here, okay? So what are the five situations? Let us look into it. First situation is your function f of x can show this nature. You can go up, stop at x1, okay? Come down, stop at x2, again go up, okay? So as you can see the cubic equation or the cubic polynomial f of x can show this characteristic. So x1 and x2 are the points where the derivative becomes zero. Remember x1 and x2 are the roots of this equation. Let me write it as two real and distinct roots. Let's say x1 and x2, okay? So at x1 and x2 it will have a maxima and a minima, okay? That's why the derivative becomes zero, okay? So this could be one situation. Another situation could be, another situation could be like this. Let's say it goes up, comes down but doesn't cut and goes back again up. That means this is your x1, this is your x2, okay? This is your second situation, okay? Remember this is the graph of f of x which I am plotting, not f dash x anymore. Cubic polynomial graph. Third situation is this.  So this is your x1, this is your x2, okay? Let me call this as a third situation. Fourth situation could be like this. It could come like this. Let's not pass through origin unnecessarily, okay? Goes up, comes down, touches here and goes back again. So this is your x1, this is your x2 point, okay? This is your fourth situation. First fifth situation will be like this. It comes, touches and goes up like this. Can you think of any sixth situation? Have I missed out any situation? Let me know. What I can think of is these five situations can happen with the f of x polynomial. So this f of x polynomial will show any one of these five characteristics. So what's the fifth situation again? This is the fifth situation. Anything that I have missed out, do let me know. So please have a good look at it and tell me if I have missed out any situation. Okay. Now in the first situation, what is happening actually, you would realize that f of x1 will have an opposite sign as f of x2. That means f of x1 into f of x2 would be negative. Check it out. Yes or no? f of x1 would be a positive value over here. f of x2 would have a negative value over here. Correct? That means f of x1 into f of x2 would be negative. Okay. Now many people ask me, sir, why do you say this is opposite in sign? Why don't we just say f of x1 is positive and f of x2 is negative? Actually I can say that but the problem is here if you divide, say it all depends upon what was your leading coefficient. Correct? Are you getting my point? So when you divide by that leading coefficient and you get this, on this you can comment but on the initial equation you cannot comment. Okay. So what I can say is this is fine or you can say f of x1 is positive and f of x2 is negative. That is fine when you know your leading coefficient of the polynomial equation was 1. But this method which I'm writing, this is equally applicable even if your graph was going the other way around. For example, if your graph was going like this, so in this case what will happen? At x1 it is negative and x2 it is positive. So still their product is negative. So I want to account for both these type of polynomial equations or both these type of polynomial graphs. That is why I'm not putting any kind of extra restriction that f of x1 should always be positive and f of x2 is always negative. No. I don't want to put that restriction. Okay. Now when you realize that the value of the function at the roots of the quadratic that you got over here, x1 and x2 is giving you opposite sign then this is an indication that the polynomial equation will have 3 real and distinct roots. Okay. As you can see one root is here, another root is here, another root is here. Okay. So please make a note of it. Situation number one is a case where you realize that the polynomial equation will give you three real distinct roots. Situation number two is a one where you realize that f of x1 and f of x2 have the same sign. That means f of x1 into f of x2 is positive. And this is a case when you have again one real root. Two other roots are imaginary. So as you can see it has only one real root over here. So now you can see one real root. There could be two situations. Either you have f dash x always positive or you have two distinct roots of f dash x, but such roots for which f of x1 and f of x2 product is positive. And let's take a little bit further here. Here also you would realize that f of x1 into f of x2 will be positive. So this also gives you only one real root criteria. So there are three real roots criteria that you can see over here. That is situation number two, three and the first situation over here. Are you getting my point? In situation number four you realize, yeah somebody is saying something. In situation number four you will realize f of x1 into f of x2 is actually zero because here f of x2 is zero. In such case, in such case, what do we conclude? We conclude that it has got three real roots. Three real roots out of which one of them is repeated. Or you can say two repeated roots out of that. In this case, which is that repeated root? X2. So one root is here, let's say call it as some alpha or something. And the other root is x2, x2. Are you getting my point? So one root is different and two other roots are repeated and that is actually x2. Fifth situation also, you will have f of x1. Let me write it in white only because I have been writing in white so far. Fifth also f of x1 into f of x2 would be zero, which means there are three real roots out of which two are repeated. And that repeated root here is actually your x1. Here it was x2. Is that clear? Any questions regarding this? Out of this, this situation is very much commonly asked. The first one, the first situation. This situation is very commonly asked. I'll put a star next to it. Please copy this and let me know if you have any concerns. Yes, Trippan. Sir, I think this is a little off topic, but then like when we have some minima or maxima occurring at the x-axis, then how many, then we say that it's only two repeated roots there or we can say any other repeated root. Now, if it is touching there, it will have always repeated roots there. How many? Like here it's cubic so only two are possible, but if it's a higher... Yeah, there you cannot comment. There you cannot comment actually. Okay. If it is touching, it can have let's say x to the power four equal to zero. It has got four repeated roots there. So it is touching the x-axis at that point. So by the fact that it is touching, it need not be only two repeated roots. It can be more than that. So maybe we can save the degree of the polynomial as well and the other roots that exist. Like for example, if we have all the other roots touching for n minus two roots touching, then we must have two roots and stuff like I think that depends on the graph itself. That depends on imaginary roots and all. That depends upon the degree. So then in that case you won't have imaginary roots possible? Why not? We can have. We can have, no. For example, a six degree polynomial can have four repeated roots and two roots are imaginary. Yes, so we can't tell those, we don't know those two imaginary roots. We can't say unless I see if I tell you that there's a graph which looks like this. Now I don't tell you the degree of this polynomial. Let's say I don't tell you the degree of this polynomial. I mean you can make out from the graph that it is a fourth degree or something like that. But let us say if I give you such a graph by which you are not able to decide how many degrees is there. So if I just ask you how many places it is touching over here. So it could be two, it could be four also. But here you know it is going to be two only because one root is here and other root is here which are imaginary. So it has got two repeated roots and two imaginary roots. But if I just give you a graph like this, no. Okay, then probably you will not be able to tell what is the degree, without the degree you will not be able to tell how many repeated roots it has. Like x to the power four, x to the power six. Yes sir, yes. So those two whenever there's some, some f dash x equal to zero occurring somewhere else then that's an imaginary root. Where, where, which one? Sir, like in the graph you drew you marked those two as imaginary roots. Yes, yes, yes. So basically two roots are imaginary because it is not able to touch. Okay. Okay, so one root is real here and two roots are imaginary. Right, similarly here also. One root is real over here and two roots are imaginary. Okay. So moving on to the third scenario where your discriminant is equal to zero. When your discriminant is equal to zero, what does it mean? It means f dash x equal to zero has repeated roots. You can say has real and equal roots, which is actually repeated roots. So let us say your f dash x is factorizable as three x minus one x one the whole square. See, let's go back. Your f dash x was a quadratic. Check this out.  Isn't it? And if you're saying this quadratic has real and equal roots and let's say that root is x one. Can I say this quadratic I can factorize like this? Okay. So let's say that repeated root is let's let's say that repeated root is x one. Okay. So can I factorize it like this? So what does this mean? It means your f of x can be written as x minus x one whole cube plus a constant. I just integrated it. Correct. Now everything depends upon this guy. Everything depends upon this guy. If if k is positive. Okay. If k is positive or k is negative, so let me write it like this. If k is equal to zero, then we say that f of x equal to zero has three repeated roots, three real and repeated roots. And what is that real and repeated root here x one itself. So if you see this is zero, then it is basically becoming x minus x one cube whole equal to zero. Right. That means it has got three repeated roots x one x one x one. A classic example of this would be your x cube equal to zero. That means it is a perfect cube. If it is a perfect cube, then it will have three real repeated roots. Am I right? Yeah, sir. Correct. And if k is not equal to zero, then again your f of x has has one real root. Are you getting my point? Why has one real root? See, if I write x minus x one cube plus k equal to zero. So what will you write? This has x minus x one cube is equal to negative k. Correct. If you take the cube root of this, remember only one will be real out of it and two will be imaginary or non real. So as to say imaginary. Correct. Why am I writing half in yellow and half in? Yeah. For example, if let's say k was one, so negative one cube root, negative one cube root, you know it's negative one minus omega minus omega squared. So two of them will be imaginary. So out of this one real root, you can actually get your one real root for the equation. The other two will be not real. So they will be imaginary in nature. Even if your k was negative quantity, let's say even if you had a minus one over here and you brought it to the other side and made a one, cube root of one will also give you two imaginary quantities. Are you getting my point? So this just tries to say that the curve x cube will just shift x one right and k up or down depending upon its sign. So if k is positive, then it would the graph would be like this. So x cube graph shifted x one to the right and k up. So in such cases, it will generate only one real root as you can see over here. By the way, I should not make it pass through origin. It gives the wrong notion that it is passing through origin. So it is actually like this. So only one real root will be created over here. So this is the root. Okay. Even if a case negative, it will be shifted. You know, anyways, I mean it could shift right down right up. Okay. Left down left up wherever it shifts. It is like shifting off x cube graph. Remember your bridge course. If you are doing something to this function like this x minus x one cube plus k. What do you what do you infer from this? It means that x cube graph has been shifted. Yeah. Shifted by x one units right and k units up. Isn't it? Okay. That depends upon sign of x one and k also. So it could be like this also. So I'm not denying it. So this point could be your x one comma k point. So anyways, it will generate one root here. Is this idea clear? So please make a note of these three situations over here. When discriminant is less than zero, discriminant is greater than zero and when discriminant is equal to zero. Let's take questions on this and then we'll take a break. Can I go on to the next page? Yes, sir. Okay. So let's take a question. Find the value of a if this cubic equation cubic polynomial equation has three distinct real roots. In fact, they should say find the values of it. So find the values of it. We'll get a range of values. So which situation will you use over here? When is the only occasion when it has got three real distinct roots? f of x one into f of x two. When discriminant is greater than zero and f of x one, f of x two is negative in sign. Yes, sir. Done. Very good. Very good. Very good. See, if you differentiate this, let's say I call this as your function f of x. Okay. So if you differentiate f dash x, you'll end up getting three x square minus three. Okay. It clearly has two roots. Minus one and one. Correct. Okay. So let's say x one and x two are the two roots. So roots are plus minus one. So, okay, let's say this is x one and this is x two.  That means a minus two into a plus two should be negative. That means a should lie between minus two to two. So this is going to be a ring interval. This is the answer. Getting my point here. Any questions? Let's take another one and after this we'll go for a break. Take the question carefully. It's a very good question. So there is a cubic polynomial equation in T. Okay. And later on this T is equated to x plus one by x. Okay. And they're saying it'll give you six real and distinct values of x. Okay. We'll see what is the right answer for that. Why you don't think so? Any reason for that? Yes. You have anything to tell me? Like by doing it, I think we get to non-cooperating intervals. Non-cooperating moment? Non-cooperating intervals. Like they don't, you know, exactly overlap. What are we mixing history and physics? Non-cooperation moment is happening, sir. Okay. Let's discuss this. Else it'll take a lot of time for us to solve the question. See, try to understand the question first. The question is saying that whatever T comes out from this equation, if you put it against the second equation. So let's say I get three values of T. Okay. So T takes three values, T1, T2 and T3. Okay. For each of these three values, if I solve this equation, let's say I solve this equation, I solve this equation, I solve this equation. Okay. Okay. I should get six real distinct answers for X. That means this should give me two answers. Okay. This should also give me two answers. This should also give me two answers because these are quadratic in X. And not only that, these answers should all be distinct. Different also. Different also. Correct. So for this to happen, first of all, T1, T2, T3 must itself be distinct. Correct. Correct. So first situation that I can say is that your T1, T2, T3 must be distinct. Must be real and distinct. And secondly, you would realize that if you want such an equation to have some roots, then your T1, T2, T3, all of them should be either be greater than two or less than minus two. Right. Secondly, your T1, T2, T3 must either be greater than two or less than minus two. Now, why is that so? Okay. If it is equal to two, you will have only one answer for it. That is one comma one. That is one and one. One is satisfied. Right. If it is less than two, you will not have any roots at all. Why? If you take a quantity x plus one by x and equate it to T, let's say. Okay. So if you just make a quadratic out of it, it will be having x square minus Tx plus one equal to zero. Right. So for you to have any roots, T square minus four must be greater than equal to zero. Correct. That means T must be greater than equal to two or T must be less than minus two. Correct. So if it is between minus two and two, there will not be any real roots at all. So forget about having six distinct roots. There will be no real roots at all coming from each of the equations. Now I've excluded two and minus two also because at two and minus two, they will have one one root each. Correct. So that means that will give me only three real distinct roots. In fact, that'll give me three real equal roots. That's why two and minus two also have excluded. Correct. Is that fine? Okay. Now taking into consideration all these things, let us say I go for the graph of this polynomial equation. So my graph should be such that my first root occurs before two. Right. So it is basically going up like this. Right. Yes or no. Yes. Second root should come after two and third root also after two. Something like this. Okay. Is this the only possibility or I could have more possibilities? Got all three be before minus two or after. Can all three be before minus two? Okay. So let's have a situation like this. What about the double derivative? Sorry, what about the derivative of this function? That will also tell you a lot of things. Let's differentiate this with respect to t. What do we get? Shankin, 60 square minus 18 t minus 18. Okay. That means your function maxima or minima should occur at zero and three. Am I right? So zero is here. Three is somewhere over here. Now, if you are claiming that all of them occurs before minus two or all of them occurs after minus two, then it will create a problem because it will not satisfy this particular condition. Yes. Can I say this is the only possibility? Sir, the root on the left can be after minus two also. It can be between, yeah, alpha one can be between minus two and two. But I think we need... Alpha one cannot be between minus two and two. These are your t values. These are your t values. Alpha one, alpha two, alpha three are your t values. Correct. So t values can be only be in this interval, no? It could be... But if you have to arrive at that of... Oh, okay. Okay. Okay. Yeah. So your root should be before minus two or after two and your function graph is such a way that it achieves a maxima and a minima at zero and three respectively. So this is the only possibility you can have. No other possibility. Correct.  So, can I say the first thing that I should adhere to is f of zero and f of three must be negative. Then only I'll have real and distinct roots. Okay. And at the same time, can I say f of minus two should be positive and f of two should also be positive? So second condition is f of minus two should be positive and f of two should also be positive. So this is my third condition. Okay. So my answer would be the intersection of all these three restrictions. So let us solve them one by one. So from the first one, what do I get? f zero into f of three should be negative. What is f zero? Thirty minus a. Okay. F three would be what? F three. Yeah. Yeah. F three would be two into 2754 minus 36 plus 30 minus a. Correct. No. Oh, wait a minute. So yes, three minus a correct. So this should be positive negative. In other words, you're saying a minus 30 a minus three should be negative. That means a should lie between three and 30. Okay. That's the first condition. Second says f of minus two. What is f of minus two? Less greater than zero. That I know. I want what is f of minus two. Excellent. Okay. I feel he is saying greater than two. F of minus two. So two into minus eight minus 36 plus 30 minus a. How much is it? Tell you are a J as parent taking so much time for doing simple calculations like this. So you're less than minus 22. That I don't want. I want the expression. I just want the expression. Are you saying minus 22 minus a you're getting? Yes. So a is less than 22. Okay. Third is f of two is greater than zero. What is f of two? A minus 10. Huh? 10 minus a. 10 minus a. 10 minus a greater than zero means a is greater than, sorry, a is less than 10. Okay. So what is the overlap of these three? Let's check it out. So the least values minus 22. Then we have three. Then we have 10 and then we have 30. First interval says between three to 30. Second interval says you have to be less than equal to 22. God. There itself it is gone. No overlap happening. Okay. Others is greater than less than 10. So there is no overlap.  So when could you are correct? No solution. Was that a guess actually? No, sir. To be honest, I'd seen this question before. I didn't understand it, but I'd seen it in my school doubt group. Good. You're honest. So now we'll take a break. Right now the time on my watch is 649 p.m. I'm giving you two seconds extra here. We'll meet at 704 p.m. There we'll talk about optimization problems. Right. So I hope everybody's back. Yes, sir. So in this segment, we are going to talk about some optimization problems which are solved by the use of maximum and minimum. Okay. Now in optimization problems. Optimization means finding the maximum or minimum of some objective function. Okay. So what will happen is that they will give you a objective function. Okay. So this is what we call as an objective function. Objective function is a function which you want to maximize or minimize. Okay.  Okay. And under this constraint, we need to maximize this objective function. Okay. So how do we do that? For that we have some simple steps that we follow. Number one, we first convert this function to a single variable function. We convert this to a single variable function by using the constraint function. Okay. After we have done that, we differentiate. We differentiate with respect to that single variable and put it zero. The roots of this equation will give you the extrema. Okay. And then the third step, which is very important and many students do not do this step in the school exams and they lose one mark or two. Is you have to verify the nature of the extrema points, whether it is actually giving you a maxima or it is giving you the minima. Let me explain this with an example. Let us say, let us say I asked you this question that if X and Y are positive real quantities that satisfy X plus Y is equal to 60. Okay. Then find the maximum value of X cubed Y. So now you can see here that your constraint function is this. So this is your constraint function. Okay. Of course, X and Y should be positive. So these are your constraints. And this is your objective function. This is what we call as the objective function. Objective function means it is your objective here to maximize or minimize this. That depends upon what is the question asked. So we have to maximize this objective function. Objective function. So what do we do? Step number one, as I've already mentioned, you convert this to a single variable. So what do we do? Step number one, as I've already mentioned, you convert this to a single variable function using the constraint. So what do I do is instead of this, I will write Y, I will write it as 60 minus X. Okay. So let's say I call this function as your function P. Okay. So now it is a single variable function. So this is now a single variable function. Single variable function. So step number two is now we differentiate the single variable function and put it to zero. So this is your step number one. Step number two, I'll add it over here. You differentiate the function with respect to that single variable and put it equal to zero. So derivative of X cubed 60 minus X, you have to find and put it to zero. If I'm not mistaken, derivative here would be 180 X squared minus 4X cubed, correct? This will be equal to zero. That means 4X squared 45 minus X will put it as zero. So there are two possibilities here. In fact, if you use your wavy curve, 45, you can write zero first here and 45. So this would be negative, positive and positive. Okay. So here we can know the nature of these critical points. So I get zero and 45 as my critical points. So 45 is the point where from positive slope it is becoming negative. This is the point of maxima. You can say local maxima. And zero is actually a point of inflection, which is a point of neither maximum or minimum. So X equal to 45 is a point where the function will achieve its maximum value. So if X is 45, Y will be 15. Okay. And then you can go ahead and find the value of this function, which is X cubed Y, which is 45 cubed into 15. That I'm not going to do. Okay. So this test is very important because when you get a point, there would be, there may be points of maxima and minima both, or there may be points of inflection also. So you need to tell the examiner that, okay, at this point a maxima will occur or at certain point minima will occur. So that test has to be done in the school exams. Sir, other than this wavy curve, the other test is double derivative. Double derivative. Yes. Yes. Okay. But if double derivative gives you zero for a point, that means you have to go for higher derivative tests. All those, or even and all that. Absolutely. Yes. Okay. So could you go to the rules for a minute? Yeah, sure. So C X comma Y is your constraint function. Done. Tirupan? Yes, sir. Thank you. Okay. So let us begin with a question. This, let's say there is a person, Venkat B. Okay. This person is trying to take a ladder from a point A to point B. Okay.  This ladder cannot be made vertical. It can only be kept horizontal. It is a heavy ladder. So you cannot make it vertical or it is, you know, such a ladder which balancing vertically is not possible. So you can only carry it horizontally like this. Okay. And this is a lane. This is the right angle lane. Okay. Now, of course, many of you would have also experienced this situation while shifting your houses or, you know, trying to buy a new furniture into your house, especially when you're living in an apartment with staircase. It becomes very difficult to maneuver the furniture about the corners. So let's say Venkat is trying to move a ladder. Let me make it a different color across this right angle street or right angle lane. My question is what is the maximum length of this ladder L that he can maneuver across this lane if this length is known to be A and this length is known to be B. So what I've drawn is basically a situation when, where he is in a deadlock position. That means his ladder is just about to cross or just about to get stuck. You can call this as a situation where it is just about to pass. That means he can now maneuver it in this way. I hope you can understand the situation. So he's standing here holding it like this and trying to take it this way. This person is Venkat. Stick diagram. Venkat stick diagram. So what is the maximum length of this ladder L that he can maneuver across this lane, which is right angular and this lane is of with A and this lane is of with B. This question can come even for your school exams. Of course, small change will be there in the question. They may not use the word Venkat there. So ladder is your objective function. So try to write ladder in terms of a thing which will keep on changing. In the interest of time, I'll also work with you because else I'll not be able to complete this chapter. So let's say this angle is theta. Okay, let's say, let's say this angle is theta. I'll just make a, okay, let's say this angle is theta. If this angle is theta, you would all admit that even this angle will be theta. Okay. So what is this length of the ladder? Let's say X and say this length of the ladder is Y. So can I say X will be A cos theta and Y will be B cos theta. This is your B. Okay.  Now, basically you have got an objective function which is in terms of a variable theta. Okay. So for your ladder length to be maximum or minimum, the derivative of L with respect to theta should be 0. Correct. Which means minus A cos theta cot theta plus B sec theta tan theta should be equal to 0. Correct. Which is nothing but saying A cos theta by sine theta should be equal to B sine theta by cos theta. In one minute. A, yeah. This will become sine theta by cos square theta. I'm sorry. And this will become sine square theta. Correct. In other words, in other words, I can say A by B. A by B will be tan cube theta. That means tan of theta is A to the power one-third B to the power one-third. Okay. So if tan theta is this, what is cos theta? Can I say cos theta would be under root of this square this square which is A to the power two by three B to the power two by three divided by A to the power one by three. Hypotenuse by perpendicular.  So this is your hypotenuse and this is your perpendicular. Okay. Similarly, what will be your sec theta? Sec theta would be again hypotenuse by base. Now why I found out cosik and sec is because in the expression of the ladder I need to use those. Added. Cosik and sec. Yeah. So multiply this with an A. So your A cosik theta plus B sec theta would be A to the power two by three under root of this plus B to the power two by three under root of. Take common. So this will become A to the power two by three B to the power two by three whole to the power of three by two. So this will be your length of the ladder. Okay. Now I'm not doing the double derivative test to check whether this was the point of maxima and minima. So I'm leaving up to you to do that. So please, please do the double derivative test. Please check that d two L by d theta square should be negative for you to have the maximum length of the lab. Okay. Any questions here? Approach is clear how it works. So in this question, one learning is there is that when you are trying to find out the objective function, you can write it in terms of a assumed parameter. For example, theta was nowhere mentioned in the question. So I assumed theta to make my life easy. So don't shy away from taking this approach. Many people, they don't like taking this approach as a result, the problem becomes super lengthy. Let's take this question. There's a small change in the function. This will be for action. So read the question everyone. Let me draw the scenario for you all so that the understanding is clear. So there is a triangle ABC. Let's say ABC and on this triangle. Okay. They're given the coordinates of these points also. So is this B is this and see is this. Okay. Now on this triangle, they're choosing some points. For example, on BC, they're choosing D. So BC, they're choosing a point D. Let's say, okay. AC, they're choosing a point E. So let's say E is here. Okay. And F, they're choosing on AB, something like this. Now, by connecting these lines, they are creating AFD. AFD would be a parallelogram. That's what they're claiming. So these points are located in such a way that AB, AFD is a parallelogram. Using calculus show that the maximum area of this parallelogram is this expression. 1 fourth P plus Q, Q plus R, Q minus R. How would you solve this question? ITJ in 1986. So if this is a parallelogram, that means this is parallel to this. And this is parallel to this. Okay. Okay. So can I say, let's say this, this. Please wait. Okay. You're trying to solve it on your own. Okay. Okay. Take your time. Next class, I'll be starting with definite integrals. That's again a lengthy chapter. So three classes minimum will go. Okay. Let's, let's proceed as it'll be too much delayed. And by naming convention of a triangle, let's say I call this as B and I call this as C. So we all know in a triangle, the side opposite to an angle is named by the small alphabet for that angle. So this is side B. And this whole thing is side C. Can I say triangle CED, CED is similar to triangle CAB, right? Because this is parallel to this. This is a common angle. So these two angles will also be the corresponding angles. So by using my basic proportionality theorem, can I say, can I say that, this length by whole length will be this length by this whole length? So can I say EC by AC? EC by AC is ED by AB. Correct? Now what is EC? EC is B minus Y. AC is B. ED is X. AB is C. Correct? So can I say from here, 1 minus Y by B is equal to X by C. That means Y is equal to B times 1 minus X by C. Okay? Let's call it as 1. Now, we all know that, let's say this is angle A. We all know that area of this parallelogram, let me call this area of the parallelogram by P. So let P be the area of the parallelogram. We all know that area of the parallelogram in this situation will be XY sine A. Am I right? Now, A angle is fixed. So this is a constant term. X and Y are two variables. So as you can see, this is my objective function. This is my objective function, which is in two variables. Now I need to convert it to one variable, which I can easily do because of 1. So by using 1 or we can say from 1, I can convert it to a single variable function. So let X be X. Let Y be B times 1 minus X by C sine A. Now if you want this function to be maximum, its derivative with respect to X should be 0. Okay? Let's do the derivative of this. B sine A will be constant. So you have X minus X square by C derivative. This should be 0. That means B sine A, 1 minus 2X by C should be 0. In other words, 1 minus 2X by C should be 0. That means X should be C by 2. If X is C by 2, Y will be... What will be Y if X is C by 2? B by 2. Correct? D by 2. Right? That means the area of the parallelogram would be X, Y, that is B, C by 4, sine A. Right? Now we all know that it is actually half of half B, C, sine A. And what is half B, C, sine A? In any triangle, what does half B, C, sine A signify? Area of the triangle. Area of the triangle. Correct? So your answer would be your maximum area would be half of the area of the triangle A, B, C. Okay? Now in order to find the area of the triangle A, B, C, I will use its coordinates which has been specified to me in the question. So these are the coordinates. We all know how to find the area of a triangle when its coordinates are specified, which is half X1, Y1, 1, X2, Y2, 1, X3, Y3, 1. Correct? So it's P square minus P1, Q square Q1, R square minus R1. And you guys have all done your determinants. I'm sure you'll be able to solve this. So let us do this operation. So you'll do this operation R2 is R2 minus R1. And let's do R3 is R3 minus R1. So let us subtract the first row with all the other rows. So it's Q square minus P square, Q plus P0.  Okay? Let's take few things common. For example, I can take Q plus P common over here. And I can take P minus R common for me. So I'll get P square minus P1. And this will be Q minus P1, 0. This will be minus R plus P1, 0. Okay? So let us expand it now. Let us expand with respect to the third. Expand with respect to the third column. What does this give you?   So area of the triangle is given by this term. So area of the parallelogram would be again, half of this will be 1 fourth, 1 fourth P plus Q. Let me write Q plus R first. And let me write this as P minus R. I think this is what we wanted to prove, right? Check it out. Yeah. 1 fourth P plus Q, Q plus R, P minus R. Yeah. Okay. Is this clear? Let us take another one. By the way, anybody who wants to copy this, please do so. I think the idea is clear now what to do. Should I go to the next page? Yes. Some easier ones. Show that the semi vertical angle of a right cone of a given total surface area, including the area of the base and maximum volume is given by sine inverse one third. Done. So let's say the semi vertical angle is theta. Now, what is given? The total surface area is given. Total surface area will be pi RL plus pi R squared. Okay. This is given as a constant. Okay. Let's call it as C. Okay. So read the question. Clearly states that the surface area is given. Given means it is fixed. You can't change it. The volume is the objective function. So volume of a cone is one third pi R squared H. So this is your objective function. And this is your constraint. This is your constraint. Now your objective function is present in two variables, R and H both. Right. So how can we change this? H is R cot theta. Right. H as R cot theta sir. R cot theta. Okay. And where will I get theta from then? The semi vertical angle. Okay. So let's say R in terms of theta. Yeah. I take this as R. Okay. Now what is given to us is pi RL. Pi RL. Now this is your L. Okay. So what is L actually? In terms of R and theta. L is R by sine theta. Sine theta. Right. So pi R squared is given to us as a constant. Correct. Yes or no? In other words. In other words. Pi R squared which you can take as a constant. 1 by sine theta plus 1 is equal to C. Okay. So you can write pi R squared as C times sine theta by sine theta plus 1. What is H? What is H? H is R cot theta. Yes sir. Okay. So what do you do in this case? One minute sir. Yeah. Theta would make it complicated. Don't you think so? Because now you have to write your R square in terms of it. And then you have to square this and then you have to differentiate it. Why don't we take a non-theta approach? Sir, isn't there any way of scanning? Can't do that sir. Go ahead. It's actually your call. See what you have to do here is you have to write R squared in terms of theta which you can place over here. But H is also having R in it. Correct. So we have to square this up. Okay. So there are multiple ways to do it. So, okay, let's follow this approach only. So now, if you see this term, V square is one ninth of pi square R to the power four H squared. And H squared is equal to R squared cot squared theta. Correct. That means H squared is equal to R squared. R squared is this term, okay? By pi. This into cot squared theta. So basically you have one third. This term is the square of this term which is C squared sin theta by sin theta plus one the whole square. Into H squared. H squared is this term which is one by pi. Or you can say C by pi. This will become cube into cot squared theta. This will become your volume square. Now you must be wondering why am I interested in volume square is because if V is max, that means V square will also be max. So I can say D by D theta of V square will be equal to zero. So right now you need to, if you collect all the constants together, this will be your constants. And the variables would be sin cube theta by sin theta plus one whole cube into cos square theta by sin square theta. So here you can cancel off, make it sin theta and differentiate this with respect to theta and put it zero. Can we write sin theta as D and differentiate with respect to D because anyways we have to find out the sin theta. Oh yes sorry, this will be one ninth. Oh sorry, what did you say? Can we use sin theta itself as a variable? Yeah, why not? That's a very good idea. So Trippan has a very good idea here. He can, he's suggesting let's write sin theta as T. So now you are differentiating D, T1 minus T squared. By the way, a lot of terms will get cancelled here also. Actually it's not a very tedious job. So you have to do this. Yeah, correct Trippan. So we can write it as DT of, this is T1 minus T1 plus T that will get cancelled with this guy. So it will be 1 plus T the whole square. Into DT by D theta it should be equal to zero. DT by D theta is equal to zero. Let's forget about it. Let's differentiate this. Oh sorry. This will be 1 minus T square whole square. Derivative of this will be 1 minus 2T. Then minus T1 minus T into derivative of this will be 2 1 plus T. Okay. Denominator I'm not considering. I'm just saying it is equal to zero. And DT by D theta which is actually cos theta. DT by D theta is cos theta cannot be zero because if it is zero that means theta will be pi by 2. That means your semi-vertical angle is pi by 2. Means your cone will be almost like flat. Okay. So let's simplify this. Out of this a lot of things will be cancelled. For example, I can cancel off 1 plus T from all the sides because I know 1 plus T cannot be zero. So if you simplify this, you'll end up getting 1 minus 2T plus T which is minus T minus 2T square. Here also you'll get plus 2T square and minus 2T equal to zero.  So you end up getting 3T equal to 1. So T is equal to one-third. T equal to one-third means sin of theta is going to be one-third. Which means theta is equal to sin inverse one-third. Okay. See actually it's your call. How you want to solve it. But many people they get muddled up in the expression of this term because they're not able to convert it to a single variable. And this is a very good question. Please mark it with a star. If it comes in your school exam, most of you would waste a lot of time on this. I'm sure. Okay. So here you can write because DT by D theta cannot be zero. So only this is possible. After this you know how to proceed. So no need to copy after that. Can I move on to the next question? Find the points on this curve whose distance from the origin is minimum. This question actually can be solved without maxima minima also. I mean the the role of maxima minima is very less in this. Any idea how to do this? Anybody? Just give him. Okay. Okay. Take your time. No worries. Okay. Let's say the point is X, Y, which is at a minimum distance from the origin. So what did I do is I took this distance as R and the angle theta here so that I can write my point as R cos theta and R sin theta. So this is the point which I'm looking for whose distance is minimum. So first of all, this must satisfy this must satisfy the given curve. This must satisfy AX square to HX Y plus AY square. Secondly, I want R to be minimum. Correct. So R is my objective function and constraint is that R cos theta comma R sin theta must satisfy this curve. In other words, you're saying A R square cos square theta plus 2H R square sin theta cos theta plus A R square sin square theta is equal to C. If you take R square common from these two terms, you'll end up getting A and from these two terms, you'll end up getting H sin 2 theta. So this is going to be a C. So R square is equal to C by A plus H sin 2 theta.  Now, the reason why I said you do not need calculus for this is because if you want your R square to be minimum, this should be maximum. Correct. And this could be maximum only when sin 2 theta is equal to 1, which anyways calculus will also give you the same thing. But why do you use calculus when it can be done without it also? That means theta should be equal to 45 degrees. Correct. That means your minimum value of R would be under root of C by A plus H. But the question is they're asking us the point. Okay. So your point would be, your points would be, can I say it would be plus minus R cos theta comma R sin theta. So R cos theta comma R sin theta is under root of C plus A plus H twice. Remember sin 45 degree is 1 by root 2. So root 2 I'm introducing here. So this will be your points which will be at a minimum distance from the origin. Sir. Sir, I think I did it in a different method. But it's not nice actually. You got the answer. Sir, I did it in different ways. I took normal and all that. Okay. The basic principle which I used was it will lie on the normal to that curve also. Like whatever curve it is, it will lie on the normal also. Like first you differentiate that A x squared and all that. So the point which is at the minimum distance from the origin should be that point where the normal drawn should pass through the origin. Yes. And then you get some big expression. But eventually the answer is the same. Okay. Let's do the last question for this topic. After that, we are going to start with definite integral enough of AOD now. Next class problem solving. Next class problem solving. For a circle x square plus y square is equal to r square. Find the value of r for which the area enclosed by the tangents drawn from the point 6,8 to the circle and the chord of contact is maximum. Okay. So it's a case of a tangent drawn from an external point to a circle. So let's say. Okay. So this is your chord of contact AB and this is your RR. This is your origin. You want the area of a PB triangle to be maximum. Okay. Okay. So for this, we don't need a lot of coordinate geometry. We can stick to geometry actually. So let's say this is theta. Then can I say this angle will also be theta? Yep. So what is the base of the triangle? What is the base of the triangle? Can I say base of the triangle will be 2r cos theta? Yes. Correct. What will be the height of the triangle? We all know that this is going to be 90 degrees. So height of the triangle, let me call it as PM. Can I say height of the triangle will be 10 minus r sin theta? Why 10? Because O2P is 10 and this guy is r sin theta. So area of the triangle, ABB will be half base into height. So half into 2r cos theta into 10 minus r sin theta. So 2 2 gone. Okay. Is this fine? Any concerns with this? No sir. Okay. Now can I say r is also given by 10 sin theta? Because this is 10. So I can write 10 sin theta for this. So 10 sin theta cos theta 10 minus r sin theta. Here also I can replace. Okay. So this is your area of the triangle. Let me call it as A for the time being. Good enough. Okay. Now the problem is we have to differentiate this. Yeah. So this will be 10 square which is 100 sin theta. Sin square will be cos cube theta. Correct. So this is your A. So dA by d theta would be equal to 0. Let's see what happens. 100 sin theta derivative will be cos theta. So I can write it as cos to the power 4 theta minus 3 cos square theta sin square theta. This will be equal to 0. Correct. From here can I say cos to the power 4 theta will be 3 cos square theta sin square theta. In other words tan square theta will be 1 by 3. So tan theta is equal to 1 by root 3. That means theta is equal to 30 degree. Correct. So your minimum value, your r value should be 10 sin theta. 10 sin theta is 10 into sin of 30 degree which is equal to 10 into half. Answer is 5. How can we solve this with conics? So first of all we know that, I'm not sure actually, I was doing it but then I lost out on it. It was taken too much time. Okay. So first we know that A B is going to be 6 x plus 8 y plus r. Equal to r. No, no, no. Okay. This is things like I derived that now. Area of that triangle is going to be rl cube by r square plus l square. What is l? The length of the tangent sorry. Okay. And then I tried differentiating but it didn't work sir. It didn't work. I think it was taking too long. Yeah, it will take too long. That's very obvious. What geometry can do here, coordinate geometry will not be able to do. It's a simple case of a geometry. Okay. Okay. So fine, we'll stop here. And next class, please do not miss it because it's going to be a very important class. We'll be talking about definite integrals. Okay. Thank you. Bye bye. Take care. Have a great day. Bye sir. Bye sir. Have a good night. Bye.


Mae'r next item of business is a statement by Kevin Stewart on suicide prevention strategy and action plan. The minister will take questions at the end of his statement and so there should be no interventions or interruptions. I call on Kevin Stewart up to 10 minutes minister. It's important that I start with a simple truth. Every suicide is a tragedy with profound and lasting effects on people's lives. It is imperative that governments set a clear course to reduce suicide and ensure that anyone affected by suicide is able to get the help that they need. Today, I'm happy to say that a few weeks ago we launched a document that I believe does just that, the Scottish Government and COSLA's new 10-year suicide prevention strategy. The work around time, space and compassion for those in suicidal crisis has emerged over the last couple of years and now forms a key foundation of our new strategy and action plan. We understand that we also need to embed these principles right across our work to build compassionate communities and services. I must acknowledge that although we launched the strategy and associated three-year action plan last month, the impact of the national period of mourning on parliamentary business meant that this statement could not be made until today. However, I am pleased that we are now able to do so because key to the strategy is an ambition for everyone to work together to prevent suicide. Members across the chamber know the devastating impact that any death by suicide has for families, for friends and for communities. Whilst the number of those taking their lives has thankfully reduced in each of the last two years, we know that there is still much to do to reduce suicide deaths in Scotland. Our aim is for any child, young person or adult who has thoughts of taking their own life or are affected by suicide to get the help they need and feel a sense of hope. Our approach needs to be rooted in the three principles of time, space and compassion across Government, across party lines and across all sectors and communities. Putting it simply, we all must do everything that we can to reduce death by suicide in Scotland. I would like to thank everyone involved in shaping the strategy, including members in the chamber here today, for the way in which they have positively engaged and supported this work. We have called our new strategy, Creating Hope Together. This is absolutely what we need to do to deliver our vision of reducing suicide deaths in Scotland. We are backing the strategy by doubling annual funding for suicide prevention to £2.8 million by 2025-26. We know that if we are to deliver our vision, we must build on our successes today, particularly the way in which our approaches have been shaped by the evidence of what works and the views of those with lived experience, including the National Suicide Prevention Leadership Panel, which has been recognised by the World Health Organisation for its work. I would like to take this opportunity to again thank the lived experience panel and all those involved in the leadership group for their work. Our new strategy continues to drive forward key actions from every life matters, our previous action plan. That includes our vital work to reduce stigma and grow public awareness of suicide and its prevention, to embed time, space and compassion into the responses that people receive when they are suicidal, both within services and within their communities, and also to support people bereaved by suicide. Crucially, we also plan to step up our approach. The strategy and action plan set out a number of new approaches, again guided by insights and evidence, and I would like to highlight a few of those. First of all, the strategy seeks to support anyone affected by suicide, whatever their age or experience. This could be their own experience of suicidal thoughts, caring for someone close to them or feeling affected by suicide more generally in our society. We know that we will only succeed by supporting everyone affected by suicide. Secondly, we are taking a whole-of-government and society approach to tackling the inequalities that contribute to suicide. Statistics tell us only too clearly that certain groups are at higher risk of suicide. Deprivation can be a key factor, so it is important that we seek to tackle the social determinants of suicide through policies such as reducing child poverty, tackling homelessness and funding debt advice. This is all the more important at a time when the cost of living crisis is impacting significantly on so many lives. The third theme within the strategy, which is a strong focus on reaching and supporting people who are at a higher risk of suicide as early as we can. In this Parliament, we have recently reflected on the value of initiatives, like men's sheds, in both promoting good mental health and reducing suicide risk. The strategy highlights our aim to take a proactive approach to supporting other groups who may have a higher risk, such as people who are LGBTI or are neurodiverse. We will work with people with lived and professional experience to understand more about how we can connect and support more marginalised groups. That includes how we can remove barriers and ensure that the responses that people receive are embodied by the principles of time, space and compassion and that they are also culturally appropriate. We also wish to focus on the different needs of population groups. For example, both older adults and children and young people may have different needs and types of responses that are suited to them. We are taking a focused approach right across our work so that we have maximum effect in our overall aim to reduce suicide. The strategy prioritises settings such as prisons and locations of concern, particular communities and parts of the workforce, where we feel our efforts can make the biggest difference. We have a very solid baseline of understandings through our work in Scotland over the last two decades and from our first-class academic research. Both of those have been fundamental in shaping the contents of the new strategy and the action plan. That research also reminds us of the importance of sensitive media reporting of suicide, so that is another priority for action. Evidence also tells us that it is crucial that people who are suicidal get the support they need from services. We know that many people who die by suicide have had contact with statutory services, often health and social care services, leading up to their death. Through that strategy, we want to make sure that, whenever someone contacts services, they will be met with compassion and given the support that they need. That support should promote their wellbeing, protect against suicide and have a primary focus on recovery. Our work encompasses primary care, mental health and unscheduled care settings. The final area that I wish to highlight is peer support. One of the strongest messages that we have heard in developing the strategy was the value of people talking to others who have shared their experience of suicide. I have been fortunate to hear first-hand accounts of the value of peer support, both at the launch of the strategy and in countless discussions that I have had with those with lived experience. Peer support clearly gives people a deep connection that not only offers them personal support but can help create a path to recovery. We will seek to embed peer support and recovery models for suicide prevention, working with key partners such as the Scottish Recovery Network. In developing the strategy, we have both deepened existing relationships and created new ones. We must continue to nurture those relationships so that we can truly create hope together in our mission to prevent suicide. The strategy and action plan sets out the Scottish Government and causeless plans to realise our vision across sectors and bring our local and national work together in a meaningful way. I will conclude by once again thanking everyone, including members here today, who have contributed to the development of the strategy and the action plan. Across all our partners, stakeholders and members of all parties, I would like to thank you all for your efforts. We all have a part to play in preventing suicide and I very much look forward to seeing the progress that we will make together as we create hope together and make a profound and lasting difference to people's lives. The minister will now take questions on the issues raised in his statement. I intend to allow around 20 minutes for questions, after which we will move to the next item of business. I would be grateful if all members wished to ask a question. We are to press the request to speak button now. I would like to thank the minister for advance sight of the statement. I am sure that everyone in the chamber will be deeply concerned that Scotland has the highest suicide rate in the UK. Last year, 753 Scots took their own lives. That is 753 people who were not given the support that they need. I welcome the new suicide prevention strategy and action plan and, like the minister, thank every organisation, family and individual that has contributed to the development of this plan. Every suicide is preventable and we owe it to those who have lost their lives and their families to make sure that Scotland's suicide prevention plan is the very best it can be. While I welcome the doubling of annual funding for the suicide prevention to £2.8 million by 2025-26, it does not matter how much money we spend on developing effective suicide prevention interventions when, ultimately, we know that the most significant barrier to improving mental health nationwide is the lack of well-trained clinicians. We may have the mental health interventions at work, but we lack the workforce or infrastructure to deliver them. A strong suicide prevention strategy requires us to intervene early. Samaritan Scotland has said that the bar is still far too high for accessing support when facing a crisis. More sustainable investment in crisis support is required to promote and deliver 24-hour open access crisis support for those in distress and crisis. Does the minister agree that the prevention strategy needs to ensure that people are supported before they face such a crisis? First of all, I would like to put on the record my thanks to Ms Weber for engaging with me and with the Government on this issue and helping us to create this strategy and action plan. I agree that we need to do more to ensure that people do not reach crisis point. I disagree with Ms Weber on some of the points that she has made around workforce, because we have the largest mental health workforce in these islands, but we still need to do better. I do not disagree with that point, but I think that we would also all agree that some of the interventions that we need to make are not at the clinical or acute stage, they are in communities. That is why this Government is investing in our children's and adults' mental health and wellbeing funds, which are making a difference in communities right across our country. What we also need to do is not just rely on those clinical interventions. All of us must do better in helping people when they are at crisis point. I know that there is an ambition from all parties that I have spoken to that we get that right. I will continue to listen right across the chamber to people's views and order that we get those interventions right. Most importantly, I will continue to listen to the voices of lived and living experience, and I know that people across the chamber will do that too. I thank the minister for advance sight of his statement. In my area of East Ayrshire, which is highlighted as one of the places most affected, the key to successful intervention is to work with people early and have properly funded training so that we can embed the concept of suicide prevention throughout every stage of care and treatment. Many professionals in primary care require further training to recognise those signs and, in doing so, bring down the backlog of people waiting for treatment, which is, of course, concerningly high. Given that it does not match previous investment in suicide strategies, I will ask the cabinet secretary directly, does he really think that the annual funding of £2.8 million and a rebranding exercise is going to cover the training requirements and the work that we need to put into making this strategy? I thank Ms Malkhan for her question. I have to say that £2.8 million for suicide prevention is not the only money that we are spending on training and ensuring that we get that right. I would say to Ms Malkhan that, for example, the trauma training that is being delivered across the country is absolutely beneficial in getting all of that right for people too. I would praise the trauma training champions in South Ayrshire that I met recently who are making a huge impact. I would say that the impact that some of those champions are making is changing the way that local authorities are creating an advancing policy. That is important as we move forward. That is something for us all to get right. It is not just that £2.8 million. It is the other investments, including, as I said in answer to Ms Weber, the £15 million for children's community mental health and wellbeing funds and £15 million for adults mental health and wellbeing funds. We must ensure that we do all that we can with the resource that is available. I wrote to the Minister the day that the strategy was published to raise concerns around the collection of GIS mapping data for suicides. Many mental health organisations warn against publishing how and where suicides are completed as it acts as a how-to on what the most successful route is to complete suicide, which is dangerous for anyone coming in. Can the minister reassure me and others that, if the data is collected, it will not be published or, for example, if it is liable to protect the public? I know the interests that Emma Roddick has in this particular topic. I will respond in more depth in writing about some of those issues. We will continue to approach our work in locations of concern with the utmost care. I can give assurance that our work will always be designed to safeguard and support individuals at risk, as well as wider communities. Any new data collection or intelligence on locations of concern is sensitive, and therefore the sharing of such data will be limited, very limited, to key partners and will not be put in the public domain. We will also be taking forward action to support responsible media reporting, as we are very conscious of the risks associated with reporting on suicide, including on this specific and important issue. Minister, early prevention is key to a robust strategy of suicide prevention. As a practising GP, I refer people all the time to mental health services, but patients face a very long wait. Many are waiting over 18 weeks for treatment in part due to staff vacancies. What steps will the minister take to increase staffing levels within mental health services? As I said in my response to Ms Webber, we have more staff in mental health services here in Scotland than other parts of the UK, and we have invested very heavily in our time and office on mental health staffing. However, we will go further, and we are at this moment looking at our mental health workforce strategy for the future, which will form an essential part of the mental health and wellbeing strategy that will be published in the spring of next year. However, it is not just about the clinical stages, it is not just about investment in clinical services, and we have to get that right too, but we also need to look at more preventative spend. That is why this Government is spending the money that it is on community services so that we can reach people at a much earlier stage before the hit crisis point. I am quite willing to share with members, Dr Gohani and others, some of the impacts that that investment is making. I encourage members to go and visit some of the projects that have benefited from that community fund because they are making a real impact and helping people from reaching those crisis points. The Minister is well aware of the on-going and long-term crisis in mental health services in NHS Tayside, which has led to two reports by David Strang, the second of which has been described to me as the worst report in Scottish public life. The conclusions of those reports led to ministers forming the independent oversight and assurance group on Tayside's mental health services. When will the Parliament see the final report of this group? Can the Minister assure us today that there will be a full debate and Government time on the content of that report to help to ensure that more lives are not needlessly lost to suicide in Tayside? I can see myself some of the difficulties that have been faced. I put in place the oversight group because I was not happy at some of the responses that we have received from the Tayside partnership. The oversight group are due to report to me very soon, Presiding Officer. I'm quite happy after they do report to me to have conversations with members around about those findings. I know that Mr Marra will probably ask for such a meeting anyway. Maybe what we can do is provide a briefing for interested members after they report, but can I assure the chamber that I recognise, without doubt, the seriousness of the Tayside situation, the fact that many patients and families feel that they have been failed, and we will do better? First, I would like to express my sincerest condolences to any family who have lost the loved one to suicide. In my constituency, there is some great work being undertaken by organisations such as Chris's House and local football teams such as Bonnery Gros, and they are seeking to address mental health and suicide prevention. Unfortunately, suicide rates in Midlothian, particularly for males, remain high, which is a great worry. What targeted help will be available for areas such as Midlothian, and if he would join me to visit these fantastic organisations in my constituency to see their work? I'm always happy to visit organisations that are doing well for their areas, and I'm more than happy to visit Bonnery Gros with Mr Beattie. A key element in all of that is the peer support element, which I mentioned in my speech. Those folks are often immense in providing that support. For men in particular, they feel more comfortable in certain environments, and football is doing extremely well in terms of helping with those issues. If I could give a couple of examples, the Changing Room's extra time project, which is being run with football clubs and Sam H, is doing wonders, and I'm sure the same thing is happening at Bonnery Gros. I was absolutely delighted to attend a day of suicide prevention work at St Myrran last year, and they have been at the forefront of some of that work, and it is quite immense to hear how well the work that they have been doing has been received by people. Also, on a football-related theme, FC United to Prevent Suicide campaign is absolutely top-notch, and I would urge Members to have a look at the work that they are doing. I'm very keen that we manage to include all Members who have requested questions, so I would appreciate short and concise responses and questions. I call Alex Cole-Hamilton and Claire Adamson. I very much welcome strategy, as does my party. If I can draw the Chamber's attention to the last line of the vision outlined in the strategy, it references children and young people, and that's important. Last month, devastating statistics from Public Health Scotland revealed that one in four deaths of five to 24-year-olds were a direct result of suicide. That is compared to just 1.2 per cent of those aged over 25. Even more shockingly, the report acknowledged that five to 24-year-olds were significantly less likely to have had contact with the healthcare service in the period before their death than people older than them. Can I therefore ask the Minister how the additional funding in the strategy outlined today is going to help specifically to give much-needed help for our young people? I think I outlined in my speech that there is going to be much more focus in certain areas, and young people is one of those areas. We must get better in terms of being able to discuss these issues and to take a nice sense of what young people have to tell us. That is one of the reasons why we have established a young people's group with lived experience to guide not only the Government but others on where we are getting it right and where we are getting it wrong. I think that they will help us to shape what we need to do in order to change our responses and to change services. I am sure, like me, that the chamber would express their gratitude to these young people for coming forward and helping us with all of this work.  Stigma has been identified as a significant factor in suicide. How will the Scottish Government place greater emphasis on the impact of stigma around mental health as part of its new suicide prevention strategy? The Scottish Government is providing £1 million of annual funding until 2026 for CME Scotland's national campaign to stop mental health stigma and discrimination. We have given them the comfort of that year-on-year funding so that they can continue to build on the good work that they have done. They, obviously, carry out a huge amount of work that includes work with employers around mentally healthy workplaces, feels FM, an app to help young people to express themselves and face-to-face work in communities across Scotland. Tomorrow, I will be attending CME's 20th birthday celebrations in Glasgow, and I would like to take this opportunity to publicly thank them for their hard work over the last two decades. How suicide is reported can make a huge difference to those left behind, not only grieving family members but others might be impacted, leading to on occasion suicide clusters. Does the minister agree that it is important that deaths from suicide when reported in the media are done so in a sensitive way without sensationalism? What discussions has the Scottish Government had with the national news journalists and newspaper proprietors regarding their approach to suicide? I absolutely agree with Mr Gibson. We need sensitive reporting of suicide by the media, both in the printed press and across social media too. Sensitive reporting is absolutely vital to protect bereaved families from further traumatisation and to avoid increased risk of suicide in the wider community. Our action plan sets out our plans to work with the national and local media sector to support responsible media coverage, and we will start engaging with all in the sector shortly on this important issue. I would also like to say that I am pleased to say that, in Scotland, the Samaritans are already working hard to support responsible media reporting of suicide and to publish guidelines for the sector, and we look forward to building on their valuable work. Gillian Mackay, to be followed by Craig Hoy. Thank you, Presiding Officer. Could the minister outline, as a result of this strategy, what will be put in place for those people who have had multiple attempts to take their lives, many with complex trauma, and how we can ensure that each of these crisis periods is taken seriously and treated without stigma? Presiding Officer, this is a really important question. What we need to do in these circumstances for the most vulnerable people is to make sure that we listen and then tailor services to meet their needs. When I talk about services, I am not just talking about mental health and wellbeing services here. We need to take a completely different approach across services to get this right for individuals. That is the most important thing that we need to do, one of the most complex things. By working together in that holistic way and creating a greater understanding of people's needs, we can do it. Craig Hoy, to be followed by Evelyn Tweet. The strategy makes passing reference to rural issues. Will Mr Stewart commit to a specific suicide prevention plan for farmers and gamekeepers, and will he support the British Association for Shooting and Conservation's proposal for a gamekeeper's task force to better understand the significant mental health challenges faced by this group? Presiding Officer, as I explained earlier, we are going to look at different sectors. We recognise that suicide is more prevalent in certain work groups. I know that there has been a fair amount of work on the issues by the likes of support in Mind Scotland. My mind is not closed to any suggestion about those particular issues. I would say to Mr Hoy and our colleagues that we do not want to fragment things too much in terms of our approach, but at the same time I am more than willing to look at anything that we need to do to create better protections and safeguards for sectors, whatever they may be. I would be happy to speak to Mr Hoy further on the issue if he wants. I recently met pupils from a local school in Stirling who raised concerns about mental health provisions for young people. Can the minister advise how the suicide prevention strategy and action plan will be rolled out to ensure that young people in my constituency are supported? The new strategy will go further in supporting children and young people affected by suicide. We know that this group needs different approaches. As I said in my answer to Mr Cole-Hamilton, that is why we have established that youth advisory group to help us to understand their needs in greater detail so that we can develop effective responses building on our existing work, such as the Better Tomorrow social media campaign that ran earlier this year. The action plan itself sets out a range of specific work to support new approaches to prevent suicidal behaviour on children and young people. That will focus on key settings such as education, health, social care and youth work. We will work with partners to develop existing and new age-appropriate resources for inclusion in the school curriculum, which will build on the understanding of mental health, self-harm and suicide prevention. All of that will be evidence informed. That concludes the ministerial statement. The next item of business is consideration of business motion 6456, in the name of George Adam on behalf of the parliamentary bureau setting out a business programme. I call on George Adam to move the motion. The next item of business is consideration of parliamentary bureau motion 6457, on approval of an SSI. I ask George Adam, on behalf of the parliamentary bureau, to move the motion. Thank you. The question on this motion will be put at decision time. There are seven questions to be put as a result of today's business. Can I remind members that if the amendment in the name of Humza Yousaf is agreed to, the amendment in the name of Sandesh Gulhane will fall? The first question is that amendment 6437.1, in the name of Humza Yousaf, which seeks to amend motion 6437, in the name of Jackie Baillie, on supporting the NHS in winter, be agreed? Are we all agreed? The Parliament is not agreed. There will be a short suspension until our members to access the digital voting system.


Back to another vlog. I don't have any product on my eyebrows. I feel like I look like Voldemort. But that is because I am getting ready to go get my eyebrows done today. I haven't, I think I've gotten my eyebrows done like two other times in my life. I have really kind of blonde, sparse eyebrows. For a while, like probably back in, I don't know, when was microblading popular? Like 2016, probably. I was like really wanting to do it but it was too expensive and I'm kind of glad I didn't because what's cute and trendy for brows is totally different now than it was back then. So I'm gonna do a lamination and a tint, I think. I've seen a couple videos on brow lamination and I kind of know what to expect. I don't know how long it's gonna take, especially with the tint. But after that, I think I might go meet my mom for lunch and that's today. So I'm excited because I, as you can see, like there's not much to work with, especially over here. So it'll be cool to like have my brows be a little bit darker, a little bit fluffier. And like the photos that I've seen on Instagram, like the transformations are wild. So before I would look into something like microblading, I'm gonna do like the lamination and the tint and all that. So I'm excited. All right, hair. That's as good as it's gonna get for today.  We're doing something big. But more about that later. Okay, gotta get dressed. I got these Nike dunks off of Poshmark and oh my God, I can't with my face. Oh my God. They're like this light pink and white color. They're so freaking cute. And the only reason I got these is because they were on Poshmark. Someone was selling them on Poshmark and I had like $300 of credits to use on Poshmark. So I got them for free. They're super cute. I'm thinking these with, hmm. I got this cute top from Abercrombie but this is kind of like a going out, out top. It's so freaking cute at like crisscrosses in the front. How cute is that? I like the idea of that. Oh, you know what I could wear. I just thrifted this top and this top is actually really cute. And then recently I got some jeans second hand. As I mentioned, I have really been loving Poshmark. So maybe like this with the dunks. Let's try it on. All right, I'm not 100% in love with this, especially with it tucked in, but like I don't really, like it doesn't look good out either. I guess I could like try to tie it up. That might be cute, actually. Okay, I tucked it up like into my bra so that it hopefully looks a little bit better.  Oh, by the way, these are the Abercrombie high rise curve love 90s jeans, I think. I got my Poshmark for like half off. So, okay, now we have to go. Okay, Jess Parked made it. I need to actually find out where it is because it's in like a salon suites type of thing and I really, really have to pee. So I'm gonna head inside. Here is the before. And we'll see you in a few for the after. We're gonna go from this to this. Oh, wow. Okay, here's the thing. They're really, really bold right now. And she told me that going into it, like they're gonna seem really, really dark for the first like 24 hours. I'm not allowed to wet them or wash my face or anything for 24 hours, I can't. They look so dark, but I swear to you they're going to be so amazing. She said that once I wash, like there's like some dye on them and there's also this like, I think it's like Vaseline or like grease or something.  So I'm actually so pleased with it and she was so sweet. She's like my new favorite person ever. I would highly recommend going to see Chelsea. Her business is called Tinted Brow Bar. It's in Scottsdale. I cannot recommend her enough. She was so kind, so sweet. We literally just chatted about everything, like being in our 30s life. So I'll leave her information linked down below like her Instagram and all of that. I would highly recommend booking with her. So I think I'll check in tomorrow and do like a little bit more of like a realistic update because she said that like sometimes when people get them done, they're like, oh no, they're too dark and she's like, trust me, it's gonna be better. So I'm so pleased.  Are you kidding me? This is all my hair. Like you saw those brows this morning, like she tinted them and then laminated them and she said that she did my tails up as well because she thought that it looked really cute. So here she is.  But what a fun way to start the morning. I'm so glad I tried this out. Like I've been wanting to try it forever. And now I can say that I did. And who knows, I might keep doing it. We'll see. Really like almost 3 p.m. now, which is crazy. This day went by so fast. My appointment was at 10 a.m. I like literally don't know what happened. Had lunch with my mom, hung out at her house for a bit and then went to hold food because my dad got me a gift card for Christmas. And I was like, I obviously need to use it. So shout out to my dad for providing me with these groceries. He knows the way to my heart. In no particular order. Actually I had to move some of this stuff around because my mom got stuff. But we got some oatmeal because why not? I was there. Oh my God. I forgot how bad the fridge smells. So last night, I made the baked by Melissa viral TikTok green goddess like dressing salad. I literally need to take it out. And it made this much. Like one head of cabbage made this much. What am I supposed to do with this? This is for a group of 12 people. Like we ate so much of it last night that it like made us sick. And I stored it in the fridge with some tin foil on it. And like it stinks every time we open the fridge. So I don't know. I might put it in like a Tupperware. Okay, wow. That was gnarly. Anyways, starting at the top I got these beet chips. I thought these looked really good. They had two different kinds. Well, actually they have like so many they have broccoli, carrot, beet, cauliflower, which I also got. And they just looked fascinating. I love veggie chips. They have kale chips. It looks like they also have watermelon slices, pineapple chips. So they have all different kinds of like dehydrated fruits and veggies. So I got buffalo ranch for the cauliflower and then just sea salt beet chips. So that should be good. I got some snap peas. What are they called? Harvest snaps. I like, they're so good. They're like, I don't know how to describe them. Green pea snap crisps is what it says. So I got some of those. I've loved those for years. Like I used to work at Whole Foods in high school and eat those. Got some tiramisu for Drew. That's from my mom. She and Drew share love for tiramisu. I love it too, but I can't have too much caffeine. So I gotta be careful. I tried this urban remedy. I got some of the bars too. I got two of the bars and one of the shakes. This is an organic chocolate banana milkshake, meal shake, I lied. And I just like love protein shakes. So it has almond milk, vanilla, banana chia, pea protein, coconut and cacao. And I just thought that sounded delicious. So this is what she looks like. We also got two of the urban remedy bars. I got this matcha one with, looks like some kind of like maybe chia seeds, sesame seeds, something like that. It's an energy bar. So I got that one for Drew. And then I got the cacao plant protein bar for myself. These look pretty good. I got a lot of snacks if you couldn't tell. This obviously isn't a grocery haul. I feel silly that I even need to clarify that. But if I ever, excuse me, if I ever like go to Trader Joe's or any, if I show any grocery haul and it's not like a complete haul of groceries, people are like, where's your veggies and fruit? I'm like, it's not, it's not my full grocery haul. I'm just, I just bought this stuff because I had a gift card. I got a nectarine, fun stuff. I got two of these get packed to human chocolate bars. I'm obsessed with this one, which is the almond butter and puffed quinoa dark chocolate. This is so luxurious and delicious. And then I also have one in the fridge. That's the, I think it's like cashew butter and vanilla. Amazing as well. So I decided to try the crunchy mint dark chocolate. So I got two of these. So one of my favorites, and then a new one. And then I have been seeing people talk. I was looking for the cookies, but I couldn't see them anywhere. And I might've been looking in the wrong place, but I saw this siete cinnamon chips. These are like churro strip chips. And siete is a really cool brand because like everything's gluten-free. I'm pretty sure it's vegan as well. Yes it is, paleo and vegan, it says on the top. And these just look delicious. And it's like made by this family, the Garza family. And there are seven of them. So that's where the siete came from. How cute. And then I've got some pistachios. I just always eat pistachios. Got some nori to make the Emily, Mariko, salmon rolls, the TikTok ones, obviously. I saw these and they looked really good. And these are called almond butter date bites. Drizzled in dark chocolate with a crunch of hazelnut croquant. I don't know French very well, but these looked amazing. Like, yeah, sign me up for that. I don't wanna eat one right now. Oh wait, I got these unreal. I've seen a lot of people talk about these and I've tried the peanut butter cups, but these are dark chocolate almond butter cups, which is actually even better. I like dark chocolate more than milk and I usually prefer almond butter things. It just like tastes better to me. So excited about those. I can't get over my eyebrows. Just realizing how dark they look. Got some of these corn snack thingies. They are like corn nuts, who loves those. I lost one, fell out. But I wanted to try this Zevia because I see everyone talking about it. And I, as we know, I love a spicy water. I love a soda, but it can be really hard to find caffeine free sodas. So I thought I would try this one. It is like artificially sweetened. I don't really mind that though. It's not really something that bothers me, but like drew. If there's even like a milligram, he'll taste it. He's like, no. So I got these and they're the strawberry flavor. And that sounded delicious. So we'll try those. I got two of my guilty pleasure, which is liquid death. I've talked about these before. It's literally just water, like actual water. Not sparkling water, not any kind of fancy water. This is from the Alps, but I don't know. There's something about this that just, I really love. And then I got these Simple Mills seed and nut flour sweet thins. And they're a chocolate brownie flavor. And I was like, that sounds amazing. I've tried their almond flour crackers. I actually have some in the pantry right now. The black pepper flavor, and they're so good. Like they're literally so flavorful and delicious. And it says that the seed and nut flour blend is made from watermelon seeds. That's insane. Caches, sunflower and flax. So they're like gluten free, vegan, all that kind of stuff. So nice. And that's it. That's literally all I got. So put all of this away and then Drew's mom's coming over for chili later. So we're gonna like probably get this place cleaned up. And I obviously need to start the chili at some point. But yeah, here with my new brows. I can't really move my brows that well cause I got Botox. Hmm. Hi. Drew's coming down. I warned him that they were really dark. They're really dark. I know. They're really dark. This is so weird. Do I look angry? You're just like really. I called him on the way home and I was like, you know, my eyebrows are really dark.  Quick update.  We're just gonna roll with it. These incredible. I'm leaving the fridge open just for like a minute to try and get the cabbage smell out. These were amazing. They actually taste exactly like the Laura bar like the chocolate chip cookie dough Laura bar. So those are amazing. We tried these simple mills. These things hit these little chocolate brownie sweet and amazing. Oh, we also tried the siete chips. I don't know where they went, but they taste like cinnamon toast crunch, like a more lightly seasoned cinnamon toast crunch. And I'm gonna try one of these unreal chocolate almond butter cups. Here we go. Yeah. Amazing. So please, but just wanted to give a quick update like everything is amazing. Just cracking a cold one after a long day. Cheers. All right, now we'll close this. Cabbage stinks. Hello, I am in the bathroom. It's the next day. It's been 24 hours. I just washed my face and did my skincare. So my face is really red. But here is the final look at the brows. Like they're so much lighter. You can still see some like leftover dye, but they don't look like as sphere. Obviously right now I don't have makeup on and I'm just doing an editing day today. So I don't plan on wearing makeup, but I really do love them. I think it's gonna look great for whenever I do my makeup I don't have to spend so much time on my brows. I think it was like a fun little thing that I tried. I'm really glad I did it. I also forgot to mention earlier in the video that like I paid my own money for this. The girl doesn't even know I have a YouTube channel. She doesn't know I was talking about it. So 100% my honest opinion, I really enjoyed it. I paid for it. I think it was $75 before tip. And I think it's great. I think they look so much better than they first did when I turned on the camera yesterday. Like, come on. They're much more like natural looking today. And I'm very pleased with them. So anyways, that's gonna do it for this vlog. I hope you enjoyed it. Thank you so much for being here. The next vlog is very exciting because I'm doing a hair transformation. I'm ready for a change. Like it's been something I've been thinking about for a while, so we're just gonna do it. And I hope you're excited and I will see you in the next vlog. Bye.