okay so um we'll be discussing paper three for this video where it is from May June 2013 23 Paper 31. okay so we'll start with question number one first they want to solve this equation that involves the exponential function to give the answer correct to three decimal places all right so um for this question right if let's say you able to see clearly you can treat actually the E power 2x are as a variable y let's say that okay you can actually treat it as a y then it will be easier for you to look at the pattern of the white okay so I let y equals to E power 2X I will have 3y minus 4 over y equals to 5. okay then I want to solve this equation I'm having 3y squared minus 5y minus 4 equals to zero so very easily you can see that it is a quadratic equation right 3y squared minus 5y minus 4 equals zero therefore when I want to get the value for y I have to use my uh the method of how we solve the quadratic equation you can factorize it if you want or you can use the quadratic formula for this question we not able to factorize it therefore we have to apply the formula so it will be negative V negative negative 5 plus minus B squared minus well AC so B squared minus 4 a is a 3 C is negative four divided by two eight so two and then a okay so by right from here you should get full value for the Y where y can be equals to A Long value okay and of course I will get another value y equals to negative 0.5 my one okay of course it is quite longer I will just keep it like this and then this is not my final answer here because our original question it is about the E power 2x therefore you need to solve for the value of x therefore I change the Y become E power 2x again so now I try to think away okay to solve for this equation and get the value for x right so simply happen for this second value here is negative 2x also E power 2 x equals to negative 0.591 okay then to solve for this 2x uh basically I need to take long for both sides when I take long both sides I'll have 2X equals to long and the long value here so then after I get the very long value for long this value 2.257 then I divide my answer by 2 and from there I'll get my answer for XR which is if the long value should be this one after dividing by 2 and of course I have to correct my answer to three significant figure right so it will be 0.407 okay so this is my one of the answer for my X and then now I have the second equation right I have to try and solve this part also and to get the if uh to get the 2X to get the X I need to take long both sides as well so when I want to take long for both sides right and if you try to press calculator you will realize that this is actually a mess area this whole term on the right hand side is non negative value is a Mass area and that's why we will ignore that because we can't get any solution from this part so we'll ignore the answer from this equation okay therefore we are we are only having one answer here which is x equals to 0.407 correct to three significant figure okay so now we come to question number two they want us to sketch the graph of y equals to modulus 2x plus 3. okay so usually before I sketch the modulus function what I do is like I will try to draw the original function first without the modular sign where I want to sketch this one first y equals two X plus three okay so to plot the graph to sketch the graph and let I will try to find the Y and X intercept I will let y x equals to zero first and then the value of y that I get is a three and then if I let y equals to zero and then the value of x that I get will be negative three over two okay then I will try to draw it so I would maybe um when you are trying to sketch the diagram here right what you can do is that you use the ruler to measure the ratio that you want to use and then for me I'm using this and this is one you need maybe what's the end for one unit then two then three I'm having negative one two and also negative three okay so my x-intercept is negative 3 over 2 which is 1.5 so should be somewhere here okay and then my when x equals to zero my Y intercept is a 3 so it should be somewhere here okay when I'm trying to join these two points together it should look something like this okay so this is the original graph this is all you know graph okay then when I put the modular sign on 2x plus 3 right the negative part of the graph I want to reflect it up become positive okay therefore my modulus function should look something like this so when I reflect it up I assume that it will be somewhere here and negative three and three then when I want to sketch the point it will be going up here so this is the basic shape for linear modulus function where it looks like a v-shape so the original one is this one okay and this is a V shape here then don't forget to label your X and also the y-axis and label your diagram y equals to modulus 2 X Plus 3. okay so for me I would encourage you to show out the X and the value for x intercept also the y-intercept in your solution okay all right so this is uh what we get for first part then now let's have a look for the second part okay so for second part they want us to solve this inequality where 3x Plus 8 9 is greater than modulus 2x plus 3. okay so to solve for this inequality before I can solve it I want to find out the intersection Point first which is the value of x or we'll call it as critical value so how to find a critical value basically we ignore the inequality sign first related equals to 2x Plus 3. and therefore a modulus function you can write into proof has write one part is the original Two X plus three another one is the reflector part where I will write it as negative 2x Plus 3. and if I try to solve the value for XR of X1 for this part I'll have x equals negative 5. and then for this part if I try to solve for the value of x um by right I will have something like this X should be equals to negative 11 over 5. okay so this is the basically the critical value that we have when I try to solve this equation with the modulus function okay and after that now I want to desire what is my what is the correct range that we should have said so that we will have our inequality fulfilled that means I want to check and see and what value of x I will have 3x Plus 8 greater than modulus 2x plus 3. okay so from here there are many many different ways that we can solve it okay for me what I will do is that I will check it part by part okay so I will try a pipe I promise that I will try to list out the value of my critical value of x that I get so I have two value here which is negative 5 and also negative 11 over 5. okay so when I draw it on number line right I'm having three parts of uh three different parts of in on a number line okay I have this one is the first part this one is the second part this one is the third part so the negative five negative 11 over five I get it from the critical value by solving the inequality letting it equal uh become the cosine okay then now how can I check it so for this part of the equation for this part of the value you can use any value that's less than negative five so as example I will let x equals to negative six and I substitute it into my inequality here where I will have three then negative 6 plus a greater than modulus to negative six plus three okay so if I try to solve the equation right left hand side will have negative 10. then greater than you take two times negative six plus three okay you have negative 9 and then negative 9 when you put the modulus you have a positive mark okay so you will get this result negative 10 greater than 9 when you substitute any value that's less than magnified for the X so you can see that this negative 10 greater than 9 actually doesn't fulfill the inequality right so that means this part is not the correct answer okay then you should try to check for the next part so for the next part this is what we have for the next part here okay then you should use a value between negative five until negative 11 over 5. so maybe for me I'm using x equals to negative three because negative 11 over 5 is negative 2.2 so between any value between negative 5 and 10 negative 2.2 I can use x equals to negative 3 and again I will apply it inside I will substitute this value inside the inequality here so I'm having negative 3 times 3 I get negative 9 negative nine plus eight okay then greater than modulus negative 3 times 2 I got negative 6 plus 3 is negative 3 then modulus negative three you get a positive 3. so an hour when you want to simplify this one you're having negative one greater than three okay so do you think that this is a correct inequality or not it is not the correct inequality right okay then you should proceed further to the third section here so I want to use any value that's more than negative 2.2 so you can let x equals to zero if you want which is an easier one right substitute into the inequality again so 0 times 3 you get a zero then plus a greater than modulus 2 times 0 and then Plus 3. okay if you simplify this inequality you are having eight greater than 6. so you can see that when I'm having a greater than 3 right it is actually a correct inequality that I get a greater than 3 it is true which means that the correct answer for this partner to fulfill this inequality will be only this one where the x is greater than negative 11 over 5. therefore don't forget to graph your final answer X will be greater than negative 11 over 5. okay so this is basically one way that you can try to consider how to check for the correct um section that we have a correct range for the X that fulfill the inequality okay and the process is the step is first you try to find out the critical value for the Express in this equation and then after that you check whether the inequality is correct or not uh part by part on the number line okay and from here you can see that we have only the correct inequality on this part of the XR therefore our final answer will be X greater than negative 11 over 5. okay there are many other ways that you can solve this question okay to decide the correct answer so this is only one of the suggestions that you can use all right okay so this is how we answer question okay so question number three they want us to find the coefficient of x power 3 in this binomial expansion so I think the Q is very obvious they will tell you that it is binomial expansion for this um two brackets the product of these two bracket okay so first of all what I want to do is I will take 3 plus X multiply with one plus four x power half okay so for the first bracket that 3 plus X it is power one right therefore you no need to expand anything it is just remain it as 3 plus X and then for this record basically we can expand it into binomial expansion by using by number expansion formula okay also I want to expand this particular bracket here with the power half one okay then again we can refer to the formula booklet the mf19 uh I think it is under the first page or second picture okay look for the formula for binomial series the first one is for as level one and for paper three we are using the second one okay so maybe I can print screen so that we can refer to this one easier okay then maybe I try to paste it here all right so this is the formula that I'm going to use up to expand the 1 plus 4X power so you can see that according to what we have here you're having XL so for X in the formula is actually our 4X in the question n is the power where our n is half okay okay all right and if you want to save time right if you want to save time basically you can just expand the term that you are interested okay so let's say I'm having three plus X I'm asking myself well all right so am I having three plus X up how can I get my final answer for X power 3 so that means all for the 3 here if I want to multiply with something this particular term should be something with X power 3. okay and then if I want to take X multiply with something here it should be X multiplied with something with X power 2. so that means later when I want to expand the bracket for One Plus One X power half right you can only focus you can just focus on getting the X power 3 the term is X power 3 and also the dummy X power 2. if you found that that is too complicated for you okay to to understand then what you can do is I expand everything here okay but for me if I want to save time I will just focus on the term with X power 2 and also the term with X power 3. all right okay so I will start now okay so to get the term for X power 2 I will take n multiply with n minus one okay so my n is half then my n minus 1 will be negative half divided by 2 factorial and then the x squared my X in this particular bracket is for X's I'm having four x and then Square then plus now I'm interested for the next term with the X power 3 okay so again n which is half n minus 1 which is negative half M minus 2 which is negative 3 over 2 divided by 3 factorial and then 4X Power Trip okay so I will stop my expansion here I will only focus on the term with x square and also the term with X power 3 because later I know that I only need these two terms to get my final coefficient for X power 3. okay so for 3 plus X I just try to copy it and then try to simplify these two terms up by using the calculator right so by right if you didn't press any uh didn't do any careless mistake then you should get 2x squared and then plus 4X Power Trip again now I want to multiply these two brackets together and I only focus want to focus on X power 3 so to get X power 3 I will take the 3 here multiply with 4 x power 3 so that I can get a term with X power 3. and then for this one the X I need to multiply with negative 2X squared so that eventually I'll get the term for X over 3 as well all right so 3 multiply before X power 2 I get 12 plus so 12 x power 3 and then for this part I'm having negative two x power 3 and now I will have 10 x power 3. okay so for this question they're only interested for the code efficient okay so to get this coefficient you need to write out again coefficient means that the value in front of the X power 3 so coefficient will be equals to 10 for this question okay so this is how you solve this question number three okay so um let's discuss question number four they want us to show that this equation can be written as finally this one okay so this is the beginning part this is the ending part all right so um before we start maybe we need to analyze a little bit okay how are we going to get the final answer here okay so we can try to observe what is the difference so basically you can see that at the starting point here we are having two data okay the trigger with two data and my final answer here the final step here the equation everything is about Theta so very easily you can see that for all the two data here you need to actually apply um double angle from the okay for the respective three goal so that you can expand it become all the term with data on it all right okay so we can start our proving part here okay so let's start with sine 2X and then plus cos 2X equals to 2 sine Square X outside Square Theta sorry this should be Theta sine two theta plus cos theta equals to 2 sine squared theta okay so first of all I want to use the double angle formula for sine two data okay so sine 2 Theta can be written as 2 sine Theta cos Theta if you can't remember you can have a look for the formula booklet provided okay so all the double angle formula will be written is written here okay all right so this is what we have therefore the sine 2 Theta I will change if become 2 sine Theta cos Theta okay after that I also need to apply cos 2 Theta okay so for the COS 2 Data here the double angle formula for cos2 data there are actually three different formula the first one is two cos Square Theta minus one the another one is one minus two sine squared theta and after that the last one can be cos Square Theta minus sine Square Theta all these are provided in the formula booklet as well okay so among these three right we need to choose maybe one of the most suitable ones so that we can save we can make our step maybe slightly shorter okay so which is the suitable um that I can apply in my step here okay so again from the origin equation to the end equation here you can see that there's no constant right I mean there's no single constant in the in the step so that means so when I want to choose the suitable The Banger formula right I will try to apply all this with one and also the negative one one where I feel that this is the most suitable one because eventually I don't have any uh constant in my in my equation okay the constant right therefore after we decide you can put cos2 data become cos Square Theta minus PSI Square Theta then the two sine squared theta I will bring it over become minus 2 sine squared theta equals to zero and I think from here right it is quite we can see the answer is here quite obvious already so I will rearrange my equation cos Square Theta and put it in front and I'm having both sine School data cos Theta in the middle so negative sine Square minus 2 sine Square I have negative 3 sine squared equals to zero so this is how we actually proof okay to from the beginning equation to the ending part here and then again we know that you have to apply the double angular formula in our step so that we can expand all the two data become everything in data all right so let's proceed to Panama B okay so for Plan B they want us to solve this particular equation for 0 or for data between 0 and 180 so 0 to 180 means our first two quadrant okay so to solve this particular equation right we'll use the result that we prove in the first part and try to solve it from there so I'll say restart from the result that we get from first partner cos Square theta plus 2 sine Theta cos Theta minus 3 sine squared theta equals to zero so usually for all this kind of questions when in the first part they asked me to prove some result or to prove something usually very likely you will need to use it in the second part okay so uh leader hesitation with straight away apply the result so this is the result that I proved in the first part right okay so how can I further solving this equation okay you can see that I'm having cos Square I'm having sine Square also basically it is a quadratic form equation and you can try to factorize that so for the COS Square Theta I can split it become cos Theta and cos Theta and then for negative three sine Square Theta right I will try to think how to get a coefficient of two so maybe for me I will need to have a sine Theta here and 3 sine Theta here so that cos multiply with three sine so I'm having three then for this one I want to have sine and cos then Y is the correct symbol correct sign here so I want to have a positive so this one should be a positive this one should be a negative value okay so when I factorize it successfully then from here I'll get cos Theta minus sine theta equals to zero and then I need to solve another equation also where cos theta plus 3 sine theta equals to zero okay so I need to solve these two equation to get my data here all right so cos Theta minus sine Theta I'll get tangent theta equals to 1. okay that means you move the PSI Theta over then divided by cos Theta therefore sine over cos you get a tangent right so tangent equals to 1 and from here I'm looking for tangent equals to 1 which is a positive value so I only want to consider the answer for Theta from first quadrant so from first quadrant it is actually 45 degree you can press calculator tangent inverse one you'll get a 45 degree from the first quadrant okay and after that for the second equation here if I try to rephrase my equation I will get tangent Theta equals to negative 1 over 3. okay then usually for me I will try to find the basic angle first where I ignore the negative sign and I try to get the basic angle for first quadrant so tangent inverse one over three you can see that I didn't put in the negative sign okay so I get 18.4 okay then after that only I try to look for the suitable uh quadrant for the data that I want okay so for this second equation here I'm looking for tangent negative so if you can ask yourself uh in which quadrant tangent will be negative so all and then sine here for all right the answer that I get is straight away the base angle and therefore the sign to get the answer for the second quadrant you have to take 180 minus the basic angle so to get this data I have to take 180 minus the basic angle which is 18.4 maybe I shouldn't use a data here so I use a base angle here this anchor so 180 minus space angle and my base angle in this um part is 18.4 so to get the data I want to consider the answer from the second part because under the second part tangent is negative all right so if I take 180 minus the base angle I'll get 161.6 correct to one decimal place for my answer in degree okay therefore I have two answer for the data eventually the first one will be 45 degree and another one is actually 161.6 okay so this is basically how we solve the question number four okay so we are at question number five now they want us to either give us this curve the equation of a curve and then from here I think you can see that it is implicit function right where a is a non-zero constant so first of all they want us to find out dydx and show that it is equal to this fraction here okay all right uh so very clearly I can see that oh this is the implicit function where I cannot separate the X and Y clearly two two sides so what am I going to do now is I have to apply implicit differentiation okay I'm having x square y then minus a y square equals to 4A power 3. okay so for the term x square y right you can treat X Y as a first bracket the Y as a second bracket it is a product of two function so we want to apply the differentiation for product of two function you have to apply product rule okay so for product rule basically what I do usually is like this I will copy the first bracket differentiate the second then differentiate the password then differentiate the first bracket copy the second okay so my first bracket is x square second bracket is wire so I want to differentiate this one I will copy the first bracket differentiate a second so differentiate the one I will get a y differentiate the Y I will get a one and when I differentiate with respect to X right I will have divided X at the back okay then Plus for this term it is copy the first bracket differentiate the second right so now I want to copy the second bracket that means I copy the Y I differentiate the first bracket which is 2X okay so this is how we differentiate the term x square plus r x squared minus x square multiplied with Y by using the product rule here okay then after that we go for this one I want to differentiate negative a y squared so differentiate a negative a y squared I copy the negative copy the a differentiate y Square I'm having two Y and again because I am differentiating something related to Y so I have to put a dividends at the back that equals to 483 they already told us that a is a constant right so when you differentiate a constant you'll get it zero Okay then if you want to simplify in a nicer way okay then what you can do is like you can rephrase it so I'm having x squared d y d x Plus 2xy then minus 2 a y d y d x equals to zero so to get the answer in terms of dividing X I want to move all my D1 DX to one side so 2 a y divided x minus x squared d y d x equals to 2xy I can factorize out my d y d x here I'm having 2 a y minus x squared equals to 2xy also and so that my DX d y will be equals to this particular fraction this is what they want us to show in the answer in the question right okay so you can see that we are applying uh implicit differentiation technique okay to get the d y d x expression foreign okay so for puppy they are saying that find the coordinate of the point where the tangent through the curve is parallel to the y-axis tangent to the curve means um d y d x parallel to the y-axis so when you are having a line that is parallel to the Y axis now that means the gradient of tangent is equals to Infinity okay all right so first let us take out the d y d exposure so you should know that d y d x is equals to the gradient of tangent right so d y d s is equals to this one so we are having two a y minus x squared this is what we get from first part okay if my d y d x is infinity right parallel to the y-axis act when you're having parallel to the Y axis the gradient should be equal to Infinity so how can I get infinitive value for dividing x a you have to make sure that the denominator here is a very very small value flows close to zero so we'll let two a y minus x square equals to zero okay then when I let two a y equals uh minus x square equals to zero right I'm having the equation where x squared is equals to 2a1 this is my first equation okay and then what's my second equation my second equation just now is the original equation that I have x squared y minus a y squared equals to 4 a power 3. so this is my second equation to get the coordinate for the intersection point right for the value for x and y r you need to solve these two equations simultaneously okay so what I'm going to do now is I will substitute the first equation into the second equation okay substitute x squared by 2 a y so that means I'm having x square here become 2 a y and then y minus a y square equals to 4 a power 3. then if I try to solve the equation here I'm having a y squared equals to 4 a power 3. okay and then y squared can be written as 4 a squared and then when I want to find the value for one I have to take square root then when I take square root by myself I have to include both plus and minus sign so I'm having two values for y here y equals to 2A plus minus 2B okay so when I already find the find out the value for y coordinate I need to continue to find the x coordinate right okay so substitute uh to find the X coordinator I will use back the equation number one since I know the y i Redeemer right okay right so I will substitute the Y into the equation number one so that I can get the 34 x okay when the Y is equals to 2A what happened to my first equation so x square equals to 2A multiplied with 2A I'm having 4 a squared Okay then if I continue to solve the value for XR the X here if I take square root both sides I will have plus minus 2A okay and then when my Y is equals to negative 2A I substitute into my first equation also I'm having x squared equals to negative 4 a squared where we cannot further Solve IT solving it there's no solution from this path okay so we only have the answer from the first part of the equation where the points that you get on the coordinates that we have is 2A x equals to 2A y equals to 2A and also x equals to negative 2A Y is still equals to 2. all right so this is how we actually get the coordinate all right where the tangent to the curve is parallel to the y-axis that means at least two particular point right when you start still into the D1 dxr you will get something like uh infinity light or Mass area Okay because when you're having any line parallel to the y-axis the gradient is equals to Infinity all right so this is how we solve question number five okay question number six we are given origin o point a point B and point C and then they give us all the position factor for a b and c here okay then they tell us that the quadrilateral ABCD is a parallelogram they want us to find the position Vector of t okay so since we know that it is a paragraph of ABCD right okay so maybe I've seen big draw a parallel here first so where this is the paragraph okay and when we label the point right this can be a b c and d okay so to find the DIA to find the coordinate for the the position factor for d right um there are many ways that of course there are many ways you can solve it so to me right to get the position factor of a certain point uh I feel that maybe we can use the coordinate geometry method to solve it which will be slightly easier okay so how am I going to use the coordinates of matching method to solve it I know that under a parallelogram the midpoint of the diagonal are the same so and damage the mid part of AC should be equals to the midpoint of DB so the coordinate for the midpoint should be the same all right so I will use this coordinate um geometry consumer to find out the position factor of the first so I will use midpoint of AC equals to the midpoint of BD okay so let's say ah when I want to convert into a coordinate right so I'm having the D which is x y z that I don't know then a I will convert the vector form into a coordinate form where I'm having two one and three c will be 3 negative 2 and also negative 4 and then the B will be 4 3 and 2. all right then now to get the midpoint of AC to get the me part of AC I will take x coordinate plus x coordinate divided by 2. so I'm having two plus three divided by 2. then y for a plus y for C divided by 2 so I'm having one minus 2 divided by 2. and then for the exact coordinate will be 3 minus 4 divided by 2. that equals to the midpoint of BD so D I will assume that it is x y z okay so X plus 4 divided by 2 and then y plus 3 divided by 2 and also that plus 2 divided by 2. finances they are having the same in point right I can directly compare the x coordinate with the x coordinate okay so from here I'll get the value for x which is equals to 1. and then for the y coordinate 1 minus 2 divided by 2 I compare it with Y plus 3 divided by 2. so from here also very easily I will get the cutting for white which is equals to negative four and then the last one of course we want to compare the Jack coordinator so Z plus two divided by 2 equals to 3 minus four divided by 2. then from here I'll get the exact which is actually uh negative 3. though since I already get the value for x y and z right it is the coordinate for the D and of course for my answer I have to rephrase it become the position factor of D so when I want to write out become the position factor of D it will look something like this either you can put it in the column Vector form 1 negative 4 negative 3 or you can put it in the igk form where I'm having I then minus 4 J then minus 3 kick okay so to me I will I will suggest to use the concept of midpoint in polynesometry so that we can get the body in the easier way to me this is quite straightforward right okay so this is how we solve the answer for part A the position factor of D okay so let's proceed to Panama B now okay for Part B they are saying that the angle between ba and also BC is Theta so find the exact value of cos Theta okay when we are looking for this one right B A and B C okay so again just now we are having this parallelogram run right and then we label it as a b c and d and now they say the angle between B A and also b c is Theta so that means this is now the data okay when I want to follow the data right in the by using Factor basically I need the two Vector that is going out from the same point so the data is actually format going out from B so I need to use the Vector a b sorry b a and also the vector BC okay to help me to calculate the value of data but now they are interested only for cos Theta okay so to find out the exact value for cos Theta I have to use the what's the formula that you can use I can use the dot product right where I have I need to have d a multiply a DOT BC sorry ba dot BC equals to modulus ba and then modulus BC then cos Theta okay so before I find out the cost data by using this formula I need to find out the vector VA and the vector BC but first okay all right so we'll start here I want to now first one get the vector of ba so how can I get a vector of B A it will be b o plus o a okay so for Bo just now what I get is negative 4 negative 3 and also negative 2. your original Bo is four three two one right where's the question okay so you can see that the b o b f is four three two therefore b o will be negative 4 negative 3 and negative 2 in the opposite direction okay right so that's why I'm having negative 4 and 3 and 92 and then for the OA you just copy what you have from the first part so it will be two one three okay then just do a simple just simplify that you have negative 2 negative 2 and also one this is your Factor B8 okay then the next one I want to find out the vector BC okay so to find the factor BC I want to use b o plus OC okay so again PO is negative four negative 3 negative 2 and then plus OC which is 3 negative 2 and also negative four you get all this value from the first part okay and then you have negative five also a negative one negative 5 and also a negative six this is your BC and now I already get B A and B C I want to apply this particular formula to find out the cost data okay so maybe I can write my solution here ba dot PC so when I have d a b a dot b c means that I have negative 2 negative two one dot negative one negative five negative six equals to the modulus so the modulus of ba will be negative 2 square plus negative 2 squared plus 1 squared and then the modulus for BC is negative 1 square plus negative 5 squared plus negative 6 squared and I'm having cos Theta okay then I try to simplify the dot product here negative 2 times negative 1 you get two okay then negative 2 times negative 5 get a ten then one multiply negative six you get negative six equals to when you try to simplify this value here you are having square root of nine and then when you try to simplify this value you have square root 62. you can just press calculator and help you and then I'm having cos Theta here okay then finally what you want us to find is to find out the value of cos Theta so to find the value of cos Theta I will get a 2 then divided by 3 I'm sorry it is not a 2 here 6 right 6 divided by three square root of 62 square root of 9 is a three right okay therefore my final value for cos Theta is 2 over square root 62. okay and they're asking for the exact value so we can just keep our answer in the sub form in this way all right so my cost data is equals to 2 divided by square root 62. and this is how we solve for part number B okay the keyword is exact value R so your answer must be in the short form or Pi form or fraction okay then let's continue to part number three for parliamentary they want us to find the area of ABCD giving your answer in this form P Square Q where p and Q are indigenous so he wants to express our answer in the third format for the area of ABCD and after that P and Q are integers all right so again uh we just have a look for this diagram again okay so when you want to find out the area okay for the parallelogram right again there are many ways for us to solve it up if I want to try and Link with what they are asked us to prove in Part B just now okay we are having better PC Vector ba and data so that means all I actually can try to use what we have just now to find out the area for this particular triangle first then after that I multiply by two okay to get the area for the whole parallelogram okay so my aim now is to try and find out this particular area for the triangle ABC okay just now I already know the length for this side the length for this side is actually the modulus of ba so what's the modulus for BA that we get just now it is a trick okay and then for this one what is the modulus for BC the vector BC the modulus for Vector BC just now I get is square root 62. okay then I can use all this result here to continue and find out the area okay so I will start the starting part here area of ABCD is actually equals to two times the area of triangle ABC okay so to follow the area for triangle ABC it is not the right angle triangle therefore I can use the formula half a b size C right okay so I have half a b that means the length for the side which is a tree the length for the other side which is square root of 62 and then sign C which is sine Theta in our equation here sine Theta in the setting our question it should be sine Theta okay so now my problem is I want to try and find out what's the value for sine Theta again to get the value for sine Theta right we look back to what we have in Part B so for publish to start we already know that we are having the value where cos Theta is 2 over 62. I want to use this result here to and try to substitute my sine data okay since I have cos theta equals to 2 over square root 62. that means if I try to draw a right angle triangle for this particular data okay so it is a opposite sorry it is adjacent over hypotenuse square root 62. okay so to get this one you need to take square root 62 Square minus the four square and then you take the square root again and then your answer should be square root 58 so this length will be square root 58. okay all right then now I try to simplify my answer further so 2 and 2 can be simplified so I still have 3 multiply with square root of 62 and then to get the value for sine Theta I will looking for the value from here so what's the value for sine Theta so sine Theta will be opposite which is square root of 58 divided by hypotenuse which is square root 62. so if I substitute my value here I am having square root 58 divided by square root 62 and then very nicely here you can see that you can simplify our answer square root of 62 and then my final answer will be 3 square root 58. okay we fulfill the request from the question because they want us to show that our answer is P Square Q okay all right so this is how we get the area for ABCD for this question number okay so um you are now at question number seven okay we have the variable X and Y with which satisfy the differential equation for this one and then they are saying that X is between 0 until pi over 4. okay then it's given that when x equals to pi over 6 and Y equals to zero they want us to solve the differential equation to obtain the value of x when y equals to pi over 6 so give our answer correct to three decimal places okay so first of all let us try to solve the differential equation first okay so to solve the differential equation basically I need to separate the X variable all to one side and then all the variable one a y at one side so if I try to separate them uh I should have okay so I would want to bring the sine Square 3y over here and then a COS 2X go down become the denominator and DX multiplied up right so if I rewrite again I will have something like this so I'm having sine Square 3y d y equals to 4 tangent 2X divided by cos 2X then DX okay so this is the first step that we do we try to separate the Y variable to one side and then all the variable X to one side so you can see that we are having clearly all right this two variable at each side here and we need to integrate the equation for both side now all right so before we can integrate it we need to analyze and check first out what is the correct integration technique or method or formula that we can use for the equation here for our equation here right so let's look for the left and side sine Square 3y I cannot integrate anything directly with PSI Tower 2. okay but I can integrate something with sine power one or maybe cos power one therefore to make the sine Square disappear in my solution here I might need to replace it by using some identity okay so maybe you can try to consider to use the com uh the data Banker formula where the original one is called and then we are having maybe two y original one I read the original one first so the original I'm having cos to y equals to 1 minus 2 sine Square y this is the original equation for double angle formula one of the formula for cos two Y is 1 minus 2 sine Square y okay if you can't remember you can refer back to the Banger formula provided in the formula booklet here so cos2a can be written as 1 minus 2 sine square a right okay so why am I choosing this sine Square it is because of in my solution here I'm having a sine square and I want to replace my side Square becomes something that is related to cos power one later why I want to replace it you uh to cos power 1 it is because of I can integrate the COS power one directly by using the simple formula okay but I cannot uh integrate the sine squared directly okay by using any simple uh any any simple and direct formula all right okay and now my problem is I'm having sine squared 3y in my solution and then in the dark angle formula here I'm having sine Square y therefore I can try to modify the uh angle here I want to change 1 minus sine Square y become 1 minus sine squared 3y so that I can replace it into my equation later right okay then what happened to the course here just now when I'm having the Y okay so I Square y then the double angle that I get is cos to Y so now if I want to change become sine Square 3y then the COS Angle now it should be double of three Y which is 6y okay and then if you try to rephrase your answer right for this one so I'm having one minus cos 6y then divided by 2. and we can actually replace our sine squared 3y by using 1 minus cos 6y a divided by 2 because um later we can integrate it directly all right okay so here I will replace it where it will become something like this maybe I prefer to take the half as a constant I take it out and inside here I'm having one minus cos 6y and then d y okay so the integration for this part should be no problem already because I can integrate it one directly I can also integrate the horsepower one directly okay then now let's proceed to the right hand side here so for the right hand side we need to have a look first okay what do we have on the left hand side and how can I integrate them so for tangent 2x over cos 2X um maybe I can try to change become tangent one of the calls is actually Secret foreign so maybe I can try to consider to integrate four tangent to x equal to X okay if you have no idea how to integrate this one maybe what we can do is try to have a look for the differentiation provided in the formula booklet okay or maybe you can we can have a look for integration if you want let's have low whether our integration have a lot of our integration don't have so maybe we can refer to differentiation okay so let's have a look for the differentiation part here okay double check and see after the differentiation under which part do I have second and tangent so you can see this one okay that means what that means so when I differentiate secant x I'll get secant x and tangent X so this is differentiation all right that means so if I want to integrate secant x tangent X I will get back the second this one is integration okay so this is differentiation okay so you can actually um fully utilize the formula provided in the formula booklet and try to find out the possible method or result or answer and also from here I know that if I integrate secant x tangent X then I will get back the second X okay I go back to my solution so sorry to integrate tangent X second X okay later I will get a second second X okay so maybe I really found the method already okay I'll continue with the next step so for the next step left hand side I will start integration integrator one I will get a y integrate cause I will have sine 6y then differentiate the thing inside the angle right so differentiate six point you get a six you have to divide it by a six after the integration then equals to and then I want to integrate tangent to X secant to X so according to why we uh refer just now right we can see that integrate tangent 2X and secant 2x equals to secant still remain S2X but because for this one is the angle with 2x right you need to differentiate the 2x you get a 2 and you need to divide by two okay then after the integration you put the plus C at the back all right okay so this part is slightly trickier so make sure that you when you have no idea how to solve it right or integrate it maybe you can have a look from the formula is there any hint given from that all right so for us we are using this one actually okay so I'm using the result from differentiation and I know that integrate secant x and tangent X will get secant x and then when you're having maybe the double angle or triple angle like secant two x tangent 2x then when you divide when you integrate you will have secant two x then you have to divide the differentiation of the angle just like what I did here okay all right then after that again before I can find out the value of C I prefer to photo simplify all my equation first okay I'll multiply the half into the equation I'm having half y then minus 1 over 12 sine 6y a that equals to simplify the 4 and the two I'm having two secant two x then plus c okay so now to continue with this uh question I need to find out the value for C first okay so how to find the value of C okay I will use the other color to show up okay so given that we we already know that when x equals to pi over 6 I'm having the value of y which is equals to zero so to find the value of C I need to substitute these two values into my equation here to get the C so half white means half zero sine six white mid sign zero okay so the left hand side will be equals to zero then equals to two secant two x so two I just copy it second and change become cos then 2x I multiply 2 multiplied by pi over 6 I get pi over three then plus C so from here if you try to simplify you're having two over cos pi over 3 is a half plus C therefore the value of C that I get is actually negative 4. okay so from here I get my particular solution where it will be something that is a half y minus half sine six y okay equals to 2 over cos 2x then minus four okay so this is actually my particular solution for this differential equation right then after I found the value for C already what else they want us to find so now they are saying what is the value for x i when y equals to pi over six okay so we'll continue further down okay now I already know that Y is equals to pi over 6 and I want to know the value of x again I will substitute into my particular solution okay to get the value of x okay so when y equals to pi over 6 when I substitute inside you are having pi over 12. okay then minus 1 over 12 sine pi okay then equals to the value of x we don't know right and we want to find out so minus 4. okay all right then if you try to um move all your values here and there right slowly okay so you're having cos 2X equals to 2 then divided by pi over 12 sine Pi is actually a zero okay so this one is a zero okay so we are having pi over 12 then last four you just try to uh rephrase Your solution okay from this step you should just try to rephrase the solutions slowly you get something like this okay or maybe I directly show you the decimal number so usually for the decimal number I'll keep it as low as possible so again something like this after that to get the value for 2x of course you need to take cos inverse okay to get the most accurate answer right I will still write it the upright of the value as long as possible okay although I write it out short not but when I press the calculator I'm using the long value here actually okay then make sure your calculator is in the radiant mode and get the answer from calculator divide by 2. so my final answer for the X should be 0.541 correct to three significant figure okay and then again a reminder when we are having all the Trigo Identity or Trigo equation uh to be solved right in our solution or whatsoever uh basically all the angles are measured in radian therefore you have to make sure that when you press the value from calculator right you have to set your calculator into the radian mode all right okay so this is how we solve question number seven okay so now we are in question number eight you have a function rational function and then they want us to express it into partial fraction okay so if you have a look here basically you can see that we're having a linear Factor and also a repeated linear Factor right okay so to express it into a partial fraction I need to write the correct pattern first okay so I'm having this is original one and two X plus one and then X plus 2 squared so when I have a linear factor I need to change it become a partial fraction with this particular linear Factor and then we are having the repeated linear Factor you will need to separate into two partial fractions the first one is without the square another one is with the square one okay all right so this is the first timer you need to write the pattern correctly and then after that to make the denominator disappear you multiply the whole equation okay so multiply the whole equation by using the original denominator which is 2x plus 1 and then X plus 2 squared so if I multiply the whole equation with this the denominator right so I'm having 3 minus 3x squared then the a that I get cancel off the 2x plus 1 and then I will have X plus 2 square on top then for the B part I'm having 2x plus 1 and then X Plus 2. because one of the X plus two I believe cancel right so I'm having still One X Plus 2. and then for the plus uh the C I multiply it cancel the X plus 2 square with X plus 2 square I'm having 2 X plus one okay then now we need to try and figure out what is the value for a b and c okay so first of all I will refer to a linear Factor first one of the linear Factor the easier one will be this one so I'm referring to this linear Factor X plus 2 and also this linear Factor X plus 2 I want to substitute a value of x so that can make it become zero so what do I do here is I will let x equals to negative 2. substitute x equals to negative 2 into the whole equation here okay so negative 2 square you get a four four times three you get negative 12 plus 3 you have negative 9 right so the 11th side is negative 9. negative two plus two you get a zero so this holding normal disappeared negative 2 plus 2 here okay also becomes zero so this whole thing also normal disappear so if I substitute negative 2 here negative 2 plus times two you get a negative four plus one you get negative three So eventually on the right side you're having negative three C okay so very easily now you can get the value for C which is actually equals to 3. all right so now I already uh find out the value of CM by using one of the linear Factor ID then now I need to refer to the second linear Factor so what's the second linear factor I can use okay so the second linear Factor here I can use this one 2X plus 1 and also 2x plus 1 here okay so to make this bracket become 0 1 I will let x equals to negative half okay so still x equals negative half into the whole equation here the left hand side by right you have nine over four if you don't do any characteristic negative one over two plus two then you square negative one over two plus two is three over two three over two U Square you have nine over four so you're having nine over four a then for this one substitute negative 1 over 2 into this bracket right you'll get a zero so this whole thing disappear okay then the simply happen for the third bracket here two X plus one so still negative half here you'll get the bracket equals to zero therefore the holding here also disappear so again very easily you can see that on the right hand side I only have the uh unknown a therefore very easily I can get the value a equals to 1. okay then after that to get the last unknown B I really know the C I already know the a I want to find the B okay so just now when I try to find a c and a right I'm using the linear Factor last I'm having X plus 2 the linear factor I have used it and then for X 2x plus 1 is another linear factor I also using e Dot and make it become zero there is no more linear factor that I can use again all right uh to make the bracket or the linear Factor becomes zero so what am I going to do now is since this is the last uh unknown already what you can do is you can use any value of x that you never used before so maybe for this case to be easier I will use x equals to 0. again substitute x equals to 0 into the whole equation here 3 minus 0 you get a 3 on the left okay and then 0 plus 2 you get a 2 2 square is a four four times a so a is a one therefore you're having a four plus you copy your Plus 2 2 x means 2 0 right 2 times 0 you get a zero plus one you get a one so zero plus two you get a two so one times two you get a two so you are having two B for the second term here I don't know what's the value for B right so but eventually when I simplify I'm having 2B then after that you substitute x equals to 0 into the last bracket here and you'll get a c so what's the value of C that you get just now it is a tree okay and again when you try to refresh everything to the right or the value to the right side your B will be equals to negative 2. okay all right so after you find out uh the finance already for a b and c usually I would prefer to write out my answer one more time in a complete in the fraction partial fraction form so 3 minus three x square divided by 2x plus 1 then X plus 2 squared equals to A is one right so 1 divided by two X plus one and then B is negative two so negative 2 over X plus 2 and also the c c will be 3 over X plus 2 squared okay so this is how we actually get the partial fraction for this particular function okay all right then let's continue to track number B okay so you can try to ever look for partner B right okay for part number B they want us to find the exact value so Hand the keyword exact value for the integration of this function giving your answer in this form a plus b Lord C where ABC are integers okay so again since our part one you already changed the function into the partial fraction right so that means not for this question when you want to file the intersection sorry integration for f x I need to use the result for partial fraction to continue for the solution okay all right so I will have 0 4 and then just now for the partial fraction that we get now we're having one over two X plus one then 2 over X plus 2 and also 3 over X plus 2 squared okay so now for the first term here when I want to integrate 1 over 2 X plus one I will get long then 2x plus 1 divided by differentiation of 2x plus 1 you get a 2 right so you divide it by the two okay so again if let's say you have no idea how to integrate this basic function then you can refer back to the integration or differentiation pattern so you can see when I want to integrate 1 over X I have long X going base integration right yeah so it can refer to the differentiation formula or you can actually have a look for the integration directly so you can see that when I want to integrate 1 over X they tell you straight away that is long X actually okay so let's go back to our solution here therefore we integrate 1 over 2x plus 1 I will get along 2x plus 1 but because the X is not 1X right so you have to differentiate the 2x plus 1 you get a 2 and you need to divide by 2. same thing happen here the 2 is uh um constant so I won't touch it during the integration so now I assume that I want to integrate 1 over X Plus 2. so integrate 1 over X plus 2 I'm having long X Plus 2. okay then of course on these two value we have to substitute in the limit 4 and 0. okay then now the last one for this last term when I want to integrate I have to use the basic formula that we learned in as earlier where trees are constant I take it out I want to integrate x plus 2 power negative 2. okay and then of course I have to I want to complete the last term here for the integration before I can find the value okay so 3 is a constant I just copy integrate the bracket minus one you're having the bracket power plus one become negative 1 divided by the negative one differentiate the thing inside the bracket you have one so divided by one then and now you have something like this okay all right then of course if you want to avoid all the careless mistake or whatsoever right usually for me that I would prefer to simplify everything first into an easier into a method or maybe into the steps that is easier to understand or safer okay so I will have negative three and then over X plus 2 power 1. then the limit is still 4 and 0. okay right so we have completed the integration part now and now we want to express our answer become a plus b Lon C so first step you just try to substitute the value for first okay so still the value for for the first term here 4 times 2 is 8 I get a 9. okay so I will have half long nine and then after that for the second term here I'm having two and then long four plus two is a six then for this third part here I'm having negative three divided by 6 also okay so after substitute the value for I want to substitute the value 0 now substitute 0 into the first term here I'm having half long one okay then into the second term having negative two long two and for the third term here I will be having uh three over two okay and now I have to think away to solve my details here like all the details here okay there are many ways to to simplify to get this answer basically all right so I will just give you a suggestion okay using my steps here so that um see whether you can understand right uh because I want to have a plus b long C right um I tends to combine all the loan together but if I want to combine all the loan together right I have to make sure that all the number in front I pull it to the back so that means all the half right I will put it to the back become nine power half so I'm having nine power hover that means square root nine square root of nine I will get a long three so for the second term here the two if I put it to the back I will have six square six power two so six power two actually I'm having long 36. so my aim now is to move back all the value okay in front of the lawn so that I can make them into a simple line like this long three minus long 36 then later I can further simplify that combine all the long together right okay so three over six is one over two long one I'm having a zero then I have a plus then long 2 square so 2 square is a lot four then plus three over two okay so let's focus on these two constant first without the learner so negative half plus three over two I already get a value which is the integer right it will be equals to one and after that I want to combine the remaining lawn so you're having three terms of long hair laundry minus long 36 and also plus long form okay so for to combine them right what I will do is I will put a plus first all the uh loan the term with the learner is positive then I will put it on top that means it belongs to numerator and I multiply them together if I'm having negative okay in front of the log basically it means that uh the value I need to make it become denominator when I combine that so long three laundry is positive number three right so the three will be on top plus long four so it is a plus that means you have to put it on top also so you multiply the three together with the four then we divided by 36. okay then again further simplify it 12 divided by 36 you're having one over three and it this for this statement it looks like we already get an answer right but if you refer again now there are requests in the question they are asking for a plus b long C where these three values should be integer so that means something wrong somewhere this is not my panel I say that that fulfill the question so I need to further simplify that okay long one minus long three okay so from here I can separate long one divided by three become long one minus long three and that was a very fold on the one so long one basically you get a zero right and if you write out your final answer it should be in one minus log three so I think this is actually the final answer that the question requires okay because from here you can see very clearly that your a is one B is negative one and then C is three okay all right so this is how we solve question number okay then now we come to question number nine the constant a is such that um it is fulfilled this equation then they want to show that a is half long four a plus two Okay then if I look at this equation here the integration part right I'm having the product of two function so when I want to apply integration I have to use integration by part okay so for integration by part so first of all I need to analyze first so the X here is belongs to algebra and then the E here is actually the exponential right according to the rules that we learned earlier so we have to use the rule l i a t e where if your group is higher then you make it become the U when the num the group of the function is lower then you make it as TV so I think very obviously a is the algebra which is a higher than e right so I will let U equals to the X in our equation here and then after that the DV will be equals to the remaining terms so I'm having E power negative 2x DX okay software you decide the correct group already for the U part you need to differentiate it for the DV part you need to integrate it to apply them in the integration by part formula later okay so when I differentiate U reflects back to X I'm having d u over DX equals to 1. integrate DV actually that's a value one here all right just that we need to write out so integrate one DV I'm having a v integral E power negative 2x I'm having E power negative 2x after that you differentiate the power you get negative two so you have to divide by negative 2. okay so after I settle the part for U and B and now I want to start applying the integration by part okay so for this iteration by Panna Theory tell us the result the integration here would be equals to one over here okay so I'm having 0 a then x e power negative 2 x DX equals to half okay so now our aim is to integrate the left hand side here okay by using integration by path and then we slowly rephrase our answer and steps up into this as a final answer okay all right so we'll start okay so we can have this one to apply integration by Pana the formula is u v minus integration of vdu so why is your u u is X y sub v v is this one right so if I rewrite it better in a better way then I will have negative half and then X and an E power negative 2X so this is UV so when I have UV I have to substitute the limit a and 0 into the X data okay so I'm having UV minus v d u okay so minus vdu why is the B now okay so for the V I'm having this which is negative half E power negative two x and then d u from here you can see d u equals to 1 DX right so you're having a DX at the back okay so maybe before I forget now I will try to write out the integration Formula First okay so this is an integration by path formula okay all right and then uh so I'm having UV minus v d u and now let it equals to half R equals to 1 over ETA on the right hand side okay then I want to slow this in before all my equation and step here okay for first term here I try to substitute the a into the equation I'm having negative half a and an E power negative 2A then I want to substitute the value 0 also when I substitute the value 0 into the X and the whole term here will become zero so this whole part I only have this after I started the limit a and zero okay then for the next part I want to integrate it so again before I integrate anything I would prefer to simplify my equation first to avoid callous mistake okay so this is what I have therefore it will be very easy for me to integrate okay so integrate E power negative 2x I'll have a negative 2x then divide it by negative 2. and now I get a and 0 that equals to 1 over 8. okay then again further simplify it bring negative one over four then I want to search into the value a inside I'm having e our negative 2 a socio 0 inside E power 0 you get a one so make sure that you put in the 1i it is not zero for E power zero okay this is a common mistake that students usually will do all right so now I'm having something like this okay then again before I rephrase it right I try to expand everything first okay so maybe I want to group all the values together like so one over four I bring the one over eight over so minus one of eight okay so one over four minus one over eight basically I'm having plus one of a okay then uh I feel that if I have the fraction like this it's very hard for me to rephrase my solution so what I prefer to do now is I try to multiply the whole equation with a so when I multiply with 8 right all the fractions will disappear so that means for the first term here half times eight I will have four therefore this second term here one over four times eight I'm getting two then eight times eight I get the one okay then what will be the next step so maybe for the next step I will want to bring my E power negative 2 a to one side first so that I get all the positive value positive sign then factorize out the e okay so I think I'm very close to the answer already because I already have this 4A plus Doula okay then for this E power negative to a right you can actually rewrite become 1 over E power to a so I want to bring it up become the numerator on the left hand side okay so in my equation I want to eat also to make the 2A come down become the number I will take long for both sides so through a log b equals to log or a plus 2. okay and then long e of course you should know that long e is equals to 1 right so from here you can get the equation that you want to show where a equals to how long for a plus two okay so this is what uh we have left for part A in question number nine okay then let's continue to Part B okay so for Part B they want you to verify by calculation that a lies between 0.5 and also the one okay so uh to prove that a lies between 0.5 and 100 there are many methods here so for me I'm using the sign change method so for scientist method right my original equation that they want us to prove the song is this one so a equals to half Lon for a plus two then to use sign change maintenance I will actually want to move all the variable to one side so that I can make this whole equation become a function at one side where I'm having half and then long for a plus 2 minus a okay then to find the sign change I need to substitute 0.5 into the equation and also I need to substitute the 1 into the equation if I started 0.5 into the equation my answer is 0.193 correct to three significant figure if I substitute the 1 into the function then I will have negative 0.104 which is a negative value so you can see very clearly that this is a positive value and this is a negative value right when you can see the sign change happened between F 0.5 and X F1 then it means that the root or the a actually lies between 0.5 and also the one okay so you can make a simple conclusion on this side so sine chain can be seen and therefore we are we can conclude that the a actually lies between 0.5 and also one okay so this one we have for Part B all right then after that we proceed to part C okay so for part C they want us to use the iterative formula based on the equation a to determine a correct to two decimal places give the result of each iteration to four decimal places so again our equation in part A is a equals to half then long for a plus 2. okay and then to use the iterative formula to find the answer right you need to use your calculator and key in the formula on the right hand side okay so this is the formula that you need to clean in your calculator so that we can continue with the iteration later I need to key in the calculator with this formula on the right hand side okay so I'm having half so I'm having long 4A plus 2 long and then for a a I will replace it by using the X then plus 2 close bracket and then um we are having divided by two okay all right then now you need the first uh the the initial value to start with so the initial value you can refer you can use any value that you like but basically we'll refer to the um number that's close to the root so we know that actually our rate is between 0.5 and 1 right so it is actually a good hint for us to use the initial value from here you can use 0.5 you can use a one or you can use any value between that all right so for me I'm setting my first initial value as one for you yourself you can try any other value basically our final answer will be the same okay all right so now we want to find out the A2 so together A2 you need to substitute the 1 into your identity formula okay then try to press calculator and see what's the value that you get so that means how you can press the calculator sign and then we'll key in the first value which is the one right so okay in one and then you press the equal sign okay so the first value that you get should be 0.8958797346 it is a long value but for this question they're asking for all decimal places therefore we are having 0.8959 so I will put in eight nine five nine eight nine five nine sorry eight nine five nine okay after that I want to continue to get my next value for the a so to get the next value for a i press calculate again key in the A2 value which is 0.8959 okay then I press equal sign one more time I will have eight five nine nine so around the answer correct to four decimal places eight five nine nine okay we'll repeat this process until we get the repeated value for two decimal places all right okay then we continue to X4 to get the X4 you press calculate again key in the value that you get just not at A3 which is 0.8599 press equal sign and you get eight four six nine so zero point eight four six nine okay so I can see that the two digit values are not repeated yet right so I will just continue the process again so calculate key in the value 0.8469 then press equal sign eight four two one and now my face value is eight four two one Okay then if you look at the A4 and A5 right although these two digit uh these two decimal number is 0.8484 right but when you correct it right uh 0.846 are correct to two decimal places you'll get 0.85 that is why you're having 0.84 so that means the two decimal places answer are not repeated yet so instead to continue further to get the next value A6 okay again press to calculate key in the value 0.8421 press the equal sign and you get 0.8403 so 0.8403 okay so if you double checking up the answer for two decimal places right this one will give you 0.84 and this one will also give you 0.84 that means you are having the repeated value already right so from here basically you can tell them that the root or the value of a should be equals to 0.84 correct to two decimal places okay but when you want to find out all the iterative iteration here you have to ensure that you are keeping it or maybe you show it in four decimal places according to the request from the question okay all right so this is how we solve question number nine okay so we are now at the last question question number 10 um we are given a polynomial denoted by p x they want us to show that X plus 3 is a factor of PX okay so um for this one to prove that X plus 3 is a factor maybe I can use the factor theorem to prove okay I know that my polynomial equals to this one okay so to apply the uh the factor theorem right I'm having the factor X Plus 3. if I let X plus 3 equals to 0 then my X will be equal to negative 3. so that means I want to substitute the value x equals negative 3 into the polynomial and I want to see what's the value that I get after I start to do it if I get a value that particular value will be the remainder which is a remainder theorem I will get a result we are using the result for remainder theorem but if we substitute the negative 3 into the polynomial and eventually I get a zero that means all uh I have no remainder when I divide it by X plus 3 okay so this is how we solve uh to how we show it okay so 32 negative three into the equation here I'm having negative 3 power 3. and then plus 5 negative 3 squared plus 31 negative 3 and also Plus 75. okay so when you are trying to simplify all the value here I'm having negative 27 and then the second term is plus 45 the third term is negative 93 and the last term is 75. okay then you just try to press calculator and that will help you a little bit so you get a zero actually so from here since I get a zero that means there's no remainder like when I divide the polynomial by using X plus three therefore I can say that the X plus 3 is basically a factor of PX okay so this is how we solve it for part a all right so let's have a look for part number B now they want us to show that uh Z equals to negative 1 plus 2 square root of 6 I is the root of p and Z equals to zero just now we are having p x so now the change it becomes PJ and they say that when p x p z equals to zero right the one of the root is this one okay so to show this one basically we are having um two different methods here or maybe more than two more than two methods are there are a few methods here the first method is you can try to substitute the deck into the polynomial also and you get a zero so if you get a zero then basically it means that the jack is the root okay all right another method is some students will carry out the long division where they take PX or pz divided by Jack Plus 3. after that they will get a quadratic Factor then they try to solve the quality Factor that's another method okay so there are a few methods that you can try to consider like if you want okay so for me I'm using the substitution so let's say I'm having p z right so p z is actually equals to uh this one okay just I need to change all the X become Z so I'm having exact power three plus five x square okay then plus 31 sec Plus 75. okay so since they said now the negative one plus two square six I is the root of this polynomial right equals to zero so what I need to do now is I substitute the Jack into my polynomial okay so I'm having power three right so that means I'm having three brackets here okay then plus five power two okay then plus 31 zag Plus 75. okay so this question right basically it is not hard but when you want to expand all the brackets for the exact you need to be very very careful try to avoid the colors mistake okay so negative one times negative one you get a one then this time this you get negative 2 Square Root 6 I then again another negative 2 Square Root 6 I so square root of 6 Times Square Root 6 you get a six times four so six times four you get 24. so 24 I Square means negative 24. okay then I'm still having the third bracket here and then for these two bracket also I'll have one minus 2 Square Root 6 I minus 2 square root of 6 I 4 times 6 is 24 times I squared is negative 24 again so minus 31 plus 62 square root 6i then plus 75 okay then now I need to simplify this I'm having negative 4 square root of 6 I then 1 minus 24 you have negative 23. then multiply with the last bracket negative 1 plus 2 Square Root 6 I okay so basically for um what we have here right we just need to make sure that we don't do any careless mistake okay the the step might be a bit tedious okay because when you want to expand everything is here you can see it's quite low okay but you just need to make sure that you try to expand everything very carefully okay so negative 31 plus 75 I will have 44 so continue here I will have for this all and then plus 62 square root 6i okay I explained these two bracket again negative four times one I get a positive right so I'm having 4 Square Root 6 I then 4 times 2 is 8 okay so I'm having negative eight negative eight times six okay I have negative 48 and then I leave 48 now by multiply with i squared which is uh negative one right so I have positive 48. okay 23 times 1 you get 23 then this one you have 23 times 2 23 times 2 you have 46 so minus 46 then Square Root 6 I then minus 20 square root of 6 I okay then you take 23 times 5 you have minus 1 1 pi okay then plus 44 plus 62 square root 6i okay so now we try to analyze it slowly yeah I'm having 4 Square Root 6 I minus 46 Square Root 6 I minus 20 square root of 6 I and then plus 62 Square Root 6 I so what do I have after I simplify of them so 4 minus 46 minus 20 then plus 62. basically I get a zero okay so you can try to simplify this part you realize that it is equal to zero and after that you simplify all the number here 48 plus 23 minus 1 1 pi plus 44 so what do I get okay if we didn't do any class basic right you also get a zero here therefore our final answer for this one is zero okay so you realize that when you substitute the Z into this equation right okay into the equation here into the polynomial I should say Okay into the polynomial then you realize that it is equals to zero that means what that means of Z equals to negative 1 plus 2 Square Root 6 I is a root okay for p z equals to zero okay so this is how we solve it and proof it then only three months is given but basically that's um it is not very hard okay it just looks a bit a little bit tedious on it all right you just need to expand everything here slowly make sure that you don't do any class mistake then by right you should get the zero after you search through the value as expanded and simplify that okay and after that for part C so for part C they want us to find the complex number Sega when we are having this one the probability of sorry the the polynomial of Z square equals to zero just now we are having p z equals to zero and now we are having p z square equals to zero okay so before I can solve this part of the question right I want to go back to part number B just now okay for part A for this polynomial I know that I'm having one real root which is x equals to negative three this is the real root right and then for the second part from here I know that I have one complex root which is negative one plus two squares uh six I okay so that means so if I try to summarize a when p z equals to zero right what is the value object that I have I have three Roots here the first rule is from first part where x equals negative three so now I'm having Z equals negative three the second rule is the one that we proved just now negative one plus two square root 6 I and what will be the third group okay so for the third right we actually have a rules where if all the coefficients for the polynomial are real numbers then the root appear in conjugate pad so you can see that my coefficient for this polynomial right all are real number 3 5 31 75 or a real number therefore I know that the other group for this polynomial will be the conjugate of Z so the conjugate offset means that it will be negative 1 then minus 2 Square Root 6 pi okay so for this particular polynomial right p z equals to zero I'm having three roots or three value object which is negative 3 negative one plus two square root 6 I negative 1 minus 2 square root of 6 I Okay then if we try to observe this holding okay if you try to observe this whole thing into part number c right okay so let's have a look at for Part B we are solving the polynomial Z equals to zero and now they want to replace the Z by using Z Square okay the polymer normally are the same but you replace all the exact using Z square that means that your jack Square Now The Roots are becomes Z square equals to negative three okay then Z square equals to negative 1 plus 2 square root of 6 I and also the Z squared equals to negative 1 minus two square root 6 I okay that for this question again I get all the value for Z squared because I replace the exact by using Z Square here right in particular C and what it was to find it wants to find the complex number Z so to find the complex number set you need to solve the three answer here one by one okay so we'll start from the simple one first Z square is negative three therefore that is basically plus minus square root negative 3. and then in the complex number right we can write this Plus or the square negative 3 become square root 3i so these are the first two answer for the exact that I can obtain okay then now to get the second value for the Z you need to solve this equation Z square is equals to negative 1 plus 2 Square Root 6 I right and now I want to find the square root of negative 1 plus 2 square root of 6 I how to find out the square root of negative 1 plus 2 square root of 6 I okay so basically I will stay together complex number which is a plus b i and then you try to bring the square to the other side square root becomes square right so I'm having a plus b i Square then I try to expand my right hand side here eventually I'm having a squared minus B squared I want to group them together so that this is the real part then plus 2 a b i so this is the imaginary part okay after that I want to compare the real part with the real part so this is the real part this is also the real part here okay so I will compare a square minus B squared it will be equal to negative one okay so a square minus B squared equals to negative one this is my first equation and the second equation I want to compare imaginary with imaginary part so 2 Square Root 6 equals to 2 a b okay so 2 Square Root 6 equals to 2 a b so from here I want to make B in terms of a la so b equals to square root of 6 over a and this is my second equation to solve the simultaneous equation I saw sub equation number two into the one okay so if I substitute into the one I'm having a square minus Square Root 6 over a square equals to negative one okay so I'm having a square minus 6 over a squared equals to negative one also okay then after that I'm having a power 4 plus a square minus 6 equals to zero so from here you can see that I'm getting a quality form equation so in terms of a square right okay so negative six I want to have positive 3 and also minus 2. okay so my a square equals to 2 therefore my a is equals to plus minus square root 2. therefore this part I will ignore it because a squared equals to negative 3 then I don't need to continue to find the Ada because if I find the aerator right when I substitute back into the equation just now the answer that I get will be actually a reputation of this one of this one so you don't need to waste time and solve that all right okay so after you get a equals to plus minus half already okay so now continue to find a b a equals to positive square root 2 What is the value of B so you substitute into the equation number two which is more straightforward so b equals to Square Root 6 divided by square root 2 and therefore your p is actually uh six divided by 2 you get a 3 right so you get square root of three and then when your a is equals to negative square root 2 then your P will be equals to negative square root 3. so if you try to list out the answer for the exact here right your deck is having this value where first one is square root 2 plus square root 3i another group will be negative square root 2 minus square root 3i okay or maybe some of the marking scheme or maybe some solution and they will show it in the shorter way where you can write become plus minus bracket square root 2 plus square root 3i these are also again okay so I get another 2 root for this equation okay so just now the first two rules was this one the first two Roots was this one exact equals to plus minus square root 3i then I solve the second equation here where I get these two answer as the root and now I need to solve the third equation so basically the steps are more or less the same for the third equation here okay so I'm having square root negative 1 minus two square root of 6 I equals to A plus b i I still get a complex number when I take the square root of complex numbers then negative 1 minus 2 square root of 6 I I will get I simplify sure so I'm having a square minus B squared plus 2 a B I like how we expand it just now still the same concept and idea compare the real Pi with the real part this is a real part right therefore you are having a square minus B squared equals to negative one so this is your third equation if you want to make it as a sign okay so this is the third equation and after that you compare right the imaginary with the imaginary part okay so if you try to compare the imaginary imaginary or you will get 2ab equals to negative 2 square root of 6. then from here we'll get b equals to negative square root of 6 over 8. and this is my fourth equation again to solve this one I will substitute the fourth equation the fourth equation into the third equation okay so I'm having a square minus negative square root of 6 over a and then I Square it equals to negative one Okay then if I try to simplify them I'm having a negative six then plus a square equals to zero okay so again I get another quadratic equation in this so I try to solve them factorize them and solve them and you can see that it is actually equals to one we get just now right it's quite similar okay so I try to solve for a equals to 2 then I ignore this part okay for a square equals to 2 you take the square root by yourself so you're having plus minus square root 2. then again I want to find out the value for B so when a goes to positive square root 2 right what's the value of B so to find the value of B you substitute into the equation number four okay so negative square root of 6 divided by square root 2 you're having negative square root 3. and I want a equals to negative square root 2 the B value that you get should be a positive value square root 3. so from here right that means you get another two value for the zaga which is the ruler it will be equals to square root 2 minus square root 3i okay then negative square root 2 plus square root 3 I or again you can rephrase it in the more simplified way where it is actually plus minus square root 2 minus square root 3 I okay right so basically you already solve all the equation here for the Jack Square so you're having six root basically you can see that yeah this is the first two roots second and the third oscillator and the fourth and also the five and the sixth so you are having six roots in total then if you double check and see when you're having P exact square right when you replace the x or the Dagger by using Z Square in the original equation uh this one okay you're having X power treatment right when you start to do exact Square power Tria you'll get Z power six when you're having the highest power the value is six right basically you're having six roots on okay so which also uh verify that our solution here is correct that means that when you try to solve p z square equals to zero you should have six answer eventually so this is the first two answer the third and the fourth and also the five and the sixth and to avoid um the possibility that maybe the examiner not able to see your answer clearly right you share our remind students again if possible or if can try to group all your answer at last line again so I'm having Z equals to plus minus I square root 3 and then plus minus square root 2 plus square root 3i and also the last one will be plus minus square root 2 minus square root 3 I okay so basically this is how we solve this question right so in my the step might be longer okay but uh to me when solving it it is not hard to understand all right so you might have other method in your Maya to solve it so you can just try and see whether you're able to get all this answer or not so this is just a suggestion of how are we going to solve this question but basically they might have more than one method on this right okay so thank you so much for watching this video and we will end the video here okay thank you very much for watching