hello everyone welcome to my YouTube channel in this video I want to solve additional MTH paper mayun 2024 paper 2 3 this is the second paper of variant three so I I I really appreciate cie for giving me opportunity to make use of the past papers from YouTube channel I'll start with question one in question one I have question coordinate geometry here I have a point 1A 4 the coordinate of a point a 1a4 coordinate of point B 5 6 the perpendicular bis sector of ab let's assume this is let's assume this is AB there's a perpendicular bis sector um intersect x axis at C let's assume let's assume this is um X AIS intersect X as C and Y AIS at D um assumption also this is y AIS X AIS this is C and Y AIS at D even that o is the origin find the area of the triangle so I'm looking for the area of this triangle this I'm looking for this I have this Tri D now what I need to do now is get the coordinate of the point of intersection of the perpendicular bis sector that's d and c and find the distance from here to here is the intercept on X AIS for the perpendicular bis sector the distance from here to here is the intercept on y AIS so first thing since it's perpendicular B sector I need the Mido the midpoint is the common point to both a b and the perpendicular bis sector so I need a midpoint so let me go for midpoint first midpoint the mid point is addition of the two axis sry addition of the corresponding axis and divide by two or the average between them so which is X1 + X2 / 2A y1 + Y 2 / 2 1 + 5 / by 2 1 + 5 / 2 comma 4 + 6/ 2 this gives 6 / 2 3A 5 this is the midpoint so the midpoint is going to be used to get the intercept on y AIS for the equation which I'm going to use y = mx + C so let's go for the gradient m y 2 - y1 Y 2 - y1 / X 2 - X1 the gradient is 6 - 4 / 5 - 1 this is 2 over 4 which is um 1 that's the gradient of ab M AB M AB the gradient of the perpendicular bcor m perpend b sector this is 1 / two this will be my 2 you don't need to apply any formula to that it's just interchanging both and denominator of the that's multiplicative inverse and addictive inverse of the multiplicative inverse minus so that's -2 find the equation of that now Y = -2x + C so you can see the essence of finding the mid point use that to get C that into so Y is five um X is 3 C 5 + 6 is = c c is 11 the equation is y = - 2x + 11 which means The Intercept here is 11 Y intercept don't forget y = mx + c y = mx + C D C is Y intercept so The Intercept of that perpendicular by sector is that so coordinate of D is 0a 11 so let me get coordinate of C I'm going to subtitute Y to be Z because at this point Y is equal Z so if Y is = 0 into this equation as equation of the line 0 is = -2x + 11 2x is = to 11 x is 5.5 that's X 11/2 so the area of triangle d c 1 / 2 * base * height 1/ 2 which is 1/ 2 the base is OC which is 11 / 2 which is same thing as 5.5 that's 11 / 2 * the he this is 11 from here to here is 11 11 this gives 1 2 1/ 4 that is the area of area required to be found now the second part of this the second question given that this equation has no roots so this is very important we're using the um discriminant find the set of possible values of K if it has no root b s - 4 a c is less than zero that means it has no Ro so from the equation a which is if I compare it with coefficient a x² + b x so comparing with quadratic equation a is equal to k a is K and B would be a is k b will be 2 Kus 1 and C is going to be k + 1 C is going to be k + 1 this is C so I'm using discriminant b s - 4 a c less than 0 2 K - 1 2 - 4 a k c is k + 1 must be less than zero expanding this this is 4 k s - 4 k + 1 expanding this also 4 K * K - 4 k 2 - 4K 2 I have - 4 k 2 - 4 K less than Z so this can see this - 8 k + 1 less than 0 so - 8 K less than -1 K is greater than -1/ - 8 because I'm dividing div by negative number I need to change the inequality sign is greater than 1 / 8 for this for this it for this equation to to have no root so moving on to the next part which is question three in question three I'm told to draw the diagram for this modulus function modulus function y = 2x - 5 and 4 AB sorry the absolute value of the modulus of 2x - 5 and 4 modul of 4 - x so in doing this this is not that difficult straight forward so don't forget you can do it by sketching the graph of 2x - 5 and reflect up and reflect up or you just subtit in values so if you subtit in values is very easy here what I'm trying to say if I I can also do it without reflection I can subtit like Y is = 2x - 5 when x is 0 Y is 5 is intersecting the graph here the graph here that's when X is0 Y is 5 that's very very important so let's look at x intercept x intercept y will be 0 2 x - 5 = 0 x X is 2.5 so which is this this is a graph 2.5 so at this point means the gra goes like this sorry the hand is not as straight I'll manage it um so I can also use the knowledge of the gradients to sketch it the gradient of this is two so which means for every two y AIS I move one horizontal if I move one to this side I'll move two down so it's going to intercept it here one to this side move to is going to inter I know where the point that is going to intercept with the knowledge of where so I can move like this the goes like this likewise this I move one two so it's going to be this it's passing through this point this point and this point so this is the graph of the first part given to me now this is GRA of this is intersecting let see seven or let me just use one line is intersecting these points these points and seven okay this is better this is the graph of y = 2x - 5 the absolute value of that the second one 4 minus the gradient of this is1 so the gradient of this is1 so if I want to look at the point it's very easy means for if um if I want to sketch this is easy if I say Y is = modul of 4 - x modulus so when X is z y is 4 is intersecting let me use another color for this is intersecting four at this point so when Y is also zero X will be 4 x is = to 4 so it's intersecting it's just a straight line going towards um so but that line since the gradient is1 so for every one axis move to every one unit move to the left and move one unit down left one down or every one unit move to the right and move one up so one unit to the left one down is going to intersect it here one unit to the left one down so drawing that will be very easy for me so this is how the graph will look like so he intersecting here also so there no nice I have something like this okay Contin yeah this is a graph now likewise this goes like this yeah so that's it this is a graph for y is = 4 - x the absolute value of that now the B part I'm told to use a graph to solve this inequality at the point where this color is less than um this color does mean if you look at the graph the point where is less equal to that is when it is below if you look at this region I use at this region the graph is below or equal to the blue one also this region the lines are below so talking about the x value is going to be the x value downward continues like this and this one also is going to be the x value from here downward so if I want to write my answer it's going to be X is less than = to 1 or X is greater than equal to 3 look at it upward the numbers and at this region down downward so those are the values of X that satisfy the inequality asked okay now moving on to the next question the next question is question of binomial expansion yeah I'm told to find find and simplify the term independent of X means the term that does not have X the term that does not have X so in doing this this is very simple to identify no need of expanding all this which means the power of this the power of X here after expanding will be equal to the power of X here after expanding so for this power to be equal to this this can be three why the power of X2 can be 3 y power of X Cub 2 x Cub can be the power of 2 x Cub can be um two so for this is not possible because addition of these two Powers must give us 10 so that's not possible also the power of x² can be 6 and the Power of 2 x 1 / 2 x Cub can be four CU 3 * 4 that's 12 this time this 12 that is the only region where they will cancel and 4 + 6 + 4 is 10 it must give you a total power so I've gotten where the term independent yeah you don't need to start expanding you can see why this question is two marks and uh no space for you to start expanding so this is going to be 10 combinations since I said the power I want to do ascending power um I want the power of this to be increasing um or whatever if the power of this is six and I want x² to take the power of six so - 1 / 2 x Cub is going to take power 4 so this is the ter is meant for me to now simplify 10 combination 6 is 210 * x^ 12 * -1 to^ 4 that's still one because when is um even power of negative number does not have the negative number does not have f 2 The Power 4 * x^ 12 because everything inside needs to take this power x^ 3^ two so let jump that step this is still 3^ 4 this is 210 x^ 12 over 16 x^ 12 this canc this 210 16 is 105 over 8 that's the answer to that now the big part of this question yeah I'm told that I should not use calculator in this part um I'm told to use binam theorem to show that this is here very simple what I need to do is just expand this first 1 + 2un 2 to the^ 4 so that's four combination Z 1 to the power of 4 2un 2^ 0 + 4 combination I'm going to have five ter here because the power is four which implies I should have 510 four combination 1 1 to the^ 3 the power of this is reducing while the power of this should be increasing so 2 to the power 1 + 4 combination 2 1^ 2 2 2 to the power of um 2 + 4 combination 3 1 to the power of 1 2 2 to the power of three and the last one let just put the last one plus four combination 4 1 the^ 0 2 2 the^ 4 now four combination Z is 1 1 to the^ anything is WR this is also one so I have one plus the next one I have um the next one I have 1+ four combination one is four this is still one 2 two is still 2 two 4 * 22 that gives 82 82 for combination two is um for for combination two is six yeah for combination 2 is 6 um 2un 2 2 2 is 4 2 2 is 4 and um 2 2 is 4 and 2K 2 2 is 2 this gives 8 so I have + 6 * 8 which is 48 the Y this gives 48 sorry um the this one gives 48 the middle one four combination three is four this is one and this is 8 * um 2 2 that's 16 < tk2 16 < tk2 * 4 that's 642 this 64 < tk2 and the last one this is one this is one 2 to the^ 4 is 16 and this gives um 4 16 * 4 64 yeah yeah this is the answer to the first expansion this expansion 1 - 2k2 what you just need to do is to replace everything here with negative value everything here with negative value of this so it's still four comination 0 1^ 4 - 2un 2 to the^ 0 plus four combination 1 1 to the 3 - 2 2^ 1 + 4 combination um 2 1^ 2 - 2k2 to the power of 2+ 4 combination 3 1^ 1 2un 2 to the power 2un 2 to the^ 3+ 4 combination 4 1 the^ 0 - 2 2^ 4 so the only part that the negative will affect to be the second one and the fourth one because negative to posi still zero this will still be the same thing as this the second one this will change to negative so this is going to be 1 - 8 < tk2 + 48 - 64 < tk2 this is going to be the same thing as this plus 64 so I'm told to subtract both of them there a question subtract both of them that is what I'm about to do so it's going to be 1 + 8 < tk2 + 48 + 64 < tk2 + 64 - 1 - 82 + 48 - 64 < tk2 642 and + 64 so if I expand that with negative 1 + 82 + 48 + 64 < tk2 + 64 - 1 + 8 < tk2 this already canc this - 48 this already can this + 64 2 - 64 this also cancel this so I have 8 < tk2 + 64 < tk2 plus another 8 < tk2 + 642 so if I add everything together this gives 144 2 this is the answer to the question this is the answer to the question I part of this question I'm told ends write this in form of this now don't forget what we did at first is this numerator is 1442 that was what we did so I have 144 < tk2 over 1 + < tk2 so the next thing to do here is to rationalize by the conjugate of the denominator which is this so 144 < tk2 multiped by this this gives 144 < tk2 minus 144 < tk2 um multiped byun2 that gives -2 8822 that gives us big denominator 1 * 1 - 2 + 2 - 2 144 < tk2 - 288 over -1 this is - 144k 2 + 288 I've divided everything by minus one that's what I've done so which gives 288 - 1442 this is the answer to this question now moving on to the next question which is question five yeah I have a question on tric ratio in question five I have question ra um it can either be Tri Identity or equation all our trigonometric ratios um I'm told to show that f of x can be written as this this is f ofx written as fraction so they don't want me to write it as fraction and change all the sign cos everything should be T the fundamental trick the see is T now looking at the numerator where 1 sin x I'm going to change one to sin s x + cos Square X we all know that is one that's the First Fundamental the first identity that you then um sin squ x + cos Square X is 1 so I have S 2 x + cos² x + 2 sin x over cos² X this is if you add this common ones like 3 sin 2 x + cos² x over cos² X which implies us I have 3 sin s x over cos² x + cos² x over cos s x so sin s x / that's T and don't for is a multiplier which is 3 have3 tan s x plus cos s x / cos x that's 1 so I've been able to write it in form of this so in form of a x + B is 3 B is is 1 a = 3 B is = 1 so I'm told to solve this equation F ofx = 4 this is f ofx which I've been written as this to to solve it when is the root ends the root ends because they want you to go back to the first thing you've done or the previous work you've done which means F ofx you're not supposed to uplift you're not supposed to uplift this this is what you will need to uplift so 3 ² x + 1 = 4 this is what I'm solving so I have 3 Time Square x = 3 it subtract four from both sides it subtract one from both sides so you have this tan s x is = to 1 if you divide divide this by three divide everything by three you have one so what is Tan x Tan x is < TK of 1 which is Plus or -1 t x is Plus or -1 you have two values of tan X and don't forget let's go back to this the region X4 this domain of the function so from < / 2 to < / 2 very important so I'll pick one by one Tan x = 1 x is tane of 1 tan inverse of 1 isun / 4 also Tan x = -1 x is tan inverse of -1 and this is also - < / 4 the two answers falls in the region X is < / 4- < before now moving on to I I part of the question n again the word n is also used so they're referring to what you've done and also find the gradient of the core at each of the points where y isal to 4 at each of the points where why we don't forget we solve the cve equal to four and we got two answers which are these so I'm to find the grent of the points at those two points so let's go back to equation of the curve first which is y is 3 t t What 3 t sare x + one I need to get the gradient function first which is the ydx the Y DX is the gradient function this is 6 you multiply the power by this which is 6 Tan x take away one from the power that's one it has the leave it and what is derivative of tan derivative of tan is SEC squ SEC s x so if you differentiate one is nothing so this is the gradient function so I'm told to find the gradient at X is = > / 4 and - p over 4 because when Y is equal to 4 we got these two answers so I'm finding grent at the two points I'm subing Pi / 4 don't forget SEC SEC has fundamental representation the 6 Tan x x is 1 / cos * 1 / cos² X so this is 6 t x 6 t x t x multi by / by cos² X this is what I have so let's put in the value is 1 by 1 at the gradient at x = < 4 let's do 1 by 1 4 which means I have M which is the gradient 6 Tan < / 4 / cos < / 4 2 so what would this give us T 4 is 1 this is going to be 6 * 1 / cos sorry the square should be outside they should be squaring everything now this is 6 * 1 because tan Pi / 4 is 1 what about cosk / 4 that's 2 2 this isk 2 / 2 2 so let me continue here so I have 6 / 2 over 2 over 4 6 / 2 4 which is 1/ 2 so this gives 12 this is 12 so the gradient is = to 12 at x = 4 the grent is 12 but x = < / 4 so the next thing is to check Pius so let's find the gradient at x = - piun / 4 so m is still the same thing 6 t - < / 4 6 * PK 4 what are we going to get - 5un 4 / cos - < / 4 cos - PK / 4 what are we going to get so I have um this will be equal to 6 * -1 6 * -1 cos - < / 4 that is 4 gives 1/ 2 so I'm going to divide that by um I'm going to divide that by so I C- / 4 squared is squared that's squared so I divide that by um < TK 2 / 2 2 which at the end of day gives one / 2 m is = - 6/ 1 / 2 this is -12 the grent is -2 they both Plus orus at 4 is + 12 at- 4 is - 12 so moving on to the next question the B part of this question I'm told to solve this quadratic equation 50 cos² Theta is = 5 sin theta plus is + 47 50 now looking at the two trig functions here this is cos this is sign they're not the same thing the best thing is to change this to sign because sin Theta cannot be changed to cos Theta the only ID we have for this is um sin s Theta + cos s Theta = 1 so cos sare Theta is 1 - sin s Theta that's I'm going to change 50 1 - sin s Theta is = 5 sin Theta + 47 this is the identity I have for this now using the identity I need to expand this 50 - 50 sin 2 Theta = 5 sin Theta + 45 so um let me take everything to the right 0 is = 5 sin Theta + 45 - 50 + 50 + 50 sin 50 sin Theta sin Square Theta so I have is 0 is equal to any like terms um just 45 - 5 5 sin Theta - 5 + 50 sin s Theta so I can write everything putting the zero on the right 50 sin s Theta + 5 sin Theta - 5 = 0 so this is what I want to solve I can divide everything by um by five I can divide everything by five oh sorry I don't know why seen a mistake this is not this is 47 not 40 um not 45 so this is 47 47 47 and this is Nega 3 -3 which means I can divide again -3 47 47 so if I multiply 50 by -3 I have a value so in factorizing this is straightforward the two factors are 5 S Theta - 1 and 10 sin Theta + 3 because are the two factors so these are the two factors that produce this from here S 5 sin Theta - 1 = 0 or 10 sin Theta + 3 = sin Theta is = 5 or sin Theta is - 3 10 have two values of theta which I'm going to pick 1 by want to get to get all two values of sin Theta I'm going to pick them one by want to get um the values of theta so let me start with sin Theta sin Theta is = 1 5 Theta is I can still do it to this side Theta is s inverse of 1 5 which means Theta is equal to the first value of this pressing calculator is 11.5 11.5 de that's Theta let's see the range or the domain of the question th to 360 so I'll use my graph I'm love with using a graph to get other values so this is graph of sin Theta this is sin Theta or Y and this is Theta the first value let's say is 11.5 on 1 over five draw your horizontal line on it is intersecting it in another place the symmetric nature of the graph from here to here is 11.5 which means the difference here is also 11.5 and this is this is 180 so 180 - 11.5 this value gives 180 - 11.5 that's 168.5 so the second value here is 16 68.5 so let me pick the second sin Theta sin Theta is = - 3 10 10 and from here I have I have Theta to be sin inverse of - 3 over 10 so which means Theta is equal to sin inverse of - 3/ 10 is -7.5 which is not in the range of the question I'll dra I can still use this graph to answer this question I can just add the negative part of the graph to it there a negative part of the graph minus 17 let's say this is let's assum this is okay let me just make it smaller -7 I say this is -7 on this is -3 - 0.3 3 10 so if I draw my horizontal line on this graph from here to here is 17.5 that's 17.5 so which means using thec nature interseting here the distance is 17.5 the distance is 17.5 and don't forget this is 360 this is 360 so 360 - .5 gives me this value also this value is 180 + 17.5 so other values the 180 + 17.5 that's 180 that's [Music] 197.4 360 - 17.5 in order to get this this is 97.5 and this is 3 42.5 360 minus 360 minus this symmetric line which is um 17.5 gives it this side 180 + 17.5 so all values of theta this Theta oh sorry the second value 3 4 so all values of theeta I have um first one 11.5 followed by 168 .5 followed by 1 97.5 and 3 42.5 T moving on to the next question question six question six I'm told that should not use calculat calculator is not allowed for this question given that x - 3 and x + are both factors of yeah in solving this they don't teach use calculator because from calculator you be able to get what the all the factors of this if you solve you inut this into calculator it gives you the solutions from the solutions you can subract CU you get X to be 3 xus one and the other one you can bring it back now since these two are factors these two are factors so there's a factor left to be multiplied to this to give us this so which implies that 2 x Cub - 3 x 2 - 8 x - 3 the same as x - 3 one of the factor the second factor and the third one then put ax + B now why I'm putting a because I don't know what this is if you look at this let's you can use the first and the last answer this question If X is um uh if this is x * X that's X2 well what will I multiply with this to give me 2 x cub that will be here be two only thing that can be here will be two and also this one let's look at the last ter minus because in order to get this you must multiply the second second second of this3 I want to get -3 * 1 is -3 so what will I multiply with -3 to give me3 that's one so this is one I don't need to this is one so now this is what I'm solving x - 3 x + 1 x + 1 = 0 is X = 3 or-1 or - 2 so be the answer to the question now the polom this is another polom which is given yeah I'm told that this polom where a b and c constants as the remainder -5 and 1 / by X - 1 the cve PX of PX act as stationary point so there is a stationary point at these two values it has remain A3 when by so let's talk about the remainder first so since that is the remainder which means if I subtitute the x value of the solution into the function it must give me5 so I'm writing the cod as P of x which is also giving us that 3 + a x 2 + B X plus C plus b x + C so that is the now since your to as a remainder which implies that P of 1 must be equal to5 very important you know that P of 1 which means I'm going subtitute one into the function so 1 cub + a 1 + b 1 + C = to I have 1 + A + B + C = to5 A + B + C = -6 equation one this is the first equation form also for the stationary part of it I have P of xal X Cub plus I can even differentiate it from the question because since it's a station stationary point Y X is equal to Z that mean p Prime of the function must be equal to Z at these two x values so let's differentiate first this gives 3x2 + 2 a x + B so P Prime of 4 3 3 4 3 2 + 2 a 4 3+ B must Beal to Z that's the statement for the first one so that must be Z simplifying this I have 3 * 4 16 9 * 3 16 16 3 + 8 a/ 3 + b = 0 mly everything by 3 16 + 8 A + B = 0 8 A + B is = -16 equation 2 also P Prime of two must be zero because you have two stationary point is is ke we so you have your two stationary point two I'm going to subtit two 3 bra 2 squared into I'm subing to the derivative of that plus 2 a * 2 + B must Beal to Z this is 412 + 4 a 12 + 4 a gives um + B = 0 so I have 4 a + b = -2 equation 3 so I have three equations now three equations which I can solve equation 2 and three simultaneously I can solve equation 2 and three simultaneously so now this is 8 A + B is = - 16 8 a + sorry 3 B this is 3B 3B 3B is 3B because when I multiplied everything by three there's no 3B when I multiply everything multip everything by three they should be 3B so I have two equations this and this and that so how do we solve the two equations let me bring them together Plus 8 8 a + 3 B is = to -16 and then 4 A + B 4 A + B is = to -2 I can multiply this equation by 3 I can multiply that equation by three to make them have the same coefficient to make B have the same coefficient or multiply this by two still the same thing so if I the first equation is still the same thing - 16 the second one if M by 3 12 a + 3 B is = - 36 so looking at the signs same sign subtract so I have - 4 a yeah - 4 a um this is 20 a is is - 5 a is - 5 let me get B in using any of the equation I use for a + b = -12 4 - 5 + B = -12 B is -2 + 20 b is 8 B is 8 since B is8 I'll use a B A and B to get C in for in equation 1 A + B + C = -6 - 5 + 8+ C is = to -6 C is -6 + 5 - 8 -4 + 5 that's - 9 C is = to - 9 so I've got in our values the values are C is -9 B is 8 and a is -5 are the values of a b and c to be found so I I question they said s use a second derivative test to show that the stationary point is a minimum which means I need to differentiate again there's a first derivative I'm differentiating this if you differentiate 3x² I give 6X if we differentiate 2 a x that gives 2 a + 2 a don't forget we know the value of a we know the value of a 6X is -5 so 2 * 5 -10 this is the second derivative I want to show that is minimum at and when it is minimum the second derivative after substituting X isal to Z be greater than zero so let's subtitute x 6 * 2 - 10 this is 12 - 10 which is 2 this is greater than Z so it is minimum at X = to 2 that's the statement you must State the statement your to to show and use it to do that now question seven in question seven I have questions on circular measure yeah in circular measure don't forget you can escape length of AR R Theta area of sector 2 r² Theta you can now use this to answer any question a of segment that's a of sector minus are of triangle so here I'm told that in the diagram a a d a to D and B to C are acts of circles with Common Center okay see OD C and O A are straight lines okay this is which means this is nine and this is also nine so Theta is the angle which the two ax which produce the two ax the are of the Shaded region of this region so in getting that Reg is area of the bigger AR minus the area of the smaller sorry area of the bigger sector minus area of the smaller sector that is what I'm going to use to get the area of sh region that's area of sector OBC that's area of the sector OBC like this minus area of this sector minus area of the sector that gives us the area of the Shaded region now in getting the area of the Shaded region I have I have um area of sector is 1/2 r² thet the biggest sector let's say the biggest sector has radius 9 so area of sector o c is = to 1 two 9 2 * 99 2 Theta and that um 81 / 2 Theta a of sector o a d see 1 / 2 5^ 2 Theta 255 2 Theta and I'm told that the area of the Shaded region that's the difference between them 81 Theta / 2 - 25 Theta / 2 is = 4 piun that's it this is equal to 4 81 / 2 -5 2 that's 2056 / 2 so I have um 56 / 2 is = 4 5 56 over 2 Theta don't forget I want to forget the Theta 56 over 2 Theta is = 4 Pi 56 over 2 th is = 4 Pi which is um 28 28 Theta is = to 4 Pi Theta is = 4 piun / 28 that's 1/ 7 that's < / 7 the the angle Theta is pi/ 7 now moving on to the B part of the question the straight line AC is added to the diagram good and um is to the DI and the region ACD is now shaded which we can see find the perimeter of the Shaded region ACD the perimeter of this is a d plus AC plus DC now this is simple if you go back to this this is five this is four this is this is four so what I need to do now a is that length of AR let me find length of AR ad length of AR ad length of r a is R Theta the radius is 5 Theta is < / 7 this is 5 pi/ 7 I have 5 pi/ 7 so don't forget DC is that because the perimeter is addition of the side that constitute the Shad region now the only thing left is AC looking at AC I'll get AC from this diagram this this is 5 this is Theta and this is 9 and I know Theta to be Pi / 7 so I want to get AC a c o so I'm using my cosine to get AC I'm going to use my cos sign which is X2 AC is represented with X X2 is = 9 2 + 5 2 - 2 9 * 5 cos Theta which is cos 5un / 7 X2 is = 81 + 25 - 45 * 2 so I said 45 is 81 yeah 45 * 2 that's 90 90 Plus 5 7 x² this is 106 -- 81.8 719 8 so X2 is = to 24 9128 24. 9128 that's X2 X is 4.99 4.99 127 so that's forget perimeter is addition of everything I've said before four the length of AR that we did here and the AC I've gotten the AC I've gotten which is 4.9 91 27 + 4 + 5 < / 7 so this gives 11235 approximately this gives 11. approximately 11.2 so this is approximate answer to three conf figures now moving on to the next question was which is question eight Yeah question eight I have question on differentiation and integration um no this is probably integration because I have the second derivative given to me and I'm told you y x is 3 at this point now since they said the Y DX is 3 4 at this which means the Y the Y with respect to X is = 3 4 when x = 3 piun 3 piun / 16 also we all know that this is X = 3 piun / 16 when I'm sorry x 3 pi/ 16 when y okay you can say when Y is 4 or Y is to 4 when X is 3 16 so since I'm told to find equation of the Cur the equation of C is going to be in form of yal to whatever now in order to do that I'm giv second derivative from the second derivative if I integrate I'm going to get the Y DX just a constant which I'm going to use this to get a constant so let's that's the sare Y the x² = cos 4x - < / 4 - < / 4 so if I integrate this with respect to X I'm going to have d y DX so if I integrate C is s so this is going to be s 4x - < / 4 ided by the derivative of what I have inside there that's reverse chain R this is derivative plus C how do I get C like I said y x 3 4 when x 3 6 I'm going sub the two 3 4 is = to sin 4 * 3 / 4 subing four here 4 * 3 4 that's 3 so I have so three 4 * 3 sorry 3 16 x is 3 16 3 < / 16 3 16 this is 4 3un 4 3 < / 4 - < 4- this Pi 4 that's 2/ 4 that's Pi / 2 so have s pi 2 after substition 4 + C so 3 4 sin / 2 is 1 that's 90 1 1 4 + C C is 3 4 - 1 4 which is 1 2 so the equation the the X the gradient is sin 4x - y 4 4 + 1 2 now in order to find the equation I need to integrate this two I integrate this integrate this I'm getting y with respect to X I'm integrating all of this so if you integrate sign is cos cos 4x - pi 4 / by 4 * 4 so I'm dividing by that way make it EAS by this four / 4 plus when you integrate 1 2 is 2x plus another constant let's disap so this is y = - cos - cos 4x - 3 piun 4x - < 4 4 over 16 + 1/ 2x + K in order to get K I'm using the point given Y is / 4 when x 3 16 so like I did the other time you substitute 3 16 here this should be 1/ two they should be one over two when I 3 < 16 rather 4 * 3un 16 4 um that's 3un / 4 - < / 4 that's < / 2 so I have y = - cosk / 2 / 16 + 1 / 2 * X is still 3 < / 16 3 < / 16+ K don't forget Y is also y has its own value Y is pi 4 why is < 4 this is < 4 so I need to find K so in getting K cos cos 90 is 0 so I have 4 = 0 plus this is 3 < / 32 + k k is equal < / 4 - 3 < / 32 so 8 Pi 8 that's 5 piun 32 this is 5 piun / 32 so the equation of the C is y = - cos 4 x - < 4/ 16 + 1 / 2X + 5 piun / 32 this is the equation of the call moving on to the next question question N I have a question on application of um um integration and combined with coordinate geometry so in this question here I'm told the diagram shows a sketch of parts of the code y = 4 + um 3x - 1 everything to the^ minus 1 one so the line X is = to 9 this line X is = to 9 UM the point a has x coordinate one so which means this point is one this is 1 comma 0 we don't sorry 1 comma K we don't know what the Y value is so which is something I can easily do that straight for it's just to subtitute Y in to this curve because the point a is where the tangent of the curve is this line is the tent to the C so both of them intersect at this point so they have the same x value the same y value at this point so that can be sorted out let me just put Y is X = One into this in order to get the Y value there 3 * 3 * 1 okay let me just say three brackets 1 - 1 everything to the^ - one this is 4 + 2^ -1 so 4 + 2^ - 1 that's 4 + 1 2 that's um 9 / 2 4.5 so K here is 9 / 2 so this point is 1 Comm 9 / 2 it has a value no so um find the area of the sh region I'm looking for this area of this region in order to get the area of this region you need to get the area enclosed by this Cur area under this Cur enclosed by x = 1 and X AIS X = to 9 and below this curve that's the area so if I get that I will not take away this area of this unshaded part this unshaded part from it that's a triangle this forms a triang right angle triangle so get the area of this right angle triangle take from Thea Clos by that so to start with this to start with this I'll find the ydx in order to get the gradient of the tent and go for the area of the triangle immediately after getting the equation of the tangent I'll go for the area of the triangle so let me start with that um so what Y is = 4 + 3x - 1 the power of1 the Y with respect to X is- braet 3x - 1^ - 2 * 3 that's the YX so this gives the Y with respect to X is = -3 3x - 1 -2 so let's get the gradient of the tarent M of the tarent is -3 bracket 3 * the tangent is at x = 1 that's point A 1 - 1^ -2 -3 3 - 1 2 2^ - 2 this is -3 over 4 that's the gradient of tangent which is the gradient of the YX at x = 1 so equation of tarent Y is = - 3 4x + C so since I know a point which is 1A 9 / I'll get my c 1 9 / 2 that I'll get my C 99 / 2 is = 3 4 * 1 + C now 9 / 2 = 3 4 + C C is 9 / 2 + 3 4 C is 9 2 + 3 4 which is 21 / 4 that's C so equation of the tent is y = 3 4 x + 21 4 that's equation of tent since I'm able to get equation of tangent therefore I can I can find The Intercept here CU with this without this intercept I can't find the area of the triangle which is this is the base from one to 1 point and I know the he is already 9 / two this is 9 /2 the he is 9 /2 so here at this point Y is = z so let me find the value of x as ADD point Y is equal to Z so let me get the x value so um X y0 is = - 3 4x + 21 4 so 3 multiply everything by 4 I have 0 is = -3x + 21 so from here I have 3x is 21 x is 7 which means this Bas is if this is 7 this is six from here to six from this point to this point is six which means this is six so I can move on with my solution so the area of the triangle the area of this triangle have this part is area is 1 / 2 the base is 6 the height is 9 / 2 area is equal to um 23 27 / 2 that's 27 / 2 the area of that part is 27 over 2 so that's not the end of the question what got in the area it Remains the area of the whole thing the are under this from one to 9 which means I need to integrate I need to integrate I need to integrate from 1 to 9 of 4 + um 1 / 3x - 1 1 3x - 1 because the power is -1 so the way you integrate that is different so 1 3x - 1 um and this if I integrate this if I integrate this I have sorry with respect to X I have 4x + 1 / 3 l 3x - 1 from 1 to 9 so if I subtitute 9 if I subtitute 9 4 bracket 9+ l 3 * 27 28 Minus when I sub one 4 + Lane 2 this is what I have 4 + Lane 2 simplify this is 36 soorry there's 1 over 3 1 over 3 36 + 1 3 l 28 -1 - 1 / 3 2 so um 36 - 1 um 36 - sorry they should be four - 4 this is 4 36 - 4 I have um 32 so I can still take it up here 32 32 2 plus um what I have here is 1 3 28 so they common logm 28 is subtraction / by 2 that's 14 so 3 l it 28 3 * 9 27 - 1 26 26 there's 26 26 so 26 6 / 2 that's 13 1 3 l 13 + 13 L is a plus orus + 13 L 13 so that's the area under the so the area of the Shaded part is 32 + 1 / 3 then 3us the area on the area of Lal 27 over 27 / 2us 27 / 2 so you can easily press your calculator this gives 19 35 4 9 8 approximately 19.4 C square unit square unit so that's the area of the ship now moving on to the next question in question 10 I have question on Vector yeah there's a diagram given to me the diagram is a parallelogram you have Point D divides OC into 2 ra3 and it stated that o c is C so this is 2 over 5 C 3 over 5 C the way you divide in ratio so since it's 2 ra3 um oh uh um Line This is line OC and this is D they both sharing this line o d and d are both sharing this line ratio 2 to three so if you want to share and you're told this is the vector is C stated the vector so you're sharing C ratio 2 to 3 so 2 over 5 of c and 3 over 5 of C so that's that now o a o a is a is a also the point P lies on a such that op is um Lambda o op is Lambda o which means there's relationship between um which means that's op B are cinear all the three points are on the same line so likewise AP also um AP is equal to Mu a a to so the three points are cinear so they said find the two expressions for op so let's let's find the two expressions for op I'll start with the first one op is Lambda o b that is already stated so this is an expression for it but I need to input I need to put the vector for OB the vector representation for OB so apart from that let me move on with the question find two expressions for each in terms of AC and scal NN show that P divides both da put da a and OB this is point P to should I divise this and this in ratio M to n so that's very important so let me start with this one here I have a op Lambda o which implies that I me continue op isal to Lambda what is OB this is Vector C because this two ab is parallel to OC to parallam they have the same Vector so I want to find I want to move from o to B I have to pass through the known Vector vectors o a a that's A+ C OB is a + C I have a + C here so op is equal to a Lambda plus C Lambda that's already established that's one expression the second expression for op I'll get that from AP look at AP it the same thing as op minus O A so from here op is equal to O A+ a o a from a and don't forget they said AP there's a relationship between AP and AD given in terms of mu op is equal to O A Plus M what is AP a so what is a let me go back to the diagram and find what a d is is very important I know what a is if I move from I want to move from a to d like thisor I want to move from a to d so I can go like this I can move like this move like this move like this cons move like this but this is shorter you must pass through the known Vector go like this that's how I want to move if I'm moving that way I have a minus a plus this is this 2 over 5 C so this is a is that so therefore op is equal to O A Plus M A + 2 over 5 C that's what I have there and what is O A I need to go back o a is a o a is a this is a small letter A so I have um op = Aus mu a + 2 over 5 mu C so this is a 1 - mu + 2 5 m c that's o p so I've gotten two Ops the two expression for Ops the next thing is to equate both of them together to find Lambda and mu I equ both of them together equate both of them together to get Lambda and M the first one is Lambda a Lambda + Lambda c equal the other one a bracket a bracket 1 - mu + 2 5 m c 2 5 m c so here we are talking about vectors equality of vectors what you do is to equate the corresponding component equate the corresponding component now how will I do that equating the corresponding component which means the component a and component a here I'm going to equate both of them together so which means I have Lambda a isal to a 1us mu can 2 Lambda = 1 mu I can call this equation one can call this equation one also this part the C component Lambda C is = to 2 m over 5 C can the C Lambda Lambda is = 2 m so I have two equations in terms of Lambda two equations 2 5 other so equate equation two equate both of them together 1 - M is = 2 5 m so 5 - 5 m = 2 m so 5 is = 7 m m is = 5 7 that is M what is Lambda don't forget Lambda is 2 5 m so sub M there 2 5 * 5 7 and this gives 2 over 7 Lambda is 2 over 7 a m is 5/ 7 this is 2 over 7 and that is 5 now this is not uh the end of the question I've gotten Lambda M so because that will help us with this expression stated with this two expression that will help us to get the ratio so the ratio in which P divides a and op a and OB so with that since I know Lambda and mu I'm going to substitute both into the expression stated there op is equal to Lambda M and A so I have op is equal to Lambda OB op is equal to Lambda what is Lambda sorry Lambda is 2 over 5 2 over 5 OB so op over OB IDE both by OB is = to 2 over 5 which means this fraction can be converted to ratio op ratio OB is equal to 2 ra 5 now let's go back to diagram op what I said there is op ratio that's op ratio o that's the part ratio the O So if that is rati 2 therefore p ratio B is 3 the ratio for this part is three the ratio for this is two so since the total ratio is five that's OB which implies that which implies that op ratio p p ratio p b is 2 ratio 5 this is 2 ra 5 2 ra 3 sorry 2 ra 3 ra that's ratio that is already established op ratio PB ratio three or any of these two any of these two will be good to answer the first part of the question so let me look at this also is that you say op ra OB op ra OB 5 or you say op ratio PB is um uh is ra three that's so let's look at the second one the second one there you have um from the diagram AP is equal to Mu a d that is what we have there so what is me AP is equal to Mu is um m is 5/ 7 7 here 5/ 7 a d so your a ratio a d is = 5 7 so which implies that if I have AP a so therefore AP sorry AP over a so AP ratio a ratio yeah I think I think the mistake I made there this should be seven sorry from the question Lambda is 2 over 7 pick up the this is Lambda 2 over 7 this is 2 over 7 sorry sorry for that 7 if that's 2 7 this should be five 7 7 7 that's 2 7 so um this I got to notice when I got here that realize this is seven that should be so yeah um AP ratio a d is = 5 ra 7 so this also let's go back to the AP a let's go back to the diagram AP a that's 5 ratio 7 which means I can say AP and PD AP ratio p d is 5 Rao 2 which is similar to this 52 so since the since p is the one dividing it and I can decide to write this as DP ratio PA a still the same thing 2 ra 5 so this is similar to this I can go with this so the ratio of the divisor is M ratio n both of them are integers so that's the answer to the question 10 that's final answer to question 10 thank you for watching and don't forget to subscribe to my channel