Oxford Mathematician Takes Irish Leaving Cert Maths Exam

[00:00:00]hello maths fans I'm Dr Tom Crawford I teach maths at the University of Oxford and the University of Cambridge but today I'm going to be taking a high school level maths exam from Ireland the leaving certificate or leaving CT as it's commonly known is a qualification taken by students in Ireland who have remained at school to study until the age of 18 they take a range of subjects however I will be focusing on mathematics and in particular this is paper one at the higher level and the exam is from the year 2023 this particular exam actually made the news back in the summer of 2023 for having some notoriously difficult questions however Beyond this I have no idea what those questions are because as usual with these videos I have not looked at the exam paper I know nothing about the syllabus nothing about the qualification I'm simply going to open the exam for the very first time try and work my way through the questions and explain my thought process and the whole thing will be captured on video for you all to enjoy right let's see what an Irish maths exam looks like two sections okay uh Concepts and skills 150 marks section B also 100 300 marks this is this feels long uh how long do I have uh two and a half hours okay two and a half hours that's a long time hopefully I won't need two and a half hours but we'll see uh 300 marks is a lot of marks all right um what else have we got use pen all right great write my answers in the book I'm clearly just going to write my answers on the iPad as usual uh diagrams are not to scale you must write the make and model of your calculator here um I don't have one so I guess na on make and model of calculator uh I really hope there aren't actual calculator questions we'll see all right okay question one um oh answer any five questions so do I I think I I assume that means I need to pick my questions one two how many have we got three four um five six oh god there there are actual questions oh but okay I've got to pick five from six this what happens when you have no idea what exam you've agreed to do on camera um all right well question one definitely looks fine so I'm going to start with question one and we'll see what I make of the others later find the two values of M which are real numbers for which the modulus of 5 + 3 m is 11 okay so 1 a we want modulus of 5 + 3 m has to be 11 now the modulus means it's always going to be positive so it could either be 11 or minus 11 so I'm simply going to solve those two cases so we're going to say case 1 5 + 3 m = 11 and then we're going to say case 2 5 + 3 m is minus 11 and I'm going to solve both of these so that means that 3 m is equal to 6 so m is 2 in the positive case and if it was minus 11 in the negative case we've got 3 m is equal to subtract another five so is - 16 so m is going to be - 16 over3 so they are my two solutions that is 1 a uh which is oh I don't know how many marks the whole thing is 30 all right I guess we'll see okay that was all right uh B for the real numbers HJ and K we have an expression all right then so B uh 1/ H is equal to K Over J + K great like that uh and what does it want me to do Express K in terms of H and J so I think it just wants me to rearrange the equation so you get k equals a function of H and J or maybe it means to solve it but either way I want k equals right is how I'm interpreting the question uh so all probably wants me to solve it it's going to be uh is it going to be a quadratic let's see um what can I do I'm going to multiply up and say J + K has to equal h k fine so J is equal to h k minus K so J is equal to K * H -1 uh so K in terms of H and J so k equal J over hus1 unless and maybe this seems too easy maybe this why it's hard unless the denominator is Z because if H is one then this doesn't work so this is true for H not equal to one otherwise H is one the equation is going to be uh J + k equals K in which case J is zero so J would be zero H is one and K is any number real number would be one possible solution I may have over complicated this uh otherwise it's j over hus one right Express K in terms of H and J yeah all right let's go with it that's one B moving on to part c um okay x^2 - PX + 1 X2 let's write that out is a factor of a cubic find the value of p and the value of R okay so that is a factor so that times something has to equal my cubic right which is X Cub - 2x X Cub - 2x uh - 3 R uh and I'm told P and R are real numbers and P is negative so that's helpful to know so P and R are real and P is less than zero all right great my plan here is I could try and do like polinomial division but that I don't really remember to do that it's been a while so I'm going to just work out what that linear factor is that must go in that bracket so that when you multiply the linear term by the quadratic I have to get the cubic that's what I'm going to try let's see um so I'm going to need an X here uh let's just use a different color because that now gives me um X cubed now I'm going to get so let's just write this out so I'm going to get x cubed minus px^ 2 now I don't want an X squ term so I'm going to need to add something to make it disappear so I think I need to add P to that because if I add P then I'm going to get plus P x^2 which is what we want then I'm going to have Min - p^2 x from there and then I'm going to have plus X from there and then plus P okay okay so uh I'm not sure if this is right which why I'm looking confused but anyway so those two cancel I'm happy with that so what I've got from my equation let's just write it out if it's x^2 - PX + 1 * x + P so I'm saying that is equal to xq - 2x - 3 R but it's also the left hand side by expanding the bracket like I just did is actually going to give me X cubed um plus how many lots of x I've got 1 - p ^ 2 lots of x uh and then I've got plus P so that okay so now if I equate coefficients so I'm going to say uh this has to be minus two that has to be that so 1 - p^ 2 has to be - 2 so p^ 2 is 3 so p is plus or minus < tk3 they tell me p is negative that makes sense cuz they're helping me out there uh so p is minus < tk3 since p is negative awesome um and then I now know that from the constant term so this was the order X term from the order one or the constant term I know that P has to be Min -3 R that's- < tk3 = -3 R so R is uh < tk3 / 3 uh and that's it yeah all right so p is uh minus root3 and I think R is < tk3 over 3 and unfortunately the denominator is already rationalized so don't need to worry about the square root question two uh now I want to work out if I want to do this don't I because I there's one of these I don't do um so I have a function uh minimums love it I'm going to differentiate limits love it like that oh draw a graph sure all right I'm going to do question two I might regret this I'm going to do question two so question two let's dive into this um so I know the first one differentiation which I like so f ofx um part A is going to be equal to X2 + BX plus C X2 + BX plus C uh where B and C are real or has a local minimum okay so minimum at um 3 - one and that's going to be X and that's going to be uh F of three has to be minus one okay uh so it's a minimum that means the derivative will be zero at that point and the second derivative will be positive the gradient increases away from the minimum um so I'm just going to make those notes so I know that F Prime is zero and F Prime because it's a minimum that has to be positive I imagine I'll need those things find the value of B and C uh in the equation right so I'm going to solve basically the two things I just said uh so let's do the derivative so F Prime uh of X is going to be 2x + B uh and that has to be zero since minimum which means turning point zero gradient um uh well it means zero uh since minimum at xal 3 is what I should say so that's actually telling you is that 6 + B is0 so B is - 6 now okay um so maybe the second derivative is just going to be two which is positive so that's automatically true so now how do I work out C that's going to come from I think the Y value um okay so let's think about this ah because I know the value of y yeah so the equation also tells me that when X is equal to 3 we must have uh f of x must be minus one so that's going to tell me that 9 uh plus b I know what that is but let's just write it as that for now * 3 + C has to be -1 B we've worked out is - 6 so that's 9 - 18 + C is - one uh - 9 + C let's do this carefully add 9 to both sides C is 8 all right so I think C is eight and B is minus 6 part B it's limit question I I feel like I picked to do this one because I saw the limit and thought it'd be fun to talk about um so question 2 b um find the value of the following limit where N is a natural number okay so we want um the limit as n goes to Infinity of n / n + 1 so that's going to have a limit of one uh plus n + a th000 Over N also has a limit one uh and then plus A3 to the power n oh which will tend to zero yes so I think as n goes to Infinity going to write this in Red so as n goes to Infinity this whole thing is just n/ n don't care about the one so this tends to one again don't care about the Thousand they're just checking you understand the size of the constant doesn't matter uh so that also tends to one and then we have a number less than one to higher and higher power so it gets smaller and smaller so this actually going to tend to zero so I I think the whole limit is two uh is going to be my answer the limit is two and that was short and sweet assuming I got it correct I made it entirely wrong but we're going to keep on moving uh to number c and graphs so uh the function G is defined in this range okay so G of X for um X between 2 and minus 2 uh less than or equal to great uh it's graph is shown each of the two two diagrams below I had a feeling I might regret the graph bit but anyway got lulled in by the limit which was actually quite easy so uh draw the graph of G of x minus 2 okay so they've given me the graph of G of X so I'm going to copy that out and we'll try and do it vaguely accurately um because I do have dots on my page so let's see if we can use these um so they reckon it's going from two to then we guard minus one one right so that's going to be minus one one is there and it's going from two three so two across and three up all right so it's quite a small scale but uh they're having it kind of do something curvy like this uh that's pretty much it all right great uh and then they've got this steep bit that's going up through three like that okay so taking the dock to be one unit space in both directions that's their original one so now we want G of X that's G of X so I want G of xus 2 so basically the whole thing just comes down by two um so that was there so that's now there that was there so it goes like that and that just comes down to two on the y- axis yes so like that okay so that right there is just G of x - 2 it's a translation in the y direction byus 2 um okay on as large a domain as possible the function is only defined betweenus 2 and two okay H I knew they were going to do this h i now want to do uh same thing but they want me to do uh this is part two should label my parts um they want me to do G of x + 3 now this means um shift three uh places to the left is what that's going to do just remembering my graph Transformations um so if I draw this one in blue it's the original one shifted three to the side so uh it's going to go like that and then it's going to go like that yes okay I'm happy with that that's G of x + 3 awesome and oh question two done these are great and I'm definitely doing question three uh is that the whole oh it's not the whole thing um but I want to do that I don't really want to do logs oh okay I'm looking at question three I really want to do part a uh because it's a cool proof but I definitely don't like logs um okay question four I like complex numbers yes big fan of complex numbers uh I want to do number five derivatives yes 100% want to do that uh yeah no I like that and six I think I want to do yeah integrals all right uh so I would say I've cheated and looked ahead but you're supposed to look ahead so I I would argue I have demonstrated good exam technique uh I was I very nearly uh started question three because I really want to prove route two is irrational it's a really cool proof um but I don't like logs uh I'm not sure I would remember all of the manipulations of logs in in B and C uh so I'm actually going to skip three I'm going to go ahead to question four you never know maybe I'll come back and do three but for now we're definitely moving on uh to question four depends how hard five and six are but four looks all right I like complex numbers so I'm going to jump in with that one right in this question I squ is minus one thank you the complex number oh okay uh is this a yes okay so zed1 is 1 + I and I'm told that's a root of the equation which is z cubed + 3 - 2 i z plus p is equal to zero okay find the value of P okay so if this is a root that means I get zero when I sub it in I could try and find the other Factor but that feels hard so I think I can just sub it in uh and it should get zero so oh but theyve got a cubit fine okay so 1+ I let's Square it * 1 + I is going to give me 1 + 2 I uh - 1 so it's just 2 I Ah that's nice okay and then 1 + I cubed is is then just 2 I * 1 + I um so that's just 2 I uh minus 2 right so if I sub in uh to the equation then I'm going to get uh z cubed so 2 i - 2 uh + 3 - 2 I * 1 + I + p is Zer right and is p complex uh yes p is complex okay good that makes it easier so I get 2 i - 2 plus uh 3 + 3 i - 2 i - 2 * i^ 2 becomes + 2 + P must be zero and so I've got 2 I and a minus 2 I and then a plus two and a minus two so I think I get 3 + 3 I uh + p is 0 so p is equal to - 3 + plus - 3 - 3 I all right let's put that in a box and keep it moving B um use deav theorem love it to find the value of w for which w^2 is -1 + < tk3 I okay so tell me to use de Mar so de theorem not sure I would have known to use it otherwise but let's write it out uh tells me that um e to the I Theta is cos Theta + I sin Theta or more generally that's probably oil theorem de Mar is e to the n i Theta is COS of n theta plus I sin of n Theta right so so I'm thinking here um if I let uh W equal r e to the I Theta then I know that r^ 2 e to 2 I Theta has to be Min -1 + < tk3 I fine um so what's that going to tell me uh but then if I take the modulus of that okay so let's work out R so the modulus of w where modulus of w^ s all right so r s um is therefore equal to the modulus of this number not sure this how you me to do it but these are the thoughts coming into my head modulus of this um yes so the modulus of that is the square root of uh -1 2ar + < tk3 2ar so that's the sare < TK of 1 + 3 which is < TK 4 which is 2 because R 2 is very clearly positive uh okay so I know that R is < tk2 because it's a distance must be positive uh now I can work out the e to the I Theta part because because because because um what do I know I know that um we need root so r s so we've got 2 * e 2 I Theta so time COS of 2 Theta + i s of 2 Theta and that whole thing must be -1 + < tk3 I so I know that COS of 2 Theta for the real part therefore because the real Parts have to be equal and the imaginary Parts have to be equal so the real part tells me cos of 2 Theta is minus a half and then the imaginary part tells me that s of 2 Theta is < tk3 /2 so what on my graphs are going to give me this um so cos is going to look like this and S is going to St a different color um so that needs to be at zero so it's going to peek there yeah and then do that right okay so bit pointy but you get the idea that's cos and S so I'm looking for a point when cos is netive a half but COS of 2 Theta is negative a half um all right so if C is going to be negative a half that's like down here okay so this is negative half but then I want s to be < tk3 /2 oh um does that work does this work uh I'm going to try a graphical approach now this is probably not what you're meant to do but I'm kind of a little bit little bit stumped so um I'm going to try an argand diagram when I say graphical approach um so I know W Squared don't I right that's yes so w^ squ is -1 + < tk3 I so I should be able to work out the angle of w s uh which will be twice the angle of w I think yeah pretty sure that's true all right so let's try and draw might have been over complicating it de Mar's theem has thrown me because I don't want to use it anyway it's probably going to say in the mark G must use it but who cares if I get the right answer I'm getting the marks um more than one way to do maths so uh I've given myself enough justification now uh we know that it's minus one uh so minus one and then we're going up root3 right so let's just suppose it's there right so this is w^2 so w^2 right yeah um so this is w^2 so I know that w^ s is oh I this is much easier I made it cuz it's clearly w^2 is 2 uh just from like Pythagoras the yeah the diagonal is going to be two is two uh e to the I Angle now figure out the angle Connor yeah um so if I know these distances it's 180 to go around so that's going to be I want to say uh let's call this Theta I know that t of theta is opposite is < tk3 one right so the graph of tan oh God the graph of tan is < tk3 at Theta equal 60 yes because I know it's I know tan of 45° is 1 that's the one I always remember yeah so it's 60 so Theta is pi over 3 probably want me to use radians I assume so Theta is pi over 3 so e to the so this is e to the I um 2/3 2 piun over 3 right so therefore W is < tk2 e to the I pi over 3 uh which is going to be < tk2 time cosine of piun / 3 + I sin of piun over 3 great so that is < tk2 right C of 60 I drew my graphs up here didn't i c of 60 is the smaller one so that has to be come in so it goes one and it goes < tk3 over 2k2 over2 1 over two so it's a half okay so I think that's a half um plus I sin pi over [Music] 3 just had a thought about Plus or minuses I'll come back to that let me just make a note to think about Plus or minuses half um plus I S of 60 is therefore um < tk3 over2 okay so that is W now I could have a minus couldn't I and it would still make sense so I was saying before that the radius R the modulus has to be positive which is true but I could take the minuses inside the e to the bit EI Theta bit so I actually think okay so I know that uh so I actually think there's a plus or minus here so let's just plug that all the way through because there should be two solutions because it's a quadratic Square equation all right so I think W is equal to either um < tk2 over 2 + I < tk3 oh god with theun on the top so that's a < tk3 over < tk2 uh yes or W is equal to - < tk2 / 2 - I < tk3 over < tk2 probably supposed to simplify that um so it's going to be root < tk2 over 2 is okay plus I top and the Bottom by < tk2 gives me aunk 6 / 2 or minus < tk2 over 2us I 6 2 now I should really check and square this and make sure I get the answer but I've already spent way too long on this question so I'm just going to move on because there is a part C uh Theon di now we're doing igon diagrams below shows the complex number U is equal to a plus b i right U equal a plus b i awesome and it's there write the complex numbers I U and I U Bar in the simplest form okay well that's complex conjugate so uh I * U therefore is a I uh minus B I think is going to be simplet form and IU of the whole thing is therefore just a i minus b or B so when you take the conjugate you just change the sign on the I so that's just going to be minus a i minus B sweet uh plot and label knew this was coming uh on the diagram as accurately as possible okay so again obviously not to scale but I can copy out what they've got uh something like this so they're saying U is up here okay so I * U is a 90° rotation I already know this so again I don't know if students taking this exam would know but multiplying by I rotates by 90° um so what you're going to get then is a 90° rotation so we were previously a across and B up and from the diagram B is clearly larger so now we are B across and a up so this is now uh B and we are just a up so we're here okay so this one should have used different colors but that one is IU now I will use blue for the second one um IU bar is you take IU and you stick it down the bottom so it's going to be down here so uh and let me just check that makes sense we've gone minus B that way yeah and now we're at minus a that way that's a minus okay yeah awesome I'm happy with that uh great yeah perfect uh is that it no is not it there is a part three okay so that one was um this was cot one this was cot 2 so there is a cot three what do they want me to do they want me to State a transformation that would send you okay so we go from U to IU bar right and I'm not allowed to include a translation very nice so what have we done we have um reflected in the line Y so this is just a reflection in the line um Y is equal to - x on that diagram so in terms of imaginary parts and real parts that would be the imaginary part is equal to negative real that's what's happened um so it's a reflection in that line um I'm going to go with that okay question five okay derivatives I'm going to clearly do this one uh so don't need to worry about three yet um so f ofx is uh 1 over 5x^2 + 7 uh find the derivative simplest form all right so F Prime is so just to help myself out this is 5 x^2 + 7 to theus1 so then when I differentiate I'm going to bring the power down uh the chain rule says I differentiate 5x^2 so that's 10 x from the chain rule uh and then I've got this whole thing to the power uh minus 2 great not as simple as that's going to go uh also uh what am I doing next I am given a function so that was part a I'm now given a function G ofx uh complicated looking thing um Okay g of X is T of X over 2 fine that's okay uh and then Times log X right sure um X is between n and Pi um okay so that means tan will go between zero and infinity right great so I'm just going to make a note to say this belongs to not infinity awesome um because T of pi/ 2 is ised infin infinity tends to it um right what am I doing find the value of G Prime okay so it wants G Prime at pi/ 2 so chain rule not chain rule well chain rle Ru and product rule product rule uh so if I differentiate tan I get sec^2 x / 2 uh dtive T of SEC squ and then a half from the chain rule times the thing I left alone and then I could have also differentiated the log and that would give me a 1 /x * T of x/ 2 okay great and it wants this uh at pi/ 2 so G Prime of Pi / 2 is um is is is Right G Prime of Pi / 2 okay so I want se s which is 1 / cosine 2un / 4 which is 45 so cosine of < 4 is cosine of 45° which is < tk2 over 2 um so when you square that you're going to get 2 over four you're going to get a half okay so that's going to be a half remember that time another half * log ofk / 2 + 1 /x so 2 over piun tan of < 4 all right so um summing in the bit I've already worked out G Prime of pi/ 2 is therefore 1 over a half yes so it's 2 * the half time log / 2 right and I can't simplify the log now we've got plus 2 over Pi tan of 45 is 1 great so just that so I think then the answer is uh 2 over pi plus log Ln of Pi / 2 uh that's G Prime of pi/ 2 which is what they wanted at least supposed to be assuming I got it correct because A and B are real they are both real log of something awesome okay part C I like this question uh might be my favorite so far feel like I know what I'm doing um dangr below shows three sets a b and c and two functions f and g f goes from A to B G goes from B to C excellent size of a is four size of G is four size of B is three fine weird and wonderful but okay so it wants me to work out G of f of3 right so F of three is the world and then when you apply G you get W so that's just W okay um feels a little bit easy but we'll go with it um part two explain why G mapping from B to C is injective but not surjective okay uh so injective uh means [Music] um each element uh what does it mean it means each element is like one to one so Maps uniquely that's the key thing yeah Maps uniquely um so EG we know that g of the plane I'm going to write in words G of the plane equals uh X we know that g of the world or the Earth is w and we know that g of rain cloud is Zed so there's only one right it's not like two things ma to zed for example so the other one you have two and three for f f Maps both two and three to the world so is not one: one I know what it means then I'm convincing all of you I deserve the marks but it's not onto right but it's not subjective because subjective means everything in your input space is mapped to something well it's mapped to everything in your output space and that's not true uh not subjective since um nothing in B maps to the output space maps to Y so you can't get to Y so it can't be onto because it has to map to every element in C but it doesn't map to Y um okay awesome right that's question five um let's move on to question six okay so I have to decide do I want to do six or go back and do three and logs which I don't really want to do um okay it's it's integral I'm going to do I'm going to do six I like calculus I teach calculus to my first years all right uh so we're going to do q6 which I think is going to be the last one in section a uh so let's go out with a bang f and g are two functions okay so F ofx is X + 4 um G of X is x^2 - 2 fabulous uh find the two values of X for which they are equal okay uh and that's part a part one right so we want f = g which means x^2 - 2 = x + 4 so that means x^2 - x - 6 is Z I can see how you're going to factorize that so that's going to be an X um - 3 and an x + 2 Z so x = 3 or minus 2 s check when X is 3 I get 7 I get seven X is minus 2 I get two I get two excellent okay there are my two answers already given myself a tick but that's to tell myself I check them they work all right find the area of the shaded region in the diagram below uh the region between these two graphs okay and I've already figured out the two Roots wait hang on yes okay fine so I'm going to draw the diagram um because as usual I'm writing on a blank page um just to help myself out so we've got the uh the quadratic doing something like this and then you've got this line F going across cross uh brilliant and this is x + 4 amazing so that means that's four uh and we want to know the the area between them between minus one and between two okay so that one's two and that's the area we want right so um how can I do this if I work out [Music] the the area between the two graphs so if I were to work out the area under f um because if I were to integrate F between min-1 and two that would give me an answer um sure but that's not going to be the extra bit that I [Music] want so how else could I do this let me think let me think um i' forgotten how to do integrals apparently so right so I've got F and I've got G um this is x^2 minus 2 right so if I integrate X2 - 2 I'm going to get the area underneath that graph but here it's negative ah so it would give me that bit plus that bit okay ah but where does it cross maybe I should work out where it crosses that might help Okay um drawing a bit of a mind blank which is interesting because I clearly do know how to do this but all right so my thought is this I do I know where that crosses um I should know where that crosses that's going to beunk two Okay so if I do all right I'm going to use another color right so if I do this particular integral I'm going to write out in a minute I think I'm going to get this area uh so let's see so I think if I do the integral of uh G so the integral of x^2 - 2 DX between -1 and < tk2 so I think this equals the green area possibly with a negative sign in fact it probably will be but let's just do it um so I think it's going to be that writing the same thing out um so if I integrate it I'm going to get um X Cub over 3 - 2x between uh -1 and < tk2 right so if I sub those values in um so that's 2 * < tk2 over 3 - 2 < tk2 um and then minus um - A3 - 2 so plus 2 right so I think we're going to get oh God 2/3 of < tk2 minus two lots of it so that's - 4/3 < tk2 um but then Min - 2 + a 3 so subtract off two and then add on 13 so God this is complicated clearly not the way to do it um so 2 - 1/3 is 5/3 so - 5/3 okay so - 4/3 < tk2 - 5/3 so then if I turn that to a positive I think that's going to be the green area so I think the green area is this is so not how you do it is 4/3 < tk2 - 5 over3 okay that's the green area now the Blue Area well let's call it the red area in fact between here so this is just simple trapezium so the red area is going to be equal to this this is so not the way to do this question um is going to be equal to minus one up to < tk2 uh of x + 4 DX so that's simply um x^2 / 2 + 4x between -1 and < tk2 um so that's going to give me x^2 so 2 over 2 1 uh + 4 < tk2 uh minus so still positive and then minus 4 great uh so it's 1 + 4 < tk2 - a half + 4 so that area is going to be 5 minus a half so 9/ 2 uh plus 4 < tk2 is the red area so um let's to help me remember how this is working let's just kind of add that red color so this is 9/ 2 + 4 < tk2 okay so there's this little bit left now that I want to work out here um so what I think I can do is actually go from < tk22 to two of F and then subtract off the G term so I can do these two together so let's write it let's write it to the right hand side uh so I think the black area is such a bad way to do this question um and I might even be doing it wrong < tk2 to2 of f x + 4 minus G so - x^2 + 2 right right DX and that's the black area um because I'm doing the area under that one minus the area under the quadratic yes yeah all right so that is now going to give me um so it gives me an x^2 over 2 we've got plus 6 so + 6 x - x Cub over 3 between uh < tk2 and 2 so I'm going to get um x^2 4 over 2 so 2 + 12 - 8/3 and then we subtract off subbing the RO t2s uh 1 + 6 < tk2 minus 2 < tk2 over3 okay so tidying it up that's going to give me 14 - 1 so 13 from those three uh minus 8 over3 plus no nearly made a mistake 6 < tk2 and then plus 2 < tk2 over3 okay so I think the total area is what has to work out then is red plus uh red plus green plus black okay so I think the answer is uh green 4/3 < tk2 - 5/3 um plus 9 /2 + 4 < tk2 um oh wait that was a plus spotted an error I made there because it was uh flipped the sign of the whole thing so that's a plus my bad yeah plus 9/2 + 4 < tk2 um and then finally this black part so + 13 - 8/3 - 6 < 2 + 2 < tk2 over 3 right so if I simplify the whole thing maybe this is one of those horrible questions mentioned at the beginning so I think the whole thing is um what have I got I've got 4 over 3 < tk2 plus another 2/3 so 4/3 + 2/3 is 6/3 is 2 < tk2 uh plus another 6 < tk2 no four so 6 < tk2 minus 6 < tk22 so all the RO tk2 terms go that's this gives me more more belief 5/3 minus 8/3 so that's uh so that one and that one give me a minus one uh and then I've got 4 and a half + 13 so that's 9 over 2 + 12 which is 16 and a half so 33 over two all right think done in definitely not the correct way but I got a nice answer I think it's 33 over2 moving as swiftly on as I possibly can because that was ridiculous um as I said maybe that's one of those crazy questions I don't know um or maybe I just didn't know what I was doing and made a mess of it quite possible uh but anyway I'm supposed to be moving on to Part B still not finished section aoo this is long uh right B is a positive constant and this is true so not to B of b e to the BX DX has to be e right so I'm just going to do the integral and hopefully this is the last question on Section a uh so the left hand side is what's the integral of that you uh if I was it's just e to the BX I think between zero and B let me just check because if I differentiate e to the BX I get b e to the BX yes uh so it's just that which is then equal to e to the b^ 2 - 1 right yes and I want that um which then we I guess I'm going to write here want that to equal e uh so we have to solve e to the B ^2 = 1 + E I mean this is a whole thing um all right sure so I can say that b^2 = log of 1 + e uh so B is square root of that yeah plus or minus the square root of log 1 + E I mean I don't think is right but that's what I'm getting and I'm definitely spent way too much time on this um so we'll go with that um oh B's positive all right fine maybe I'm right then who knows B is greater than not so I think it's the square root of log of 1 + e all right question six was hard but uh there you go uh let's move on I think now section B right section B let's see how section B compares to section a uh it says answer any three questions from this section and there're apparently each worth 50 marks which is a very daunting number of marks for a question to be worth um all right so how many have I got I've got seven question seven question eight um what else got question eight long question nine um I'm assuming there'll be at least one more question 10 is that going to be it yes okay so I've got to do three out of four um right well let's just start by looking at seven do I like the look of seven um acceleration sure um average speed derivatives yeah I like this stuff minimum time okay I think I'm going to do seven um it just kind of looks okay so we'll dive on in number seven uh right f is driving on a Motorway uh passes a point a her speed is given by this right so we have an equation for Speed as a function of time is 2/3 T cubed - 6 T ^ 2 + 13 t plus 109 all right so that's the speed V is her speed t minutes after okay passing the point a okay so this is from a onwards okay great uh and T is between Z and five just going to note all of this stuff down CU feel like these questions might possibly a bit more involved work out her speed when she passes the point a okay so V is her speed after passing it so it's when T is zero so uh her speed speed at a is simply V at tal not which is 109 probably have to give units uh kilometers per hour okay so that I was about say that's speeding but it's not I think that's actually within the speed limit isn't it my my lack of knowledge there of Irish speed limits but I think we're okay F I think you're okay um right uh okay so it great work out Fiona's acceleration that is the rate at which her speed is increasing 5 minutes after she passes a so the acceleration is the derivative B Prime um so that's simply 2T ^2 - 12T + 13 and we want to know her acceleration 5 minutes after and T was the time in minutes so we want V Prime of 5 so that's 2 lots of 25 - 12 lots of 5 + 13 so that's 50 - 60 - 10 so that's 3 so her acceleration in kilometers per hour per kilometers per hour per minute um km per hour uh over minute I guess so she's increasing her speed by 3 km/ hour so it's going up by three every minute uh is the acceleration great okay uh C uh find the time at which Fiona reaches her maximum speed during the first four minutes okay so that is simply solving the derivative the acceleration being zero right yes maybe okay well it ask me to derivatives which I've done so the derivative is 2 T ^2 - 12T + 13 so I'm interested in could that be zero so let's solve it equal to zero 2T ^2 - 12T uh + 13 equal 0 all right now is might have to use a quadratic formula I'll use the formula just to be safe um so T is equal to minus B 12 plus or- < TK of b^ 2 144 - 4 * a which is 2 2 * C which is 13 all / 2 a which is 4 so T is equal to 12 actually that's just uh keep it simple is a quarter 12 plus or minus the square < TK of 144 minus 8 lots of 13 five of them is 65 plus another 39 so that's 94 does that sound right uh yes 94 no what am I on about 8 lots of 13 is 65 + 39 is 104 yes it's 104 okay so T is equal to um 3 plus or minus 1/4 * < TK 40 okay now I can take out so T is 3 + or minus bring out a four and left with aunk 10 um so I can actually bring out an8 no got a root eight so I'm got a 2 < tk2 * aunk 5 all over four maybe not that helpful all right so it's three plus or minus um a half < TK 5 yes not root 5un 10 my badunk 10 yeah okay plus orus < TK 10 / 2 now < TK 10 is like three so that's going to be bigger than four so the plus Roots bigger than four so therefore the maximum Max is at um T = 3us < TK 10 / 2 Which is less than or equal to four because it says in those first four minutes now um so that's the time it says give it decimal places don't have a calculator deal with it um and that will be okay that will definitely be positive is it a Max that's the question isn't it uh so the second derivative I think means it will be a Max let me just check so the second derivative V Prime of T is going to be 4T - 12 and for this to be a Max this should be negative so as long as T is less than three which it is so this is less than not at uh let's call it t star so therefore it is a Max okay awesome there we go probably didn't need to do that but I wanted to check for my own sake uh great part D um use integration to work out F's average speed over the five minutes after she passes point a okay um so what do we know I know the so if I work out the area oh God I have to think about speed time graph stuff okay so what have we got we've got a function which uh looks something like this I don't know between Z and five okay probably doesn't look like that but it's doesn't really matter something of this nature um and we're trying to say what we're trying yeah cuz it's going to carry on like that and then go up like that isn't it okay so it does look something like that I think because it's cubic positive cubic um and this is time and this is the speed so if I integrate speed with respect to time that's going to give me the distance right is that correct because speed times time which is kind of what I'm doing is going to give me a distance yes so I don't really care about a distance I want to know the average speed um ah but if I know the distance covered and then I know the time I can work it out the average speed okay yes gotcha so uh so I reckon the distance covered is equal to the integral from not to 5 of the function um which is where is the function written up there 2/3 T cubed - 6 t^ 2 uh + 13 t uh + 109 DT right so that's the distance and then if I divide by the time which is going to be divide by five give me the average speed that's my Approach so um let's integrate uh so it's two increase power divide by the new power so it's over 12 T 4 minus increase the power divide by that so 2 T cubed + 13 / 2 T ^2 + 109 t between 0 and five so they've all got T's so I have to worry about the zero one so it's just when it's five why we need a calculator but here we are 2 over2 * 25 sared oh Jesus Christ 25 squared is uh 25 is squared is 625 isn't it yeah 625 okay uh minus two lots of um o 5 cubed 5 Cub is 125 um plus 13 / 2 Sol 25+ 109 * 5 is 1090 half is 545 545 yes so is this going to simplify probably not um so that's 1350 over 12 minus uh 250 plus 13 * 25 bloody L uh 250 plus another 3 3 25 over 2+ 545 right great so then the that's the distance so the average [Music] speed is just simply distance divided by time and here time is five so it gives me 1350 over 60 - 50 plus 325 / 5 is going to be 50 wait no I'm doing this wrong 13 * 5 is uh 65 over two yes 65 over 2 I agree plus 545 divid by 5 will back to 109 okay all right so it's whatever that number is which I can tidy up a little bit and say is um 135 / 6 + 65 / 2 uh- 50 so plus 59 all right that's looks about right maybe 59 235 over 6 is like 20 so it's like 79 plus another 30 yeah a bit faster than she was I'm happy with that ballpark feels correct don't have a calculator deal with it all right party it is what it is at this point as at all of these you know could get a calculator but by this this point I'm just being uh I'm just rebelling taking V Prime to be the derivative of V and V double Prime to be the second derivative okay so we're told that V Prime of one is positive so that means um the change in so acceleration is positive yes uh but you're slowing down so this means acceleration is uh greater than not but the D I ative of it so but um slowing down rate of increase slowing down rate so you're still speeding up you're accelerating but the rate at which you are is starting to tail off um is what that second derivative is telling me uh being negative okay four graphs are shown below okay close to where t equal 1 the graph of yal V of T Al look like one of the four graphs given above I kind of Drew V of t uh So based on my picture it's got to be B yeah so I'm almost certain it's B but let's see okay right now which graph this is okay so I think it's graph B just based on the picture that I drew but it wants me to give a justification so the point is um V Prime of one is positive so we know that at one it must have a positive gradient yeah so V Prime of 1 positive means um positive slope positive [Music] gradient right so that way so it has to be b or d simple as that um and then I see now why they want me to explain V Prime of this being negative means um gradient is decreasing so it's getting shallower so B is getting shallower D is increasing getting steeper so that tells you it's B so decreasing meaning shallower there we go all right I got it from the picture I sketched out earlier H but I have hopefully satisfied what they want me to do all right this question is never ending tell these are 50 markers all right there is an average speed zone on the motorway starting at the point a and ending at the point B this is from A to B is 10 km okay so A to B and that's 10 km marvelous uh cameras record the time taken the cars travel from A to B each car's average speed is then calculated work out God the minimum time in minutes a driver could get from A to B while not driving a above okay so um okay so you're allowed to go so if V is 100 kilm hour so let's suppose you're driving at exactly the speed limit 100 km/ hour and you need to cover 10 km so your distance uh is 10 km then your time uh so speed equals d/ T just to make it really clear to myself don't make any silly errors V speed is distance over time so time is distance divided by speed so time equal 10 over 100 which is 1110th of an hour which equals 6 minutes so I think the fastest you can do it is going to be six minutes ah yes minimum time Perfect all right moving on to Rohan okay I've been watching a lot of Lord of the Rings rings of power recently so I'm immediately picturing some kind of Rohan plane and someone on a horse but anyway Rohan nice as side is driving from A to B passes the point a at constant speed okay so at a um B is 120 km per hour great uh after 2 minutes he starts to decelerate um okay so he's going so if I'm drawing a speed time graph he's going 120 for for 0 to two um okay um at a constant rate until he reaches B but by the end he has to have an average of 100 okay so he then does this so this is going at 120 yes so he's going at 120 and then he goes until he reaches the point B he decelerates all the way through okay so this is a this is B right and then I know that this new height here is 100 okay and uh his average has to be 100 km per hour okay um so can work this out um so if he average so he's going 120 * 2 minutes means it is equal to 120 * 1/ 30th 130th of an hour so he's covering there uh 4 km okay so in those two minutes he's covered four kilm so he has six kilm left so there are six kilom remaining and he clearly needs to be sufficiently slow that his average is 100 so if he's gone above that there okay okay okay okay okay um so I know his speed and I know his time and I know that the area is 4km yes great so remaining we know that this has an area of 6 km down to some unknown no he's at 100 exactly B um right so I think I think I can work this out through areas because I think I've got this kind of shape like this uh where I know that this bit here is 20 I know this bit is 100 um because he's going to be at 100 no ha that's incorrect Thomas he's not ending at 100 he's at some unknown speed right this is just some like uh you who really knows or cares right it's to start again that's not right but his average has to be 100 okay right okay so so I can maybe still use my area approach so I can say we have a shape like this and I know that this is 120 this is some U that I don't know that's his ending speed but I do know that um it has to cover 6 kilm uh so the total area here is 6 km okay and then this is like some fractional amount sure ah maybe this isn't the way to do it um okay fine uh how else can I do this um okay so he's got to slow down sufficiently um so he's got to be going slower for some point of time so I'm this is not a very methodical way to do this but I'm thinking if I just throw in like what if uh he's going 100 halfway through work out what that means I can probably just SL that like adjustment it makes sense in my head I'll try and explain now what I mean so I'm thinking um what if um he's going to pass through 100 partway through so this is going to be 100 and this is going to be having done three there's got to be a wased this as an area hasn't there yes there has has the has the has so the area Okay so this is to do with what I just worked out isn't it so I just worked out that the average okay I'm going rewrite this so I worked out in the earlier part the average speed is just equal to the the total distance divided by the time taken right gotcha that's it so he has to cover 10km because that's fixed we don't know T but we know that his average has to be 100 right uh so that tells me I worked out T didn't I um so the time taken yes so 100 km per hour time T has to equal 10 km so T = 1/10 of an hour so T equals 6 minutes okay so he's got to travel there it is got to travel for another four minutes because his total time yes okay so that's at two which means this is now at four six sorry total yes so he's traveling between 2 minutes and 6 minutes 4 minutes across the bottom okay and he's got a constant slope down in those four minutes right but what does he have to end up on so that the area is equal to 6 km okay so what I know then is I know that if I put this as 0 to 0 to4 right and that's the speed has I've got to work that out have I workout is deceleration okay right I I have an idea in my head so um I don't know the value he's going to end up at no so what's the area of a trapezium oh god um so I know the area must equal six but it's also equal to uh the square so for youu uh and then plus the bit at the top which is I don't know the height the height is 120 minus U uh times the width times a half okay I think that's right so six must be 4 U plus two lots plus 240 - 2 U so that's 2 u+ 240 now that was kmph that was minutes which is not helpful um so maybe shouldn't have done it like that that should have been in hours um so 4 minutes equals 115th of an hour okay so let's change that number this is fiddly this is fiddly um so rather than being four it's 1 over 15 okay so it's U over 15 plus that plus 1 over 15 okay so it's U over 15 plus I'm just going to have to redo this on I okay um that's 1 over 30 so 1 over 30 is 4 120 ID 30 is 4 minus U over 30 okay so I think what it actually should be tidying this up is we'll get there eventually um 6 equals U over 50 minus a 30th so this gives me a u over 30+ 4 so U over 30 is 2 so U equal 60 right this makes sense so he ends on 60 here he's going 60 um right so he's starting at 120 so he's going from 120 to 60 down to 60 in exactly 4 minutes in 115th of an hour so he's slowing down so every minute he slows down so 60 so minus 60 km pH in 4 minutes so his deceleration therefore is um every minute he slows down by 15 minus 15 kmph um km/ hour per minute so every minute he slows down by 15 uh kilom per hour okay I think at this point I've spent way too much time so that's my answer and is that the end thank God that's the end of question seven that one was long all right two to go this is already feels like a slog and I've still got 100 marks left of the 300 um kind of running out of steam this is this is long this is long mainly because I feel like seven was pretty awkward uh or you know maybe I made a mess of it who knows anymore who even knows okay um so eight is oh interest maybe I'll do interest financy stuff um okay nine is oh factors of numbers ah I don't really like factors of numbers it's like number Theory esque to me H gosh okay and then 10 is what's 10 geometry not a huge fan of geometry oh dear okay well I feel like I'm probably going to do finance and then one of those other two that I don't really like the look of so let's get eight out of the way um so question eight let's get stuck in to some economics if you will okay alga CH Fiona F and ran back along with their friends Olga in chat all right so Olga puts €3,000 um I like obviously it makes sense it's euros but you know I always like it when you have different currencies in these exams from different countries 20 in savings interest is added at a rate of 2.4% um per year okay yeah please be a compound interest question work out the amount after 5 years yay it is a compound interest so after 5 years you'll have 3,000 um uh but it's that times each year you add 1.024 and you take that to the power at five yes right it's 3,000 times I have to think about this now because it's 3,000 time 1.024 gives you an answer and then it's times another 1.024 and then dot dot dot to the last one yeah okay yeah so whatever that is in a calculator I'm not I'm not dealing with it it's that that's the answer I can anyone can put that into a calculator all right b um explain what is meant by the present value gosh of a payment of,000 whoa who explain what is meant by the present value of a payment of,000 Eur in one year's time at a particular interest rate what okay so a th000 EUR in one year's time will have increased okay um so the € 1,000 in one year's time will have increased to um 1,00 * I where I is the interest rate right where I is the interest rate okay this is a weird question uh not sure what I'm supposed to say but um what is meant by the present value of it um so therefore the present value oh it's a th in a year okay okay so we have okay so if you let P equals present value I get it then um P * I where I is the interest rate will be a th000 so P the present value is actually going to be a th000 / I where I is the interest rate so again because I have no clue what they actually want me to do here I am just going to add an example so EG in the first part I was equal to 1.24 so P would be 1,00 / 1.024 all right so it's kind of saying th000 in the future is actually what you know could be like 970 now or something of that nature right that's the idea that I think it's getting up okay um is that what I'm supposed to do um yeah all right whatever who knows number two let's keep it moving B part two Chen puts a different amount in a savings account with the same interest rate after 6 years Chen has €4,000 look at how much money CH put in initially okay so we have an unknown amount X and we know that X um and then is multiplied by the interest rate so 1 time 1.024 and that's happened six times because there were six years has to equal uh 4,000 so therefore uh X is simply 4,000 / by 1.024 to the^ 6 again I'm not putting that in a calculator deal with it uh cool okay part C fer is taking out a loan at the same annual interest rate 2.4% Fiona makes payments quarterly well done Fiona nice work uh four times per year work out the quarterly end oh God the quarterly interest rate will be equivalent to an APR which I assume means annual I actually don't know what APR means of 2.4% give your answer as a percentage to two decimal places okay all right so what are we trying to work out here we need to work out the to okay the total interest okay so the interest rate over the year I think I get the idea the question is trying to get out so I think the interest rate over the year is 1.024 right that's the the 2.4% now it's not as simple as saying 2.4% divided by 4 would give you 0.6 sorry would give you 0.6% but you don't pay that four times because it's a power compound right so that's I think what it's trying to test the understanding of so it's not that it's not 0.6 but what you would have to solve is we need um 1 plus some number to the power 4 has to give us the same as if we paid 2.4% right so how would I solve that so I would say 1 + I is 1.2 4 to the power of a quarter which again I would put in a calculator if I had one so I the interest rate is 1.24 to the power of a quarter minus one again I could put it in a calculator deal with it right okay D Rohan wants to put the same amount of money in a savings account at the start I feel like I'm learing so much Finance at the start of each month for 36 months so at the end of three years he will have a total of 12,000 and his interest rate is that all right okay think about this again same amount start of each month for 36 months okay so after 36 months we want the total to be 12,000 okay yeah like it and the interest rate is 0.1 1% so that's 1 Point uh not uh not one1 uh per month that's the increase every month okay taking a to be the amount he puts in at the start of each month write down a geometric series to show the total amount of money in the account at the end of three years okay so he has a and then after the second month he has 1. no11 time a then oh wait but then he adds another a but he had that at the beginning yes so it's okay all right I think it's okay because I have to so I was just thinking to myself the month one he has a the beginning he has a the month one he's got the percentage times a cuz that's grown plus the a he puts in so it's okay this is the total for two months and then after month three he's got the one from the beginning has this squared time a so now I see why it's a getic series uh plus the one from last month and plus the one he's just put in right now which is just a so that's going to go all the way up to uh so this is month one just to label this so this is like uh month 36 this is month 35 this is month 34 so two months have passed all the way down to month one so it's going to be 1 one1 to the 34 do you want the first two and the first two and the last two yeah right times a and then the last term because this one would be from month two and then from well from month one because it's wait is that right oh oh um no that would be month two yes yeah that's correct I think that's correct all right plus uh 1.11 to the 35 a so that was the one added in month one so yeah 30 that many months have passed I agree and then 36 months have passed so the end of the 36 month yes the end of three years okay so I'm happy with that so he would not therefore have added would he have added month 36 no because that would be the start right so I don't I'm just going to put this in red I don't think if I've understood this correctly there will be 36 terms because he's made 36 deposits right so this is like at the start okay uh was that all I had to actually do yes it's going to make me work this out though isn't it hence find the value yeah so we want right so I have a geometric series um and I want the total is 12,000 so it's a geometric series um first term is um one 1.11 a yes 1.11 A the common ratio is 1.11 and there's the last term which is 1.11 a to the 35 No 8 to the 36 my bad yeah okay so the sum of a geometric Series so the sum which we want to be 12,000 is um to think about this one now uh it is so you put the first term out the front so that's common to all of them and then it's oh God what is it one plus the next term along um so 1.11 a to the 37 um I think it's the first term times 1 plus the ratio to a higher it's not a to 37 to the next term along I think it's that I think it's 1.11 to the 37 ah no because it would be uh the whole thing should be the next term along so that's 36 okay slightly confused here I think that's right I'm worried about balancing my indices I'm going to go with that and it's over um Rus one or 1 - R um 1 - R yeah is what I would put on the bottom but that's not going to work work ah because one of them is that a minus that might be a minus and then it's one minus 1.11 trying to remember my geometric series formula um I'm going to prove the geometric series for me just to make I don't want to make a silly error right so if s is it's a plus and this is why I was getting a bit confused because it's a r plus plus a r to the n then you say well if I did R * S I get a r + do dot plus a r to the n + one okay great so I'm pulling out the a so if I pull out the a what I'm left with is a 1.011 to the 35 having pulled out the a factor because then you do that top one minus the bottom one yes which is what I've written and then um so you would do r s - s over R -1 yeah would equal a r n +1 - A so you say s * R -1 equals this yeah so s is that over Rus one right so it's actually that okay glad I checked um so I think it's it's that yeah okay I think it's that so if I put that into a calculator so I'm going to again I'm going to be like I'm not bothering with that um so 12,000 has to equal 1.11 * a time 1.11 to 36 -1 all divided by 011 okay so I would substitute that in so a is simply equal to 12,000 *.11 divided by uh 1.11 multiplied by 1.11 to 36- one okay I have no idea what that's going to give me but I think that's my answer I am going to move on because again taking a while all right e uh Park cells three types of ticket child student adult the table below gives information on the price of each ticket and the percentage of tickets sold for example 15% of all tickets sold are student uh yes okay the expected value of the price of the ticket is 13 this is this is a whole other question now isn't it no more banking is at 1385 okay so e of X expected value is uh 13.85 sure um work out the value of x the price of an adult ticket okay so I think e of X is also equal to um 11 * the probability which is.52 um plus the student price which is x - 5 yep times 0.15 uh and then plus 33 0.33 sorry * X all right solve for x so we would say 13.85 is um God 11 * 0.52 is 5.72 plus 0.15x minus 5 lots of that is 0.75 plus 0.33 X okay so 13.8 five um and then on that side I've got um 4.97 so - 4.97 has to equal 0.48 x so I think X is equal to uh it's going to be 8.85 8.88 over.48 so it's kind of doubling it it's like 189 something like that okay cool F again I'm not pointing to a calculator can't be bothered f um when an item is being sold the markup is the profit as a percentage of the cost and the margin is the profit as a percentage of the selling price okay a Shop sells an item with a margin okay margin of 18% uh work out the markup give your answer as a percentage but I have no clue of the price is selling it okay okay so the margin is the profit as a percentage of the okay so let's suppose the selling price I'm just going to Define my variables a lot going on selling price equals P right P for price um okay and then we know the margin is the profit as a percentage of the selling price so if I'm selling for p then 18% of that is profit so 0.1 8 p is profit right I'm selling it at P I make 18% is the profit so that's my profit okay the markup is also the profit but as a percentage of the cost okay so I don't know that so.18 p is the profit um let's call the cost price C and I want to know the profit as a percentage of the cost o sells an item with a margin of 18% okay now work out the markup for this item but I don't know the cost price do I not did not do any kind of economics and getting rather confused by the definitions okay let's think about this again told you it's a slug my brain isn't working the margin is the profit as a percentage of the selling price yes now the markup let's just write this out the mark up is the profit equals uh profit as a percentage of the cost oh okay oh okay so I want to say if the margin's 18% then does that mean that the cost is 0.82 p I think right because it's the cost it costs you right so you're selling it let's Okay let me just think about this so let P equal 100 so then I'm saying the cost equal it cost 80 wrote pounds 82 but I don't need a currency it costs 82 and then I'm saying um so if I sell at 100 then my margin is 18 % so I make a profit of 18 okay so then the markup is the profit as a percentage of the cost so I think the markup is 18 out of 82 that makes sense to me because the markup is the profit as a percentage so it's a proportion of the cost yes so 18 out of 82 is going to give me a number less than one so then I Times by 100 okay to get percentage all right had to I still not sure I understand their definitions but that makes sense to me uh looking at the numbers right I think that was I had a light bulb Moment In My Head where I was like if that's 18 then I think the cost must be 82% and then that allowed me to kind of go from there anyway we're going with that we're going with that and that is thankfully the end of question eight uh so now I just have to decide am I going to do nine or am I going to do 10 so I'm going to obviously look through both of them and make my choice so um I don't really want to do the number theory part of nine but then it looks like derivatives and I like derivatives as we've seen um and 10 looks like a lot of geometry and I don't I really I don't think I've got the brain power for geometry uh so yeah I don't really like either but I'm probably going to pick nine let's let's see what happens question nine all right okay AA is investigating factors of different numbers looks at numbers written as powers of a prime okay list the five different factors of 2 to 4 okay well that's okay because it's prime factorization and it's already written like that so a part one the factors of 2 to 4 are simply uh the other powers of of two so 2 to the 0 2 to the 1 2 to the 2 2 cubed and 2 to the 4 they are five factors how many different factors uh 3 to the 7 has uh same idea um so factors of 3 to 7 uh so it's going to have eight factors uh so number is eight uh I don't think I have to write them out but right I'll just write out the pattern you're going to get three to the 3 to the one all the way up to at 3 to the 7 so there'll be eight different powers of three okay maybe this one wasn't as bad as I thought how many different factors uh two oh okay 2 to the 10times I think I know what they're getting at here 2 10 * 3 to 12 now yeah because right so 2 to the 10 has 11 factors 3 to the 12 has 13 factors so um you can pick a different number of them to combine um and you can yes so you pick one from each so you can take them all by themselves or you can pick them as pair so I think it's 11 * 13 so I think it's 143 cuz it's all multiplied so I think the total factors is just 11 * 13 uh which is going to be 143 so I think it's just as simple as that actually um all right uh B right what's AA doing now um I can get the relationship between Pairs of factors she makes a table to show the pairs of factors of 12 so the pairs of natural numbers X and Y where X Y = 12 great complete the table sure uh X and Y okay so X goes 1 2 3 4 6 12 1 2 3 4 6 and 12 uh and then y would have to be 12 it would have to be 6 3 * 4 4 * 3 6 * 2 12 * 1 okay plot the six points above on the coordinate diagram below one of the points is shown so showing me when X is 3 and Y is four uh so it's showing me that one okay uh it's going to be nice and symmetric that's for sure so um positive X and Y so when X is 3 Y is four and then when X is four Y is three um and then when we go down to like one it's like up here at 12 yeah and then down here this is 12 this is one and then two and then six in the middle all right and then two and then six and that's all of them 1 2 3 4 6 12 yes okay right great uh do I want to join them up I don't know if I'm going to join them up I'm going to join them up because you're going to get this nice kind of pattern like this between them okay oh I'm Feathering sorry Ben Sparks for those of you that have seen My GCSE video uh I I am calling that correct uh screw you Ben okay three AA realizes the relationship is this yes see that's what I that's what I drew right hyperbola okay so AA has worked out very cleverly that y equals 12x awesome on the coordinate diagram draw the graph in the domain oh I kind of did all right I already did uh so yeah so this is now just this is literally y = 12x and it wants me doing the domain between one and 12 great that's what I've done uh perfect um C awesome okay so far I'm happy with my choice it's going to get worse now I've said that a tangent to the curve is drawn at a point P so we have a tangent at um P for x and so Y is 12 over P excellent I agree with that um p is positive and a real number awesome show the equation of this tangent is this right so um that's a straight line yeah okay so the gradient is dy by DX um Dy by DX which is um so it's 12 12 * x -1 so it's going to be -2 x^2 uh so therefore the equation is y equal MX uh plus c x + C where the gradient is the x value so - 12 over p^2 yes because you evaluate that at x = p so that's the first bit right and then we have to solve so we know that y y = 12 P when X is p so that's going to give me c so 12 over P has to be -2 over p ^ 2 * p + C uh yeah that does work so that gives me 24 over p = c which is the answer okay great awesome tick that's what we got that's your intercept uh okay awesome uh show that the equation of tent is this done liking it C2 um the area of the triangle formed by the X the Y and the tangent is always k s units wa wait wait okay hang on the area of the triangle okay so area of triangle uh is equal to K what K where K is a natural number okay work out the value of K right so it's saying it's always the same no matter the choice of P so um right so let's just think about this so my graph is y equals this so I am just going to take so what have I got it's y = -2 p^ 2 um X+ C where C is 24 P so if I take P to be 12 right so I'm going to let P equal 12 to make my life hopefully a bit easier because it's telling me it's always the same so therefore my line is um - x + 2 so now I know that it's going through two and it has a gradient of Min - x so it's going to look like this so therefore it's a triangle of side lengths two so the total area is two so the area equals 2 which implies k equal 2 all right cool I don't think I would have immediately realized they were always the same kind of does make sense though I think but yeah I'm happy with that CU it's saying pick a value of P wait that's not correct h SP my I think I SP my own mistake because it's P Squared isn't it that's that's that's wrong my bad my bad that was lucky um it's p^ squ it's p^ squ so I don't want P to be 12 um let's take P equals 2 let's take PS 2 my idea will still work um because I can pick any value of P to make my job easy I just did a silly uh because I'm going to let P equal 2 which I know is a point on the curve um because now I can say Y is = to -12 over 4 * x + 12 so actually Y is equal to - 3x + 12 right that makes more sense it seemed too small the answer I got so what you're actually going to look like over here redraw this is you're going to go through 12 and it's going to come down quite steep so when Y is zero X has to be four so it's going to look like that let's dra more like that so this is four so the area is actually 24 right because the area of that triangle is a half * 12 * 4 so it's 24 which is K okay so K is actually 24 um which I find more believable thinking about the picture in my head yeah all right I'm going to go with that so I think K is 24 uh and let's keep it moving oh no that's it oh well that one was short is that really it that can't be it okay now that is it all right well okay so a surprising end there that one was so much easier number nine there was so much easier than seven and eight so maybe I picked maybe eight was one of those other awkward questions I feel like seven and eight were both actually a little bit tricky they might have been the hard questions uh who knows I can look up the uh I can actually look up the newspaper articles now and find out but before I do that as usual I've stopped I've finished done the exam it's taken a while I've exhausted I do need to mark it or at least check that I've got the right answers the right working so take a little break I'm going to go and get myself a drink at least and I'll be back in just a moment to do the marking and then I'll kind of rank where does Island sit in my list of International High School maths exams right I'm feeling refreshed and ready to do some marking so let's see how I got on uh I have the mark scheme in front of me I will also as usual put the mark scheme and the exam paper in the video description so you can have a go at these questions yourself and also check that I have marked my own answers correctly apart from you know not having a calculator we're just going to ignore that uh but otherwise uh I'm feeling really confident there are some hard questions so you know we'll see my confidence might be misplaced um but I I feel like most of my answers seemed reasonable so as usual I'm not going to focus on exactly the very specifics of the mark scheme if I have the correct answer I'm giving myself the marks that's how we're going to do it otherwise this will be a whole other hour of marking and at this point I'm tired you're tired we just want to know how I did question one uh m is 2 or - 16 over 3 boom uh B uh K is J over H minus one k is J of hus one it doesn't mention the zero thing but that's fine I just did more right so that's fine I'm claiming that's okay um p is minus < tk3 yes R is < tk3 over3 amazing uh that's question one okay so that's 30 out of 30 Good Start I was expecting that one uh to have been okay okay uh C is 8 B is minus 6 that sounds like what I got amazing the limit that was quite a nice one that was two pretty straightforwards for me at least uh the graphs um so it's been shifted down to amazing and it's been shifted left amazing and that looks perfect yeah yep yep yep all right awesome and that was it all right another 30 out of 30 forgot how nice these early questions were uh number three uh oh no I didn't do number three I left out number three uh I went to four yeah okay p is minus 5 minus three oh oh oh wait I might have dropped marks ah oh dear and this is out of live oh gosh okay so what have I done wrong they reckon that I should have what one oh wait I've done the wrong question have I have I I might have written out the wrong question let me look at this I think I one of us me or the mark SCH answered the wrong question it's probably me um oh gosh okay so I've done Cube which I thought was kind of annoyingly difficult I think it might have meant to have been squared um it is it's Z squared right so I have answered the wrong question good one for me but I've lost marks it's fine that right there should be z s and I have written z cubed so I have solved the wrong question now I've got to give myself 02 three or five um what am I going to get uh let's see low partial credit um some correct substitution and multiplication I definitely got that high partial credit is fully correct substitution and multiplication so I am going to argue that I deserve three because here it's that's high partial credit the markim says basically if you've done everything correct but just not got the answer then you deserve full credit and I'm pretty sure what I've done is correct I've just messed up at the end because uh I'm an idiot and wrote out the wrong question so I'm going to say that's three out of five um so I have got a big old minus two on that one uh and you know lesson learned read the question double check the question after you've read it I answered the question I wanted to answer instead of the question in front of me a okay B oh this one was a nightmare um what is w what is w um okay it's < tk2 over2 plusun 6 6 over 2 I that kind of rings a bell < tk2 over 2 plus root six oh get in I'm actually really happy I got that I've not used their method at all but who cares minus oh amazing get it all right good and there's a solid 15 so I get 15 for that awesome and then the last 10 I'm going to be uh down here um right so I need to get I * U is oh gosh what does it say is a i minus B yes and then when I bar it it's minus B minus that yes uh minus B plus a i for the other one yes yeah cool so they're all correct number two uh yeah that looks correct that looks correct um okay oh this bit at the end okay few it says axial symmetry in the line through the origin with slope minus one uh which is exactly what I said I said it's a reflection in that line amazing so I get 28 out of 30 question four ah answer the wrong question don't mind that it is what it is silly error I shouldn't have made and none of you should make but uh yeah whatever my MK was fine I think okay - 10x over 5x^2 + 7^ 2 awesome that was fif that can't be 15 marks wow okay next one is 2 over pi plus log pi over two awesome um and then we've got uh W yep okay this injected bit h no element of C is used more than once means each element Maps uniquely yeah so I have said that perfect and it's not surjective because one element of C is not used uh nothing in B Maps why awesome yeah weird question but I think that's a 30 out of 30 for that right 50 markers still sounds like a crazy high number for a question uh number six 33 over2 units that rings a bell that does ring well wait wait wait there was more before that I've scrolled too far got too excited X is three and X isus two yes already tick that one and then 33 over 2 unit squar get in that one again I don't think I did the right method but got right answer it's all that matters um and is that it no oh and somehow I got that one correct the square root of log that awesome this wasn't a 50 marker was it this was another 30 getting ahead of myself right because I skipped question four or three or whichever one it was okay now we get to the 50 markers okay um 109 yes 3 km per hour per minute yes um right I didn't use a calculator 1.42 I'm pretty sure 1.42 and I did check gradient amazing um that is going to be 1.42 I I'm I'm certain it is so that's all good awesome and then Part D uh oh yeah this horrible number I didn't want to work out I'm pretty sure that's what I've got because they've integrated and then divided by five yeah okay I'm going to trust my algebra there it looks pretty much like it's going to be 11233 to be honest um yes wait uh oh you know what it is it's near enough it's it's clearly going to be that or very very close to that um I did the right thing it's fine okay um question seven answer B uh because the function is increasing and the rate of increase is slowing yes amazing now the horrible one um oh no time was six minutes yep figured that out come on deceleration 15 km per hour per minute get in h and there was such an easier way to do it than than how I did but you know what it's kind of nice when you figure it out from a very convoluted confusing way um there's like I get a lot of satisfaction the fact I was still able to figure it out there is a much easier way to do it given in the mark scheme which you can take a look at uh my job here is not to teach you anymore it's just to see what it's good so I did get the right answer thankfully um right economics one number eight uh have I got that formula I do have that formula so therefore that's correct um have I got that formula divided by 1.024 to the six yes awesome that's correct did it explain the amount that should be invested today to give you a th000 in that time I mean yes weird question um okay great that would be C uh oh that to the quarter minus one awesome just didn't put it in the calculator uh amazing geometric series okay so I got lazy again but I've got 12,000 times not point that yeah and then one minus that but I've got that minus one because of the flip get it yes it's the right formula again I've not put it in it would come out the same answer € 326 and6 cents a month um would bought you would be investing okay X is 8.88 over 0.48 brilliant uh oh God 9 out of 41 time 100 that's why I written written is 18 over 82 same thing Wicked that is correct all right question nine nearly there is the last one isn't it cuz I didn't do 10 I didn't have to didn't even really read it properly but didn't look nice okay question nine factors are uh yeah just the powers of two amazing okay it just wanted me to say there were eight awesome 143 factors brilliant uh that is my graph yes this is looking good this is looking good uh okay and then finds d y by DX and then works out the intercept by suming in the Y value yep brilliant and K is K is what come on 24 unit squared uh and that's it 50 out 50 and the one above was also 50 out 50 wasn't it somehow that was 50 out 50 uh and then the other one was that also I think it was you know I got so excited that I got the deceleration correct forgot to write the mark So I think I've actually everything I'm going to claim all of the maths I wrote down was somehow correct this doesn't happen often so I'm going to going to make the most emphasize this everything I wrote down was correct but I misread one of the questions so on the complex number question I misread it as Zed cubed so I actually got the wrong answer so I dropped two marks so I've scored there a final total of 298 out of 300 which I am extremely pleased with because that was a long slog of an exam there you go 298 out of 300 uh obviously a very good Mark as I absolutely should be getting but I think the challenge there for me taking that was um it got a bit puzzled on the economic stuff right that that sort of margin versus markup I've just never studied it so like I was kind of tired having slogged through all of those questions and it my brain was getting fatigued so I did struggle a little bit to kind of get my head around that first and I sort of guessed fortunately guessed correctly using a bit of common sense um so that I had to sort of puzzled about a little bit um and I think on some of the the integration question for the area I definitely did that not at all the best way to do it but you know I managed to get there so I think it's an example showing you that um you know there are whilst there are methods to follow if you can't remember them or figure them out like you can still just do sensible things just do maths to get the answer right there's more than one way to solve these problems and even in the mark scheme it was giving method 1 2 three I didn't look in detail because I again wanted to be quick just work out my total uh but I feel like I've definitely used some slightly different methods especially on those integration questions and that deceleration question my method there was so different to what was in the mark scheme but I don't care I got it right that's all that matters uh from my perspective at least sure teachers will tell me something different but nonetheless um now difficult so I did uh after finishing the exam I did look up before I marked it uh the hard questions uh and it was the stuff to do with injective and surjective the question about showing route two was irrational came up as one of the hard questions so actually I know that proof I didn't do that question because I didn't like the rest of it about logs but I do know that proof so to me that wasn't hard but if as a student if you don't remember that proof I didn't pay attention to it that's going to look like a horrible question um and then the subjective and injective again if you don't recognize those words you're going to be like what the hell is going on so um I can see why some students came out being like what are they testing here this is like nothing I've seen before I I am sympathetic to your cause but I actually for me again University level of course I found the more computational ones where i' just forgotten the the techniques I found those a bit harder uh and the time it was a slug it was absolutely a slug right 300 marks I I I need to lie down right I'm exhausted so I think it was all right overall where does it fit on my list um I felt like an a-level exam is sort of I've forgotten all of the ones I have on the list you can see them on screen but I can't remember right now I've done that many now but it feels like it wasn't quite as hard hard as further maths it feels sort of similar to a level um I feel like I have the Arbiter from Germany is a bit harder than a level this felt very very similar to a level so when I'm editing this as I'm currently now recording you'll be able to see I'll slot it in somewhere around a level but how it compares to the others you can see on screen uh cuz I just can't remember all the others I've done you can tell my brain has just it's gone right just done a two and a half hour exam I'm pretty exhausted uh so my brain's melted I can't remember the full list but it felt like a level so therefore it's going to be harder than the American ones that I've been doing but not there were more advanced concepts in AP Calculus so maybe it's not as hard as AP Calculus just because at that one had more advanced concepts right there was like sequences and series and harder limit stuff in AP Calculus so I think it's not as bad as that don't think it's quite as hard as the German Arbiter so it's somewhere between a level and arbiter I think is where it's going to fit on the list but again you can all see it on screen that's the ranking right so I'm going to go take that lie down I keep going on about pretty exhausted um thank you for suggesting exams please do keep suggesting exams I want to do as many countries as possible I I am slowly trying to diversify away from just uh English speaking so I've done Germany it would be nice to get a few more translations so if you have an exam you'd like me to do maybe from your home country and you are able to find an English version or send me a translation then please do I'm trying to do as many different countries as I can but thank you as always for watching and thank you for supporting the channel if you would like to increase your support for the channel then you can either sign up for my patreon more details are in the video description or completely free option is you can subscribe and then you'll get notifications the next time I release a video but thank you for watching and I'll see you all soon take care