Comprehensive Guide to IGCSE Maths Paper 2 (2025 Edition)

hello and welcome to my all of IGCSC maths paper 2 the 2025 edition That's right So this is really focused on

non-calculated topics Those topics that are much more likely now to appear on the paper two rather than the paper 4

And there are four key ideas to this So first of all as I just mentioned here the non-calculator aspect It is based on

the specimen papers So what's been coming up in the specimen papers third my prediction videos So I do these

prediction videos every year for the last three or four years Again you can check out the most recent one above me

here And then number four which is really important to yourselves here which is the harder topics Now not every

topic is really really hard but certainly things like differentiation and vectors and things like this I'll be

talking about in this video So get yourself a coffee get yourself a tea sit down for the next couple of hours or so

and work through these topics so you're really prepared for those harder topics on IGCC math paper 2 in 2025 Right let's

get started Okay so let's go through some typical paper two questions and all

these questions that you see today come from the last three years of papers So find the xcoordinates of the point not

the coordinates of the points on the graph of y = x ^ 5 - 5x ^ 4 where the gradient is zero So usually give some

indication of either a turning point or where the gradient is zero So this is the thing to look out for when looking

for a differentiation question So what we're going to do is we're going to take our function and we're going to work out

the derivative dy by dx Now the process that we do here to do this is we take the five we bring it to the front So I'm

going to bring the five to the front and now we reduce the index by one So what do I mean by this well we're going to

reduce this from a five to a four And this is the standard process for differentiating IGCSE this one here Well

we're going to take the four We're going to bring it to the front So what does that mean here well we need to do five *

4 and that's equal to 20 And then we reduce the index by one So we get x to the power of three Now in the question

we want to find out where the gradient is zero Notice dy by dx This is what's called the gradient function So we want

to take the answer we just made and make it equal to zero So we need to solve this equation here Now this is actually

an equation which we we can put into a single bracket So what do these two things have in common well they have a

five in common and they have an x cubed in common because notice x * x * x also fits into this as well Then we open up

our brackets We say to ourel well what do we multiply 5x cubed by to get 5x ^ of 4 just x What do we multiply 5x cubed

by to get - 20x cubed well that's just -4 And now we use the null law So either this has to be equal to zero or this has

to be equal to zero Because you have two numbers or two things that multiply to give you zero then one has to be zero So

either this is the way I like to write it either 5x cubed is equal to 0 which means that x then has to be zero It's

the only thing that works or x - 4 is equal to zero And therefore we solve the equation x= 4 And notice all they wanted

for this question are the x coordinates We don't need to work out the y-coordinates in this case So our

answers here are x is equal to zero and x is equal to 4 So we write that in However in our next question we're going

to look at on a paper too Uh they sometimes ask you for the actual coordinate itself Let's look at that now

So we want to differentiate 6 + 4x - x^2 So we need to differentiate this function So let's go through that

process now Now if you have a number and differentiate it it disappears Okay so we're going to ignore that If we

differentiate this well we still use the power law as normal Notice x on its own is the same as x the^ of 1 So we bring

that one and bring it to the back 1* 4 is 4 Reduce the index by one So x ^ of 0 Now some of you already know what the

answer is but I'm just going to write it out in full And if we differentiate the minus x^2 well the two comes to the back

So we get min -2 And then we reduce the index by one Well let's reduce that down from 2 to 1

Now at the moment it's in kind of ugly algebra So let's tidy this up So remember anything to the power of 0 is

equal to 1 1 * 4 is 4 And remember x^ of 1 We generally like to write that as x So our answer for this question here is

4 - 2x like so And now we need to work out the coordinates of the turning point So

that's when the gradient is equal to zero the other way around to the previous question of this function here

So we want to set our gradient function 4 - 2x our dy by dx from the previous question equal to zero So this is a

straightforward equation to solve I'm going to -4 on both sides of the equation This cancels we get -2x is

equal to minus4 And if we divide by minus2 on both sides of the equation this cancels and we're left with x is

equal to 2 Now notice that's not our final answer That is just the xcoordinate of the turning point To work

out the y-coordinate we need to sub x = 2 into here So we need to then do y is

equal to 6 + 4 lots of 2 - 2^ 2 And if we work that out well 6 + 8 - 4 remember your bid mass 6 + 8 is

14 14 - 4 well that's equal to 10 So our final answer for this question is then 210 So you need both the x coordinate

but also the ycoordinate there as well And you can have a look at the mark scheme This is where you pick up your

method marks if you go a little bit wrong right at the end And so we're going to look at this one here As you

can see it's out of eight marks So it's going to take a little bit more work So first of all we want to work out dy by

dx of this function here So the way we do that is just like we did on the paper two questions So we take the three here

we bring it to the back So 3 * 2 is equal to 6 We reduce the index by one So it was an x cubed And now it's going to

become an x^2 Likewise for this part here we're going to take the two there We're going

to bring it to the back -4 * 2 is equal to -8 We reduce the index by 1 So this becomes x to the power of one or as we

like to write just x and that's all we need for our derived function Remember if you differentiate the six then that

gradient of that is zero So that will disappear And now this is very typical in the part two We want to work out the

gradient at a specific point So we want to work out the gradient when x is equal to 4 So we take

our initial answer here So 6 x^2 - 8 x and replace the x with 4 So we get 6 lots of 4^ 2 - 8 * 4 And remember our

bid mass So 4^ 2 is equal to 16 8 * 4 here is equal to 32 And now we just need to do this calculation here So

16 * 6 Now again we got a calculator but again you can always do it manually this way as well This gives you 96 So then 96

minus 32 Again any of this can be done on your calculator as well And this gives you the final answer of 64 So the

gradient at that particular point is equal to 64 And this is a good question to go through because it goes through a

lot of the typical kind of paper 4 style differentiation questions And we want to work out the coordinates of the two

stationary points on the curve So as soon as you see the phrase stationary point or turning point you need to be

thinking that the gradient function that's a t gradient function is equal to zero Well what does

that mean so we take our initial function here our 6x^2 - 8x and make it equal to zero So you need to have good

quadratic skills as well in order to be able to do these questions Now here this is an example of where we factoriize

into a single bracket So what do these two things have in common well they've certainly got a two in common and they

have an x in common as well So then we open up our bracket What do we multiply 2x to get 6x^2 well that's going to be 3

3 * 2 is 6 and an x What do we multiply 2x by to get - 8 x that's going to be -4 is equal to zero And just like we did

in one of the paper two questions we use our null law So when 2x is equal to zero so we know one

of these things has to be zero We don't know which one Well x is equal to zero And

when 3x - 4 is = 0 well 3x is = 4 x is equal Oh no So near the bottom of the page is equal to 4 It doesn't want me to

write it I'm going to write it over here See there we are 4/3 That is going to be 4/3 There we are I'm determined to write

it There we are So x is equal to 4/3 So that's our xcoordinates of the turning points So 0 and 4/3 And now we need to

work out the y So we need to take our function right up here So this is equal to y = 2x cubed - 4x^2 + 6 And now we

need to replace the x in here with these values of 0 and 4/3 So let's start with zero So when x is equal to zero y is

equal to well that disappears That's cool This also disappears That's even cooler And we got a plus six So these

will disappear and we're left with just six So our first turning point or our first stationary point is equal to

06 Next one's going to be a little harder So we're going to go to when x is equal to

4/3 I'm going to write this out for the examiner So we got 2 lots of 4/3 cubed - 4 lots of

4/3 squared + 6 to work out these more complex calculations with fractions And that gives me the

y-coordinate of the other turning point like so So there's the mark scheme So you can

have a quick look through and where you actually get the marks and what some of the mark uh method marks are dependent

on as well Okay on to the next question So again this is very very typical of the paper 4 and the 0580 course They

don't start necessarily with a differentiation problem It's usually embedded in another question So in this

case the equation is cubic xub - 4x^2 + 4x can be written in this form Find the value of a So cubics generally speaking

will be written in such a way so you can factoriize this quite easily So notice they've all got a factor of x here So

we're going to take a factor of x out What do you multiply x by to get x cubed x^2 What do you multiply x by to get -

4x^2 - 4x What do you multiply x by to get 4x is just + 4 At this point we got a quadratic within the bracket So at

this point we can put this into two brackets And we put an x at the front an x at the front and we want two numbers

that multiply to give me four but add to give me minus4 Well that's going to be -2 and -2 And notice because you got two

x - 2's this can be simplified further as x - 2^ 2 Now that's not the answer that they're looking for The actual

answer is what's this missing value a here and notice it's the value of two Now while the reason they've done

this in the first part is now we can now draw a very nice sketch So if I was to solve this equation so let me pop this

over here So x /x - 2 all^ 2 * x - 2 all^ 2 is equal to zero Then using the null law I know then that x must be

either equal to zero or equal to two However this is a cubic So remember the shape of a cubic generally speaking

looks something like this So it crosses the axis three times This is a very special cubic where only crosses or

touches the x-axis twice So let's mark in those uh turning points So got zero here and two over here We have a

positive cubic Notice this is an x cubed And so we've got all the information we need to draw a good sketch So watch

carefully We're going to go up up up up We go through the point and this is the crucial part because two is a repeated

route We're just gonna touch the graph there and then we go off That's a little bit of a bend there Try and make sure

that's a nice smooth curve as well Okay So notice that I've followed the instructions of the question where the

graph meets the axis where it meets it at 0 0 meets at 2 0 as well There is some sort of local maximum but that's

not necessary for this question you'll see in the mark scheme if I go forward to the M scheme So here's your graph So

it's going through the origin going through 0 uh the positive cubic and it's the right shape as well So you can see I

pick up all four marks Now comes to the differentiation part So we take that cubic and we want

to find the equation of the tangent to the graph at x= 4 So what that means in real English is if I take um go green x

is four is probably up here somewhere Okay So if we read that off that's four So basically at that

particular point we want to find the tangent line So the line that just touches the curve and then disappears

off So this really really steep line and we want to find that equation of that particular line So the way we do this is

we use some differentiation This is where it's embedded So first of all we need to differentiate the function So we

get dy by dx So remember we take the three bring it to the front reduce it by one So we

get 3x^2 We do the same procedure here I'm speeding up slightly now So -4 * 2 is equal to - 8 Reduce the index by 1 is

8x And don't forget because you got a 4x here we do the same procedure because this is a 1 1 * 4 is 4 And then reduce

the index by 1 That's x ^ of 0 which is just 1 So we get the function of dy by dx is equal to 3x^2 - 8x + 4 Now at this

point we actually want to work out the gradient at the specific point We don't want to work out turning points or

anything So we need to work out dydx where x is equal to 4 What does that mean in real English well we're

going to put four into this equation So we get three lots of 4^ 2 minus 8 * 4 + 4 Well 4 2 is 16 16 * 3 is 48 8 * 4 is

32 + 4 And that simplifies to 20 Okay So the gradient of that green line that I drew over here let me just

mark this in The gradient of this line is equal to 20 So we've got that information Now that means that the

tangent line so I'm going to write the word tangent has the form y = 20

x + c And now we need to work out the y intercept This c Now the only point that we know on this particular tangent line

is this point here Okay and we know the x coordinate is four but we currently don't know what the ycoordinate is The

way we can work that out is actually substitute x = 4 into this equation So when x is equal to 4 y is equal to 4

cubed minus 4 lots of 4^ 2 + 4 * 4 That's a lot of fours That is definitely a lot of fours So we get 4 cubed is

equal to 64 Well 4^2 is 16 16 * 4 is 64 Oh that's nice That cancels And then we get 4 * 4 is 16 So when x is 4 y is 16

So this coordinate is actually 416 And now we have enough information to work out C So when X is= 4 Y is equal

to 16 You can see why this is a seven mark question A lot of work involved So we've got 16 = 20 lots of 4 + C So 16 is

equal to 80 + C And then C is equal to 16 - 80 This is really where you don't want

to make a silly mistake by the way So 16 - 80 that's equal to - 64 So our final answer is equal to y =

20x - 64 and you'll see that agrees with the final answer at the end for all seven

marks So just be very careful of that kind of calculation right at the end You're going to be tied at the end of a

question like this Make sure you're accurate all the way until the end Hello and welcome to another ginger

mathematician video and today I'm going to go through IGCSC advanced vectors is I did a similar video on meneration with

advanced meneration and I thought I'd do exactly the same for vectors as well Now if you're really not sure about vectors

again what a vector is and doing those basic calculations with vectors then please do check out my UDMI course on

vectors it's completely free and that would get you up to speed on what you need to know for IGCSC vectors in

general but I thought I'd take some more advanced questions have had a lot of requests for a video like this So you

can go through this nice and carefully Right let's get started So we have a tri a diagram that shows a triangle O A B So

got O here A and B And a parallelogram over here So O K A L or O A L K And the position vector of A is A and the

position vector of B is B So that's already marked on here What it's basically saying here this word position

vector we're going to talk about this a lot in this video Often confuses students but all a position vector is

it's the vector from O to A So O to A is just vector A and the vector O to B is just B The O here is just the origin

It's the center point where we're working from Now K is a point on AB So this straight line here such that the a

k uh the ratio a k to kb is 1:2 So what that means is remember with ratios if we do 1 plus 2 we get three So that means a

third is represented by this section and 2/3 is represented by this section from b to k So putting in all that key

information just to help us when we actually work out what the question wants us to do And what it wants us to

do is work out the position vector of L So remember the position vector just like I mentioned here we're starting

from O So we're looking for O L Now in order to go from O to L we have a couple of options Uh one option is to go from O

to A and then go A to L So we could do that or we could go O to K Okay Plus K L So we can take either of

these roots up here in order to work out the problem Now because this is a parallelogram so this is an important

word in the question It basically means that the vector O A is going to be the same as KL Okay they're the same length

and direction So I can actually write this in as well that this vector is also going to be A as well So it means I can

take either route and it's not really going to make much difference Okay So this is the where we try and use this

sort of retrograde thinking Um I want to work out okay that's really important for the question before I can work out

okay I need to work out well to go from O to K I need to go O to B and B to K for example or O to A and A to K So I

need to work out this vector B A or A B So I'm going to have to in some shape or form work out this vector I'll put it in

blue here so you can see this nice and carefully And I need to work this out Whichever direction that we go I'll just

take this direction So from B to A So how do I go from B to A well to go from B to A I can go from B to O and O to A

So I'm going to write that in B to O and O to A Now the vector B to O because I'm going against the arrow is going to be

minus B and O to A I already know this This is plus A However in order to work out this O to K

I need this section of my vector here in order to do O to B and B to K So if I want to work out the vector B K in that

direction remember I wrote this down earlier It's 2/3 of the entire vector 2/3 of BA So it's going to be equal to

2/3 of - B + A And so that's equal to minus 2/3 b + 23 a Okay So we're working through now

we're working forwards rather than working backwards Now I can actually use this to work out Okay which was

important from what I wrote earlier So to go from O to K Okay Sorry that joke is so bad Um then

we go from O to B and then we go B to K Now O to B I already know that's just equal to B And B to K is what we worked

out earlier which was - 2/3 B + 2/3 A Now we're going to collect up some like terms Remember this is one lot of B So 1

- 2/3 that's going to be equal to 1/3 B And then 2/3 A stays as it is We collect up A's separately and B's separately as

well Now okay is not the answer we're looking for Remember we're looking for the position vector We're looking for O

L here So to work out O L just like I did above we need to go from O to K which we just worked out plus K L which

we already know So we have then 2/3 A + 1/3 B And then we've got this vector K to L That's just equal to A Now

remember we've got 2/3 A here We've got a whole A here So 1 + 2/3 is 1 and 2/3 or

5/3 a +/3 b which is going to be our final answer Make sure you got your squiggles

on your vectors That's really really important to make sure you're indicating it's a vector quantity and not a number

or scalar quantity So you can have a look at the mark scheme here again for all four marks We've got our answer and

you can see where you pick up those method marks as well Right on to question two here So they like their

position vectors Yeah O A and O B So we're given that X and Y and O D So the vector O to D is equal to 37 X + 47 Y

And we need to calculate the ratio A to D and D to B Kind of backwards to some of the questions we're going to see

today Okay So before we can really discuss this I think we just need to work out our vector A to B Notice that's

always a very standard thing to do the co soal long way to go around and find this vector So A to B it's A to O and O

to B So that's equal to minus X + Y You can write down min - X plus Y straight away once you're used to these kinds of

questions So in order to work out O to D here we then just go O to A and then some fraction of this line here some

fraction of AB So let's just give that a number Let's just call it K So K lots of AB We don't know how long we need to go

along the line That's what we're trying to work out here So we have O to A that's equal to just the vector X plus K

lots of what we just worked out which is min - X plus Y Now we can expand the bracket here So we just get x

+ minus minus kx

plus k y Again we can rewrite this slightly So we can factoriize out an x So we get

uh 1 - k x + k y and we're told this vector od as you is seen in the question here is

equal to 3 over7 x + 4 over 7 y So in order for this to be true again the k must be

equal to 4 over 7 uh in order to get this as 3 over 7 So our ve uh our ratio here so this is four the whole thing

needs to total to give us seven So therefore this needs to be three because 4 + 3 is 7 So we have our ratio here of

4 to 3 Careful not 3 to four but here four to three because we have this extra position vector in here as well Okay So

you can check the answer here Again quick two mark question Again it's working backwards to what we've actually

kind of seen already Okay And now we've got a bit of work with magnitude and subtracting

vectors Okay A bit more on the easier side So first of all we want to work out P minus Q If you've watched my demi

course I go through this in quite a lot of detail So we've got two minus min -1 and then we have 8

-4 So remember two minuses here This is what you have to watch out for to not miss a mark out So minus minus is a plus

So we get three at the top and four at the bottom And now it wants us to work out the magnitude of this vector So this

is something we'll talk about in later parts of the video And remember if you want to work out the

magnitude of the vector you want to do Pythagoras So we put this into the Pythagoras formula And again this is

quite nice numbers if you spot the answer straight away Really well done So we get square<unk> of 9 + 16 that's

equal to<unk> 25 and that's equal to 5 Again if you notice that Pythagorean triple or 345 really well done You can

pop that straight in as you'll see in the M scheme here Again get one mark for some idea of putting three and four

together Right onto our next question So we have a trapezium not a parallelogram this time So a trapezium So we have to

be a little bit more careful O is the origin O R is A and OP is B Okay so we've got our position vectors here and

the magnitude of RQ So let's look at this vector at the top here is 3 fifths of the vector OP Now we know OP is equal

to a whole B So the top here must be equal to 3 fifths B This is a very common technique that I'm doing these

questions Before I even look at the question I try and fill in as much information as I have So then when I

come to the question I've already got that information there ready for me Now we need to find the vector P cub in

terms of A and B So we want to go and let's put this in a different color There we are So going from P to Q and

we're going to have to take the long route round I'm afraid So if you want to work out

PQ then that's equal to P to O So we're taking the long way around here P to O to R and then R to Q And by putting in

that extra information we can just put in all the information we have So P to O going against the arrow minus B And

we're going with the arrow O to R So plus A And then we've got 3 fifths B Remember this is minus one B So

this simplifies then to A And we've got -1 + 35ths That's going to give us -25s b You need to be good with fractions

when you're working with these kinds of vector question So let's pop that in We have a min -

2s b Nicely done Now this next question a lot of people ask me about this kind of question So do pay careful attention

to this So when pq and o r are extended they intersect at some random point called w And we want to find that

position vector of W So we want to find O W whatever that's going to be So what does this look like so P to Q is

extended So imagine going up like this And what else is extended o R is extended So at some point and you can

probably agree with this it's going to meet at ah resume that slideshow It's going to meet Oh great I have great fun

with this Okay It's going to meet at some point W at the top Now you probably agree here that if I do this in a

different color it must be just some vector just with a in it because we're just going along this line Again we're

just making the vector bigger or smaller So we're changing the magnitude but not the direction So this vector must equal

to something a We just don't know what that something is Now the way to think about these kind of problems is actually

think of it in two different ways So we know that OW the vector O to W is equal to some K * A We know that from what I

just described However we could also take a longer route We could take the route O to P and then P to W So we could

take the vector O to P and P to W This is where part A is very important for part B Now the vector

O to P is equal to B And so this vector pw is some multiple along this line multip multiple of this pq that we

worked out before So let's call this like some kind of t We don't know what this t is of that vector pq So let's

write that in So some t * pq So pq was a minus 2s b And let's just multiply out and collect stuff So

we get B + T lots of A minus 2 T of B Thinking why are we doing this but you'll see in a moment

And then we collect our A's and B's together So we have T A and then we have we've got one here and we've got minus

twoths T We don't know what T is at the moment B So there's two ways of describing OW either the first thing

just some multiple of a and that means zero b there's no b involved or t a a different letter plus 1 - 2 tb if these

two things are true then that means that the k must equal t first of all and that zero most importantly is equal to what I

have here we're doing basically equating coefficients we use this a lot at a level as well so what I can say is this

zero at the top here is equal to whatever's here So 1 - 2 t That gives us a really easy equation to solve So we

get 25ths t is equal to 1 And therefore t must equal 5 over 2 Again this is advanced video I'm not going to go

through that in much detail And remember I also said that t is equal to k which is what we wanted to find So our answer

here will just be the position vector ow is just equal to 5 over 2 a like so these questions come up a lot at the top

end where there's a a star questions if you want to get those a stars specifically these are the kind of

questions you need to kind of get your head around and practice So there's the M scheme for the two things that we have

already done so far and you can check the M scheme as well What I'd like to talk about is a

very similar question to the specimen paper two and that is a cubic function However if they give you a cubic

function on IGCSC0580 maths it will be in this kind of factorized form So don't feel like

you'll have to use a factor theorem remainder theorem to do these So let's talk about first of all the shape of the

graph Now notice these x's So if I multiplied all of this out I would get x * x * x which is equal to positive x

cubed or 1 x cubed Now because it's a positive x cub so a is equal to 1 the graph goes uphill So what do I mean by

this a general cubic has two different shapes Either it goes this way or for a

negative it goes this way And because a is equal to positive one it's going to have this general shape here Okay so

that's what I mean by going uphill Imagine you start here and you're hiking and you're going all the way up to here

Then you're essentially going uphill You have got a little dip here but the general trend is that you're going

uphill So we know from the from the a= 1 that the graph goes uphill Perfect So we can tick this off Now we

need to find the y intercept So set x0 Again we've looked at this before So when x is equal to 0 y is equal to well

0 - one 0 + one And anything times by 0 is just going to equal zero So we've got

our y intercept at 0 0 and now we need to find the roots Now notice it's in a very nice form which it will be for

IGCSC 0580 maths So we have it in this form and that means then that either this x is zero which we already

knew from the y intercept anyway x -1 is 0 So x is pos1 And we also know x + 1 is zero So x

is equal to -1 So we know it's going to cross at zero at one and at minus1 as well So that is useful information to

know Now I know a few people have asked about this about how do you actually work out the maximum and minimum points

Now if I put here it's not usually required for IGCSE but if you're moving on to other courses and you're looking

really on that extension those unseen kind of questions Let's at least talk about this So if I expand all this out

notice we have it in the difference of two squares form So if I leave this x alone but just expand the brackets well

we've got x^2 + x - x -1 These two cancel and so we're left with x brackets x^2 minus one So this is a difference of

two squares You should recognize that from the IGCSE course Now we can just multiply out this bracket So we get x *

x^2 is x cubed and then x * -1 is equal to x So if we expand all of this out then we get this particular fairly

simple cubic here Now to work out the maximum minimum point we have this method called differentiation which some

of you have talked about in the chat already So how do I differentiate this function how do I work out dy by dx well

we take the three we bring it to the front So that gives us 3x and then we reduce the index by one So we go from

three to two just like that For this one we do the same process So we take the one bring it to the front So we have

minus one x we reduce the index by one So that gives us x to the power of zero Now you should all know this is a really

important fact Anything to the power of zero x the^ of 0 is equal to one 7 to the^ of 0 is equal to one My favorite

football team Derby County They're in the second league Hopefully they'll survive this season To the power of zero

is also one as well So anything to the power of zero is one And so this term here just becomes

minus1 like so So we've now found the derivative of the function Now why do we care about such things why do we care

the reason for that is to find any stationary points So minimum or maximum So I'll use min or

max We set our derivative now to zero So we take the 3x^2 minus 1 and we make it equal to zero So we're going to solve

that equation So I'm going to give you a moment in the chat Can you solve that equation for me 3x^2 - 1 = 0 Please be

careful That's all I'm going to say to you so far Let's see if anyone can answer

this Take a little break and take a little coffee Okay so we've got our square root

of 1/3 Can we be a little bit more specific here so I agree 3x^2 is equal to 1 x^2 is equal to

a3 Good Plus and minus Thank you Hassan Really well done So we've got the plus and minus the square root That makes a

big difference for this kind of question Um I'm sure your teacher has said when you find the square root of a number

like 64 is equal to 8 That's the principal root is what we like to call it But when it comes to equation solving

specifically then you get two answers So there's a difference between just knowing a square root and square rooting

in an equation There are two different things going on Nice So we've got plus and

minus the square root of 1 over3 Now just to test your third knowledge because that is also important for

uh paper two So we've got plus and minus If we do the square root top and bottom we get 1 over<unk>3 Can anyone tell me

in the chat how I rationalize the denominator here this is a phrase you need to know from s How do I rationalize

the denominator here let's see if anyone can do that in the chat and know what I'm talking about because it's really

important It comes up now or will do a ton So how do I rationalize one over root3 what answer do I get trench has

got the right idea Can you give me an answer good Yeah exactly Well done So we get

the plus and minus the square root of 3 over3 Generally speaking you should rationalize the denominator if you get

the chance to do it Whether they take a mark off IGCSE for this kind of question I doubt it but they certainly could do

that Now once I know the plus and minus <unk>3 over3 once I know those two answers I can substitute that back into

my function to work out the y-coordinate Now I'm not going to do that now but if you wanted to work out the y value we

take plus <unk>3 over3 and then we can work out y And then when x is minus <unk>3

over3 then we can work out the y-coordinate there So notice we're going to have a maximum point and a minimum

point Now I haven't talked about the second derivative yet about working out if it's a maximum or minimum But we can

be a bit lazy in this question and there's a reason why Can anyone tell me why I don't need to use the second

derivative test to work out which is a maximum which is a minimum so can anyone tell me that whilst I start drawing our

sketch here let me just duplicate this slide Can anyone tell me why I don't need to

use the second derivative test okay one is negative already Yeah I

think you got the right ideas So let me just draw in my sketch So this is why the y-coordinate doesn't

matter too much I just want to show knowledge here So we've got goes through 0 0 goes through one and minus one

wasn't it yeah one and minus one one here minus one here now if you take the general shape

of a cubic function general shape notice it will always have a shape when it's positive like

this so I've drawn this best should be symmetrical by the way but that's a point for a different video so notice

the first point always has to be the maximum here and the second point always has to be the minimum because we know

the shape of the graph It's our very first thing that we did So the go minus roo<unk>3

over3 and whatever the y value is comes here and then roo<unk>3 over3 and whatever the y value is then we'll come

here So you can use the shape of the graph to actually work out if it's a maximum or a minimum

Okay But yes you can use the second derivative test as well But notice I'm trying to save you time in the exam If

they say "Okay I want you to draw a function and give the minimum maximum points." You don't want to be sitting

there doing another derivative substituting values in seeing if it's less than zero or greater than zero This

is a much quicker and refined way of doing it So I'm trying to give you lots of tips in this video So just to speed

you up in the exam so you can get that extra time to really work on those difficult questions

Yeah Um does this apply for 0607 paper uh I don't think so Um because you got the graphical calculator for four and

six anyway But this is good knowledge generally to have when you go into IB or A level Okay So nice little trick to

know for cubics You can use the general shape to know where the minimum is and where the maximum is Okay Any questions

in the chat before we do a actual genuine specimen paper question based on what we've talked about so

far Could I do more menstruation videos i've done two I've done an ordinary meneration I've also done an advanced

menstruation video So again type in to YouTube IGCSC menstruation ginger They should come up Hey not related but what

questions are more likely to come in paper two yes something like exact trig values domain the exact trig values SS

has to come in non-cal There are certain topics that have to be non-cal based Did I ask minimum maximum IGCSC um

probably to calculate it like I just did in this question Probably not but it's good knowledge to have anyway And if

they want to try and extend those A star questions which they always do some twist every year like I predict the

exams and I predict it with like 60% accuracy 60 65% Uh but they'll always put something in there to try and really

differentiate between a candidates and a star Uh ATMAS video there will be one coming out fairly shortly I'm not going

to tell you which topic it is but it will be coming out Do ask to find maximum in sketching

grass IGCSE Not that I've seen yet but there's no reason why not You have the skills for it Uh prediction papers are

already on my website So you go ginger mathematician prediction papers into Google You will find those for IGCSE

both paper two and paper four and Alevel papers one and five Okay So to finish this video today

we're going to do part 26A using what I've literally just said here So on the diagram sketch the graph of y= f ofx

between our domain So notice this word we should be more familiar with now So what x is allowed to be Now it's given

obviously on our diagram We're going between minus3 and four So can anyone tell me what the first stage is it's a

three mark question Probably takes you four minutes in the exam What should I do so far what should I

do yeah the prediction papers are paid for but it's a way of supporting the channel sports all the other content

that I do and there's tons of content out there Uh expand No don't expand Please don't expand Please don't

expand Don't need to expand Definitely don't need to expand Okay first thing shape of the

graph First thing shape of the graph So we've got x * x * x So we've got x * x x Yeah if expand it just takes up too much

time It doesn't really help you too much It's an x cubed So it's going to be uphill So we go like this So it's going

to be this shape of graph So that's the first thing that we should be looking at Put in calculator Yeah Yeah It's paper

two not paper four question Um what do we look for next equate each to zero Find x roots Yes Again try and

systematize this So first next thing would be y intercept There's the next thing I do and that

means set x to zero If x equal to zero using the f notation which is fine So 0 2 - 3

Remember anything times by 0 is just zero So we know it's going to go through 0 0 just like the example we did

before And yes now we've got the roots Now we can talk about the roots here So the roots called lots of different

things roots x intercepts solutions Um there's another word I've forgotten So we had x x + 2 x - 3 is

equal to zero So either x is equal to zero x is equal to minus2 or x= pos3 So we know it goes through minus2 So minus2

here We'll put three over here And now we've got everything that we need in order to draw this function Yeah Next x

intercepts Well done Hello Muhammad So we go up through the

function Now notice we always cross the axes We don't just touch there There's a reason for that which I'll talk about in

one of my Alevel videos And then we go up like this So we've got our sketch here If we wanted to we could try and

work out these maximum points But for this question here it's only a three mark question it certainly would give

you more marks to work with if they wanted those points Notice it says very clearly in the question here show the

values of the intersections with the axes and that's what I've done I've shown here here and here If you look at

the M scheme here so you'll see so correct sketch without any labels So the shape of the

graph we've got minus 2 0 and three indicated So here here and here If you have the general shape of the graph it

goes through 0 0 There's a maximum here a minimum here and again just continues in both directions So that's where we

pick up the three marks here using what we've talked about in this video because we have a given uh true because got a

given domain It doesn't keep going This is true Again I'm usually taking it where it goes That is a good point there

I will cross out my arrows because it just stops at minus three and four Yeah that's true

Okay welcome to another Ginger Mathematician video here and we're going to go through everything to do with

similarity that a couple requests for this kind of video So let's get started on this part D and we're going to

combine some other topics in here Things like circle theorems and meneration as well So GKM light on the circle center O

FGH is a tangent to the circle at G So notice this line that goes across just touches the circle and then goes off And

mg is parallel to O That's why we've got these arrows here and here And we need to show that triangle GKM So it's worth

identifying which triangle this is So this one here So G KM So G KM So those letters there um is mathematically

similar to triangle OG So when you do a question like this it's quite useful just to label the two triangles that

we're focusing on That makes it easier then to work out what is shared between the two triangles Give a geometrical

reason for each statement you make So this is where this kind of questions a bit more like science because you see

there are four marks available for this So we're going to put down four bits of information as we go along So the first

thing to notice here and this is a circle theorem Again if you're not sure about circle theorems here then do check

out the video above Is that if we take G notice this angle here is a right angle A tangent meets a radius of a circle at

a right angle And in a very similar way if you take a diameter of the circle then two lines that come out from the

diameter also meet at a right angle as well So notice they've integrated some circle theorems into this What this

means then is that angle G M K Now how do I know to use this notation g M K So M is where the angle is Then G and K are

the points surrounding it Okay and we're going to get lots of practice on this And then we look at O G H Notice G is

where the angle is Our surrounding And if you're thinking well why not put down HG O well that's also absolutely fine

too So these two angles are the same and both equal to 90° and we need to write some reasons So

tangent meets radius at 90 degrees and then the other reason

here the diameter So lines from diameter again not much right room to

put all this in but it's important from diameter meet

circumference at 90° I'll show you the mark scheme afterwards to see yeah you can make this a little bit shorter in

terms of the information So we have those two angles that are shared which is very very important Now what we're

going to do is show that two other angles are the same And the angles I'm going to pick on here are this one here

So this angle here So this is G O H I'm going to show this is equal to this angle down here So that's

KGM And the reason they're equal is another angle fact you need to know Again a very useful angle fact that

comes up a lot So if I do this now with my two triangles notice there is what's called informally a Z angle And so we

need to give the proper reason for this We need to use the right language here And the reason is that alternate So the

Z angle is known as alternate angles are equal So it's really important you put

these very clear reasons down as you go So if two angles are the same then it stands to reason if we look at the very

last angle that we haven't considered here these two must also be the same as well So G K M is equal to O HG There's a

couple of reasons you can put for this You can say third angle in triangle That is absolutely fine But

hopefully some of you have spotted out there This is also the alternate segment theorem So this angle here must be the

same as this angle here for the alternate segment theorem Again I really encourage you check out my circle

theorem video if you're not sure about that particular idea And because then the three angles

are the same So all three angles the same You can write down a aa as well if you so wish are the same

then uh they are similar triangles So they are similar

triangles So we've got all the information there for the four marks Let me break that down So the first part is

showing their 90 due to those two circle theorems one mark Then alternate angles are equal therefore that's second mark

The third angle in the triangle therefore they're equal and therefore some context sentence at the end saying

because all three angles are the same the two triangles must be similar and you can read through the mark scheme

very very carefully you can see where I've picked up all those four different marks again we had some alternatives

here about not using alternate seance just using the idea it's the third angle in the triangle that is also fine as

well okay and onto a more classical kind of question here so question 11 Again whenever it's a similar a similarity

kind of question here they are going to write down that phrase They're going to write down mathematically similar So

that's a great trigger word to use to go ah I need to remember this video that you're currently watching on similarity

And what we need to do here is calculate the side QR So again it's always good to know exactly what side we're looking for

As soon as I see a similarity question like this I'm thinking scale factor Now what do I mean by the scale factor well

the way that I would work this out is I take two corresponding sides of the shape Now what do I mean by

corresponding sides well the same side in both shapes So AB here and P Q like so Again they're both the bottom side of

the shape so they're corresponding to each other And then I take one and divide it by the other So eight divided

by six Now you could do it the other way round as well I'm going to stick with big divided by small Now if I do that

again we're just going to divide both by two and we get 4 over3 You don't need to rush to a calculator with this kind of

question Try and be happy to leave it as a fraction Now what we need to do here is work out our Q So QR that's what

we're looking for And notice we're going from big to small So this is the big triangle This is the small triangle So

if I'm going this direction I want to be dividing by that scale factor because I want to go from big to small That's

going the other way It would be multiplying So all we're going to do is take the corresponding side to RQ in

this shape which is BC here So therefore six And all we're going to do is we're just going to do

six divided by 4 over3 Now of course you can use your calculator for this That is not a problem at all I'm going to do

some keep it switch it flip it because that's good fun So keep this switch this to a times and then flip the fraction

Again that is also absolutely fine too And if I work this out this is 18 over4 Again new to calculator it's a

calculator question And that's going to lead to my final answer here of 4.5 You're wondering where that comes

from but 18 / 4 You can always use the bus stop method That's always good fun and we get four there Carry the two

Fours into 20 go five Again like using non-calculated methods but of course you can just use a calculator at that point

And a very normal follow-up question to this is they'll do something to do with prisms here So mathematically similar

prisms And the way I read this question is the volume So that's really important here of the larger prism is 320 We want

the volume of the smaller prism So as soon as I think of volume I'm going to be thinking of volume scale

factor Now you're probably thinking well how do I work out volume scale factor i just have the scale factor 4 over3 from

the previous question Well all we do is we take that 4 over3 and we're going to cube that number Okay because notice

we've got centimeters cubed or cubic centimeters So we want to take that and then cube it Now again we can use a

calculator here but if I cube this we're going to get the same as 64 over 27 4 cubed 64 3 cubed is 27 Now we're going

to do the same logic here We are looking for the smaller prism So we're going from big for small So we're going to

divide by this new scale factor So to work out the new volume of small just to remind me what I'm doing here I'm going

to do 320 divided by 64 over 27 cubic cm for the two

marks Again you can check the mark scheme here Again you get a method mark for cubing that scale factor Again

either way around Notice I used 8 over six but you could also do the same process with 6 over 8 as long as your

times you're multiplying correctly too Okay And onto the next question here Again very similar kind of questions

This is why you watch these videos to really see the patterns between the questions that come up And in triangle

PQR so that's the whole triangle So that's the whole triangle that goes around the

outside Again M lies on QR and S lies on PR Okay Probably don't need to put the ink there Explain giving reasons why

triangle PMR So let's identify that Okay Well we've got this This is our

PMR Little bit on the wobbly side It's fine Saved it There we are Uh it's similar to triangle MSR Okay So that's

the smaller triangle inside And we need to find out why they are

similar Okay You can't just put they're contained within each other That doesn't

work Right so PMR is equal to MSR Right so notice that this is a right angle here It's on the same straight

line And we've got a right angle here as well So they share this right angle So we need to write this in nice and

carefully So P MR is equal to M SR and they're both equal to

90° Again I think that's quite clear why that's the case I don't think we need to write any more reasons on that Uh the

next thing to realize is they share an angle So notice that this angle is the same in both triangles So again we need

to identify this carefully So angle P R M is equal to S RM Uh so shared

angle So that's my second piece of information Okay So we've got two angles Now I think that makes it actually quite

easy now because the last angle that we haven't mentioned yet where did my pen go my pen decided to

disappear There we are So if this angle is the same in both we've got a right angle that's shared That must mean the

last angle we haven't talked about So this one and this one So M PS must be equal to S Mr Don't worry

won't do any SMR videos here Um and that's because it's the third angle in the triangle you know two are

the same and the other one must be the same as well add up to 180 So therefore and we can write down a aa or

all three angles Um so that means the triangles are

similar You don't need to over complicate these questions As long as you find two angles then you can find

the third angle It's going to be the same and therefore the two triangles here will be similar So we'll go through

part B and then we'll come up You can just see the answers quickly here So you can see putting AA gives you that third

mark which is important So let's get started on question one here where we need to

simplify the following expression Now the first thing that I notice here is notice there is a common factor of x in

the numerator So when I'm copying this across I'm going to take a factor of x out here And then I'm going to work

backwards So what do I multiply x by to get 5x a five What do I multiply an x by to get - x^2 that's going to be a - x

Now usually in these algebraic fraction questions there will be a factor that looks like this in the denominator So I

can cancel through So that's kind of my little cheat method here I'm looking out for a 5 - x down below Now if we take

the 25 - x^2 this is a very special form of factorizing known as dots ds the difference of two squares

difference because we have a negative here and squares because we have a square number here and a square number

here Now this is a special number 25 If we find the square root of 25 that is equal to 5 And of course the square root

of x^2 will then just be x So I'm going to use these here and I can write this now in a special factorized form Taking

these numbers five and x I write here a five minus x and a five + x Now if I'm not sure that that's actually going to

work one thing I can do here is expand my brackets to check if it's correct 5 * 5 is 25 5 * x is + 5 x - x * 5 is - 5x -

x * + x is + x^2 And you may be thinking here well that's not the same as denominator It's quite different Notice

there's a cancellation We have a + 5x here a minus 5x here That then gives us cancels This should be a minus by the

way Well spotted if you notice that because - x * + x is - x^ 2 Notice that leaves us with 25 - x^ 2 So we do get

the denominator Now why do we actually factoriize these things in the first place well notice we have a 5 - x in the

numerator 5 - x in the denominator We can cancel those out giving us x over 5 + x as our final answer I can pop that

in here Now if you're finding this quite tricky at this point don't worry We're going to do a ton of different examples

so you're up to speed on this tricky topic There's the mark scheme and where you pick up the method marks over here

And on to question two In this case we need to write as a single fraction in its simplest form So the first thing to

notice here we have different denominators We have an x - 5 and we have a 2x So one way to make one

fraction here is to multiply the two denominators together So if we do that we just get x - 5 * 2x Notice in algebra

for multiplying we just put the two things next to each other Now how do we work out what goes in the numerator we

essentially do an idea of cross multiplying So we do a 2x * 3 here and an x - 5 * - 7 Notice sometimes it's

called the butterfly method Okay this idea of multiplying crossways to work out what goes in the numerator So we

have a 3 * 2x that gives us 6x and then we have a minus 7 Be very careful there's a minus here Time x - 5 Once

we've got this we just now need to tidy it up So how do we do this we keep the denominator the same I'm just going to

switch this over It looks a bit more neat So 2x brackets x - 5 And at this point we have 6x we'll leave alone for

the time being We have a single bracket which we multiply using the hook That's what I like to call it here So we get -7

* x is -7 x - 7 * -5 Be careful here That's equal to + 35 And now if we simplify this we get 6x - 7 x So I have

six bananas and then seven bananas disappear minus one banana and then we have + 35 over 2x brackets x - 5 Notice

I didn't expand at the bottom here You could I would still get all the marks but I prefer to leave it in this

factorized form because it is its simplest form essentially and leave it like this to get me all three marks

Again notice in the M scheme you can also put it the other way round This is maybe a little bit more tidy Notice - x

+ 35 is exactly the same thing On to question three here Again we want to simplify We're going to go

through a very similar method to what we did in question one I look at the top of my fraction and go what do they have in

common so notice there's an x in common And then let's look at the numbers So what number goes into four and 16 that's

going to be four And now we work backwards So what do I multiply four sorry 4x to get

4x^2 that's going to be equal to x What do I multiply 4x to get -6x it's going to be -4

Just like in the previous question number one I'm expecting an xus 4 in the

denominator because we need to have some cancellation going on Now notice we have another difference of two squares here

with x^2 - 16 because remember the square root of 16 So something times something gives you 16 and it's the same

What is it that's four and the square root of x^2 is just x So we can factoriize it in that very very special

form that we've seen before where we have an x -4 and an x + 4 We achieve what we're looking for here Notice we

have an xus 4 above and below We can cancel those out giving us 4x over x + 4 And that is our final answer for a

whopping three marks just for that fairly short process There's the mark scheme here And

you can see where you pick up the method marks On to question four here This is slightly different Again lots of marks

available here Five marks And we need to solve the equation Now however we do start it out like we did with question

two So in question two we found a common denominator And the best way of doing that is just multiplying the two

denominators together So I can just write them next to each other X + 1 * X + 9 The way that we work out the

numerator we do this cross multiplying method here So that gives us 1 * X + 9 is just X + 9 9 * X + 1 Okay 9 lots of x

+ one and that's equal to one here Let's tidy up that numerator So if we look at the numerator here we still got the x +

9 Let me pop in the bottom bit here So x + 9 and then we have 9 * x is 9 x 9 * 1 that's equal to 9 and this

is equal to 1 So we've kind of done this the method in question two that's about two or three marks worth Now we actually

have to solve this thing Now first thing is I do not like fractions here This fraction is irritating me So what's the

opposite of dividing by x + 1 * x + 9 Let's multiply by x + 1 x + 9 on both sides Quite hard to write all this out

This cancels giving us the numerator We can simplify that at the same time So we get 9x + x that's equal to 10 x 9 + 9 is

equal to 18 And this is equal to 1 * this So it's just going to be x + 1 * x + 9 We've still got work to do here So

first of all we need to expand out the bracket So remember our first outside inside last or I prefer to call it the

double hook Give it a big double hook So if we work out the right hand side here we have x * x is x^ 2 x * 9 9 x 1 * x is

just x and 1 * 9 is 9 We are making progress I assure you So let's continue this up here So I'm going to tidy up the

right hand side So that's x^2 + 10 x + 9 is = 10 x + 18 Notice I've also switched the sides but that's absolutely fine

because we have an equal sign in the middle So I can swap the sides That isn't a problem Now let's get rid of the

10 x So the opposite of adding 10 x is minusing 10 x from both sides Get a nice massive calc cancellation here So this

and this disappear giving us x^2 + 9 This cancels with this giving us 18 We're almost there Keep with it So we've

got an add nine here So the opposite of that is minus 9 So this cancels leaving us

x^2 18 - 9 is equal to 9 And then final step here we're almost there What's the opposite of squaring something we're

going to square root on both sides That leaves us with x on its own The square root of nine This is where we have to be

very very very careful Notice there are two answers it's looking for So it's not just three but it can also be minus3 as

well Why is that the case well -3 * -3 - * a minus is a plus and 3 * 3 is 9 So whenever we're square rooting there are

two possible answers to consider So our very final answer here is x is equal to 3 and x is equal to -3 like so Again you

can have a look through the mark scheme and see where you pick up all these method marks And if this video has been

really really helpful to you so far then do think about liking and subscribing That's the best way to keep up with all

the different IGCSE content that is coming out And I'm going to talk to you about the sponsor of this video which is

me with my predicted papers So these predicted papers are built by ourselves here making sure we're using that 12

plus years of experience that I have as an IGCC maths teacher to give you the very best authentic experience for that

IGCC math paper 2 that's coming up in May 2025 So if you want to check that out again go to my website predicted

papers the link will also be in the description Okay welcome to another Ginger

Mathematician video where I'm going to go through the exact trigonometry values that you need to succeed at IGCSE paper

2 but also very useful for IB learners as well If you like the content I'm doing then please do think about liking

and subscribing Right let's get started So you should know first of all sin 0 co0 and tan zero These are worth

learning off by heart and you'll see them just there on the screen there But now let's work out these things like sin

30 or cos 45 and how to know those off by heart And the way I like to do this is by using two very special triangles

So my first triangle I'm going to draw over here So this is a right angle triangle Weird kind of angle there but

there we are And then we have a right angle here And I'm going to make this side one and this side one as well Now

if I know these two sides I can work out the hypotenuse here by using a bit of trig bit of Pythagoras So remember that

x^2 is equal to well 1^2 + 1^2 So using pythagoras that means then x^2 is equal to well 1 2 is 1 1 2 is 1 1 + 1 is equal

to 2 Therefore x the hypotenuse here well I square root both sides and that gives me the square root

of two So this is how I know this is the square root of two And because this is one and this is one this is an isosles