I'm an IGCSE Maths Teacher - Post your questions here.

[StudentXYZ]

The only nth terms that you need to be able to find are for 1) arithmetic series and 2) simple geometric series.

1) E.g. 3, 7, 11, 15, 19, ... These have nth terms of the form an+b

2) E.g. 3, 9, 27, 81, 243, 729,... These have nth terms of the form a^n

The nth terms of all other types of sequences can usually be found either by using previous sequences or by substituting values like in this question. Some sequences e.g. 1, 4, 16, 25, 36, ... you are expected to recognise.


d)

You could use algebra here, but you will end up with a quadratic equation which seems a but unnecessarily just for one mark!

Have a look at the sequence again: 0, 2, 5, 9, 14, 20, 27, 35, 43, ...

The question wants you to find two consecutive terms in this sequence where the difference is 30.

E.g. n=5, n+1 = 6. A polygon with 5 sides has 5 diagonals and a polygon with 6 sides has 9 diagonals so the difference is 4.

You need to find where the difference is 30. Remember, the difference goes up by 1 each time.


For sequences questions like this, it's always best to look for patterns and use logic before using algebra.

OK, thank you very much.

 

[Relon]

How to solve question 9 , 0580-w07-qp-2 ?

 

[notnek01]

How to solve question 9 , 0580-w07-qp-2 ?

If you've learnt the 'completing the square' method for solving quadratic equations then you just need to complete the square.

If you haven't learnt it, expand the form given:

(x+p)^2 + q = x^2 + 2px + p^2 + q

Now compare this with x^2+4x-8 and you can see that 2p=4 and (p^2+q)=-8.

Does this make sense?

 

[Relon]

If you've learnt the 'completing the square' method for solving quadratic equations then you just need to complete the square.

If you haven't learnt it, expand the form given:

(x+p)^2 + q = x^2 + 2px + p^2 + q

Now compare this with x^2+4x-8 and you can see that 2p=4 and (p^2+q)=-8.

Does this make sense?

I got it , Thank You

 

[AmazynGal]

how to get the position vector?? help please!! T^T~ that was on MJ 2009 question8 paper 2 variant 2 (b)

 

[notnek01]

how to get the position vector?? help please!! T^T~ that was on MJ 2009 question8 paper 2 variant 2 (b)

Call the centre of BCDE, M. Then the position vector of M is O->M.

Starting from B, the point M is halfway along the vector a and halfway along the vector g.

So O->M = O->B + 1/2 a + 1/2 bg = 2a + 1/2 a + 1/2 g = 5/2 a + 1/2 g.

 

[Ni19Mo]

I need help with no 11

 

[mak kings]

There are a few ways that you can do this question. If you don't like the method below, please tell me and I'll give you another one.

I'm going to find a relationship between the area of a sector and the arc length:

Area = a/360 x pi x radius^2

Arc length = a/360 x 2 x pi x radius

You can see that that for area compared to arc length, there is an extra radius and a missing 2.

So radius x Arc length / 2 = Area


For this question, Arc length = 4r so

Area = (5r) x 4r / 2 = 20r^2 / 2 = 10r^2


Does that make sense?

nopes sorrry i dint get it!! .... :S

 

[mak kings]

Your method is correct - parallel lines have equal gradients.

But first you need to write both equations in the form y=mx+c:

5x = 4y + 10 --> 4y = 5x - 10 --> y = (5/4) x - 10/4

2y = kx - 4 --> y = (k/2) x - 2

So (5/4) = (k/2) ...

so is the answer 5?

 

[notnek01]

I need help with no 11
View attachment 24935

Which part specifically are you having trouble with? Do you understand what property they are talking about?

 

[notnek01]

nopes sorrry i dint get it!! .... :S

I'll write a as the angle of the sector. The formula for arc length is: a/360 * 2 * pi * radius.

The radius is 5r and the arc length is 4r so we have:

a/360 * 2 * pi * (5r) = 4r

Next make a/360 the subject of the equation:

a/360 = 4r / (2*pi*5r) --> a/360 = 4/(10*pi)


The question asks for the area of the sector. The formula is: a/360 * pi * radius^2

The radius is 5r again and we can use the a/360 formula that we found earlier. Substituting all this in gives:

Area = (4 / (10*pi) ) * pi * (5r)^2

If you simplify this (try this yourself) you get:

Area = 10r^2


If you're still stuck, please tell me exactly which parts you do understand and which parts you don't.

 

[notnek01]

so is the answer 5?

Not 5.

(5/4) = (k/2) --> k = 10/4 = 5/2

 

[notnek01]

Oct/nov 2009 paper 4 . I am having trouble with question 6 b (iii) . How can we find it when the line y=f(x) is not a straight line.

Also, in the same paper Q2 a(iv). My answer was shear with a factor of 1 , but the mark scheme says the factor is -1, I don't understand why.

6biii) f(x)<g(x) when the graph of f(x) is below the graph of g(x). Can you see where this is?

2aiv) I showed this example here: https://www.xtremepapers.com/community/threads/mathematics-post-your-doubts-here.2565/page-148

Scroll down until you see the image.

 

[StudentXYZ]

6biii) f(x)<g(x) when the graph of f(x) is below the graph of g(x). Can you see where this is?

2aiv) I showed this example here: https://www.xtremepapers.com/community/threads/mathematics-post-your-doubts-here.2565/page-148

Scroll down until you see the image.

2a(iv) yes, I got it , thank you.

6 (b)yes, it is the wavy part of the line below g(x), but how do I work out the answer.

 

[notnek01]

2a(iv) yes, I got it , thank you.

6 (b)yes, it is the wavy part of the line below g(x), but how do I work out the answer.

The answer is the range of x values where f(x) is below g(x) i.e. where you see this 'wavy line'.

The answer is written like a<x<b and the two values a and b are where the two graph intersect.

E.g. a is approximately -4.4.

 

[StudentXYZ]

ok, I got it, thank you very much.

 

[Ni19Mo]

Which part specifically are you having trouble with? Do you understand what property they are talking about?

I understand that we get the same answer in both multiplications and that the tens and units swap places but from part b) onwards i just dont get it

 

[notnek01]

I understand that we get the same answer in both multiplications and that the tens and units swap places but from part b) onwards i just dont get it

Once you know the pattern for part (c), it's easy to find an example that works for (b). But to be honest, I'm not sure how they expect you to do (b) without knowing the pattern in (c).

Since you haven't found the pattern yet, I may as well tell you: pr=qs. E.g. for 26 x 93, 2 x 9 = 6 x 3.

So pr=qs is the answer to (c). Have a go at (d) now you know this and don't worry if you get stuck - it's hard!

Is this from an IGCSE paper? Which one? If it is, it's one of the hardest questions I've seen (except part (a), which is one of the easiest )

 

[StudentXYZ]

May/June 2012 paper 41.
I am having trouble with question 9 (b).
g(x)=7 -2x (we multiply by -2 and add 7)
so inverse we will divide by 2 and subtract 7 so it will be x-7/2 but the mark scheme says 7-x/2 . I don't understand why.

 

[notnek01]

May/June 2012 paper 41.
I am having trouble with question 9 (b).
g(x)=7 -2x (we multiply by -2 and add 7)
so inverse we will divide by 2 and subtract 7 so it will be x-7/2 but the mark scheme says 7-x/2 . I don't understand why.

I thought I replied to this already but it got deleted.

'we multiply by -2' so you need to divide by -2 (you divided by 2) then you'd get: (x-7)/-2 which is the same as (7-x)/2.

The method that you use can easily lead to mistakes. I recommend this method: make x the subject then switch x and g(x) and relabel g(x):

g(x)=7-2x --> g(x)+2x = 7 --> x = (7-g(x))/2

Now switch x and g(x) and relabel g(x) as g^(-1)(x):

g^(-1)(x) = (7-x)/2