Mathematics: Post your doubts here!

[MustafaMotani]

So I can call on you for help if I'm stuck in any question?

like i would say no ..

 

[zainuxx]

i would be really grateful if anyone uploads or gives me the link to download the 2001 p3 and p5 paper?

 

[Esme]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_33.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_ms_33.pdf

Q10 ii) b) Why do we have to take the square root of z that we've found in a) ? Since they're asking for z^2, shouldn't we square the values of z that we got in the previous bit ?

MustafaMotani ?
Or anyone else ?

 

[Shreeram]

please help me with s12 qp 51 q 3. thanks please explain into details im dying here

 

[50sum]

can someone tell me how to find the least/greatest value of arg z or mod z. I have no idea about what these are.

 

[Esme]

can someone tell me how to find the least/greatest value of arg z or mod z. I have no idea about what these are.

This might be helpful
https://www.xtremepapers.com/community/threads/complex-no-max-min-izi-and-arg-z.9862/

 

[Khalid Mohammed Okiely]

Hi, I want to ask about question 8 (full one, including part ii) in this paper.

How would I go about that? I realize I have to perform a double angle formula in part (i), but I just can't seem to get it right. Any help would be much appreciated!

 

[Esme]

Hi, I want to ask about question 8 (full one, including part ii) in this paper.

How would I go about that? I realize I have to perform a double angle formula in part (i), but I just can't seem to get it right. Any help would be much appreciated!

It was too tedious to type all the workings , so here is the one that I solved
And umm.. sorry, it's too light. You might have to zoom in a little.

 

[iFuz]

It was too tedious to type all the workings , so here is the one that I solved
And umm.. sorry, it's too light. You might have to zoom in a little.

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf
Question 8(ii)
Question 9(ii)
and Question 10.

Anyone?

 

[Esme]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_32.pdf
Question 8(ii)
Question 9(ii)
and Question 10.

Anyone?

I've totally reached my saturation point right now, I'll help you tomorrow first thing in the morning. Sorry

 

[Khalid Mohammed Okiely]

It was too tedious to type all the workings , so here is the one that I solved
And umm.. sorry, it's too light. You might have to zoom in a little.

Thank you! I actually think I will stop solving for now because what I thought I never knew how to solve was a VERY silly identity mistake from my side that messed up the whole question.

Again, thanks a lot!

P.S. You have a great handwriting...

 

[Esme]

Thank you! I actually think I will stop solving for now because what I thought I never knew how to solve was a VERY silly identity mistake from my side that messed up the whole question.

Again, thanks a lot!

P.S. You have a great handwriting...

Yeah.. same is with me right now. Been at it the entire day, can't take anymore! :/

Lol you can't be serious...everyone tells me my writing sucks !!

 

[iFuz]

I've totally reached my saturation point right now, I'll help you tomorrow first thing in the morning. Sorry

Sure Im waiting
and prior Thankyou

 

[MustafaMotani]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_33.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_ms_33.pdf

Q10 ii) b) Why do we have to take the square root of z that we've found in a) ? Since they're asking for z^2, shouldn't we square the values of z that we got in the previous bit ?

MustafaMotani ?
Or anyone else ?

aah yu got confuse in functions ....
p(z) = z^3 + 6z^2 +24z +32
p(1) = 1^3 + 6(1)^2 +24(1) +32 =63
similarly
p(z^2) = (z^2)^3 + 6(z^2)^2 + 24z^2 + 32 (just replaced z by z^2)
now if yu make substitution x=z^2 to solve the equation ... u shall get the same equation as in part ii)a
x^3 + 6x^2 +24x + 32

now the solutions of this equation (values of x) will be same as that of values of z in previous parts

now z^2 = x
so z = x^(1/2)
ans so we square root all the values of x to get z ....
Hope yu get it ..

 

[MustafaMotani]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_32.pdf
Question 8(ii)
Question 9(ii)
and Question 10.

Anyone?

in 8ii) yu just have to rearrange the equation so that it resembles your equation in question .. ( graph equation)
when you will be don rearranging... yu shall see that in place of "y" u will get 0.5 .. ( u have to try it to get it)

in 9ii) you got the roots in first part .. from those roots yu have to form an quadratic equation
Formula :
x^2 - (sum of roots)x + (product of roots)
once you got the equation just perform long division and yu shall get another quadratic which will give you another two roots..

 

[MustafaMotani]

I've totally reached my saturation point right now, I'll help you tomorrow first thing in the morning. Sorry

Q.10
Crossing OC and AB shall give perpendicular vector ...
AB = (-1,1,3)
AB x OC = (12,0,4)

Equation :
r = (x,y,z)
r.n=a.n
12x + 4z = 52
3x + z = 13

ii) its quite difficult to explain this part on forum ... there are several ways you can solve this question

First we have to finD CF

make an eqution for line through AB
it will be
r = (3,-2,4)+k(-1,1,3)
since OF lies on the line
OF = (3-k, -2+k, 4+3k)

CF = OF - OC
CF = (2-k, 3+k, 7+3k)

Since CF is perpendicular to AB.. dot product of CF and direction vector of AB should be zero

(2-k , 3+k, 7+3k) . (-1, 1, 3) = 0

k = -2

CF = (4, 1, 1)
simply find lenght
(4^2+1^2+1^2)^1/2

I know i have missed some steps ... but u might get it if yu go through it thoroughly..

hope it helps

 

[Hirdayesh.B.Shrestha]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_31.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_ms_31.pdf
In question number 3ii)b), I didn't understand what the marking scheme says. I got the roots +-squareroot(2) and +-2i. Is there supposed to be six roots or just the ones that I mentioned?

 

[SararaIH]

Can someone please help me with #4 of http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_33.pdf ?

 

[Hirdayesh.B.Shrestha]

Can someone please help me with #4 of http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_33.pdf ?

The answer in the marking scheme is INCORRECT for this question. The examiner report shows the correct answer which is {(x^2+4)^3}/2
I'm too lazy to type out the solution so here you go.
I used limits to for integration. You could also integrate and find a constant instead of using limits, which I find easier.

 

[MustafaMotani]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_31.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_ms_31.pdf
In question number 3ii)b), I didn't understand what the marking scheme says. I got the roots +-squareroot(2) and +-2i. Is there supposed to be six roots or just the ones that I mentioned?

u just have to squre root all three ansers in ii)a ... i did similar question just few posts above...