Mathematics: Post your doubts here!

[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...

 

[asexamskillme111]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_31.pdf
Can someone please explain why the circle has to pass through the origin in 4 ii? I understand that there's a circle. What I did is I made the radius of the circle equal to the modulus of u since it made sense to me. Help please?

 

[WayneRooney10]

Why does (cosx)^4=5/8 have 4 answers? Can someone explain the logic behind it? Thanks!

 

[kiara15]

hey plz give me link for how to draw argand diagrams, i m totally messed up with this

 

[kiara15]

can anyone solve permut n comb question from 2011 p.paper, i had an ambiguity solving it.. paper 3

 

[kiara15]

c an u solve 2-3 quest of perm n comb from p.papr? i had ambiguity or anyone?

 

[Esme]

hey plz give me link for how to draw argand diagrams, i m totally messed up with this

These notes were uploaded by Phyzac

 

[kiara15]

These notes were uploaded by Phyzac

thank u

 

[Esme]

Why does (cosx)^4=5/8 have 4 answers? Can someone explain the logic behind it? Thanks!

That's because you'll write it as cos x = +- (5/8)^1/4
just like for square root, you write +-(square root)...
So you'll have x= cos^-1 ( positive (5/8)^1/4 ) and x=cos^-1 (negative (5/8)^1/4)
This should give you 4 different values of x.

 

[Esme]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_31.pdf
Can someone please explain why the circle has to pass through the origin in 4 ii? I understand that there's a circle. What I did is I made the radius of the circle equal to the modulus of u since it made sense to me. Help please?

Yeah you're right. The modulus of u is the radius of the circle. Automatically the circle will pass through the origin because the modulus is the distance from the origin to the vector i.e. your radius of cirlce

 

[kiara15]

Which questions do you need help with?

p63- q 3 and p62 - q5

 

[biba]

The complex number u is defined by u = (1 + 2i)^2 / (2+i)

(i) Without using a calculator and showing your working, express u in the form x + iy, where x and

y are real. [4]

(ii) Sketch an Argand diagram showing the locus of the complex number s such that |s − u| = |u|.
[3]


can sum1 help me with the argand diagram ?

anyone??? [^__^]

 

[SararaIH]

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.

No wonder I couldn't understand the Mark Scheme -_-
Thanks so much

I'm confused with one thing
When can I use this formula---> (1/2a)*ln(a+b/a-b) ?

 

[Esme]

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 ..

Yeah I got it thanks

 

[talalz]

does any body have MATHS P3 VECTORS notes ?? please i need help urgent ??

 

[SararaIH]

does any body have MATHS P3 VECTORS notes ?? please i need help urgent ??

This might be really useful
Vectors Study Guide