Mathematics: Post your doubts here!

[Fiery987]

In second part i dont get it from 3rd step onwards. After squaring on both sides why is (z-2i)^2 equal to (z-2i)(z-2i)*. I dont get the conjuguate part (*)

DON'T read the first part where I crossed it. The answer starts from |z-2i| = 4

Well, I used the relationship given in part (i). As they made us prove it, there must be some use of it, right? But I did it this way (as I proved in reverse) :

because |z|^2 = zz* {Part (i) question}
then |z-2i|^2 = (z-2i)(z-2i)* {z is a complex no. so z-2i must also be a complex no. because add/subt of a complx no. = another complx no.}

 

[Fiery987]

Um, I just posted the Part (ii) explanation. Refresh your browser.

 

[farhan141]

Um, I just posted the Part (ii) explanation. Refresh your browser.

Ohh got it! Thanks man .

How's ur prep overall? How is S1 going? What abt Physics and Chem?

 

[areeba240]

why do we put this negative sign????
thanks in advance

 

[i_try9621]

View attachment 60224
why do we put this negative sign????
thanks in advance

what negative sign?

 

[areeba240]

what negative sign?

 

[i_try9621]

oh okay.
So the question says "reflex" angle which means it's greater than 180 degree and less than 360 degree, which falls in 3rd and 4th quadrant where sin(theta) is negative. So you use a negative sign to indicate that.

 

[qwertypoiu]

qwertypoiu

Waiting for ur magic

Hey idk why I didn't receive this tag at all!
Please quote questions if there are still any left. I think others have helped you out?

 

[qwertypoiu]

doubt in 7) iv) .
It's from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_41.pdf

Take a look at this:
https://www.desmos.com/calculator/my8jguvy5q

 

[farhan141]

Ple

Hey idk why I didn't receive this tag at all!
Please quote questions if there are still any left. I think others have helped you out?

please do the first complex no. question, the vector one and the 2 integration ones

 

[qwertypoiu]

Ple

please do the first complex no. question, the vector one and the 2 integration ones

Quote them plz

 

[farhan141]

View attachment 60174 View attachment 60172 View attachment 60173

View attachment 60178 View attachment 60179

 

[farhan141]

Quote them plz

First question (complex), vector part b and both integration ones

 

[qwertypoiu]

The complex one.
Note that for the diagram question, the origin of my circle should be at (0,2) but it looks like it's slightly higher. I only realised that later so yeah just imagine it like that

Spoiler

farhan141

 

[Fiery987]

The complex one.
Note that for the diagram question, the origin of my circle should be at (0,2) but it looks like it's slightly higher. I only realised that later so yeah just imagine it like that

Spoiler

View attachment 60246

farhan141

Um, why did you eliminate x if x>0? x can't be 1 only.

 

[qwertypoiu]

The vector one

Spoiler

 

[qwertypoiu]

Um, why did you eliminate x if x>0? x can't be 1 only.

No I divided by x on both sides.
I justified this by reminding that x>0 so it cannot be equal to zero.
If you divide zero on both sides you can run into all sorts of trouble. But as long as x is not zero you can divide on both sides like that.
Or you can think of it as just factoring out x. Then the second bracket can be equated to zero as well.

 

[Fiery987]

No I divided by x on both sides.
I justified this by reminding that x>0 so it cannot be equal to zero.
If you divide zero on both sides you can run into all sorts of trouble. But as long as x is not zero you can divide on both sides like that.
Or you can think of it as just factoring out x. Then the second bracket can be equated to zero as well.

Ohhh, right. Thanks a lot!

 

[qwertypoiu]

One of the integration questions...

Spoiler

 

[farhan141]

No I divided by x on both sides.
I justified this by reminding that x>0 so it cannot be equal to zero.
If you divide zero on both sides you can run into all sorts of trouble. But as long as x is not zero you can divide on both sides like that.
Or you can think of it as just factoring out x. Then the second bracket can be equated to zero as well.

Whole 5 and 6 question.

Also please do both integration ones from the previous ones i uploaded. Thanks for the others, will look into it in a while then will let u know