Trial Revision Thread (1 Viewer)

[XcarvengerX]

I have a couple of questions:

1. Solve cos (x-3) + sin (3-x) = 1 for 2pi ≥ x ≥ 0
2. tan b = (gy) / [a (g + a) + y²]
Find db/dy and hence write an expression for y which would maximise b given that g and a are constants.

This is where I stuck.
1. So I substitute A = 3 – x, and as cos A = cos (-A), then cos A + sin A = 1
Then how to solve it from there? Substitute back, then??? I got confuse because it is solve, not prove…
2. I try to differentiate tan b implicitly because that is the only way I can think of, but mess up so badly.

I realise I am wasting time trying to type questions online. The next set of questions (that is if I stuck again) will be in pdf form - handwritten. If you have scanner, please write your answer in paper and scan it because I am so bad at reading Math things on the forum. Or use paint (but must be readable).

Thank you in advance.

 

[vafa]

(1) cos(x-3)+sin(3-x)=1

let 3-x=Alpha
cos(-alpha)=cos(alpha)
e.g. cos(-45)=cos(45) and etc

cos(alpha)+sin(alpha)=1 where alpha=3-x

and then use subsidary angles

(2) sec^2b*db/dy=(g(a(g+a)+y^2)-2gy^2)/(a(g+a)+y^2)^2

db/dy=(ag(g+a)-gy^2)/(a(g+a)+y^2)^2

then the rest is so obvious.

 

[vafa]

i do not have catholic2004. if i get that, then i will answer you, riviet.

 

[Riviet]

And here's lots and lots.

http://www.geocities.com/ext2papers

http://www.geocities.com/fourunitmaths

http://www.angelfire.com/ab7/fourunit

The first one has all catholic trials down to the 1980's.

 

[Rax]

Hey Riviet, how do you do 6)b of the 2004 CSSA

I can't remember the methods for these things.

I will have a dig at 6)a) now

 

[YBK]

arggg!
2002 is REALLY annoying me.

 

[Riviet]

Hey Riviet, how do you do 6)b of the 2004 CSSA

I can't remember the methods for these things.

I will have a dig at 6)a) now

Let the roots be a, b and c.
a + b + c = -b/a
=0 (since there the coefficient of x² is zero!)
required roots are:
(a+b)/c², (b+c)/a², (a+c)/b²
= (sum of roots - c)/c², (sum of roots - a)/a², (sum of roots - b)/b²
=-1/c, -1/a, -1/b
=-1/x

Substitute -1/x into original polynomial and simplify into the polynomial form.

 

[YBK]

Hey Riviet, how do you do 6)b of the 2004 CSSA

I can't remember the methods for these things.

I will have a dig at 6)a) now

lol i answered that above

 

[Riviet]

arggg!
2002 is REALLY annoying me.

Already doing '02? =O

I still have to finish off '03 tonight, none of these past exams have been close to moderate in difficulty!

 

[XcarvengerX]

(1) cos(x-3)+sin(3-x)=1

let 3-x=Alpha
cos(-alpha)=cos(alpha)
e.g. cos(-45)=cos(45) and etc

cos(alpha)+sin(alpha)=1 where alpha=3-x

and then use subsidary angles

What are subsidary angles?

 

[YBK]

What are subsidary angles?

auxillary method...

 

[YBK]

Already doing '02? =O

I still have to finish off '03 tonight, none of these past exams have been close to moderate in difficulty!

oh haven't done 03 yet... lol

 

[bboyelement]

solution to cssa 2004 question 6a part (ii)

 

[XcarvengerX]

And what is auxiliary method?

db/dy=(ag(g+a)-gy^2)/(a(g+a)+y^2)^2

then the rest is so obvious.

Where do you put the sec² b? I still can't see it...

 

[Riviet]

Cool, thanks for that bboyelement, the last bit was a bit confusing when you had 1+cos²x become sin²x but it was meant to be a 1-cos²x.

 

[YBK]

And what is auxiliary method?



Where do you put the sec² b? I still can't see it...

Hmmm.. do you have cambridge yr 12 3u maths book?

If you do, it's in there. It's a 3u thing...

 

[XcarvengerX]

Hmmm.. do you have cambridge yr 12 3u maths book?

If you do, it's in there. It's a 3u thing...

No,

and I think I suppose to know it already, but maybe stress has got into me...
can you please explain?

 

[vafa]

I just started to do question 7 a first because i thought it is more interesting and it i guess that there should be an easier way than mine however i attach my solution. i was delighted to answer all of these questions but believe me i have trial exams but i try to answer as much as i can.
good luck

my solution is in 3 pages i attach 2 of them by this post and the third one by next post.View attachment 13469

View attachment 13470

 

[Riviet]

Going off to finish '03 and a few last 04/05 questions.

 

[vafa]

this is the third pageView attachment 13471