Need Answers from Random Textbook (1 Viewer)

[Dreamerish*~]

Does anyone know what the textbook is called? We don't use it at school so I only remember vaguely what the cover looks like.

I think the 2U book is blue. it has vertically symmetrical, dark-and-light blue diagonal stripes, and the "2U Mathematics" thing is written in the top half, in the middle.

If anyone happens to have it, can they please save me a trip to school and scan the answers to the uh... 300 review questions at the back?

 

[insert-username]

"Blue cover with diagonal lines" - that sounds like Maths in Focus, by Margaret Grove. I've got the 3 unit books 1 and 2, which are basically the equivalent 2 unit books except with more stuff. The only thing is, they don't have a big review section at the back of the textbook - just revision exercises scattered through the chapters and Challenge exercises after each chapter. Is that the one? I've also got the Fitzpatrick New Senior Mathematics 2 unit text (which is yellow), but that has a longish "Test Paper" section at the end with 5 separate text papers. Is that what you're looking for?


I_F

 

[Dreamerish*~]

"Blue cover with diagonal lines" - that sounds like Maths in Focus, by Margaret Grove. I've got the 3 unit books 1 and 2, which are basically the equivalent 2 unit books except with more stuff. The only thing is, they don't have a big review section at the back of the textbook - just revision exercises scattered through the chapters and Challenge exercises after each chapter. Is that the one? I've also got the Fitzpatrick New Senior Mathematics 2 unit text (which is yellow), but that has a longish "Test Paper" section at the end with 5 separate text papers. Is that what you're looking for?


I_F

I think it could be Maths in Focus, but it does have 300 review questions because I did them! *Cries* I've been waiting to mark them but forgot to ask my teacher for the book before I left school.

 

[insert-username]

Can you tell me what the Review questions look like, i.e. the title on each page, the format, etc. Then I can confirm it and scan the right one.


I_F

 

[Dreamerish*~]

Can you tell me what the Review questions look like, i.e. the title on each page, the format, etc. Then I can confirm it and scan the right one.


I_F

Umm, two columns of questions, simple font, probably Times. No graphs in any of the questions, they're quite short at the beginning and get longer towards the end with more sections. The title "Revision Exercises" is not very big, and sits above the first page of questions.

Thank you so much.

 

[piyo_extreme]

help me

can sum1 answer my question pleasee

A chord PQ of a parabola x^2=4ay passes through the focus S(0,a). The co-ordinates of P are (2ap,ap^2) and the gradient of the line PQ is c

given from the 1st part that p-1/p = 2c

Show that the midpoint M of the chord PQ has the coordinates (2ac, a(1+2c^2))

thanx guyz

 

[brett86]

 

[piyo_extreme]

thank you!! hmm sori can sum1 answer another question for me thank you so much!

prove by induction that, for all integral n, n ≥ 0


cosθ.cos2θ.cos4θ......cos2^n θ = sin (2^n+1 θ)/2^n+1 sinθ

 

[insert-username]

For induction:

1. Prove true for lowest value of n (0)

cos2^0 θ = cos θ

sin (2^0+1 θ)/2^0+1 sinθ

= sin (2θ)/2 sinθ

= 2 sinθcosθ/2sinθ

= cos θ

= LHS


2. Assume true for some value n = k

3. Prove true for n = k + 1 (this section gets a bit yukky, anyone else want to have a go?)

4. State that since it is true for 0 and for k + 1, then it must be true for 1, 2, etc.


Dreamerish, I've checked the textbooks I've got, and none of them have a review exercise section in that format. One has Revision Exercises at the end of each chapter in a single column and Test Papers, the other just Test Papers. Sorry I couldn't help you beyond that.


I_F

 

[Dreamerish*~]

For induction:

1. Prove true for lowest value of n (0)

cos2^0 θ = cos θ

sin (2^0+1 θ)/2^0+1 sinθ

= sin (2θ)/2 sinθ

= 2 sinθcosθ/2sinθ

= cos θ

= LHS


2. Assume true for some value n = k

3. Prove true for n = k + 1 (this section gets a bit yukky, anyone else want to have a go?)

4. State that since it is true for 0 and for k + 1, then it must be true for 1, 2, etc.


Dreamerish, I've checked the textbooks I've got, and none of them have a review exercise section in that format. One has Revision Exercises at the end of each chapter in a single column and Test Papers, the other just Test Papers. Sorry I couldn't help you beyond that.


I_F

It's ok, I really appreciate you trying to find it. Thanks.

Oh and I'm curious about that question too, can you actually do it? I got up to here (see attachment) and from there on it kind of went like nuclear fission.

 

[insert-username]

Dreamerish: I got up to about there too and gave up. I tried rearranging the exponents, but that didn't help. Is there some trig rule you can use or is it just stupid indexes?


I_F

 

[Dreamerish*~]

Dreamerish: I got up to about there too and gave up. I tried rearranging the exponents, but that didn't help. Is there some trig rule you can use or is it just stupid indexes?


I_F

It's a mixture of indexes, trig, algebra and induction. What a nighmare!

I was actually quite determined. I wrote out three pages of crap and then crawled back to chemistry.

 

[fantasy27]

Dreamerish, is this the book you're after?

http://img.photobucket.com/albums/v429/X_wildberry_X/Photo0076.jpg

the only thing different to you description is that it is 3u not 2u, so i'm not sure, but its got the 300 review questions!

 

[Dreamerish*~]

Dreamerish, is this the book you're after?

http://img.photobucket.com/albums/v429/X_wildberry_X/Photo0076.jpg

the only thing different to you description is that it is 3u not 2u, so i'm not sure, but its got the 300 review questions!

Yes! That's it!

 

[word.]

That induction question you guys are trying to prove does not work with n > 0.

This is why brackets are essential:
cosθ.cos2θ.cos4θ......cos2^n θ = sin (2^(n+1)θ)/2^(n+1)sinθ

try again:

cosθ * cos2θ * cos4θ * ... * cos2ⁿθ = ^sin(2n+1θ)/_{2ⁿ⁺¹sinθ}

 

[insert-username]

2 to the power of 0 = 1.

cos^1 θ = cos θ


Using your brackets:

sin (2^(n+1)θ)/2^(n+1)sinθ

= sin (2^1.θ) / 2^1.sin θ

= sin 2θ / 2 sin θ

= 2 sinθcosθ/ 2 sin θ

= cos θ


I don't see any problem. Where is it?


I_F

 

[Dreamerish*~]

Okay, people, how about next time when we post complicated questions, we scan a handwritten one? :uhhuh:

 

[Trev]

Dreamerish, you're writing is so neat... =\

 

[word.]

2 to the power of 0 = 1.

cos^1 θ = cos θ


Using your brackets:

sin (2^(n+1)θ)/2^(n+1)sinθ

= sin (2^1.θ) / 2^1.sin θ

= sin 2θ / 2 sin θ

= 2 sinθcosθ/ 2 sin θ

= cos θ


I don't see any problem. Where is it?


I_F

I don't understand what you're getting at...
It was a coincidence that the way you interpreted the question worked with n = 0

However it doesn't work with n > 0 so you're trying to prove an incorrect statement.

cosθ * cos2θ * cos4θ......cos(2ⁿθ) = sin[(2ⁿ+1)θ]/(2ⁿ+1)sinθ

when n = 0
LHS = sin[(1+1)θ]/(1+1)sinθ
= sin2θ/2sinθ
= cosθ = RHS

when n = 1
LHS = cosθ * cos2θ
RHS = sin3θ/3sinθ

you can sub values for θ and realise that LHS <> RHS

With the brackets:
cosθ * cos2θ * cos4θ * ... * cos2ⁿθ = sin(2ⁿ⁺¹θ)/2ⁿ⁺¹sinθ

when n = 0

LHS = cosθ
RHS = sin2θ/2sinθ
= cosθ

when n = 1

LHS = cosθ * cos2θ
RHS = sin4θ/4sinθ

you can sub values for θ and realise that LHS = RHS

hence the question contains brackets around n + 1

 

[香港!]

^ I agree with "word."!!
hehe
Nice hand writing Dreamerish