Extension One Revising Game (1 Viewer)

[tommykins]

-_- editted.

 

[kaz1]

heres a question

Solve:

5/[(2 - x)(x + 2)] >1

Is it -2 < x < 2?
cbf typing the working out.

 

[conics2008]

Is it -2 < x < 2?
cbf typing the working out.

yeap your correct xD

here ill post up some level 2 questions, people posting too many simple question.

Q]

Let P(x) be a polynomial give by

P(x)= ax^4+bx^3+cx^2+dx+e

Suppose the remainder is px+q when P(x) is divided by x^2-m^2

i) show that p= 1/2m *{P(m)-P(-m)} and q=1/2 *{ P(m)+p(-m) }

this was also a 4unit question relating to complex numbers, but its from a 3unit paper.

 

[vds700]

Is it -2 < x < 2?
cbf typing the working out.

yeah thats correct

 

[Mark576]

Let P(x) be a polynomial give by

P(x)= ax^4+bx^3+cx^2+dx+e

Suppose the remainder is px+q when P(x) is divided by x^2-m^2

i) show that p= 1/2m *{P(m)-P(-m)} and q=1/2 *{ P(m)+p(-m) }

this was also a 4unit question relating to complex numbers, but its from a 3unit paper.

By the polynomial division transformation:

P(x) = (x² - m²)Q(x) + px + q

P(m) = pm + q <---- (1)

P(-m) = -pm + q <---- (2)

(1) + (2):

P(m) + P(-m) = 2q => q = [P(m) + P(-m)]/2

(1) - (2):

P(m) - P(-m) = 2pm => p = 1/(2m)[P(m) - P(-m)]

Someone else can post up a new question.

 

[bored of sc]

There are three identical blue marbles and four identical yellow marbles arranged in a row.

How many different arrangements are possible if...
- There are no restrictions?
- Just five marbles are used?

 

[omniscience]

There are three identical blue marbles and four identical yellow marbles arranged in a row.

How many different arrangements are possible if...
- There are no restrictions?
- Just five marbles are used?

1. 7!/3!4! = 35

2. don't understand the question. Elaborate on it please. Wait, I think I understand what you intended in your question...bear with me for a while. I will post my solution.

7 marbles

1 B marbles + 4 Y marbles = 5!/1!4!
2 B marbles + 3 Y marbles = 5!/2!3!
3 B marbles + 2 Y marbles = 5!/3!2!

Add them up, so you get: 5 + 10 + 10 = 25

 

[bored of sc]

1. 7!/3!4! = 35

2. don't understand the question. Elaborate on it please.

How many different arrangements of just five of these (the seven) marbles are possible?

 

[tommykins]

回复: Re: Extension One Revising Game

You're correct omniscience.

 

[bored of sc]

Factorise a³ + b³.

Hence or otherwise, show that

(2sin³A + 2cos³A) / (sinA + cosA) = 2 - sin2A, if sinA + cosA < 0 and sinA + cosA > 0

 

[clintmyster]

Factorise a³ + b³.

Hence or otherwise, show that

(2sin³A + 2cos³A) / (sinA + cosA) = 2 - sin2A, if sinA + cosA < 0 and sinA + cosA > 0

a³ + b³ = (a+b)(a²-ab+b²)

therefore

[2(Sin A + Cos A)(Sin² A - Sin A Cos A + Cos² A)] / Sin A + Cos A
= 2(1 - Sin A Cos A)
= 2 - Sin2A
= RHS

 

[tommykins]

回复: Re: Extension One Revising Game

It's 2-sin2A, not sin^2 A.

 

[clintmyster]

Re: 回复: Re: Extension One Revising Game

It's 2-sin2A, not sin^2 A.

yeah i fixed it up..daym i had it right when i did it on paper.

 

[conics2008]

Re: 回复: Re: Extension One Revising Game

com'on guys some good question can u post up.

 

[midifile]

Re: 回复: Re: Extension One Revising Game

com'on guys some good question can u post up.

okay..

From the top of a cliff an observer spots two ships out at sea. One is on a true bearing of 042^o with an angle of depression of 6^o, while the other is on a true bearing of 312^o with an angle of depression of 4^o. If the two ships are 200m apart, find the height of the cliff to the nearest metre.

 

[midifile]

Re: 回复: Re: Extension One Revising Game

My working is everywhere :S Using scrap paper and I've got triangles everywhere. I got a diagram and broke it apart to simplify it xD

I got 12m for the height.

Yeah thats right. or 11.6m to 1 dp

 

[Trebla]

Here's one to think about. What is the flaw in the following:
-1 = (-1)³
= [(-1)⁶]^0.5
= 1^0.5
= 1
Hence 1 = -1

 

[Pwnage101]

ur taking the positive square root, when u could also take the negative square root????

like saying (-1)^2=1, therefore (-1) = 1^0.5 = 1

 

[bored of sc]

= [(-1)⁶]^0.5

[(-1)⁶]^0.5
= (root (-1))⁶
= does not exist in the real number system (i.e. root (-1))

 

[lolokay]

ur taking the positive square root, when u could also take the negative square root????

like saying (-1)^2=1, therefore (-1) = 1^0.5 = 1

yeah I'd say that's it. you could also take the negative square root.
The square root is defined to be positive, but the indice law you're using is then only true for numbers >0