Maths Game! (1 Viewer)

[Trebla]

(c)Let x= [maths]\frac{pi}{\22}[/maths].
Then from (b), as n --> ∞,
(sin x) / x = cos (x/2).cos (x/4).cos (x/8)................
=[maths]\frac{\sqrt{2}}{2} . \frac{\sqrt{2 + \sqrt{2}}}{2} . \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} .......[/maths]
using exact values of trigonometric expressions.

Explain how you got each of the values: [maths]\frac{\sqrt{2}}{2} . \frac{\sqrt{2 + \sqrt{2}}}{2} . \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} .......[/maths]

 

[Trebla]

(1+ax)^n is given by 1-24x+252x^2-....

Find the values for a and n.

[maths](1 + ax)^{n} = \sum_{k=0}^{n}{^{n}C_{k}\:a^{k}x^{k}}\\$When k = 1, we have the term\\$^{n}C_{1}ax = - 24x\\\Rightarrow an = - 24\\\Rightarrow n = -\frac{24}{a}\\$When k = 2, we have the term$\\^{n}C_{2}\,a^{2}x^{2}}} = 252x^{2}\\\Rightarrow a^{2} \frac{n!}{2!(n-2)!} = 252\\\Rightarrow a^{2}n(n-1) = 504\\\Rightarrow -a^{2}\frac{24}{a}(-\frac{24}{a}-1) = 504\\\Rightarrow -24a(-\frac{24}{a}-1) = 504\\\Rightarrow 576 + 24a = 504\\\Rightarrow a = - 3 \\$But$\\n = -\frac{24}{a}\\\Rightarrow n = 8[/maths]

 

[jet]

Just curious - Why are you doing 17 units for your HSC?

Why not???

 

[shaon0]

or you could just realise its a semi circle radius a
therefore S [limits:a to -a] sqrt(a^2-x^2) dx
= pi.a^2/2

Yeah i know that. But do it using integration.

 

[study-freak]

Explain how you got each of the values: [maths]\frac{\sqrt{2}}{2} . \frac{\sqrt{2 + \sqrt{2}}}{2} . \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} .......[/maths]

Well I left it out coz I thought it was straight forward..
well sinx/x=cos(pi/4)cos(pi/8)cos(pi/16)...
=[maths]\frac{\sqrt{2}}{2} . \frac{\sqrt{2 + \sqrt{2}}}{2} . \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} .......[/maths]

since cos(pi/4)=1/sqrt(2)=[maths]\frac{\sqrt{2}}{2}[/maths]
and cos(pi/4)=2cos^2(pi/8)-1= [maths]\frac{\sqrt{2}}{2}[/maths] [double angle formula]
which when rearranged gives: cos(pi/8)=[maths]\frac{\sqrt{2 + \sqrt{2}}}{2}[/maths] [since cos(pi/8)>0]
and etc

 

[bored.of.u]

Did you teach yourself this stuff? Or dd you have a Private Tutor?

im not an accelerant i have just finished ext1 course from doing the maths in focus n cambridge books i want to be ranked first till yr 12 for maths =D...n no better to learn by yourself (at least thats wat i think)

 

[study-freak]

im not an accelerant i have just finished ext1 course from doing the maths in focus n cambridge books i want to be ranked first till yr 12 for maths =D...n no better to learn by yourself (at least thats wat i think)

lol go on and finish 4unit once before you go into yr12 like i did

 

[bored.of.u]

lol go on and finish 4unit once before you go into yr12 like i did

hehe yep i plan too =D i had a look at 4u n really different to 3u i was really surprised =O...btw sumone post a question to keep this thread going thnx =D

 

[study-freak]

hehe yep i plan too =D i had a look at 4u n really different to 3u i was really surprised =O...btw sumone post a question to keep this thread going thnx =D

Yea I'd say in terms of difficulty, 4u=3u*20

anyway, can't think of a question..sorry

 

[kurt.physics]

lol go on and finish 4unit once before you go into yr12 like i did

Hey, sorry to butt in and such, but im kind of doing the same thing, i have done all of prelim adv and ext (learnt by myself), should i start learning 4 unit when i start doing my prelim adv and ext 1 classes this year? And should i start year 12 aswell?

What are the main advantages with starting extension 2? My school hasnt got the greatest amount of tallent, and so i would probably be the only one in my ext 2 class when i start it. Should i organise with the teacher to start it early and then just continue when i "officially" start class? What are the advantages?

 

[kurt.physics]

Question: Circle geometry

The side AB of a triangle ABC is a diameter of a circle of radius R, and C lies on this circle. The angle bisector of angle BAC meets BC at E and then the circle at D. When produced, AC meets the circumcircle of triangle CED at F. If BC = a, express CF in terms of R and a.

 

[bored.of.u]

Hey, sorry to butt in and such, but im kind of doing the same thing, i have done all of prelim adv and ext (learnt by myself), should i start learning 4 unit when i start doing my prelim adv and ext 1 classes this year? And should i start year 12 aswell?

What are the main advantages with starting extension 2? My school hasnt got the greatest amount of tallent, and so i would probably be the only one in my ext 2 class when i start it. Should i organise with the teacher to start it early and then just continue when i "officially" start class? What are the advantages?

your going into yr11 next year correct?? im prob not the best person...but i would not suggest you begin ext2 together with ext1...you could get an idea of the work if you have the cambridge 4u book...currently im doing the maths in focus hsc course (maths in focus is good for a foundation but otherwise a joke) alongside cambridge 3u. good luck kurt.physics =D

too keep this thread going:

Find all values of x:

tan^2 (x) + 1/cot (x) - 20 = 0 for 0<x<360

 

[study-freak]

Hey, sorry to butt in and such, but im kind of doing the same thing, i have done all of prelim adv and ext (learnt by myself), should i start learning 4 unit when i start doing my prelim adv and ext 1 classes this year? And should i start year 12 aswell?

What are the main advantages with starting extension 2? My school hasnt got the greatest amount of tallent, and so i would probably be the only one in my ext 2 class when i start it. Should i organise with the teacher to start it early and then just continue when i "officially" start class? What are the advantages?

You should definitely do 4u (yet after you finish all others in 2u and 3u) once either by yourself or with a tutor before going into yr12 (i personally like doing it on my own).

Well, the advantages are: you get more time and experience and 4unit is much harder compared to 3unit so it gets you prepared to face harder problems in question 8's of trial or the HSC.

Also, yr11 marks do not go towards anything and so I recommend you to keep a certain satisfactory (that is, good enough in your opinion) standard in other subjects than maths in yr11 and concentrating on learning ahead in maths in preparation for yr12. Alternatively, you can accelerate (yourself) in other subjects yet I reckon doing so in maths is the most valuable when you later face increased workload of yr12.

And btw, the fact (or at least an anticipation) that you'll be the only one doing ext2 is a good thing on one hand. This means you are definitely the first of the year in the subject at your school, which then gives you a chance to get a higher mark than what you would have gained when you were 2nd or 3rd.

 

[random-1005]

Question: Circle geometry

The side AB of a triangle ABC is a diameter of a circle of radius R, and C lies on this circle. The angle bisector of angle BAC meets BC at E and then the circle at D. When produced, AC meets the circumcircle of triangle CED at F. If BC = a, express CF in terms of R and a.

" circumcircle", whats that?, i drew up a diagram and i dnt no what your on about

 

[gurmies]

too keep this thread going:

Find all values of x:

tan^2 (x) + 1/cot (x) - 20 = 0 for 0<x<360

tan^2 (x) + tan (x) - 20 = 0

[tan (x) - 4][tan (x) + 5] = 0

tan (x) = 4 or/ tan (x) = -5

x = 75' 58", 255' 58", 101' 19", 258' 41"

Solve 3x² - 2x - 2 ≤ |3x|

 

[bored.of.u]

Solve 3x² - 2x - 2 ≤ |3x|

Hmmm...so i'll do this is two parts coz it would be weird doing them next to each other (postive values above the negative values):

positive values of 3x:
3x^2 - 2x - 2 < 3x
3x^2 - 5x - 2 < 0
(3x+1)(x-2) < 0
.'. -1/3 < x < 2
OR
negative values of 3x:
3x^2 - 2x - 2 < -3x
3x^2 + x - 2 < 0
(3x-2)(x+1) < 0
.'. -1 < x < 7/3

Solve for n where,

2^(4n+1) - 10[2^(2n)] + 8 = 0

 

[gurmies]

Hmmm...so i'll do this is two parts coz it would be weird doing them next to each other (postive values above the negative values):

positive values of 3x:
3x^2 - 2x - 2 < 3x
3x^2 - 5x - 2 < 0
(3x+1)(x-2) < 0
.'. -1/3 < x < 2
OR
negative values of 3x:
3x^2 - 2x - 2 < -3x
3x^2 + x - 2 < 0
(3x-2)(x+1) < 0
.'. -1 < x < 7/3

Solve for n where,

2^(4n+1) - 10[2^(2n)] + 8 = 0

Correct, however it's safer to combine the two inequalities and say that -1 < x < 2

For your question:

2[2^(4n)] - 10[2^(2n)] + 8 = 0

(2^n)^4 - 5(2^n)^2 + 4 = 0

Let 2^n = u

u^4 - 5u^2 + 4 = 0

(u^2 - 1)(u^2 - 4) = 0

u = ±1, ±2

So, 2^n = ±1, ±2

BUT, since k^n will always be positive

2^n = 1, 2

n = 0, 1

Find all real x such that

|2x - 1/2| > √(x - x^2)

 

[kurt.physics]

" circumcircle", whats that?, i drew up a diagram and i dnt no what your on about

The circumcircle is the circle that touches all 3 points of the triangle. wikipedia (Circumscribed circle - Wikipedia, the free encyclopedia) has more info.

 

[bored.of.u]

Correct, however it's safer to combine the two inequalities and say that -1 < x < 2

For your question:

2[2^(4n)] - 10[2^(2n)] + 8 = 0

(2^n)^4 - 5(2^n)^2 + 4 = 0

Let 2^n = u

u^4 - 5u^2 + 4 = 0

(u^2 - 1)(u^2 - 4) = 0

u = ±1, ±2

So, 2^n = ±1, ±2

BUT, since k^n will always be positive

2^n = 1, 2

n = 0, 1

Find all real x such that

|2x - 1/2| > √(x - x^2)

how and why can u add the inequalities??

to your question:

because of the square root both the positive and negative values r equal

2x - 1/2 > √x - x^2
4x^2 - 2x - 1/4 > x - x^2
5x^2 - 3x - 1/4 > 0
20x^2 - 12x - 1 > 0
(20x - 10)(20x - 2) > 0
(2x - 1)(10x - 1) > 0
.'. x > 1/2 , x<1/10

Is this correct?? cbb posting up a question sum1 pls do it for me =D

 

[gurmies]

Not quite, i'll leave it up to you to find your error