HSC 2013 MX2 Marathon (archive) (1 Viewer)

[deadpatch]

Re: HSC 2013 4U Marathon

integral of xlogx

let u = logx then use base conversion
let dv=xdx

Use IBP

 

[asianese]

Re: HSC 2013 4U Marathon

let u = logx then use base conversion
let dv=xdx

Use IBP

Why would you need to base convert? It is understood that logx means log to the base e of ex = lnx.

 

[Sy123]

Re: HSC 2013 4U Marathon

The polynomial with roots:





So we have a polynomial with roots +alpha and a polynomial with roots -alpha

So how to make a polynomial that is a combination of these roots, we have to multiply them, we can observe this by simply writing each polynomial as (x-a_1)(x-a_2)...(x-a_2n)

Then if we multiply them and we get a polynomial with both roots!

So the polynomial T(y) has roots

 

[Sy123]

Re: HSC 2013 4U Marathon

 

[HeroicPandas]

Re: HSC 2013 4U Marathon

The polynomial with roots:





So we have a polynomial with roots +alpha and a polynomial with roots -alpha

So how to make a polynomial that is a combination of these roots, we have to multiply them, we can observe this by simply writing each polynomial as (x-a_1)(x-a_2)...(x-a_2n)

Then if we multiply them and we get a polynomial with both roots!

So the polynomial T(y) has roots

How do i combine this product?

 

[darkfenrir]

Re: HSC 2013 4U Marathon

The polynomial with roots:





So we have a polynomial with roots +alpha and a polynomial with roots -alpha

So how to make a polynomial that is a combination of these roots, we have to multiply them, we can observe this by simply writing each polynomial as (x-a_1)(x-a_2)...(x-a_2n)

Then if we multiply them and we get a polynomial with both roots!

So the polynomial T(y) has roots

I see what you did there.....

 

[HeroicPandas]

Re: HSC 2013 4U Marathon

x = arccost ..[1]
y = arcsint ..[2]

[1] + [2]
x + y = pi/2
y = pi/2 - x

x = arccost, with 0 < x < pi (x=pi)



THerefore 0 < x < \pi/2 , |x| = |arccost|

y= pi/2 -|x|

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

Let

Find a recurrence formula for

 

[Sy123]

Re: HSC 2013 4U Marathon

x = arccost ..[1]
y = arcsint ..[2]

[1] + [2]
x + y = pi/2
y = pi/2 - x

x = arccost, with 0 < x < pi (x=pi)



THerefore 0 < x < \pi/2 , |x| = |arccost|

y= pi/2 -|x|

I made a mistake in my solution (by interchanging doing something with (sin^-1, cos^-1) in the middle of it hence muddling my solution up), it is indeed supposed to be:



So your first part is right. And my restriction on x is wrong (because for a while I was treating it as sin^-1)

Sorry for the hassle.

 

[bootybitches]

Re: HSC 2013 4U Marathon



Realisenothing i think this is your answer


Sandeep - out

 

[asianese]

Re: HSC 2013 4U Marathon

Let

Find a recurrence formula for

Pretty sure its wrong but worth a stab:

 

[sandeeptheG]

Re: HSC 2013 4U Marathon

A good induction problem for the ones out there close to my level


The sequence of numbers x0, x1, x2, . . . begins with x0 = 1 and x1 = 1 and is then recursively determined by the
equations
xn+2 = 4xn+1 − 3xn + 3
n
for n ≥ 0.
(a) Find the values of x2, x3, x4 and x5.
(b) Can you find a solution of the form
xn = A + B × 3
n
+ C × n3
n
which agrees with the values of x0, . . . , x5 that you have found?
(c) Use induction to prove that this is the correct formula for xn for all n ≥ 0.

 

[sandeeptheG]

Re: HSC 2013 4U Marathon

Oh damn the subscripts aren't working... well hey see if you can get it anyway

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

A good induction problem for the ones out there close to my level


The sequence of numbers x0, x1, x2, . . . begins with x0 = 1 and x1 = 1 and is then recursively determined by the
equations
xn+2 = 4xn+1 − 3xn + 3
n
for n ≥ 0.
(a) Find the values of x2, x3, x4 and x5.
(b) Can you find a solution of the form
xn = A + B × 3
n
+ C × n3
n
which agrees with the values of x0, . . . , x5 that you have found?
(c) Use induction to prove that this is the correct formula for xn for all n ≥ 0.

That's a very standard induction question that appears in 4U a lot. If only the subscripts were non-ambiguous though.....

 

[sandeeptheG]

Re: HSC 2013 4U Marathon

If only the subscripts were non-ambiguous though.....

Here's the link, sorry bout the copy horse person.
http://www.maths.ox.ac.uk/files/imp...graduate/practice-problems/pdf/induction2.pdf

 

[Sy123]

Re: HSC 2013 4U Marathon

Here's the link, sorry bout the copy horse person.
http://www.maths.ox.ac.uk/files/imp...graduate/practice-problems/pdf/induction2.pdf

Interesting question:

By simply using the fact that



We can make 3 equations in A B and C, we solve for all of them and we arrive at the conjecture:



However when trying to prove by induction this result I found it really frustrating to use it in that form, so I did this:



After this to prove by induction, establish the 2 base cases of n=0 and n=1
Then assume the results for n=k and n=k+1

Then use the expressions that you gain from the assumption:





Then construct using the above equations:



By the recurrence relation this turns out to be:



So we just sub it in, do some manipulation, algebra and we arrive at the answer that Step 3: n=k+2 yields

Hence proved by mathematical induction.

 

[harryharper]

Re: HSC 2013 4U Marathon

a) Find

b) Hence find

c) Finally, for those top notch students, prove that, if

then, for all positive integers n,

 

[psychotropic]

Re: HSC 2013 4U Marathon

a classic Andrei Kiselyov problem. very good mr harper, very good

 

[Lieutenant_21]

Re: HSC 2013 4U Marathon

Very popular conics question (don't tell which paper it is from if you know, please):

 

[darkfenrir]

Re: HSC 2013 4U Marathon

sandeep, your name should be sakdeep, thats a more common indian name isnt it? LOL

and William Sidis I can't think of a fast way to do your question! I can only think of a rather tedious one.... -_-

what shortcuts r there?