3u Mathematics Marathon v2.0 (1 Viewer)

[pLuvia]

07er's you are so inactive in the maths forums unlike the previous years So to stop that (I hope)

Time to start this again, point of this thread is to test each other's maths skills, general revision etc.

General rule: A person posts up a question (not homework-help questions) and then another person answers it in spoilers and after they answer they put up another question for the next person to answer (thank you webby!)

How to do spoilers : [spoiler ]Like this[/spoiler ], but without the spaces in between the brackets so without it, it would be something like this

Spoiler

Like this

I'll start off to get the ball rolling, last question from the previous marathon

Question

If a>b and c>d prove that ac+bd>ad+bc

Edit: Scrapped that question. don't think it was possible with 3u methods

 

[webby234]

Re: 3u Mathematics Marathon 2.0

Just something you didn't mention - the person who answers the question posts a new one for others to solve.

 

[pLuvia]

Re: 3u Mathematics Marathon 2.0

That too

 

[onebytwo]

hey nice first question pluvia, looks familiar, its the one we encountered in mondays tute! lol

 

[pLuvia]

Re: 3u Mathematics Marathon 2.0

There has already been a question posted so please do not post another one until the previous one has been answered, that's what happened in the last 2u marathon and it got messy

hey nice first question pluvia, looks familiar, its the one we encountered in mondays tute! lol

You in my tute? dammit I got to find out who you are lol

Looks like the one in our tute but I just randomly made one up using that question we had

 

[Mark576]

I'm only in year 11 btw:

Spoiler

ac + bd > ad + bc

ac - ad > bc - bd

a(c-d) > b(c-d) (able to divide by (c-d) since c > d and therefore (c-d) is positive)

a>b, which is true, therefore ac +bd > ad + bc

Not sure if that's correct, probably isn't lol. From what has been said above this method seems too easy.

Here's the next question: (mind you, it's year 11)

Prove that x^2 +xy + y^2 > 0 for any non-zero values of x and y.

Remember to put in the spoilers!

 

[LoneShadow]

here's a "solution"...did not spend time thinking. It might be wrong. I just wanted to do it like most year 11 kids would do:

Spoiler

Let f(x) = x^2+yx+y^2
Using the quadratic formula: x = (-y+/-Sqrt[-3y^2])/2.
But -3y^2<0 always. hence f(x) has no roots. and since it's a positive quadratic, it's always greater than zero.

someone post the right answer!

 

[SoulSearcher]

Here's the next question: (mind you, it's year 11)

Prove that x^2 +xy + y^2 > 0 for any non-zero values of x and y.

Long time I've done one of these marathons, gonna have a go at this:

Spoiler

Consider:
x³-y³ = (x-y)(x²+xy+y²)

When x > y,
x³-y³ is positive,

x-y is positive, and thus

x²+xy+y² is positive.

When y > x,
x³-y³ is negative,

x-y is negative, and thus

x²+xy+y² is positive.

Therefore x²+xy+y² > 0 for non-zero values of x and y.

I'm not completely sure that is correct, but if there's an error, then anybody else is free to correct it.

Next question:
When a polynomial is divided by x-p, the remainder is p³. WHen the polynomial is divided by x-q, the remainder is q³. Find the remainder when the polynomial is divided by (x-p)(x-q).

 

[Mark576]

Very nice solution soulsearcher, but you could also say this:

Spoiler

we know that x^2 and y^2 are greater than 0, and that x^2 +2xy + y^2 > 0, so:
(x^2 +y^2)/2 +((x+y)^2)/2 > 0, simplifying this, it becomes:
x^2 + xy + y^2 > 0

 

[bos1234]

Next question:
When a polynomial is divided by x-p, the remainder is p³. WHen the polynomial is divided by x-q, the remainder is q³. Find the remainder when the polynomial is divided by (x-p)(x-q).

Spoiler

P(x)=(x-p)(x-q).Q(x) + ax + b
P(p)=ap+ b=p^3
p(q)=aq+b=q^3

equating simlutaneously,
ap-aq=p^3-q^3
a=(p-q)(p^2+pq+q^2)/(p-q)
a=p^2+pq+q^2
b=-p^2q-pq^2

therefore ax+b = R(x) = p^2+pq+q^2-p^2q-pq^2

Is this right or wrong? if wrong please correct!

question:

A finance company has agreed to pay a retired couple a pension of $15,000 per year for the next twenty years, indexed to inflation which is 3 1/2 percent per annum.


(a)How much will the company have paid the couple at the end of the twenty years?


(b) Immediately after the tenth annual pension payment is made, the finance company increases the indexewd rate to 4 percent per annum to mathc the increased inflation rate.
Given these new cond., how much will the company have paid the coiple at the end of twenty years?

year 12 - 3 unit qn.

 

[S1M0]

We need one of these for Yr.11.

 

[pLuvia]

You are allowed to post any level of questions in here except they just have to be within the 3u level, year 11 or year 12 are permitted

 

[idling fire]

Well... now I just feel like hanging myself... can't even think about solving any one of those...