Maths Methods 3/4 Help Thread 2011

[thushan]

Hmm, the thing is that sum doesn't have anything to do with combinatorics...

The fifth term is basically that expression you got, with k = 7 substituted. Why does it start with k = 3? Well, k can start anywhere really, but that sum expression you have stipulates that the first term is where k = 3.

The 5th term is 7/2 a^4 b^-49.

Am I right?

EDIT: That 100C3 thing might have been there as a trick?

[nacho]

Hmm, the thing is that sum doesn't have anything to do with combinatorics...

The fifth term is basically that expression you got, with k = 7 substituted. Why does it start with k = 3? Well, k can start anywhere really, but that sum expression you have stipulates that the first term is where k = 3.

The 5th term is 7/2 a^4 b^-49.

Am I right?

EDIT: That 100C3 thing might have been there as a trick?

Sorry, I'm not too good at this, but isn't the summation sign another way of saying 100C3 ?

[pi]

Sorry, I'm not too good at this, but isn't the summation sign another way of saying 100C3 ?

No

[thushan]

Your fifth term is:

[pi]

^^ How long did that take you to type out?

[onur369]

http://vce.atarnotes.com/forum/index.php/topic,37731.msg400624/topicseen.html#msg400624

Can you clarify or solve please. Dont know the solution.

[thushan]

http://vce.atarnotes.com/forum/index.php/topic,37731.msg400670.html#msg400670

This solution is correct.

[yabbaboo]

hi,

Question:

P(x) = x^5 - 3x^4 + 2x^3 -2x^2 +3x + 1

Given that P(x) can be written in the form (x^2 -1)Q(x) + ax + b where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the reminder when P(x) is divided by x^2 - 1

[pi]

I would go for 'otherwise' and just long divide straight away (if the remainder is all you have to find).

[thushan]

Haha that would fall under "hence" actually

[pi]

Haha that would fall under "hence" actually

Haha, it would actually!

[thushan]

Just to clarify, do polynomial division, then you can get the remainder.

[Rairiko]

First time post. Just wondering how do you sketch the modulus function y=|x|^2 - 2∣x∣ and the steps involving how to do it

[xZero]

you can split them up into 2 scenario, where x is negative and x is positive
when x is negative, y= (-x)^2 - 2(-x), -infinite<x<0
when x is positive, y=x^2-2x 0<x<infinite

rest should be straight forward

[thushan]

Well, take f(x) = x^2 - 2x. Sketch that first. Now consider f(IxI) = IxI^2 - 2IxI, which is what you want. To get that graph, the part of f(x) to the right of the y-axis is the same as f(IxI) = f(x) when x>=0. The section of f(IxI) to the left of the y-axis is simply a reflection of the right-hand part of the y=f(IxI) graph in the y-axis, as f(IxI) = f(-x) when x < 0 and hence -x > 0.

Your final graph should look a little like a rounded W-shape.

Does that make sense?