HSC 2016 MX1 Marathon (archive) (3 Viewers)

davidgoes4wce · Feb 8, 2016

Re: HSC 2016 3U Marathon

Maybe I should have done the whole induction process instead of just substituting the numbers in? Step above only does the first step of induction.

Paradoxica · Feb 8, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
Here's a question before I forget :

$Solve:$

$y=x^3+x^2-4x+3---(1)$

$(y+1)(y^2-4)=x-7 ---(2)$

$$\noindent Equation 1 rearranges into $(x+1)(x^2-4) = y-7 \\ $It is now obvious that the only possible solutions occur iff $ x=y \\ $This reduces the problem to finding the zeroes of $x^3 + x^2 - 5x + 3 = 0 \\ x^3 + x^2 - 5x + 3 \equiv (x+3)(x-1)^2 \\ $Hence, the two distinct solutions are $x=1,\,x=-3$

leehuan · Feb 9, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
There's no use of it in any HSC level. (except maybe general notsure about the syllabus.)

General has no use for it either just to clarify.

InteGrand · Feb 9, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Can an arithmetic progression have a difference of 0?

For part a)

I looked at the difference of the second term and first term D= x-(-1)=x+1
Then I looked at the difference of the 3rd and 2nd term D=x^3-x

equating that

x+1=x^3-x

x^3-2x-1=0

In solving for x, I get x=-1.

You can have an A.P. with common difference of 0, but Paradoxica said in the question that that was a trivial one, implying that he wanted you to find the other solutions.

InteGrand · Feb 9, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
b) Since we know the solution of part a), by equating n=1, it returns an integer of 3.

$\noindent There are simpler ways to do this, for example, recall that $\phi$ is the positive solution to $x^2- x -1=0$. Then via products of roots, the other solution is $\psi := -\phi ^{-1}$. So$

$\begin{align*} \phi^2 + \phi ^{-2} &= \phi^2 + \psi ^2 \\ &= \left(\phi +\psi \right)^2 -2 \phi \psi \\ &= 1^2 -2\cdot (-1) \quad (\text{sum and product of roots}) \\ &= 3.\end{align*}$

Flop21 · Feb 9, 2016

Re: HSC 2016 3U Marathon

Oh... so my thread got killed and put in here?

Nailgun · Feb 9, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$$\noindent$ f(x) = \frac{x^2-6x+6}{2x-4} \,\,\, g(x) = \frac{ax^2+bx+c}{x-d} \\ $The following information is known about $g(x). \\ g(x) $ and $ f(x) $ have the same vertical asymptotes.$ \\ $The oblique asymptotes are perpendicular, and intersect on the $y$-axis. $ \\ f(x) = g(x) $ has a solution at $x = -2. \\ $Find $g(x)$ and hence find the other solution to $f(x) = g(x)$

Lololol

InteGrand · Feb 9, 2016

Re: HSC 2016 3U Marathon

Nailgun said:
Lololol

Haha, why is that question funny?

Nailgun · Feb 9, 2016

Re: HSC 2016 3U Marathon

He posted it in the 2U marathon first, and then when he realised that oblique asymptotes were in 3U only, he just posted it here instead.

idk i find it amusing

Paradoxica if it makes you feel better, I tried to learn out to find the oblique asymptotes, but then it involved long division so I ceebsed lel

davidgoes4wce · Feb 9, 2016

Re: HSC 2016 3U Marathon

From the 2002 Cherrybrook 3U Exam paper:

Paradoxica · Feb 9, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
From the 2002 Cherrybrook 3U Exam paper:

"outside times whole equals outside times whole"

thus, x = 7

It would be infinitely easier if the stupid teachers would just let us write

"power of a point theorem"

davidgoes4wce · Feb 9, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
"outside times whole equals outside times whole"

thus, x = 7

It would be infinitely easier if the stupid teachers would just let us write

"power of a point theorem"

First time I have heard of that term.

InteGrand · Feb 9, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
First time I have heard of that term.

"Power of a point theorem" is a real name, but isn't used in the HSC as far as I know. So don't write that in the actual HSC. You can read more about it here: http://www.cut-the-knot.org/pythagoras/PPower.shtml .

davidgoes4wce · Feb 9, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
"Power of a point theorem" is a real name, but isn't used in the HSC as far as I know. So don't write that in the actual HSC. You can read more about it here: http://www.cut-the-knot.org/pythagoras/PPower.shtml .

Interesting, thanks I'm going to read it now.

leehuan · Feb 10, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
"Power of a point theorem" is a real name, but isn't used in the HSC as far as I know. So don't write that in the actual HSC. You can read more about it here: http://www.cut-the-knot.org/pythagoras/PPower.shtml .

Products of distances from intersecting secants at point of contact are equal.

Something like that.

InteGrand · Feb 10, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Interesting, thanks I'm going to read it now.

The theorem itself is in the HSC, but somehow they don't seem to give it a name (in the HSC). So instead of referring to its name (Power of a Point Theorem) in the HSC exam, you should just write something like "product of intercepts of intersecting secants are equal" or something (basically writing out the content of the theorem in words).

Flop21 · Feb 11, 2016

Re: HSC 2016 3U Marathon

Find the roots of the following polynomials:

x^3 -5x^2 + 4

Now I'm not really sure how to do this. I remember the lecturer saying just guess any number, so I found +1 makes it equal zero, hence (x-1) is a factor. What do I do now to find the other roots?

davidgoes4wce · Feb 11, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Find the roots of the following polynomials:

x&3 -5x^2 + 4

Now I'm not really sure how to do this. I remember the lecturer saying just guess any number, so I found +1 makes it equal zero, hence (x-1) is a factor. What do I do now to find the other roots?

Im assuming you mean x^3-5x^2+4?

Flop21 · Feb 11, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Im assuming you mean x^3-5x^2+4?

Yes my bad.

InteGrand · Feb 11, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Find the roots of the following polynomials:

x^3 -5x^2 + 4

Now I'm not really sure how to do this. I remember the lecturer saying just guess any number, so I found +1 makes it equal zero, hence (x-1) is a factor. What do I do now to find the other roots?

$\noindent To find the other roots, we can use polynomial long division (or synthetic division, or inspection) to find $\left(x^3 -5x^2 +4\right)\div \left(x-1\right)$. This will come out to be a quadratic (it should be $x^2 -4x-4$). Then we can find the roots of this quadratic using the quadratic formula or completing the square etc. The roots of this quadratic are then the other roots of the original cubic.$

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