Preliminary mathematics marathon (10 Viewers)

hscishard · Jun 2, 2011

New question:

Is this prelim only?

SpiralFlex · Jun 2, 2011

hscishard said:
New question:

Is this prelim only?

Probably, I hope not. Does James Ruse Preliminary count? I see little Year 12 Year 11s.

ohexploitable · Jun 2, 2011

hscishard said:
New question:

Is this prelim only?

Anything goes.

kooliskool · Jun 6, 2011

random-1006 said:
lol what expand ( p -a) ( p-b) (p-c) ( p-d), where a, b, c, d are roots

its a little less work i guess

There's actually another way that only takes really a few lines only,

If you look at the $\left ( p-\alpha \right )\left ( p-\beta \right )\left ( p-\gamma \right )\left ( p-\delta \right )$ as product of roots, we can create a new equation with $\left ( p-\alpha \right ) , \left ( p-\beta \right ),\left ( p-\gamma \right ),\left ( p-\delta \right )$ as roots like follows:

The relationship between the new roots $y$ and the old one is $y=p-x$

Hence $x=p-y$

Substitute this back into the old equation which will be:

$\left ( p-y \right )^{4}-p\left ( p-y \right )^{3}+q\left ( p-y \right )^{2}-pq\left ( p-y \right )+1=0$

and expand to get the leading coefficient and the constant term will give you the product of roots then.

Leading coefficient will be: $y^{4}$

Constant Term will be: $p^{4}-p^{4}+p^{2}q-p^{2}q+1$

So if I'm not mistaken, is the answer just 1?

ohexploitable · Jul 7, 2011

shameless revive

spiralflex or someone, get in here

SpiralFlex · Jul 7, 2011

ohexploitable said:
shameless revive

spiralflex or someone, get in here

Yes Sir!

a) Since it is Preliminary, Preliminary student's only answer this!

I will start with a basic question.

Express the following fraction in the partial form,

$\frac{p}{x+2} + \frac{q}{1-x}$ . Where $p$ and $q$ are constants.

The fraction is $\frac{5x+6}{2-x-x^2}$

For Preliminary students,

b) The roots of the equation $ax^2 +bx+c=0$ are in the ratio of $p:q$ . Show that $(p^2+q^2)ac=(b^2-2ac)pq$

For HSC students, [Sorry this question was on the thread already. Didn't know. Attempt it if you want anyway. No cheating.

]

c) Differentiate with respect to $x$ ,

$x^x$ . Hence find the differential coefficient of $x^x^^^x$

[Hint: Don't differentiate normally.]

For HSC students,

d) Show that $\frac{e^x}{e^2^x-1}=\frac{1}{2}{(\frac{e^x}{e^x-1}-\frac{e^x}{e^x+1})}$ and hence evaluate $\int_{1}^{2} \frac{e^x}{e^2^x-1}$

Remarks: HSC students! Don't answer the Preliminary sections! Preliminary students! Attempt HSC questions if you can!

Drongoski · Jul 7, 2011

c) is a very interesting question.

AAEldar · Jul 7, 2011

I think I've still got the working for b) on my pc. I shall let someone else answer it though.

Well, it's question c now.

SpiralFlex · Jul 7, 2011

AAEldar said:
I think I've still got the working for b) on my pc. I shall let someone else answer it though.

Well, it's question c now.

Good times...

ohexploitable · Jul 7, 2011

i think question c has already been asked in this thread

SpiralFlex · Jul 7, 2011

ohexploitable said:
i think question c has already been asked in this thread

Really? What are the odds. In Preliminary too! I see little Ohexploitable on the first page of this thread!

ohexploitable · Jul 7, 2011

SpiralFlex said:
Really? What are the odds. In Preliminary too! I see little Ohexploitable on the first page of this thread!

ohexploitable said:
New Question:

Differentiate...

$y=x^{x^{x}}$

SpiralFlex · Jul 7, 2011

ohexploitable said:

What?! What are the odds!

ohexploitable · Jul 7, 2011

was it khorne?

powlmao · Jul 7, 2011

I Want to know!

ohexploitable · Jul 7, 2011

spasticaznkid said:
Find the value of k:
The remainder is 25 when:
$2x^{4}-3x^{2}+5$ is divided by $x-k$
I know P(k) = 25 but im unsure what to do after.

$2k^4-3k^2+5=25$

solve for k

Hermes1 · Jul 7, 2011

spasticaznkid said:
Find the value of k:
The remainder is 25 when:
$2x^{4}-3x^{2}+5$ is divided by $x-k$
I know P(k) = 25 but im unsure what to do after.

sub in x = k, then get a quadratic equation in terms of k^2 and solve it, then you should get 4 solutions for k.

Hermes1 · Jul 7, 2011

ohexploitable said:
$2k^4-3k^2+5=25$

solve for k

yeh like this

Drongoski · Jul 7, 2011

$\therefore (2k^2 +5)(k^2-4) = 0$

$\therefore k^2 = 4 \implies k = \pm 2$

Note: for real k (no complex for MX1):

$(2k^2+5) \neq 0$

powlmao · Jul 7, 2011

Not sure which program everyone uses, but solve it normally

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