HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Mar 31, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$tan(x)tan(2x)tan(3x)=tan(3x)-tan(x)-tan(2x)$

Yeah oops :s

Sy123 · Apr 1, 2013

Re: HSC 2013 4U Marathon

$Prove that the polynomial has no real roots$

$p(x) = \frac{x^6}{6} + \frac{x^5}{5} + \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + x + c$

$(c is a constant)$

$$If$ \ \ \ c > \frac{37}{60}$

Sy123 · Apr 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
This was a good one (even though I said I won't post more series questions)

$$\sum_{k=1}^{n} \tan^{-1} \left (\frac{1}{1+k(k+1)} \right )$

First lets establish what we know about the tan inverse function, due to:

$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$

It then follows that:

$\tan^{-1} A + \tan^{-1}B = \tan^{-1} \frac{A+B}{1-AB}$

So lets try and split up the tan inverse in the summand. We need to find A and B so that A+B = 1 1-AB = 1+k+k^2
We find that A = n+1, B = -n

Therefore $\tan^{-1} (n+1) - \tan^{-1} n = \tan^{-1} \frac{1}{1+n(n+1)}$

So the summation is a telescoping sum, we end up with,

$-\frac{\pi}{4} + \tan^{-1}(n+1)$

Which simplifies to:

$\tan^{-1} \frac{n}{n+2}$

psychotropic · Apr 2, 2013

Re: HSC 2013 4U Marathon

prove that the curves x^2 -y^2 = c^2 and xy = c^2 cross at right angles

RealiseNothing · Apr 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that the polynomial has no real roots$

$p(x) = \frac{x^6}{6} + \frac{x^5}{5} + \frac{x^4}{4} + \frac{x^3}{3} + \frac{x^2}{2} + x + c$

$(c is a constant)$

$$If$ \ \ \ c > \frac{37}{60}$

Differentiating:

$x^5+x^4+x^3+x^2+x+1$

Obviously -1 is a root of this, so when x=-1 there is a stationary point. Subbing this into P(x) gives $\frac{-37}{60}$

Hence to have no real roots P(x) needs to be shifted up by $\frac{37}{60}$ , that is $c > \frac{37}{60}$

RealiseNothing · Apr 2, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Differentiating:

$x^5+x^4+x^3+x^2+x+1$

Obviously -1 is a root of this, so when x=-1 there is a stationary point. Subbing this into P(x) gives $\frac{-37}{60}$

Hence to have no real roots P(x) needs to be shifted up by $\frac{37}{60}$ , that is $c > \frac{37}{60}$

I might note that if you factorise P'(x) you get:

$(x+1)(x^4+x^2+1)$

And the second bracket has no real roots, hence why I took x=-1 as my minimum point on the graph.

Killerb47 · Apr 2, 2013

Re: HSC 2013 4U Marathon

Let x= z+iy and its conjugate be x-iy.
After multiplying by the conjuagtes of each fraction, combine the two to get 2x/(x^2+y^2)= 1
Hence 2x = x^2 + y^2.
Moving the 2x over and completing the square: (x-1)^2 + y^2 =1 with (0,0) excluded from the locus as it would create a zero denominator in the original question.

Sy123 · Apr 2, 2013

Re: HSC 2013 4U Marathon

$Evaluate$

$I=\int\frac{1}{x^2\sqrt{1+x^2}} \ dx$

asianese · Apr 2, 2013

Re: HSC 2013 4U Marathon

Evaluate?

Sy123 · Apr 2, 2013

Re: HSC 2013 4U Marathon

asianese said:
Evaluate?

Eh, not sure what word to put there, Compute, Find?

asianese · Apr 2, 2013

Re: HSC 2013 4U Marathon

I think the HSC uses 'Find' in indefinite integrals.

Would students be expected to know integrals of cotxcosecx, though? It is easy to find this out if you know what to differentiate.

Edit: nvm it can be done.

Trebla · Apr 2, 2013

Re: HSC 2013 4U Marathon

Consider three unknown numbers. It is known that the sum of these numbers is 4, the sum of the squares of these numbers is 10 and the sum of the cubes of these numbers is 16. Find the product of these three numbers.

RealiseNothing · Apr 2, 2013

Re: HSC 2013 4U Marathon

Trebla said:
Consider three unknown numbers. It is known that the sum of these numbers is 4, the sum of the squares of these numbers is 10 and the sum of the cubes of these numbers is 16. Find the product of these three numbers.

I've seen this question before...

I just can't remember where, and that's annoying lol.

Sy123 · Apr 2, 2013

Re: HSC 2013 4U Marathon

Trebla said:
Consider three unknown numbers. It is known that the sum of these numbers is 4, the sum of the squares of these numbers is 10 and the sum of the cubes of these numbers is 16. Find the product of these three numbers.

$\int \frac{dx}{1+e^x} = \int \left(1-\frac{e^x}{1+e^x} \right ) \ dx = x-\ln(1+e^x) + c$

asianese · Apr 2, 2013

Re: HSC 2013 4U Marathon

-4?

Trebla · Apr 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$\int \frac{dx}{1+e^x} = \int \left(1-\frac{e^x}{1+e^x} \right ) \ dx = x-\ln(1+e^x) + c$

Sorry I changed my question because the original one was too easy

seanieg89 · Apr 2, 2013

Re: HSC 2013 4U Marathon

$abc=\frac{1}{6}((a+b+c)^3-3(a^2+b^2+c^2)(a+b+c)+2(a^3+b^3+c^3))$

too lazy to put the numbers in

.

edit: -4.

asianese · Apr 2, 2013

Re: HSC 2013 4U Marathon

Dat trinomial expansion.

Sy123 · Apr 2, 2013

Re: HSC 2013 4U Marathon

Trebla said:
Sorry I changed my question because the original one was too easy

Yeah I noticed, your new one isn't much harder its just a bit long.

Is there any cool trick we can do to make it much faster (except for trinomial expansion

)?

RealiseNothing · Apr 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yeah I noticed, your new one isn't much harder its just a bit long.

Is there any cool trick we can do to make it much faster (except for trinomial expansion )?

What I did:

Construct a polynomial in the form:

$x^3-ax^2+bx-c$

Now let the roots of this polynomial be our three unknown integers, denoted by $\alpha,~\beta,~\gamma$

$a=\alpha +\beta +\gamma = 4$ as it is the sum of the roots.

$b=\alpha^2 +\beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma) = 4^2-2b = 16-2b = 10$

So $b=10$

$\alpha^3+\beta^3+\gamma^3=4(\alpha^2+\beta^2+ \gamma^2) - 3(\alpha + \beta + \gamma) + 3c$

$16 = 4(10) - 3(4) + 3c$

$-12=3c$

$c=-4$

So we have the polynomial $x^3-4x^2+3x+4$

Now we want $\alpha\beta\gamma$ which is just the product of the roots, which is just -4 by observing the polynomial.

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