# Thread: HSC 2012 Marathon :)

1. ## Re: HSC 2012 Marathon :)

Originally Posted by barbernator
~ between words
thanks bruh

3. ## Re: HSC 2012 Marathon :)

dulip, that question is more suited for an Extension 2 student under the topic "Polynomials". It is a very commonly asked question, and is essentially the 'Discriminant' of a cubic.

Here is one for the mean time:

By letting x = tan(u), evaluate the integral: (Medium/Hard)

$\int \frac{dx}{\left (1+x^2 \right )^2}$

Or:

By letting u = cos(x), show that: (Easy)

$\int \tan(x) dx = - \ln(\cos(x)) + C$

4. ## Re: HSC 2012 Marathon :)

Originally Posted by Carrotsticks
dulip, that question is more suited for an Extension 2 student under the topic "Polynomials". It is a very commonly asked question, and is essentially the 'Discriminant' of a cubic.

Here is one for the mean time:

By letting x = tan(u), evaluate the integral:

$\int \frac{dx}{\left (1+x^2 \right )^2}$
$\frac{1}{2}(tan^{-1}x + \frac{x}{1 + x^2})$

Hopefully i did it right in my head lol.

5. ## Re: HSC 2012 Marathon :)

Originally Posted by nightweaver066
$tan^{-1}x + \frac{x}{1 + x^2}$

Hopefully i did it right in my head lol.
Nope, but close.

7. ## Re: HSC 2012 Marathon :)

Correct!

I'll put up another question since you forgot to post one up yourself.

Find the condition such that the two lines:

ax+by+c=0 and dx+ey+f=0

Are:

a) Perpendicular.

b) Parallel.

8. ## Re: HSC 2012 Marathon :)

Cmon Carrot.

ax+by+c=0 => y=[-ax-c]/b has gradient -a/b
dx+ey+f=0 => y=[-dx-f]/e has gradient -d/e
a) If the two lines are parallel then their gradient are equal,
i.e. -a/b=-d/e
Hence, bd=ae
b) If the two lines are parallel then the product of their gradient is -1,
i.e. (-a/b)(-d/e)=-1

9. ## Re: HSC 2012 Marathon :)

Originally Posted by bleakarcher
Cmon Carrot.

ax+by+c=0 => y=[-ax-c]/b has gradient -a/b
dx+ey+f=0 => y=[-dx-f]/e has gradient -d/e
a) If the two lines are parallel then their gradient are equal,
i.e. -a/b=-d/e
Hence, bd=ae
b) If the two lines are parallel then the product of their gradient is -1,
i.e. (-a/b)(-d/e)=-1
For the parallel one though, you also have to put in the condition that they aren't the same line.

10. ## Re: HSC 2012 Marathon :)

Originally Posted by RealiseNothing
For the parallel one though, you also have to put in the condition that they aren't the same line.
oh right completely forgot about that one. so you just add c/b =/= f/e
so bf=/=ec

11. ## Re: HSC 2012 Marathon :)

A polynomials question.

CodeCogsEqn.gif

12. ## Re: HSC 2012 Marathon :)

^ a lot of you will probably recognise that one. If you do, please leave it for someone else lol.

13. ## Re: HSC 2012 Marathon :)

hmm, if you write the polynomial as a product of n terms then take the derivative using the special nth product rule, then use a trig sub...you will get no where
so i will leave it for the hsc kids to figure out what they are properly meant to do

14. ## Re: HSC 2012 Marathon :)

alright

anyone gonna give the question a go?

15. ## Re: HSC 2012 Marathon :)

ok ill give it a go in words.

Express z^n -1 as a procduct of its n roots in factored form, divide both sides by z-1, which is a factor of z^n-1. now the hsc kids finish it for me... i i i...forgot what to do next.

16. ## Re: HSC 2012 Marathon :)

-cotx / 2 using the chain rule twice

17. ## Re: HSC 2012 Marathon :)

You could have used log laws to simplify the expression before continuing =)

$\frac{d}{dx} \left ( \ln \left ( \frac{1}{\sqrt{\sin x}} \right ) \right ) = \frac{d}{dx} \left ( - \frac{1}{2} \ln \left ( \sin x \right )} \right ) = - \frac{1}{2} \cot x$

20. ## Re: HSC 2012 Marathon :)

Originally Posted by deswa1
This isn't the answer- maybe try again or someone else can have a go.
i am sure that is the answer

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