HSC 2012 Marathon :)

**bleakarcher** · 25 Apr 2012, 2:22 PM

Originally Posted by barbernator

~ between words

thanks bruh

**dulip** · 25 Apr 2012, 3:04 PM

$\frac{9!}{4!\times 2!}$

new question:
If two of the roots of the equation $x^{3}+qx+r = 0$ are equal show that $4q^{3}+27r^{2}=0$
[easy/medium]

**Carrotsticks** · 26 Apr 2012, 5:16 PM

Originally Posted by dulip

$\frac{9!}{4!\times 2!}$

new question:
If two of the roots of the equation $x^{3}+qx+r = 0$ are equal show that $4q^{3}+27r^{2}=0$
[easy/medium]

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)

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

**nightweaver066** · 26 Apr 2012, 5:19 PM

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.

**Carrotsticks** · 26 Apr 2012, 5:21 PM

Originally Posted by nightweaver066

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

Hopefully i did it right in my head lol.

Nope, but close.

**Timske** · 26 Apr 2012, 6:23 PM

$\frac{1}{2}~(\frac{x}{x^2+1} ~ + ~ \tan^{-1})$

**Carrotsticks** · 27 Apr 2012, 1:06 PM

Originally Posted by Timske

$\frac{1}{2}~(\frac{x}{x^2+1} ~ + ~ \tan^{-1})$

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.

**bleakarcher** · 27 Apr 2012, 1:18 PM

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
Hence, ad=-eb

**RealiseNothing** · 27 Apr 2012, 1:20 PM

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
Hence, ad=-eb

For the parallel one though, you also have to put in the condition that they aren't the same line.

**bleakarcher** · 27 Apr 2012, 1:23 PM

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

**bleakarcher** · 27 Apr 2012, 1:31 PM

A polynomials question.

CodeCogsEqn.gif

**bleakarcher** · 27 Apr 2012, 1:32 PM

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

**math man** · 27 Apr 2012, 10:55 PM

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

**bleakarcher** · 28 Apr 2012, 8:55 PM

alright

anyone gonna give the question a go?

**math man** · 29 Apr 2012, 12:15 AM

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.

**barbernator** · 2 May 2012, 6:11 PM

$z^n-1=0\\ \\ LHS=z^n-1=(z-1)(z-z_{1})(z-z_{2})...(z-z_{n-1})\\ \\ divide~both~sides~by~(z-1)\\ \\ z_{n-1}+z_{n-2}+z^{_{n-3}}+...+z+1=(z-z_{1})(z-z_{2})...(z-z_{n-1})\\ \\ let~n=1\\ \\ n=(1-z_{1})(1-z_{2})(1-z_{3})...(1-z_{n-1})~as~required\\ \\ new~question\\ find~the~first~derivative~of~ln(\frac{1}{\sqrt{sin (x)}})$

edit: difficulty, medium

**deswa1** · 2 May 2012, 6:20 PM

$\frac{d}{dx}ln(\frac{1}{\sqrt{sinx}})=(\sqrt{sinx})(\frac{-cosx}{2(sinx)^{3/2}})\\ =\frac{-cosx}{2sinx}\\ =\frac{-cotx}{2}$

$\textup{Evaluate }\int_{0}^{1}2^xdx$

Medium-hard

**zeebobDD** · 2 May 2012, 6:22 PM

-cotx / 2 using the chain rule twice

**Carrotsticks** · 2 May 2012, 6:29 PM

Originally Posted by deswa1

$\frac{d}{dx}ln(\frac{1}{\sqrt{sinx}})=(\sqrt{sinx})(\frac{-cosx}{2(sinx)^{3/2}})\\ =\frac{-cosx}{2sinx}\\ =\frac{-cotx}{2}$

$\textup{Evaluate }\int_{0}^{1}2^xdx$

Medium-hard

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

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

**Timske** · 2 May 2012, 6:39 PM

$\frac{d}{dx} ~ \ln(\tfrac{1}{\sqrt{sinx}}) , using ~ \ln f(x) = \frac{f'(x)}{f(x)} \\ \frac{d}{dx}~\tfrac{1}{\sqrt{sinx}} = \frac{d}{dx} (sinx)^{-1/2} \\\\ = -\frac{1}{2}(sinx)^{-3/2}*cosx = \frac{1}{2}*\frac{-cosx}{sin^{3/2}x} \\\\\therefore \frac{d}{dx} ~ \ln(\tfrac{1}{\sqrt{sinx}}) = \frac{1}{2}*\frac{-cosx}{sin^{3/2}x} ~ * \sqrt{sinx} \\\\ = \frac{1}{2} * \frac{-cosx}{sinx} \\\\ = \frac{-cotx}{2}$

**barbernator** · 2 May 2012, 6:52 PM

$\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)~/~ln(2)~dunno~how~to~do~numbered~box~brackets\\ =1/ln(2)$

whoops fail answer lol, it should be 1/ln(2)

new question, at a guess for not having done the question, medium to hard

$2sin(\theta -\frac{\pi}{3})=cos(\theta +\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a+b\sqrt{3},~where~a~and~b~are~integ ers.$

**Timske** · 2 May 2012, 6:56 PM

Originally Posted by barbernator

$\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)$

$\left [(x)\ln(2) \right ]_{0}^{1}$

best i could get it

\left(\right) x_a^b

**deswa1** · 2 May 2012, 7:19 PM

Originally Posted by barbernator

$\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)~/~ln(2)~dunno~how~to~do~numbered~box~brackets\\ =1/ln(2)$

whoops fail answer lol, it should be 1/ln(2)

new question, at a guess for not having done the question, medium to hard

$2sin(\theta -\frac{\pi}{3})=cos(\theta +\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a+b\sqrt{3},~where~a~and~b~are~integ ers.$

This isn't the answer- maybe try again or someone else can have a go.

**Timske** · 2 May 2012, 7:20 PM

Originally Posted by barbernator

$\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)~/~ln(2)~dunno~how~to~do~numbered~box~brackets\\ =1/ln(2)$

whoops fail answer lol, it should be 1/ln(2)

new question, at a guess for not having done the question, medium to hard

$2sin(\theta -\frac{\pi}{3})=cos(\theta +\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a+b\sqrt{3},~where~a~and~b~are~integ ers.$

$tan(\theta ) = -3-\sqrt{3}$

right??

**barbernator** · 2 May 2012, 7:22 PM

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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