HSC 2012 MX2 Marathon (archive) (1 Viewer)

barbernator · May 6, 2012

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{x^2}{\sqrt{1-x^2}}dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{x^2}{\sqrt{1-x^2}}dx" title="\int \frac{x^2}{\sqrt{1-x^2}}dx" /></a>

juantheron · May 6, 2012

Re: 2012 HSC MX2 Marathon

$\hspace{-16}\bf{\int \frac{x^2}{\sqrt{1-x^2}}dx}$\\\\\\ Put $\bf{x =\sin \theta \Leftrightarrow dx=\cos \theta d\theta}$\\\\\\ $\bf{=\int \frac{\sin^2 \theta.\cos \theta.d\theta}{\cos\theta}=\frac{1}{2}\int 2\sin ^2 \theta d\theta }$\\\\\\ $\bf{=\frac{1}{2}\int \left(1+\cos 2\theta\right)d\theta}$\\\\\\ $\bf{=\frac{1}{2}\left(\theta+\frac{\sin 2\theta}{2}\right)+\mathbb{C}}$\\\\\\ $\bf{=\frac{1}{2}\sin^{-1}(x)+\frac{1}{2}.x.\sqrt{1-x^2}+\mathbb{C}}$$

juantheron · May 6, 2012

Re: 2012 HSC MX2 Marathon

$\hspace{-16}\bf{\int \frac{(4x^5+5x^4)}{(x^5+x+1)^2}dx}$

smartalec · May 6, 2012

Re: 2012 HSC MX2 Marathon

hey guys,
so this is totally an easy question for all of you, but I figured this would be the best place to get the quickest response. Its a measly prelim yr 11 mathematics questions. Its about domain nd range and I'm a bit confused :S

f(x) Square root 2x -6

smartalec · May 6, 2012

Re: 2012 HSC MX2 Marathon

rep for best answer!!

SpiralFlex · May 6, 2012

Re: 2012 HSC MX2 Marathon

$f(x) =\sqrt{2x -6}$

We know that the expression inside square root is always positive, (can be equal to zero too)

$\therefore 2x-6 \geq 0$

$x\geq 3$

And for f(x)

$f(x) \geq 0$

As it is always positive (can be equal to zero)

smartalec · May 6, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
$f(x) =\sqrt{2x -6}$

We know that the expression inside square root is always positive, (can be equal to zero too)

$\therefore 2x-6 \geq 0$

thanks for the answer spiral! I should have mentioned I got that far, but I am not sure how to determine the domain and function and how i would d

$x\geq 3$

And for f(x)

$f(x) \geq 0$

As it is always positive (can be equal to zero)

And how would I determine the domain and range and draw it on a graph?

barbernator · May 6, 2012

Re: 2012 HSC MX2 Marathon

smartalec said:
And how would I determine the domain and range and draw it on a graph?

he just solved for domain and range, x=>3, y=>0. the graph is the positive axis of a sideways parabola

smartalec · May 6, 2012

Re: 2012 HSC MX2 Marathon

smartalec said:
And how would I determine the domain and range and draw it on a graph?

ohhh ok! thanks so much!

smartalec · May 6, 2012

Re: 2012 HSC MX2 Marathon

What is the congruent of 2 root 3 + 3 root 7?
Thanks so much guys!! Sorry for the stupid questions!

Carrotsticks · May 6, 2012

Re: 2012 HSC MX2 Marathon

smartalec said:
What is the congruent of 2 root 3 + 3 root 7?
Thanks so much guys!! Sorry for the stupid questions!

This thread is for MX2 questions.

If you have your own, make your own thread so you can ask as many questions as you like. Best not to hijack this one.

smartalec · May 6, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
This thread is for MX2 questions.

If you have your own, make your own thread so you can ask as many questions as you like. Best not to hijack this one.

its my last question

haha

SpiralFlex · May 6, 2012

Re: 2012 HSC MX2 Marathon

Congruent? Lol

smartalec · May 6, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Congruent? Lol

whoops auto correct hahahaha.

SpiralFlex · May 6, 2012

Re: 2012 HSC MX2 Marathon

Graphically if you plot some points (or if you already know what the square root function looks like)

Then you can immediately tell the range/domain. Try it.

Fus Ro Dah · May 6, 2012

Re: 2012 HSC MX2 Marathon

Did you mean 'Conjugate' instead of 'Congruent'?

A congruent number is an integer that is the area of the triangle with 3 rational sides.

A cool theorem related to this is "There does NOT exist a square number that is congruent". It is called Fermat's Right Triangle Theorem.

barbernator · May 7, 2012

Re: 2012 HSC MX2 Marathon

to get the ball rolling again, <a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{2x^2@plus;13x@plus;16}{(2x@plus;1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{2x^2+13x+16}{(2x+1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" title="\int \frac{2x^2+13x+16}{(2x+1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" /></a>

Fus Ro Dah · May 7, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
to get the ball rolling again, <a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{2x^2@plus;13x@plus;16}{(2x@plus;1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{2x^2+13x+16}{(2x+1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" title="\int \frac{2x^2+13x+16}{(2x+1)^2(x-2)}~show~all~working.~This~definitely~tests~skill~in~partial~fractions" /></a>

$\frac{2}{2x+1} + 2 \ln(x-2) - \frac{3}{2} \ln(2x+1) + C$

Find the area between the curve $y=x e^{-x^2}$ and the x axis over the span $[- \infty,+\infty]$

barbernator · May 7, 2012

Re: 2012 HSC MX2 Marathon

Fus Ro Dah said:
$\frac{2}{2x+1} + 2 \ln(x-2) - \frac{3}{2} \ln(2x+1) + C$

Find the area between the curve $y=x e^{-x^2}$ and the x axis over the span $[- \infty,+\infty]$

1 units squared. find the inverse function of <a href="http://www.codecogs.com/eqnedit.php?latex=y=2e^{2x} -14e^x @plus;6" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=2e^{2x} -14e^x +6" title="y=2e^{2x} -14e^x +6" /></a> and state its new domain and range.

ismeta · May 7, 2012

Re: 2012 HSC MX2 Marathon

Fus Ro Dah said:
$\frac{2}{2x+1} + 2 \ln(x-2) - \frac{3}{2} \ln(2x+1) + C$

Would it be possible to see your working?

I'm having trouble with the question.

HSC 2012 MX2 Marathon (archive) (1 Viewer)

Active Member

Active Member

Active Member

Member

Member

Well-Known Member

Member

Active Member

Member

Member

Retired

Member

Well-Known Member

Member

Well-Known Member

Member

Active Member

Member

Active Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)