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What the hell goes on during O-Week?

-1\leq \sin x\leq1 \\\\ -\frac{1}{x}\leq \frac{\sin x}{x}\leq \frac{1}{x} \\\\ \lim_{x\to \infty}-\frac{1}{x}\leq \lim_{x\to \infty}\frac{\sin x}{x}\leq \lim_{x\to \infty}\frac{1}{x} \\\\ 0\leq \lim_{x\to \infty}\frac{\sin x}{x}\leq 0 \\\\ \therefore \lim_{x\to \infty}\frac{\sin x}{x}=0

I didn't do any work those holidays...=/

Re: 2 Unit Revising Marathon HSC '10 Quotient rule you mean xD

Yup

Re: 2 Unit Revising Marathon HSC '10 Please read the guidelines of this exercise, you must post a question upon answering one correctly (which you did).

Ty Sir =]

Exactly this.

Any idea how to put a solution in a box?

A great tip for "irregular solids" would be using the fact that the relationship between the variables is always linear. This saves the hassle of drawing diagrams and utilising similar triangles (if they can be found). PS: A very wise man shared this pearl of wisdom with me.

P\left ( z \right )=z^{4}+bz^{2}+d \\\\ P'\left ( z \right )=4z^{3}+2bz \\\\ If \,\, 3i \,\, is \,\, a \,\, double \,\, root \,\, of \,\, P\left ( z \right ), \,\, then \,\, P'\left ( 3i \right )=0 \\\\ P'\left ( 3i \right )=-108i+6bi=0\Rightarrow b=18 Just an alternative solution.

The method I have used since the very beginning - this can't be beaten.

Come on man, I was blessed with piss-easy exams and I only did decently in them =/ Yours were far harder!

This is why I got the marks I did in HSC Mathematics.

u^{2}=x+2\Rightarrow 2u\frac{du}{dx}=1\Rightarrow dx=2udu \\\\ \int \frac{x-2}{\sqrt{x+2}}dx=2\int \frac{u^{2}-4}{u}du \\\\ = 2\left ( \frac{u^{2}}{2}-4\ln|u| \right )+C\Rightarrow u^{2}-8\ln|u|+C \\\\ Since \,\, u^{2}=x+2, \,\, \int \frac{x-2}{\sqrt{x+2}}dx=x+2-8\ln\left ( \sqrt{x+2} \right )+C

Use \,\, x^{2}-\left ( \sum \alpha_{i} \right )x+\prod \alpha_{i}=0

Quartic - not quintic xD

Indeed I did. I had a super-long quadratic formula expression which, unless I made mistakes on paper, could be boiled down to y = a/b and z = b/a - however I tried completing the square and it seemed to give a less tedious solution.

Haha - it wasn't so much elegant, as it was me skipping steps :haha:

\sum a_{i}=a+b=-m \\\\ \prod a_{i}=ab=n \\\\ \sum y_{i}=y+z=\frac{m^{2}-2n}{n}=\frac{\left ( a+b \right )^{2}-2ab}{ab}=\frac{a^{2}+b^{2}}{ab} \\\\ \prod y_{i}=yz=1 \\\\ \therefore y^{2}-\left ( \frac{a^{2}+b^{2}}{ab} \right )y+1=0 \\\\ \left ( y-\frac{a^{2}+b^{2}}{2ab} \right )^{2}=\left (...