Search results

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent We rearrange and integrate using the Fundamental Theorem of Calculus, from which we know that $\int _a ^b \frac{\mathrm{d}y}{\mathrm{d}x}\text{ d}x = \Big{[}y(x) \Big{]}_{x=a} ^{x=b}$, where $y(x)$ is the function...

$In order to get an intuitive feel for why $\int _a ^b f(t) \text{ d}t= F(b) -F(a)$, it is probably best to think of things in terms of rates rather than areas.$ $Above, $f(t)$ is the derivative of $F(t)$, so $f(t)$ is the \textsl{instantaneous rate of change} of $F(t)$ (which could represent...

You're right that in order to calculate the area, we'd have to do "height times width". But we'd need to take infinitesimally small widths in order to get the exact area. As your book should show, we do this by taking rectangles to approximate the area and then take the limit as the no. of...

Non-mathematical simple version: maybe just get one of the subjects (English or History) (see my above post for reason why).

$\noindent Maybe you could just get the results for the \textsl{one} subject you're most curious about. The difference in utility/joy gained from getting one subject's results compared to no subjects is probably greater than the difference between getting two subjects' and one's (in other words...

Just said it to reduce confusion for those reading this thread who did not know the right spelling. :)

Re: HSC 2016 4U Marathon $Note that $0$ is not a root. So we can divide through by $x^2$ to obtain$ $$3x^2 - x + 4 - \frac{1}{x}+\frac{3}{x^2} = 0$.$ $This can be written as $3\left(x^2+\frac{1}{x^2} \right)-\left(x+\frac{1}{x}\right) + 4 = 0$.$ $Now, note that $x^2+\frac{1}{x^2} =...

Actual spelling is ''guarantee'', as leehuan said.

$\noindent ATAR Calculators typically crash in the days leading up to ATAR release (especially the day before, since then the marks are released and everyone is frantically trying to calculate their ATAR), so expect long load times / crashes for ATAR Calculator sites.$

Re: HSC 2016 4U Marathon It seems to take too long to prove to make it part of that question. So we can make it the next question. \noindent $\textbf{NEW QUESTION}$ $\noindent Let $\mathcal{C}_1$ and $\mathcal{C}_2$ be two circles lying in the plane without any intersection. Let $X$ be a...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Let $f(x) = \cos (2x) -\left(2\sin ^2 x -1\right)$, so the inequation is $f(x) \geq 0$. Show that this holds at $x=0$ by substituting $x=0$ into the inequation. Then since the first (non-negative) root is at $x =\frac{\pi}{4}$...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Both methods are valid, so you must have made a substitution error or something in the one where you got a negative answer (the answer should be positive).

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $It's not $y = \ln |x|$ you should be subbing in to; it's the equation of a tangent. So subbing $x=0$ into the equation of either tangent gives $y=-1$. So the intersection point is $(0,-1)$ and the height is thus $1$. So the...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $You got the wrong slope for the tangent at $x=-1$. The tangent is downward sloping there, so you should have got a negative gradient for it.$

I think Blitz_N7 was referring to the Pender textbook (the Year 11 3U one) (publisher: Cambridge).

Re: HSC 2016 3U Marathon Use the following facts: • We can find angle CBE using the fact that adjacent angles on a straight line add up to 180 degrees (that is, are supplementary). • Then we can find angle BCE isn't angle sum of triangle BCE. • But angle DCE equals angle BCE as we are told...

Here's the BOS thread about it for anyone who wants to see: http://community.boredofstudies.org/214/news-current-affairs-politics/282143/student-challenges-ruling-her-hsc.html

Yes, that's the one.

I can't remember exactly but I think that person did their HSC in 2008. There was a thread on BOS about it I think, that person received quite a bit of flame on it…

That person a few years back whinged about something else iirc. They claimed iirc that they should have received some special provision for something and didn't receive it, and this contributed to their failure to get 100 UAI (it was UAI back then). Iirc that person still got 99.95 UAI with at...