Search results

Yeah, you may want to pre-derive that formula before setting out to tackle an integral like this so that you have it handy if you need it.

Oh sorry lol, I was thinking of a sec rather than sec^2 being the thing expanded. $\noindent If we have $\sec^m x\tan ^n x$, where $m$ is an odd positive integer and $n$ is an even positive integer, then $n =2k$ for some positive integer $k$ and so $\tan ^n x = \left(\tan ^2 x\right)^k =...

$\noindent Say $I = \int \sec ^5 x \tan^ 4 x \text{ d}x$. Then $\tan^4 x = \left(\sec^2 x-1\right)^2$. So $I = \int \sec ^5 x \left(\sec^2 x-1\right)^2\text{ d}x$. Expanding this via the binomial theorem will give us some terms with an even (and some with odd) power of $\sec x$.$

I'm afraid this case is the most tedious of the bunch.

If you plan on studying something that requires maths at university, you should probably not do General (if possible).

Yeah it won't hold if M can be any real matrix at all (edited in an example). If they meant for M to have integer entries, then you can show the n = 2 case using Cayley-Hamilton theorem for instance. The characteristic polynomial will be pM(z) = z2 + bz + c, where b and c are integers (you can...

$\noindent Hint: It can be done quite easily by induction (show it starting from $n=1$, since this is an easy base case, and you can use strong induction if needed). :-). Doing it this way will also be able to give you a recursive formula for the $a_n$ and $b_n$, assuming you are able to find...

$\noindent For the upper half of a circle of radius $a$ ($a$ some constant greater than $0$), we can write it as $y=\sqrt{a^2-x^2}$, $-a<x<a$. Then $y^\prime = \frac{-x}{\sqrt{a^2-x^2}}$, for $-a<x<a$. You can do a similar computation for the lower half, where $y=-\sqrt{a^2-x^2}$ $\Big{(}$well...

Probably the definition of ''definition''.

If the midpoint lies on the unit circle, then P must be 2 units from O. Conversely if P is 2 units from O, the midpoint lies on the unit circle. Hence the desired locus is the circle of radius 2 centred at the origin, i.e. x2+ y2= 4.

$\noindent The discriminant is $\Delta = \left(k+1\right)^2 - 4\left(2k-3\right)\cdot (-1) = k^2 + 2k + 1 + 8k - 12 = k^2 + 10 k\, \color{blue}{ - 11}$ (rather than your $-10$). This factors into $\Delta = \left(k+11\right)\left(k-1\right)$, and so the given answers are right.$

Well the HSC definition for "Assess" is "Make a judgement of value, quality, outcomes, results or size" ( http://www.boardofstudies.nsw.edu.au/syllabus_hsc/glossary_keywords.html ), so you should give a final judgement of whether it is overall a positive impact or negative impact. So probably...

Now that really is quite strange (but probably typical of HSC Science).

It's pretty common in HSC Science to lose marks for things like this (not closing tables, or not drawing the lines in ruler, etc.).

One can find full details of the quadratic interpolation procedure used for the moderation of school marks in the attachment in this comment: .

I was intentionally using the word "scaling" to refer to a general adjustment procedure.

The scaling process isn't that simple though is it (I.e. your internal mark that'll get used isn't in general simply P% of E1)?

$\noindent Except if $x = -1$, when the answer is $\ln |e|+ \mathcal{C}$. :$\mathcal{P}$$

Re: HSC 2016 3U Marathon $\noindent Let $P$ be the point $\left(a,b\right)$ in the Cartesian plane. Let $P^{*}$ be the point obtained by reflecting $P$ about the line $y=x$. Show that $P^{*}$ has coordinates $\left(b,a\right)$.$

Re: International Baccalaureate Marathon 2016 So basically (making the approximating assumption that these countries all have roughly same proportion of their total population be of Year 12 (or equivalent) age), the US has a larger ratio of its students doing the IB compared to UK (and also...