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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent We can do each of these (except the last one) using the fact that $\frac{\mathrm{d}}{\mathrm{d}x} \left(a^{mx+b}\right) = a^{mx+b}\cdot m \ln a$.$ $\noindent To prove this fact, we can let $y = a^{mx+b}$, and then we...

Re: MX2 2016 Integration Marathon Can never be too sure… ;)

Re: MX2 2016 Integration Marathon When you used do integrals here by inspection, were they usually actually by inspection, or did you mainly say that to save having to type out lots of tedious working out / LaTeX code?

$\noindent 1. The integral of $\ln x$ is $x \ln x -x +c$, but you don't need to know how to derive this result unless you do HSC 4U Maths.$ $\noindent 2. The volume is given by $ $$\begin{align*}V &= \pi \int_{1} ^{5} y^2 \text{ d}x \\ &= \pi \int _{1} ^{5} \frac{2^2}{2x-1} \text{ d}x \\ &=...

Re: HSC 2016 4U Marathon $\noindent Since adjacent angles of a rhombus add up to $180^\circ$, the value of $\theta $ is $\pi -\angle AOB = \pi - \frac{\pi}{6} =\frac{5\pi}{6}$, or in degrees, $\theta = 150^\circ$.$

Re: MX2 2016 Integration Marathon $\noindent \textbf{NEW QUESTION}$ $\noindent Find $\int \frac{\sin x}{\sin x+ \cos x}\text{ d}x$.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent If we want to factorise it, we have$ $$\begin{align*} y^\prime &= x^2 (x+1)^3 \left(3(x+1) + 4x\right) \\ &= x^2 (x+1)^3 (7x+1). \end{align*}$$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent The answer is $y^\prime = 3x^2 (x+1)^4 + x^3 \cdot 4\left(x+1\right)^3$.$

Re: Extracurricular Integration Marathon Yeah, it can be proved that way: https://en.wikipedia.org/wiki/Sophomore%27s_dream#Proof .

That was the case before the current syllabus. In the current HSC Physics syllabus (2001 onwards), no knowledge of 2U maths is assumed or required.

Re: HSC 2016 3U Marathon Blast1 posted a method using complex number vectors in the 4U marathon: http://community.boredofstudies.org/14/mathematics-extension-2/344755/hsc-2016-4u-marathon-11.html#post7100224 (Check his quote of a Paradoxica post there too, as it contains the question with a...

$\noindent Let $f(x) = x \ln x -x^2$, $x>0$. Then $f^\prime (x) = \ln x + 1 -2x$ and $f'' (x) = \frac{1}{x} -2$. So $f''(x) =0$ if and only if $x=\frac{1}{2}$, and it is clear that $f'' (x)$ indeed changes sign here. So the inflection point is at $x =\frac{1}{2}$. At this point, $y =...

Re: HSC 2016 4U Marathon $\noindent This is an ellipse by the focus-directrix definition of ellipse $\Big{(}$its eccentricity is $\frac{3}{5}$ and a focus is $(0,3)$, with corresponding directrix $y=\frac{25}{3}$$\Big{)}$. So in terms of visualisation, observe that any point in an ellipse...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent He meant $y=2x$.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Observe that the summand is identically equal to $\frac{\frac{1}{2}}{r} + \frac{-1}{r+1} + \frac{\frac{1}{2}}{r+2}$, and use telescoping.$

Yeah, it probably is a General Maths Q.

He might be talking about 2U Maths, since this looks like a typical 2U question (calculating arc lengths of circles). Edit: oh wait, never mind, there's a formula for when the angle is in degrees in the General Maths formula sheet. So it is probably a General Maths question.

$\noindent The formula $\ell = r\theta$ is for when $\theta$ is in radians. If the angle is in degrees, say $\phi ^\circ$, the arc length would be $\frac{\phi}{360}\times 2\pi r$.$

There isn't any linear algebra in the NSW HSC maths courses now, though there used to be some a few decades ago.

$\noindent For the second case, we have $-1 \leq x \leq 2$, so in this interval, $x+1 \geq 0\Rightarrow |x+1| = x+1$. Also, in this interval, $x-2\leq 0$, so $|x-2| = 2-x$ again.$