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Move the negative to the other side (in other words, multiply both sides by -1).

$\noindent When $x<-1$, we have $x+1<0 \Rightarrow |x+1| = -(x+1) = -x-1$ and $x-2<0 \Rightarrow |x-2| = -(x-2) \Rightarrow -|x-2| = x-2$. So the equation for this case becomes: $-x-1 + x-2 =2 \Longleftrightarrow -3=2$. This is a contradiction. Hence the given equation is never satisfied if...

$\noindent Split it up into three intervals (cases): $x<-1$, $-1\leq x \leq 2$, and $x\geq 2$. Find how to simplify the absolute value signs in each interval using the definition of the absolute value sign as $\left | \xi \right |=\begin{cases}\xi & \text{ if } \xi \geq 0 \\ -\xi & \text{ if }...

Re: MX2 2016 Integration Marathon $\textbf{NEW QUESTION}$ $\noindent Find $\int \frac{1}{\left(x^2 -1 \right)^2}\text{ d}x$.$

Re: HSC 2016 4U Marathon - Advanced Level It's the point of intersection of the altitudes. An altitude is the perpendicular from one vertex to the opposite (possibly extended) side (doesn't necessarily meet it at the midpoint).

Re: HSC 2016 4U Marathon - Advanced Level Do you want to post your GeoGebra picture?

Try putting the maths stuff between dollar signs, i.e. write $\frac{2}{5}$.

$\noindent By part (i), an anti-derivative of $\frac{2}{\left(9+x^2\right)^2}$ is $F(x):=\frac{1}{27} \left(\frac{3x}{9+x^2}+ \tan^{-1} \frac{x}{3}\right)$. Hence the integral evaluates to $F\left(\sqrt{3} \right) -F(0) = F\left(\sqrt{3}\right)$, as $F(0)=0$. I'll leave you to evaluate...

The special allowance is a one-off for having at least one emergency call-out in the week.

$\noindent The main trick is that the double-time emergency payments are for at least 4 hours. So if he did emergency work for fewer than 4 hours, it gets treated as though it was for 4 hours. So his emergency work for the second time was worth a payment of $\boxed{4\times 2\times \$ 22.35}$...

Is this HSC General Maths material (or is it from somewhere else)? Also, we don't seem to be told whether Wendy has dependants, or her partner's income. So do we have enough information to do the question?

$\noindent The equation is $2x^{\frac{3}{2}}-5x + 2\sqrt{x} = 0$, so we'll assume $x$ needs to be non-negative. If we substitute $u=\sqrt{x}$, the equation becomes $2u^3 -5u^2 + 2u=0 \Longleftrightarrow u\left(2u^2 - 5u +2\right)=0$. Factorising the quadratic term: $2u^2 - 5u +2...

Have you learnt how to rationalise the denominator of a single fraction? If so, these questions are just based on that. Rationalise each denominator, and then add the fractions with rational denominators by taking a common denominator. If you haven't yet learnt how to rationalise denominators...

Re: HSC 2016 4U Marathon $\noindent Note that $z =\frac{1}{u-1}$, $u\neq 1$ (so we lose the solution of $u=1$). Also, you may find it helpful to show that $\frac{1}{\mathrm{cis}\, \theta -1} = -\frac{1}{2} \left(1+i\cot \frac{\theta}{2}\right)$, where $\theta \in \mathbb{R}$ and $\sin...

Re: HSC 2016 3U Marathon $\noindent It could still be done via the chain rule if one only knew the rule for $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin^{-1} x \right)$.$

Re: HSC 2016 3U Marathon It's a standard differentiation formula as davidgoes4wce showed in his post.

Re: HSC 2016 3U Marathon $$\frac{1}{\sqrt{9-x^2}}$$

Re: HSC 2016 3U Marathon $\noindent What do you mean about how to find the initial one? The initial one is given to us. Just keep iterating that process (do it three times) so that you obtain the next three approximations, $x_1, x_2$ and $x_3$. So once you find the first approximation $x_1$...

$\noindent I didn't mention the case where precisely one of the altitudes is outside, but this isn't too important. I mentioned the case where both were outside (i.e., $P$ on minor arc) because it was slightly different. Even in the case where exactly one altitude is outside (which is covered by...

I assumed they were allowed to, as I mentioned for one of the cases.