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In the mythical land of 5U :p

Yeah. For most questions in nU Carrotsticks trials, they would probably only appear in (n+1)U (or higher) papers :p

I think this question was from a Carrotsticks Trial, designed to trip people up. If it came up in the 2U HSC paper, it'd probably have a few sub-parts to it.

Also, when we get the x <= 0 case in case i), we can immediately say this gives us just x = 0, since case i) had x >= 0. In other words, for each case, we take the intersection of the solution sets with the condition for the case (e.g. for case i), we take the intersection of ''x <= 0 or x >=...

$\noindent This is when $x\geq 0$ as the Case 1 line says, so $x=-2$ isn't under consideration. In other words, we're considering the case where $|x|$ is just the same as $x$, and we're saying this is equivalent to $x$ being non-negative (which is true, since the definition of the absolute value...

$\noindent If we sketch the graph correctly, we don't need to actually `test' the solution of $x=0$ (i.e. it will come straight from our sketch rather than having to do a separate test to `find' it), because it's clear that the `middle' of the `W' will be at $x=0,y=1$ (of course, it's good to...

$\noindent Yes, $x=0$ is a solution, since we get equality in that case, as seen by direct substitution into the L.H.S. Here is an algebraic method to solve this inequation.$ $\noindent \underline{\large Case 1. $x \geq 0\Longleftrightarrow |x| = x$}$ $\noindent Now the inequation is...

$\noindent I think it's decently easy to do graphically too (probably easier than algebraic method), as long as you're careful not to miss out on a solution. Basically the graph comes out to be this sort of W shape. We will see that the solutions are: $x\leq -2$, $x=0$ (this is the one easy to...

The hardest part is visualising the diagram correctly, unless you are accustomed to spherical geometry (which may well be the case for General Maths students). I'll try my best to describe my diagrams. $\noindent Consider this diagram:$...

Is that the Sydney Morning Herald?

Why don't the unofficial school ranks get based on ATAR instead (e.g. median ATAR)? With the Band 6 system, school A could have 100% band 6's in the form of a 90 mark from every paper, whilst school B could have less than 100% Band 6 rate but with a much higher ATAR median, yet School A would be...

Yeah, if you're trying to find for which values of lambda the equations have no solution, just try and make diagonal entries 0, and investigate what happens (it's still possible that there solutions, because the right hand column may also be 0 where there's a zero row, depending on the situation...

Always try the ones that make diagonal elements 0. If a diagonal element is 0, it means the matrix is either not yet in row echelon form (e.g. Plugging lambda = 2 makes the matrix require further operations to get into row echelon form, since then the third column needs to be cleared up), or we...

Because then the denominator of the RHS would be 0, and we can't divide by 0. Edit: oh you said why *can*. Well, it can't be -1 in fact, as you had.

What do you mean? Plug. in lambda = 2 to the matrix you posted. Then interpret the rows as equations as usual.

Don't need to know about determinants for this. Sub. lambda = -2, then equation 2 (row 2) says that -z = 1, whilst row 3 says that -12z = 3, a contradiction.

Everyone here has explained that y needs to be positive, but to truly claim the range is y>0, we would also need to show that is indeed every positive y value can be attained (e.g. for the function y = 1/sqrt(1+x^2), we also have y > 0 always, but the range is not just y > 0, because in fact y...

I think it is also common for HSC students to think something like d/dx ∫f(x) dx = f(x) is an application of FTC.

Did he get the marks for the Pascal's theorem one? Since he basically assumed the result if he just quoted the theorem as his proof.

Ah yeah, that was the Q. I was thinking of (the irrationality of pi one, I think it's from 2003).