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Re: HSC 2018 MX2 Integration Marathon In the case of a misprinted question, I think the best thing to do here would just be to give everyone the marks.

(Draw a diagram if you haven't already - this makes it much easier) From the given axis ( x = 0) and latus rectum ( y = -3), the focus of the parabola must be (0, -3) . (The latus rectum must pass through the focus) Since the vertex is the origin, that means the equation of the parabola is...

Re: HSC 2018 MX2 Integration Marathon Are you sure the integrand is correct?

My school's math department writes their own papers. Apparently the difficulty is on par with the HSC.

87% Maths Ext. 2 82% Maths Ext. 1 87% Software Design 57% Chemistry 60% English Adv. Chem mark looks bad but was slightly above average because our marking is really harsh. 3U was terrible and my worst result in trials by a long shot.

Given that D(k) - kD(k-1) = -[D(k-1) - (k-1)D(k-2)] for some k > 2 . Set k =n to get (1):\, D(n) - nD(n-1) = -[\color{red}{D(n-1) - (n-1)D(n-2)}\color{black}] That red expression on the RHS looks very similar to the LHS. Set k = n-1 to get (2):\, \color{red}{D(n-1) -...

If \arg(z+z^2) = \pi/3 , then \tan \frac\pi 3 = \frac{\mathrm{Im}(z^2+z)}{\mathrm{Re}(z^2+z)} But since \tan\pi/3 = \tan -2\pi/3 , \tan \frac\pi 3 = \frac{\mathrm{Im}(z^2+z)}{\mathrm{Re}(z^2+z)} also gives the locus of \arg(z+z^2) =-2\pi/3 . To get around this we can require z^2...

Using your equations for time, make v the subject. Integrate again to get x as a function of t , and then solve x_1 + x_2 = h. Note that C = g/k \implies k = g/C.

A parachute maybe? Or perhaps glue a bunch of springs to it.

First the 20% discount is applied, then the 10% discount is applied onto the new price. So Zoe saves 20% initially. That means the price is now 80% of what it was. She receives a further 10% discount off this new price. 10% of 80% is another 8% saved, with respect to the original price...

Let l = 1/k . \begin{aligned} \Delta &= (4m)^2\left(k-\frac 1k\right)^2 - 4(4m^2)\left(k^2 + \frac{1}{k^2}\right) \\ &= 16m^2\left(k^2 - 2 + \frac{1}{k^2} - k^2 - \frac{1}{k^2}\right) \\ &= 16m^2(-2) \\ &= -32m^2\end{aligned} which is negative for all nonzero real m .

Have you tried expanding (k+1)^3 + 5(k+1) and rearranging?

You only need the substitution l = 1/ k and it is possible to prove the resulting expression is always negative regardless of the value of k .

It is given that the y co-ordinate of P increases at one unit per second. That is, \frac{d(ap^2)}{dt} = 1 We want to use part iii) to answer the question. That means we need to find dp/dt. \frac{dp}{dt} = \frac{d(ap^2)}{dt} \cdot \frac{dp}{d(ap^2)} And, \frac{dp}{d(ap^2)} =...

You have to find a balance between rushing through a question too fast and not leaving enough time for yourself, which can be hard to do if you're prone to screwing up. First, always read the entire question properly. Then after answering the question do a quick skim over it and make sure your...

Not really. Although ellipses are covered in depth in 4 unit, it mainly deals with their properties like the directrices, foci and eccentricity. None of that knowledge is required to answer the question. This is a perfectly reasonable question for a 2 unit student to be asked. The equation...

Actually forget what I said, I misread the question. The easiest way to solve this would probably be graphing. You could: i) graph | x + 2 | = y and | x -3 | = y , add their graphs and check where it equals 5, or ii) square both sides (which doesn't introduce new solutions in this case...

1. \frac{1}{1-\sin^2x} = \frac{1}{\cos^2 x} Therefore, this expression is undefined when \cos x = 0 . 2. You can either take cases, or use the definition |x| = \sqrt {x^2} . (you will end up with x^2 - x - 6 = 0)

Actually in this case, since the initial velocity is negative, a positive acceleration means that the particle would be doing a U-turn, as the velocity goes from negative to zero (and eventually to positive). The magnitude of velocity (speed) would therefore be decreasing.

Conventionally, the right is taken as the positive direction (like on the Cartesian plane), a negative velocity means the particle is moving left. Acceleration is the change in velocity. Since it is negative, the velocity is decreasing over time, i.e. moving further in the negative direction...