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Maybe you expanded the \left(\frac{12}{y} - 1\right)^2 as \left(\frac{12}{y}\right)^2 + 1 instead of as \left(\frac{12}{y}\right)^2 - \frac{24}{y} + 1? This would get the 800pi but miss the log term...

Also, 11e is easy by sketching y = sin x and y = cos x and adding... the solution x = 0, pi/2, 2pi drops immediately.

Simplification is not required / mandatory outside of "show" questions. Both one over root 2 and root 2 over 2 would be accepted, for example. I am confident that both + ln(1/3) and - ln3 would be accepted. Not changing 12 / 4 to 3 is careless, and leaving 4 / 12 instead of 1 / 3 isn't good...

For 14(c)(ii), the product rule can be avoided. Working towards the form of A that I have here may be quicker, too. \begin{align*} A &= Rr|\sin{t}(1 + \cos{t})| = Rr|\sin{t} + \sin{t}\cos{t}| = \frac{Rr}{2}\left|2\sin{t} + 2\sin{t}\cos{t}\right| = \frac{Rr}{2}\left|2\sin{t} + \sin{2t}\right|...

For 14(c), Let p be the projection of a onto b. Then the vector a - p is the the altitude of the triangle with base as b, and so area of triangle, A = |b||a - p| / 2 It's another "otherwise" approach...

Yes, 10 is B. The students must be arranged as two pairs plus a single student. Seat the first teacher in 1 way. There are three ways to seat the other two teachers, and 2 ways that they can be arranged. The five students can then be seated in 5! ways Thus, arrangements = 1 x 3 x 2 x 5! =...

For the volume question, the bottom part is just a cylinder with V = \pi r^2h=\pi(10)^2 \times 4 = 400\pi\ \text{cu units}, so skip integration. For the top part, we can make the numbers a lot easier to deal with... \begin{align*} V &= \int_4^{12} \pi x^2\ \!dy \qquad \qquad \qquad \text{where...

I get 8 is A and 10 is B

Q9 is definitely D

If we consider a cubic, the polynomial P(x) = Ax^3 + Bx^2 + Cx + D being monic requires that A = 1. However, if we have a cubic polynomial in an equation, like 2x^2 - x^3 + 7x - 4 = 0 then it could be described as monic because it can be re-written as x^3 - 2x^2 - 7x + 4 = 0 and hence...

The x-axis here is wrong. It is not the [FeCl2] and if it was, that would be in mg/L rather than in ppm. The value being plotted is actually the concentration of iron(II) cations (in which case the units mg/L and ppm are equivalent): [Fe2+] / ppm.

To your question opening the thread: You can, but it would be unwise... it is much better to tailor your response to the situation and the specific chemicals involved. In this question, the question is: State ONE safety concern associated with organic liquids and suggest ONE way to address...

I think that many of these difficult MCQs are more easily approached by eliminating incorrect answers rather than identifying the correct one. This can be a case of applying examination technique strategically, which can be of questionable educational value... but it can also be an application...

I am sure that many people here have attempted the North Sydney Boys Extension 1 Trial from this year. I was looking at it earlier today and noticed that none of options provided are correct. I am wondering if anyone else has noticed that this MCQ does not offer a correct answer. The question...

No, the answer B is correct. Sketch the inverse by reflecting the given graph in y = x. That's g(x) g(|x|) is the same as g(x) for x > 0... this starts at (0, a) and decreases to (d, b)... but g(|x|) is also even, so g(|x|) starts at (-d, b), rises to a y-intercept of (0, a), then decreases...

A function and its inverse must meet on y = x, but if a function is its own inverse (like for xy = 1 or y = -x) it will meet itself at every point in its domain. f(x) = -x is a function that meets all the criteria specified and it is clear for it that (A), (B), and (C) are false.

The magnitudes of the two tension vectors are the same, but the two vectors are different.

Yes, I don't use it.

A faster way is to recognise that x = 2 + 4\sin^2{t} is an equation of simple harmonic motion. Since 0 \le \sin^2{t} \le 1 \quad \implies \quad 2 \le x \le 6, it follows that the amplitude of motion is A = 2 m. The period of the motion is \pi, and over each complete period, the particle...

There are some markers who passionately hate cis... though that's more likely to be an issue in trials than in the HSC, where the markers will be required to conform to the same standards. Making "cis" and "cos" clearly different is important, though. This can be dealt with by writing...