Search results

The cis notation is actually not explicitly in the old syllabus. It seems to have been introduced by teachers and textbooks as their own abbreviation. The problem I have using Euler’s formula in the HSC is that you have to introduce it without proof - otherwise you have to go beyond the...

The change of sign alone doesn’t tell you which way around the inequality is. You need to know the behaviour of the function in that domain to be able make that conclusion. You are slightly saved by the fact that x2 can be explictly shown to be greater than x1 in the particular example.

The question specifies a particular term. Therefore the order matters and because the equation is linear, there can only be one value for k, not two. If you did (5x)^k*2^(n-k) the solution you find for k identifies the coefficient of x^k. If you did (5x)^(n-k)*2^k then the solution you have...

Not really, because the x is attached to the 5. The former is the coefficient of x^k and the latter is the coefficient of x^(n-k).

For i) since the two graphs intersect only once then there is only one real solution where the y-values are equal -then rearrange to get the result. For iii), I came up with this: a^3 - ka^2 - 1 = 0 a^2(a - k) = 1 a = k + 1/a^2 Since a^2 is positive then a > k and it follows that k + 1/a^2 < k...

Depends on what it is. If it’s just “hey just look at the graph therefore it is true” then no.

Not really, NESA does it because the HSC exam is meant to cover all key topics across the syllabus. It would make no sense for them to suddenly skip an entire topic in the HSC exam.

The equation of the tangent is in the reference sheet. Therefore you are allowed to quote it.

It’s not really a “short cut” tbh. The only difference is you either simplify the products separately then multiply or you multiply them first then simplify. If anything, the latter is more prone to error.

So if you had V = f(r) By “implicitly differentiating” you get dV/dt = f’(r)*dr/dt The marker will read this as equivalent to using dV/dt = (dV/dr)*(dr/dt) anyway because you’re not really doing anything special outside the 3u syllabus. You are literally just using that very chain rule the...

...and here is the latest version of Mathematics Extension 1 paper, solutions (now with colour in the graphs!) and results. And that's wrap so thanks to everyone who helped and participated this year! :) To the students, best of luck for the Mathematics Extension 1 HSC exam!

Implicit differentiation IS the chain rule. It's not exactly a new concept but rather just applying the chain rule in a different context. If you asked a 3u student to differentiate x[f(x)]^2, it's pretty much the same as differentiating implicitly xy^2 where y=f(x). Both approaches explicitly...

The questions are usually designed so that you don’t need to use 4u techniques at all. Otherwise it disadvantages those who only do 3u.

Depending on what your goals are you could try the BoS trials

Further feedback for students after marking Q12 of the Extension 1 paper. • For the inverse trigonometric functions question, almost every student got the derivative of f(x)=\sin^{-1}(2x^2-1) incorrect. The most common mistake was asserting that...

Just a few remarks on the parametrics question after marking Q11 for the Mathematics Extension 1 paper: In (b)(ii) far too many students derived the coordinates of P' (intersection of tangents at P and R) by deriving the equations of tangents at P and R and then finding their intersection. This...

Yes everything in 2u is definitely in scope for the ext1 exam. gg if they do cover that in the exam lol

You shouldn’t be using your actual school results in an ATAR calculator, let alone ones from Year 11. Every school has different standards of marking whereas the actual HSC puts everything on the same standard. The inputs for most ATAR calculators are the HSC marks you will actually get. So for...

When you solve that quadratic equation in terms of t, you are effectively finding the times where y=h

Yes