Final Tips for 4U + some questions

[fkkiwi]

Hey guys,

2 days to go until Extension 2!!! Just wondering what some final tips are for the last 2 days of prep? I'm comfortable with most topics (except combinatorics, occasionally inequality and miscellaneous questions like the ones attached).

In regards to the questions attached, what topic would they be under? I'm thinking harder 3U but I'm not sure since they don't really seem to belong to any topic. If there are questions like this in the exam which aren't really classified within a certain 4U topic, how would you suggest approaching it?

[RuiAce]

The first one, whilst not necessarily obvious to do, is certainly a usual question in the graphs topic. The course required you to understand how implicit differentiation would be useful in curve sketching. The course also taught how absolute values cause reflections in the resulting graph.

The second counts as harder 3U and whilst it isn’t exactly an inequality proof, should be treated like one. I don’t imagine that the answer would be immediately obvious.

Whereas the third is just a 2U question however with considerably greater difficulty.

There’s not much more that can be said for the first, unless you require a full solution. The second one would probably have been a question I left until the very end to try out, but I believe that the third should be doable using only rules taught in the 3U/2U course.
(I might make an attempt at the second one later myself if nobody else beats me to it)

[fkkiwi]

Hi Rui,

On your first note, how is implicit differentiation useful in curve sketching because I was never taught that connection in class.

Do you happen to have a solution for the 2nd question?

I will give the 3rd one another crack later

[RuiAce]

I had an attempt at the second one and also checked with some other sources. From what I've seen, the intended method was actually just to use techniques from the 2U quadratics topic, which was something I wanted to avoid. (This is more or less NESA's given solution, but with more details)






I'll return to anything else here in a bit

[RuiAce]

To answer the original question briefly:
Given that there's only 2 days left, there's at most 4 past papers that you can realistically do (i.e. 2 a day), and optimistically at most 6. That's assuming that you temporarily have enough time to focus studying on MX2. So out of the remaining past papers, think about which ones you would want to do. Any paper from 2001 onwards is viable, but as a rule of thumb they get increasingly harder the further back you go.

Ensure that you're set on all of the comparatively basic stuff (e.g. usual integration by parts) and have a firm idea of where your silly mistakes are, so that you know what to watch out for when checking your responses during the exam. Also remind yourself to frequently look at the clock.
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This is, of course, a point where the curve 'flattens out' and the tangent to the curve becomes horizontal. As opposed to the next case.

And thus as \(x\to 0^+\), the curve slops further and further upwards.

[3.14159265359]

this is probably really dumb because its only 1 mark but this is what I did

dy/dx= ay(1-y)

max value when dy/dx=0

dy/dx=ay(1-y)=0
y=o or y=1

so where the freak does 1/2 come in the story??

[RuiAce]

this is probably really dumb because its only 1 mark but this is what I did

dy/dx= ay(1-y)

max value when dy/dx=0

dy/dx=ay(1-y)=0
y=o or y=1

so where the freak does 1/2 come in the story??

You're maximising the wrong thing. You've set \(\frac{dy}{dx} = 0\), but that maximises \( \boxed{y}\).

We actually want to maximise \( \boxed{\frac{dy}{dx}} \) itself. In theory, we would set \( \frac{d^2y}{dx^2} = 0\) to do this. But because we can't compute the corresponding second derivative, we just use our knowledge of quadratics to maximise it instead.

[3.14159265359]

You're maximising the wrong thing. You've set \(\frac{dy}{dx} = 0\), but that maximises \( \boxed{y}\).

We actually want to maximise \( \boxed{\frac{dy}{dx}} \) itself. In theory, we would set \( \frac{d^2y}{dx^2} = 0\) to do this. But because we can't compute the corresponding second derivative, we just use our knowledge of quadratics to maximise it instead.

ohhhhhhhhhhhh

we just use our knowledge of quadratics to maximise it instead.

wdym?? and how?

I'm sorry I know this is dumb because its only 1mark

[RuiAce]

ohhhhhhhhhhhh
wdym?? and how?

I'm sorry I know this is dumb because its only 1mark

The maximum of a quadratic that concaves down occurs at its axis of symmetry, which is halfway between the intercepts (in this case at \(y=0\) and \(y=1\)).

[3.14159265359]

The maximum of a quadratic that concaves down occurs at its axis of symmetry, which is halfway between the intercepts (in this case at \(y=0\) and \(y=1\)).

ohhhhhhhhh that makes sense, thank you!!!