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jamonwindeyer

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Mathematics Extension 1: A Complete Guide to the Course!
« on: June 21, 2015, 02:35:43 pm »
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For 2016 Half-Yearlies, we have a thread for questions here and we also offer free online HSC tutoring!

Hey all! Here are all the guides I have written for Extension linked in one place!! If you have any questions, ask away in the respective thread!!

Note: Some of these guides are under maintenance following a site upgrade. If you spot something that doesn't look right, let me know and I'll move that guide to the front of the list

Extension Trigonometry : A separate guide on the additional 3 unit content!

Extension Calculus
Spoiler
Hey everyone!

This guide is going to cover the extra parts of calculus in the Extension 1 course. These are usually some of the hardest parts of an exam, so read carefully, think about the examples, and of course, ask questions! You can register for an account here. There are awesome notes here . And finally, be sure to check out the 2 unit guide on differentiation .

So there are a few, fairly distinct areas to cover here, so we'll go one by one. First, rate of change questions. These have the potential to be quite nasty, but really, all you need to know is this basic idea:

$\frac { dy }{ dx } =\frac { dy }{ da } \times \frac { da }{ dx } \\$

The rest of the work comes with tricky proofs, usually linking volume, radius, and height of a container. The best way to prepare for these is practice, so let's look at one from 2013

EG:A spherical raindrop of radius r metres loses water through evaporation at a rate that depends on its surface area. The rate of change of the volume V of the raindrop is given by
$\frac { dV }{ dt } ={ -10 }^{ -4 }A\\$
where t is time in seconds and A is the surface area of the raindrop. The surface area and the volume of the raindrop are given by $A=4\pi { r }^{ 2 }\quad \\$ and $V=\frac { 4 }{ 3 } \pi { r }^{ 3 }\quad \\$ respectively.

(i) Show that $\frac { dr }{ dt }$ is constant.

(ii) How long does it take for a raindrop of volume ${10}^{–6} {m}^{3}$ to completely evaporate?

This example is fairly straightforward, but a good way to introduce the concept. We require the change in r with respect to time. We have the change in volume with respect to time. So we can use the formula above to get what we need as shown:

$\frac { dV }{ dt } =\quad { -10 }^{ -4 }A\\ \\ \quad \quad \quad \quad =\quad { -10 }^{ -4 }.4{ \pi r }^{ 2 }\\ \\ V=\frac { 4 }{ 3 } \pi { r }^{ 3 }\\ \\ \frac { dV }{ dr } =4{ \pi r }^{ 2 }\\ \\ \therefore \quad \frac { dr }{ dt } =\frac { dr }{ dV } \times \frac { dV }{ dt } \\ \\ \quad \quad \quad \quad \quad \quad =\frac { 1 }{ 4\pi { r }^{ 2 } } \times \frac { -4\pi { r }^{ 2 } }{ { 10 }^{ 4 } } \\ \\ \quad \quad \quad \quad \quad \quad =\quad -1{ 0 }^{ -4 }$
which is a constant.

For part (b), the easiest way to consider this is when the radius is zero. We first need to know the initial radius. In the interest of brevity, I'll skip the algebra:

$\frac { 4 }{ 3 } { \pi r }^{ 3 }\quad =\quad 1{ 0 }^{ -6 }\\ \\ \therefore \quad r=\frac { 1 }{ 100 } \sqrt [ 3 ]{ \frac { 3 }{ 4\pi } } \\ \\$

From part (a), we know that the radius decreases at the constant rate found. This makes the next part easy:

$\frac { dr }{ dt } =-1{ 0 }^{ -4 }\\ \\ \therefore \quad r=-1{ 0 }^{ -4 }t\\ \\ t=-1{ 0 }^{ 4 }r\\ \\ \quad \approx \quad 62\\ \\$

So the answer is 62 seconds. The hardest part of these questions is definitely knowing what to do with the data. Do some derivatives, write down everything you know, and look for the magic pairing. It will be there.

Next is projectile motion. These are usually one of the last questions in the exam, and there is no other way to say it... They are brutal. These are the questions which will earn you a Band E4, if you can do them. Now there is an almost infinite number of things they can ask, but one thing that pops up a lot of deriving the standard formulae you learn. In fact, it's normally worth a good 4-5 marks! Let's make sure you can do it:

Remember that we can find velocity by integrating with respect to time, and position by integrating with respect to velocity. The hardest part is considering the constants of integration. Make sure you memorise this process. First the y (vertical) equations:

${ a }_{ y }=\quad -g\\ \\ { v }_{ y }=\quad -gt\quad +\quad { C }_{ 1 }\\ \\ t=0,\quad { v }_{ y }=Vsin\theta \\ \\ \therefore \quad { C }_{ 1 }=Vsin\theta \\ \\ { v }_{ y }=Vsin\theta -gt\\ \\ { s }_{ y }=Vtsin\theta -\frac { 1 }{ 2 } g{ t }^{ 2 }+{ C }_{ 2 }\\ t=0,\quad { s }_{ y }=0,\quad \therefore \quad { C }_{ 2 }=0\\ { \therefore \quad s }_{ y }=Vtsin\theta -\frac { 1 }{ 2 } g{ t }^{ 2 }$

The x equations are much easier in comparison:

${ a }_{ x }=\quad 0\\ \\ { v }_{ x }=\quad { C }_{ 1 }\\ \\ t=0,\quad { v }_{ y }=Vcos\theta \\ \\ \therefore \quad { C }_{ 1 }=Vcos\theta \\ \\ { v }_{ x }=Vcos\theta \\ \\ { s }_{ x }=Vtcos\theta +{ C }_{ 2 }\\ t=0,\quad { s }_{ x }=0,\quad \therefore \quad { C }_{ 2 }=0\\ { \therefore \quad s }_{ x }=Vtcos\theta \\ \\ \\ \\$

Remember that V is the initial speed, and it is separated into it's xy components with some simple trigonometry. These questions are varied in what they ask, and the best way to prepare is to practice as many past questions as possible. However, some tips for these questions:

1- Remember that maximum range is achieved when the angle of projection is 45 degrees!
2- Examine the equation you need to derive. The trig functions are a big clue. How do you get to tan from sin and cos? Never just start manipulating, think about where you are heading and how you might get there.
3- Remember that, all other things constant, two different launch angles give the same landing position. This factors into lots of questions.
4- Draw a diagram! I cannot recommend this enough, it will give you best idea of what's going on, and make your logic clearer for the marker.

The final big area is simple harmonic motion. These are easier than projectile questions, but they can still be tricky! Be sure to remember the few key formulas:

$a=-{ n }^{ 2 }x\\ \\ T=\frac { 2\pi }{ n } \\ \\ x=Asin(nt+\emptyset )\quad /\quad Acos(nt+\emptyset )\\ \\ \\$

Simple harmonic motion occurs when the particle oscillates about an equilibrium position. It is a consequence of the properties of the acceleration; the first formula shows this, and is what you must prove to prove SHM is occurring. A is the amplitude of the motion, it indicates how far the particle moves from the equilibrium position, indicated by the phase constant. If the formula is a sine formula, the particle starts at this equilibrium position, if it is cos, it starts at it's max/min position.

Knowing these formulas and their consequences makes most questions fairly easy. Take this question I did in my HSC:

EG: A particle is moving in simple harmonic motion about the origin, with displacement x metres. The displacement is given by x = 2 sin 3t, where t is time in seconds. The motion starts when t = 0.
(i) What is the total distance travelled by the particle when it first returns to the origin?
(ii) What is the acceleration of the particle when it is first at rest?

For part (i), we can immediately read the amplitude of 2 from the equation, so the answer is 4. BE CAREFUL HERE GUYS. It travels to the end point, then back, so the answer is 4, not 2.

For part (ii), if you know your SHM, we know that the particle is first at rest when it reaches it's maxima (2 units from the origin). We use the standard formulae for a simple 2 marks:

$a=-{ n }^{ 2 }x\\ \\ x=2,\quad n=3\\ \\ \therefore \quad a=-18\\ \\$

The actual worked solution provided by BOSTES for this question is 15 lines long, 15! Just by knowing a little bit more about SHM, these questions become easy, so remember these two other key facts:

1- Velocity is maximum at the equilibrium, and zero at the endpoints
2- Acceleration is maximum at the endpoints, and zero at the equilibrium

Of course, these are not the be all and end all of potential questions. Miscellaneous questions on velocity and acceleration pop up a fair bit, and be prepared for more complex applications of the chain, product and quotient rules. And of course, you have a few extra derivatives to remember as well. The trick is knowing your formulas like the back of your hand... If you have this, you can focus on interpretation and getting logically developed solutions. In terms of approaching the really tricky projectile questions, practice makes perfect! And of course, if you get a specific question you want examined or explained, post it here and I will post a worked solution for everyone to benefit from!

That's about it for this guide. I hope these are proving beneficial, feel free to post some feedback below, check out the notes, and talk to each other! You have an amazing community to capitalise on

GUIDE BY JAMON WINDEYER

Circle Geometry
Spoiler
Hello all! I've just finished writing a guide to 2 unit geometry, check it out here (INSERT HYPERLINK. In this guide, I want to add to that by giving a brief run down of what to expect in terms of circle geometry in a HSC/Trial Exam. It's the dark horse of extension exams... It's one of the first things you cover in Year 11 (usually), and lots of students think it won't be covered much in an exam. Wrong . In 2012 it was worth nearly 10% of the marks for the exam. That's the difference between a Band E3 and a Band E4. Read this guide, and it will brush you up on everything you should know. As always, remember to register and ask any questions you have below, and check out the awesome notes available for both 2 Unit and Extension 1.

In Extension One, circle geometry questions are best approached armed with a detailed knowledge of the list of rules (and there is a lot). Look at the question, and ask yourself, what is being shown? If there is tangents, think alternate segments. Are there chords? Think of your chord formulas. Are there distances involved? Think of your ratio formulas. Don't look at the whole thing with every formula in mind; break it down, and consider which formulas COULD be used, and go from there.

Let's introduce the concepts with a few questions.

Example One (HSC 2014) : The points A, B and C lie on a circle with centre O, as shown in the diagram. The size of ∠ACB is 40°. What is the size of ∠AOB? [/b]

This should bring to mind the following ideas:
• Chords which are equidistant from the centre of the circle are the same length and make the same angle with the centre
• The perpendicular from the centre of the circle bisects a chord
• The angle made by a chord at the centre of the circle is twice that made at the circumference (be careful of reflex angles here)

Only that last idea is relevant here, and from this idea, it is easy to see the answer is 80 degrees.

Example Two (HSC 2012): The diagram shows a large semicircle with diameter AB and two smaller semicircles with diameters AC and BC, respectively, where C is a point on the diameter AB. The point M is the centre of the semicircle with diameter AC.
The line perpendicular to AB through C meets the largest semicircle at the point D. The points S and T are the intersections of the lines AD and BD with the smaller semicircles. The point X is the intersection of the lines CD and ST.

a) Explain why CTDS is a rectangle.

Looking at this diagram, the three semicircles should immediately ring alarm bells for the idea that the angle in a semicircle is 90 degrees . I'll save myself the trouble of writing it three times in the proof below, but this fact along with basic straight angle ideas is all you need:

$\angle ADB=90°\\ \angle ASC=90°,\quad \therefore \quad \angle CSD=90°\\ \angle CTB=90°,\quad \therefore \quad \angle CTD=90°\\ \therefore \quad \angle ADB=90°\quad$

Since all angles are 90 degrees, CTDS is a rectangle. I'll skip the rest of that question, it goes more into easy 2U geometry from there.

Example Three (HSC 2014) : In the diagram, AB is a diameter of a circle with centre O. The point C is chosen such that Triangle ABC is acute-angled. The circle intersects AC and BC at P and Q respectively.

a) Why is ∠BAC = ∠CQP?

The answer here is the first of the two key properties of cyclic quadrilaterals, both of which should spring to mind when you spot this diagram.
• The external angle of a cyclic quadrilateral is equal to the opposite internal angle
• The opposite internal angles of a cyclic quadrilateral are supplementary

b) Show that the line OP is a tangent to the circle through P, Q and C.

This one is a little trickier, but as soon as the word tangent is mentioned, you should think angle in alternate segment . BOSTES loves to use it for some reason. The reasoning is as follows.

Let ∠APO=x degrees
So ∠OAP=x degrees (since Triangle APO is isosceles, equal radii, and the base angles of an isosceles triangle are equal)
So, from part(a), ∠CQP=x degrees
Now if we extend OP beyond the edge of the circle to some point N, the angle ∠CPN=x (vertically opposite at Point P). The solution diagram provided in the marking guidelines shows this nicely.

Now, since ∠CPN (∠CPT in the diagram), is equal to ∠CQP, then the angle in alternate segment theorem is satisfied (the acute angle formed between the tangent and a chord is equal to the acute angle formed by that chord at the circumference). Thus, the line OP is a tangent.

Example Four (HSC 2013): The circles C1 and C2 touch at the point T. The points A and P are on C1. The 3 line AT intersects C2 at B. The point Q on C2 is chosen so that BQ is parallel to PA. Prove that the points Q, T and P are collinear.

This one is again, a little trickier. It doesn't relate to any new ideas (and there are more, which I will summarise in a bit). Instead, it shows that just like the last question, you sometimes have to be proactive, and add in your own information to answer the question. Again, I think the diagram provided by BOSTES does the best job of explaining it, so I'll walk you through it.

We draw a common tangent between the two circles, which remember, allows us to use the angle in alternate segment theorem. Here is the proof.

We let $\angle PTY=\alpha ,\quad \angle QTX=\beta$

By angles in alternate segments, $\angle PAT=\alpha \quad \angle QBT=\beta$

But $\angle PAT=\angle QBT$, alternate angles on PA//BQ

So $\therefore \quad \alpha =\beta$, which means they are vertically opposite at Point T. As a consequence, Q, T and P must be collinear.

This isn't all of the rules there is. The main ones missed:

• Angles made at the circumference, standing on the same arc are equal. Thus, so are angles made at the centre
• Chords intersect in proportion, meaning that if chords AB and CD intersect at point X: $\frac { AX }{ BX } =\frac { CX }{ DX }$
• A similar formula works for the intersection of a tangent at Point T and an extended chord AB, according to the formula: ${ T }^{ 2 }=AX.BX$
• Tangents to a circle from a common, exterior point are equal

These appear less often than the earlier theorems, but they may appear, so be ready for them regardless.

That's it for this guide! Feel free to leave some feedback below, drop a question, give us some tips on how you approach these questions. And another reminder to check out the notes , they are seriously awesome. Have fun!

A GUIDE BY JAMON WINDEYER

Binomial Theorem and Induction

Spoiler
Hello once agin everyone! This will be the last of Extension guides  but we get to finish with the fun stuff! Induction and binomial theorem. These are perhaps the strangest parts of the Extension 1 course; they don't really fit in with anything else (the binomial theorem does have applications in more advanced probability questions), and they offer some of the hardest questions you'll face in an exam scenario. While it is impossible to go through everything that can be asked, I'll go through some common examples, and this guide will hopefully act as a nice refresher before Trials!

I highly recommend the notes available for HSC Math , free! They are awesome and super handy to have! You can also register for an account to post in these forums and share in each others expertise! Share the love

Okay, we'll begin with induction. Induction is a cool process in mathematics which is used to prove a variety of results. In the HSC, it is usually restricted to equality, inequality, and divisibility proofs. For all three, the process is reduced to three steps.

Step One: Prove the result for the smallest integer in the domain. For example, if it is asked to prove for $n\ge 1\\$, then you would prove for $n=1$.

Step Two: Assume the result true for $n=k$, where $k$ is an integer in the domain. Use this assumption to prove the result for $n=k+1$.

Step Three: Make a statement resembling the following: Therefore the result is true for $n=k+1$. Since it is true for $n=1$ (from Step 1), it is true for $n=1+1=2$, $n=2+1=3$, etc, for all $n\ge 1\\$.

Again, there are three main types of induction. The first is equality proofs of sum results.

Example One: Prove that $\sum _{ r=1 }^{ n }{ \left( 4r+1 \right) } =n\left( 2n+3 \right)$, for all positive integers.

We first prove the result for the smallest positive integer. When $n=1$:
$\sum _{ r=1 }^{ 1 }{ \left( 4r+1 \right) } =5\\\\ \quad \quad \quad \quad \quad \quad \quad \quad =1\left( 2+3 \right)$
$\therefore$ the result is true for $n=1$

Next we assume that: $\sum _{ r=1 }^{ k }{ \left( 4r+1 \right) } =k\left( 2k+3 \right)$, ie, the result is true for $n=k$. We then tie this into the proof for $n=k+1$. Remember, we seek to prove that: $\sum _{ r=1 }^{ k+1 }{ \left( 4r+1 \right) } =\left( k+1 \right) \left[ 2\left( k+1 \right) +3 \right]$.

$\sum _{ r=1 }^{ k+1 }{ \left( 4r+1 \right) } =\sum _{ r=1 }^{ k }{ \left( 4r+1 \right) } +\left[ 4\left( k+1 \right) +1 \right] \\ \\ =\quad k\left( 2k+3 \right) +\left( 4k+5 \right) \\ \\ =2{ k }^{ 2 }+7k+5\\ \\ =\left( k+1 \right) \left( 2k+5 \right) \\ \\ =\left[ k+1 \right] \left[ 2\left( k+1 \right) +3 \right]$

So the result is proven! And we just copy the statement from above as appropriate. No tricky math here! Just a bit abstract, lots of people struggle with proving a result with an assumption. But it is allowed, and extremely useful. Besides sum results, another common one is divisibility results. The trick with these is coming up with the condition. This is easy. If we want to prove something is divisible by 5 (for example), we are trying to prove it is equal to $5M$, where M is some integer. This proves divisibility by five. Check out the example below.

Example Two: Show that ${ 3 }^{ 2n }-1$ is divisible by eight for all positive integers n.

So, we want to prove that ${ 3 }^{ 2n }-1=8M$. Subbing in n=1 yields a correct result, so lets get to the good stuff.

Assume that ${ 3 }^{ 2k }-1=8M$ where k is some positive integer.

Now let $n=k+1$, and using the assumption above:
${ 3 }^{ 2k+2 }-1={ 9\left( { 3 }^{ 2k } \right) }-1\\ =9\left( 8M+1 \right) -1\\ =72M+8\\ =8(9M+1)$
which is divisible by 8. Thus, the result is true by mathematical induction (again, in an exam scenario, insert a concluding statement like the one above).

The final type are inequality proofs, which operate on the same principles as above. They aren't common in the HSC, but I'll quickly run through one.

Example Three: Prove that ${ \left( a+b \right) }^{ n }\ge { a }^{ n }+{ b }^{ n }$ for all positive integers n. $a>0$ and $b>0$.

Again, the result can be proven for $n=1$, and we assume true for $n=k$:

${ \left( a+b \right) }^{ k }\ge { a }^{ k }+{ b }^{ k }$

Now using this result:

${ \left( a+b \right) }^{ k+1 }=\left( a+b \right) { \left( a+b \right) }^{ k }\\ \ge \left( a+b \right) \left( { a }^{ k }+{ b }^{ k } \right) \\ \ge { a }^{ k+1 }+a{ b }^{ k }+{ a }^{ k }b+{ b }^{ k+1 }\\ \ge { a }^{ k+1 }+{ b }^{ k+1 }$

This proof is actually a little strange, check you understand it. We tie in the inequality from the assumption. In the last line, we omit terms, which we can do since both a and b are positive (since, if the result is greater than the result in the third line, it is definitely bigger than the result in the fourth, which is what is required). Let me know if there is any issues with this, because it is a bit weird. But again, the result is proven.

Induction is a cool process which is really, assuming the question isn't absolutely terrible, fairly easy marks! Treat the process like an essay, be clear and logical. Most of the marks lost will be lost from the marker not following your working.

Moving on now to binomial theorem, the general form of which looks like this (excuse the poor formatting):

${ \left( a+b \right) }^{ n }={ a }^{ n }+\overset { n }{ \underset { 1 }{ C } } { a }^{ n-1 }b+\overset { n }{ \underset { 2 }{ C } } { a }^{ n-2 }{ b }^{ 2 }...+\overset { n }{ \underset { n-1 }{ C } } { a }{ b }^{ n-1 }+{ b }^{ n }$.

This is an extremely useful theorem with ties into probability and other aspects of mathematics. Questions in the HSC will either be linked to probability, or abstract number proofs. In probability questions, a is the probability of one event occurring, and b is the probability of the other. Thus, $a+b=1$. Say for example, you were looking at basketball shots. You want the probability of getting 6 shots out of 10 when the chance of getting one is 60%. So, you set a as 0.6, b as 0.4, and n as 10, in the binomial expansion above. The probability of getting six shots is defined by the term where a has a power of six. Easy.

$P=\overset { 10 }{ \underset { 4 }{ C } } .\left( { 0.6 }^{ 6 } \right) \left( { 0.4 }^{ 4 } \right) \\\\ \approx 0.25$

The more difficult questions come in the form of tricky proofs. Now, there are literally infinite numbers of things they could ask you here, here is one of the slightly easier ones to get the feel of what the questions look like:

Example Four: Use the binomial theorem to show that $0=\overset { n }{ \underset { 0 }{ C } } -\overset { n }{ \underset { 1 }{ C } } +\overset { n }{ \underset { 2 }{ C } } -...+{ \left( -1 \right) }^{ n }\overset { n }{ \underset { n }{ C } }$.

This seems unusual, but actually very easy. The terms on the right represent a binomial expansion of ${ \left( 1-1 \right) }^{ n }=\overset { n }{ \underset { 0 }{ C } } -\overset { n }{ \underset { 1 }{ C } } +\overset { n }{ \underset { 2 }{ C } } -...+{ \left( -1 \right) }^{ n }\overset { n }{ \underset { n }{ C } }$

And of course, $1-1=0$, and ${ 0 }^{ n }=0$, therefore the result is proven!

Once the theorem makes its way to the back of the paper, things get nastier. 2013, for example, required a complicated binomial expansion proof in multiple steps. However, you are lead through, and given the opportunity to continue with given results even if you can't prove them yourself. Do this if you need too! It is way better to get some marks than none at all, don't skip any questions! Some tips for tackling these terrible, monstrosities of questions:
• Having to differentiate or integrate both sides is common. Check if this may be what is needed in the proof (you may be able to tell by examining what would happen to the powers)
• Linking to this, binomial proofs are one area where you may have to manipulate both sides to make a proof work. Using calculus as listed above is an example of such a case.
• Remember that $\overset { n }{ \underset { a }{ C } } =\overset { n }{ \underset { n-a }{ C } }$, it may come in handy with some proofs!

The best way to prep for these sorts of questions is to practice. Feel free to post any specific questions you wanted me to go over in the forums below! The examples in these guides are definitely not as tough as you could get, they are there as a quick refresher of questions of average difficulty. Make sure you are attempting the harder questions if you want to aim high, and again, let me know if I can help!

Well, that wraps up these series of guides. I hope they are helpful! I'm looking forward to doing material for other subjects, stay tuned!

Remember, if you need extra revision, check out the notes , which go into more detail than I do, designed for more detailed revision rather than refreshers/tips like these guides are. Be sure to register to get access to the notes, and be able to ask me questions! I'm here to help!

Ciao

Integration : Integration by substitution.

Probability: Combinatorics

Functions : Polynomials, root estimation and inverse functions!

Plane Geometry : Locus, and interval division
« Last Edit: April 06, 2016, 12:03:52 pm by jamonwindeyer »

IkeaandOfficeworks

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Re: Mathematics Extension 1: A Complete Guide to the Course!
« Reply #1 on: March 26, 2016, 10:12:36 am »
0
Hey Jamon! I'm just confused when to switch to radians or degrees in my calculator as i'm doing trig functions at the moment. Thanks!

Happy Physics Land

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Re: Mathematics Extension 1: A Complete Guide to the Course!
« Reply #2 on: March 26, 2016, 11:10:27 am »
+1
Hey Jamon! I'm just confused when to switch to radians or degrees in my calculator as i'm doing trig functions at the moment. Thanks!

Hey Ikea and Officeworks!

As a general rule of thumb, once you hit year 12, all of the 2u maths, 3u maths and 4th maths contents would encompass radians. If the question requires you to put your answer in degrees, then you will have to convert into degrees. You can tell when the question needs you to put your angles in radian forms: if the question expresses angles in terms of pi, you need to use radians, if the question puts angles in terms of small numbers such as 1.34 without a degree sign, its also radians. So in year 12, generally we use radians instead of degrees. One place in 3u maths that we do need to use degrees is projectile motion. Sometimes they can ask you for the direction of the resultant velocity. In cases such as these you would be required to convert into degrees (e.g. North 45 degrees West). Other than this, you barely need to use degrees.

Hope my answer helped in one way or another (hehe), if Jamon comes around he can express his opinion on this issue which might make it clearer for you!

Best Regards
Happy Physics Land
HSC Subject Choices:

Mathematics: 96
Maths Extension 2: 93
Maths Extension 1: 97
Physics: 95
Chemistry: 92
Engineering Studies: 90
Studies of Religion I: 98

Vyrocious

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Re: Mathematics Extension 1: A Complete Guide to the Course!
« Reply #3 on: April 02, 2016, 12:36:30 pm »
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The examples on Trigonometry are missing

jamonwindeyer

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Re: Mathematics Extension 1: A Complete Guide to the Course!
« Reply #4 on: April 06, 2016, 12:06:23 pm »
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@Jamon feel free to just use the quote tool and copy/paste the LaTeX onto your post to fix it

Cheers Rui! I actually already had this one fixed, but I didn't get the new version everywhere it should have been, thanks for covering in the mean time!!

fizzy.123

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Re: Mathematics Extension 1: A Complete Guide to the Course!
« Reply #5 on: October 05, 2016, 01:34:43 am »
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To get an E4 in MX1, what hsc mark do u need to get?

RuiAce

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Re: Mathematics Extension 1: A Complete Guide to the Course!
« Reply #6 on: October 05, 2016, 08:01:33 am »
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To get an E4 in MX1, what hsc mark do u need to get?
Refer to the raw marks database for values in the past.

The actual number will vary every year as every year's paper will be of differing difficulty.
Interested in Mathematics Tutoring? Message for details!

ATAR: 98.60
Currently studying: Bachelor of Science (Advanced Mathematics)/Bachelor of Science (Computer Science) @ UNSW
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fizzy.123

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Re: Mathematics Extension 1: A Complete Guide to the Course!
« Reply #7 on: October 22, 2016, 09:58:21 am »
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what mark out of 70 do you need for MX1 to get scaled up to a 85+?

RuiAce

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Re: Mathematics Extension 1: A Complete Guide to the Course!
« Reply #8 on: October 22, 2016, 10:06:48 am »
0

what mark out of 70 do you need for MX1 to get scaled up to a 85+?
If it's not in the database; then the answer is we don't know. Because the alignment algorithm isn't fixed every year
Interested in Mathematics Tutoring? Message for details!

ATAR: 98.60
Currently studying: Bachelor of Science (Advanced Mathematics)/Bachelor of Science (Computer Science) @ UNSW
Formerly studying: Bachelor of Actuarial Studies/Bachelor of Science (Advanced Mathematics)