# What raw marks in extension 2 generally correspond to state ranks? (1 Viewer)

#### TheOnePheeph

##### Active Member
Speaking of BoS trials....

*cough* *cough*
I do kind of want to do the BOS trial this year, but I am unfortunately outside of Sydney so it is a hassle to go up to do it. The past papers I have done from it have been quite a good challenge though (I think I've done 2018, 2017, 2015).

#### sharky564

##### Member
Did you end up making IMO?
no, unfortunately, i simply didnt have consistent enough performance :/

#### blyatman

##### Well-Known Member
There was a really good resisted motion problem in one of the BOS trials though where there was resisted SHM and you had to solve a second order linear differential equation (but it gave you the integrating factor and basically told you how to do it). Makes me wish differential equations were in the 4u syllabus, would make mechanics so much better. I reckon taylor series could also be in the syllabus instead of something like conics or graphs.
I think the main reason why Taylor series wouldn't be taught is because of the baggage that comes with it. Deriving the basic Taylor series is straightforward, but things like the remainder term, error bounds, radius of convergence (in the context of complex analysis) etc is more indepth that requires more than a surface understanding, and teaching that would feel somewhat out-of-place compared to the straightforwardness of all the other topics. Students don't even know the epsilon-delta definition of a limit, which is typically where you start with undergrad calculus. Likewise with ODE's: solving linear second-order ODE's is straightforward, but typically ODE's are paired with linear algebra, which students don't know.

Also, do you remember the SHM ODE? I'm curious, since I've never really seen integrating factors used to solve second-order ODE's, only in first-order ODE's.

#### TheOnePheeph

##### Active Member
I think the main reason why Taylor series wouldn't be taught is because of the baggage that comes with it. Deriving the basic Taylor series is straightforward, but things like the remainder term, error bounds, radius of convergence (in the context of complex analysis) etc is more indepth that requires more than a surface understanding, and teaching that would feel somewhat out-of-place compared to the straightforwardness of all the other topics. Students don't even know the epsilon-delta definition of a limit, which is typically where you start with undergrad calculus. Likewise with ODE's: solving linear second-order ODE's is straightforward, but typically ODE's are paired with linear algebra, which students don't know.

Also, do you remember the SHM ODE? I'm curious, since I've never really seen integrating factors used to solve second-order ODE's, only in first-order ODE's.
Here is the question:
I remembered it slightly wrong, it basically involves making a substitution of y=v + nx to make it a first order seperable equation and then solving the remaining first order differential equation in x using an integrating factor, which it basically gives to you. I guess its basically like how you would solve a second order homogeneous linear equation without knowing how to use the characteristic equation. I still thought it was a pretty cool way to incorporate differential equations.

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#### blyatman

##### Well-Known Member
Here is the question:
I remembered it slightly wrong, it basically involves making a substitution of y=v + nx to make it a first order seperable equation and then solving the remaining first order differential equation in x using an integrating factor, which it basically gives to you. I guess its basically like how you would solve a second order homogeneous linear equation without knowing how to use the characteristic equation. I still thought it was a pretty cool way to incorporate differential equations.
Ah right, yeh so you break down the second-order ODE to a first-order ODE before using the integrating factor - that makes more sense. Is the integrating factor necessary though? The solution to y' + ny = 0 is simply just y = y0*exp(-nt). Or am I missing something here...

#### TheOnePheeph

##### Active Member
Ah right, yeh so you break down the second-order ODE to a first-order ODE before using the integrating factor - that makes more sense. Is the integrating factor necessary though? The solution to y' + ny = 0 is simply just y = y0*exp(-nt).
The equation in y doesn't need an integrating factor, its just seperable, but when you subsitute v + nx back into it you get dx/dt + nx = y0exp(-nt), which does need an integrating factor.

#### blyatman

##### Well-Known Member
The equation in y doesn't need an integrating factor, its just seperable, but when you subsitute v + nx back into it you get dx/dt + nx = y0exp(-nt), which does need an integrating factor.
Ah right, I didn't work through the next step, but I see what you mean.

#### Trebla

Here is the question:
I remembered it slightly wrong, it basically involves making a substitution of y=v + nx to make it a first order seperable equation and then solving the remaining first order differential equation in x using an integrating factor, which it basically gives to you. I guess its basically like how you would solve a second order homogeneous linear equation without knowing how to use the characteristic equation. I still thought it was a pretty cool way to incorporate differential equations.
The resisted SHM question given there was a special case where the roots of the characteristic polynomial were equal. This lends itself to a more 'first principles' approach to solving it in a nice way without calling upon any ODE specific theory.

#### TheOnePheeph

##### Active Member
The resisted SHM question given there was a special case where the roots of the characteristic polynomial were equal. This lends itself to a more 'first principles' approach to solving it in a nice way without calling upon any ODE specific theory.
Yeah I would love to see something like this in the HSC this year. Without the hint to multiply by e^nt (this might be taking it too far though), I think something similar could be a really nice, challenging last question, much better than something like the geo question from last year.

#### UStoleMyBike

##### New Member
We all know what happened the last time they tried that *2014 HSC*. All the top students/state ranks literally did it in two steps because it could be done by inspection.

So na after that disaster I don't think they will haha

Which one? The last question? How could it be done by inspection (i had to use u sub)

#### TheOnePheeph

##### Active Member
Which one? The last question? How could it be done by inspection (i had to use u sub)
It was basically just reverse quotient rule

#### UStoleMyBike

##### New Member
It was basically just reverse quotient rule
Is this taught in the syllabus?

#### DrEuler

##### Member
Which one? The last question? How could it be done by inspection (i had to use u sub)
Yup what TheOnePheeph said.

Reverse quotient rule is a dodgy inspection method of getting the answer. I'll explain

So we have

$\bg_white I = \int \dfrac{lnx}{(1+ lnx)^2}dx$

Observe that the denominator is of the form (something)^2. The only way we can get this is when you differentiate something of the form u/v. As this will give you

$\bg_white \dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{vu' - uv'}{v^2}\hspace{2mm}\text{(Standard Quotient rule)}$

Notice the denominator is of the form (something)^2.

We can therefore say the Anti derivative takes the form of

$\bg_white \dfrac{f(x)}{1 + lnx}$

Now comes the dodgy part. You guess the numerator. Usually it's something simple and you should guess it as something that is similar to the integral. For example if the integral contained a bunch of sines and coses, you might guess sin(x) as the numerator.

If you test 1 and -1 it doesn't work but if you test x which is the second most simplest thing it checks out

And no this isn't taught in the syllabus. This is just intuition you develop over time or by a tutor or someone telling you

#### UStoleMyBike

##### New Member
Yup what TheOnePheeph said.

Reverse quotient rule is a dodgy inspection method of getting the answer. I'll explain

So we have

$\bg_white I = \int \dfrac{lnx}{(1+ lnx)^2}dx$

Observe that the denominator is of the form (something)^2. The only way we can get this is when you differentiate something of the form u/v. As this will give you

$\bg_white \dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{vu' - uv'}{v^2}\hspace{2mm}\text{(Standard Quotient rule)}$

Notice the denominator is of the form (something)^2.

We can therefore say the Anti derivative takes the form of

$\bg_white \dfrac{f(x)}{1 + lnx}$

Now comes the dodgy part. You guess the numerator. Usually it's something simple and you should guess it as something that is similar to the integral. For example if the integral contained a bunch of sines and coses, you might guess sin(x) as the numerator.

If you test 1 and -1 it doesn't work but if you test x which is the second most simplest thing it checks out

And no this isn't taught in the syllabus. This is just intuition you develop over time or by a tutor or someone telling you
Ahh, that's so simple I'm surprised I didn't notice it. Thanks for the explanation. Good to know that if I'm struggling on an integral with a power of 2 in the denominator I'll consider this.

Any other 'tricks' like this to watch out for? Most of the integrals in the HSC aren't too hard (usual by-parts or partial fractions or u-sub)

#### TheOnePheeph

##### Active Member
Ahh, that's so simple I'm surprised I didn't notice it. Thanks for the explanation. Good to know that if I'm struggling on an integral with a power of 2 in the denominator I'll consider this.

Any other 'tricks' like this to watch out for? Most of the integrals in the HSC aren't too hard (usual by-parts or partial fractions or u-sub)
They generally make it pretty clear when you have to do this, but using the property:

$\bg_white \int_a^{b}f(x)dx=\int_a^{b}f(a+b-x)dx$

Usually when it is done it is for something like trigonometric integrals, where you substitute pi/2-x, and change a sinx to a cosx. When you add this to the original integral, I , it simplifies it a lot and allows for you to solve for 2I. On the extremely rare chance they didn't make it clear they wanted you to use this method, look for bounds involving pi/2 and 0, with trig integrals, as it will normally be what its used for. There's also sometimes substitutions of u=-x, usually when you have a complicated function involving a lot of exponentials. Look out for bounds in the form of +/-a for these types of questions, and functions that you could simplify by adding this new form on to the original.

#### stupid_girl

##### Active Member
They generally make it pretty clear when you have to do this, but using the property:

$\bg_white \int_a^{b}f(x)dx=\int_a^{b}f(a+b-x)dx$

Usually when it is done it is for something like trigonometric integrals, where you substitute pi/2-x, and change a sinx to a cosx. When you add this to the original integral, I , it simplifies it a lot and allows for you to solve for 2I. On the extremely rare chance they didn't make it clear they wanted you to use this method, look for bounds involving pi/2 and 0, with trig integrals, as it will normally be what its used for. There's also sometimes substitutions of u=-x, usually when you have a complicated function involving a lot of exponentials. Look out for bounds in the form of +/-a for these types of questions, and functions that you could simplify by adding this new form on to the original.
Why not have a crossover of trig and exponential?

Here, I'm using re-arranged version that I finds more convenient.
$\bg_white \int_{b}^{a}f\left(x\right)dx=\int_{b}^{\frac{a+b}{2}}f\left(x\right)+f\left(a+b-x\right)dx=\int_{\frac{a+b}{2}}^{a}f\left(x\right)+f\left(a+b-x\right)dx$

$\bg_white \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin x\cos x\ln\left(1+e^{x}\right)}{\sin^{4}x+\cos^{4}x}dx$
$\bg_white =\int_{0}^{\frac{\pi}{2}}\left(\ln\left(1+e^{x}\right)-\ln\left(1+e^{-x}\right)\right)\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$\bg_white =\int_{0}^{\frac{\pi}{2}}x\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$\bg_white =\int_{0}^{\frac{\pi}{4}}\left(x+\frac{\pi}{2}-x\right)\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$\bg_white =\frac{\pi}{2}\int_{0}^{\frac{\pi}{4}}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}dx$
$\bg_white =\frac{\pi}{4}\int_{0}^{\frac{\pi}{4}}\frac{2\tan x\sec^{2}x}{1+\left(\tan^{2}x\right)^{2}}dx$
$\bg_white =\frac{\pi}{4}\int_{0}^{1}\frac{1}{1+u^{2}}du$
$\bg_white =\frac{\pi^{2}}{16}$

#### stupid_girl

##### Active Member
Observe that the denominator is of the form (something)^2. The only way we can get this is when you differentiate something of the form u/v. As this will give you

$image=https://latex.codecogs.com/png.latex?\bg_white+\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg)+=+\dfrac{vu'+-+uv'}{v^2}\hspace{2mm}\text{(Standard+Quotient+rule)}&hash=20daa0746796e01c22f550f6ecffcf92$
haha...
There are other possibilities but it's good to enrich the toolbox. When reverse quotient rule works, it is really amazing.

This is an example with denominator of the form (something)^2...but reverse quotient rule may not be a good idea.
$\bg_white \int_{0}^{1}\left(\frac{x^{2}}{x+\sqrt{x^{2}+1}}\right)^{2}dx$
$\bg_white =\int_{0}^{1}x^{4}\left(x-\sqrt{x^{2}+1}\right)^{2}dx$
$\bg_white =\int_{0}^{1}\left(2x^{6}+x^{4}-2x^{5}\sqrt{x^{2}+1}\right)dx$
$\bg_white =\int_{0}^{1}\left(2x^{6}+x^{4}\right)dx-\int_{1}^{2}\left(u-1\right)^{2}\sqrt{u}du$
$\bg_white =\frac{67-44\sqrt{2}}{105}$

#### Jacobagel

##### New Member
Hey guys,

Would any of you happen to know what the marks were looking like for the 1995 HSC (i.e. highest, average, minimum mark for E4, etc.). I have nothing to compare my own mark with for that paper.

#### TheOnePheeph

##### Active Member
Hey guys,

Would any of you happen to know what the marks were looking like for the 1995 HSC (i.e. highest, average, minimum mark for E4, etc.). I have nothing to compare my own mark with for that paper.
Not sure, but everyone says 1993 was super hard, and the average for that was like in the 40s out of 120, and imo 1995 is of similar difficulty to that, possibly a bit easier.

#### DrEuler

##### Member
Not sure, but everyone says 1993 was super hard, and the average for that was like in the 40s out of 120, and imo 1995 is of similar difficulty to that, possibly a bit easier.
Ironically that year Anthony Henderson got 100 raw, the only one so far.