HSC 2015 MX1 Marathon (archive) (3 Viewers)

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
the 2nd line on the right should be sin (x) / x

Can you use Latex D: The images are to big

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

In the solution says 'Draw the perpendicular OA from O to DE and the perpendicular PB from P to FC.' Now in pencil, I have drawn the line and shown point A there, should i also draw a line from centre P to line FG instead of FC as mentioned in the solution?

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

How do you upload photos i have the solution

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
How do you upload photos i have the solution

go to photobucket

Carrotsticks · Jul 5, 2015

Re: HSC 2015 3U Marathon

Use imgur, faster.

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

PW= OA because its equal chord length and equal circle radii.

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

I get the feeling with answering Circle Geometry answers, there isn't necessarily only 1 solution. There are a number of ways you can get to an answer.

Crisium · Jul 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I get the feeling with answering Circle Geometry answers, there isn't necessarily only 1 solution. There are a number of ways you can get to an answer.

That tends to be the case a majority of the time

People also find memorising the 17 or so theories relatively simple, but struggle to apply them to exam questions

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

Crisium said:
That tends to be the case a majority of the time

People also find memorising the 17 or so theories relatively simple, but struggle to apply them to exam questions

All the questions you should be thinking : Why do they give the information that they do. I have an arsenal of techniques. X amount is eliminated. Then go through possible ones

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

Is there a way students try to learn this rule? I for one dont want to memorize. Any tips from the experienced in here?

I think for me is, the best way is to draw the '4 Quadrants' and work at deriving those formulas.

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

I had a question tonight, find the 'General solution of tan 2x=sqrt(3)' whilst its not a hard question. I still had to refer to the formula book to get the right answer. Maybe its a case of me needing to do many more questions similar to that for those formulas to be reinforced in my head?

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Is there a way students try to learn this rule? I for one dont want to memorize. Any tips from the experienced in here?

I think for me is, the best way is to draw the '4 Quadrants' and work at deriving those formulas.

Memorise circle geo.

davidgoes4wce · Jul 5, 2015

Re: HSC 2015 3U Marathon

Well I think I have an easier way of remembering it they all have " a + n Pi" in common and work things out from there

Drsoccerball · Jul 5, 2015

Re: HSC 2015 3U Marathon

I didnt see the image sorry. Derive those :L

InteGrand · Jul 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Is there a way students try to learn this rule? I for one dont want to memorize. Any tips from the experienced in here?

I think for me is, the best way is to draw the '4 Quadrants' and work at deriving those formulas.

$The second are third are the easiest to etch into your memory. They reason for those two are as follows (they are based on the odd/even properties of the functions and their periodicity).$

$Since the cosine function is $2\pi$-periodic, it means that $\cos t = \cos(t+2n\pi)$ for any integer $n$ (in other words, if we change the argument by a multiple of $2\pi$, the function value doesn't change, which is precisely what $2\pi$-periodic means). Now, clearly one solution to $\cos x = \cos \alpha$ is $x=\alpha$. Therefore, $x=\alpha + 2n\pi$ is also a solution for any integer $n$, since $x=\alpha + 2n\pi$ just means we start at $x=\alpha$ and then move forward or back by a multiple of $2\pi$, and we just said that this doesn't change the function value, so $\cos (\alpha + 2n\pi)=\cos \alpha$, which is precisely what we mean by saying $x=\alpha + 2n\pi$ is a solution. This is one set of solutions.$

$We can also see that $x=-\alpha$ is a solution, because $cos(-\alpha)=\cos \alpha$, as the cosine function is even. (E.g. if I tell you $\cos x = \cos \frac{\pi}{3}$, then you know that $x=-\frac{\pi}{3}$ will satisfy the equation.) Applying the same logic as before, this means that $-\alpha + 2n\pi$ is the other set of solutions. Combining this with the solution set above, we can write the general solution as $x=\pm \alpha + 2n\pi$.$

$We can do a similar thing with tan, except that this is $\pi$-periodic rather than $2\pi$-periodic. If $\tan x = \tan \alpha$, then clearly $x=\alpha$ is a solution, but then so will $x=\alpha + n\pi$ be, since the period is $\pi$. This means the overall solution is $x=\alpha + n\pi$. (We don't need to go further because there is one and only one solution to $\tan x = \tan \alpha$ in one `branch' or period of the tan function, since the tan function is one-to-one in each branch; for the cosine function, you will see that in each period of $2\pi$, the cosine function attains each value twice (except for the special values of $\pm 1$), which is the reason there is a $\pm \alpha$ (to account for \emph{two} solutions per period).$ $Also, we only need to consider solutions within one period of each function, and then can just add on the $n\cdot period$, since the behaviour is the same in any other branch, and to get to the other branches from a point in one branch, we just add a multiple of the period.)$

$The reason for the sine formula is that the general solution is: $x=\alpha + 2n\pi$ or $x=(\pi - \alpha)+2n\pi$ (this formula comes from the $2\pi$ periodicity of the sine function, the fact that, like the cosine function, there will generally be two solutions per period, and the identity $\sin x = \sin (\pi - x)$ (as opposed to $\cos x = cos (-x)$), hence the $\pi - \alpha$ instead of $-\alpha$.)$

$Rewriting this, the general solution is $x=\alpha + 2n\pi$ or $x=-\alpha + (2n+1)\pi$. What this says is that the solution is either $\alpha$ plus an even multiple of $\pi$ (as $2n$ is even), or $-\alpha$ plus an odd multiple of $\pi$ (as $2n+1$ is odd). So we have a $+\alpha$ when the multiple is even and a $-\alpha$ when the multiple is odd. This can be expressed in the form $x=(-1)^m \alpha + m\pi$, where $m$ is an integer, because when $m$ is even, $(-1)^m=+1$, and when $m$ is odd, $(-1)^m=-1$, as required.$

davidgoes4wce · Jul 6, 2015

Re: HSC 2015 3U Marathon

The question . Simplify, where alpha is an acute angle.

I'm a bit hesitant as to where to start from this one. I know the domain of cos^(-1) x is between -1 and 1. (cos 0 =1 and also cos 180=-1)

I set Pi + Alpha between 0 and Pi but am coming up with negative values. My thinking is we have to possibly change the domain from say y from Pi to 2 Pi.

Sy123 · Jul 6, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
The question . Simplify, where alpha is an acute angle.

I'm a bit hesitant as to where to start from this one. I know the domain of cos^(-1) x is between -1 and 1. (cos 0 =1 and also cos 180=-1)

I set Pi + Alpha between 0 and Pi but am coming up with negative values. My thinking is we have to possibly change the domain from say y from Pi to 2 Pi.

The first helpful thing to notice, is that the range of the cosine inverse function is between $0$ and $\pi$ . This means we can't do the old method of cancelling out the inverse cosine and the cosine because this would give us $\pi + \alpha$ which since $\alpha$ is acute would not be valid since it is outside the range.

However, we can still try to apply this by manipulating the expression. First, to give a proper definition of the trick, remember that:

$\cos^{-1}(\cos x) = x \ $for$ \ 0 \leq x \leq \pi \ (*)$

This is true precisely because of the definition of an inverse function, and precisely because that domain $0 \leq x \leq \pi$ is the domain of the original $y = \cos x$ function that we wish to 'invert'.

So proceeding from this, we want to manipulate the given expression into one in which we can use $(*)$ .

Remember that, $\cos (\pi + \alpha) = -\cos \alpha$ and $\cos (\pi - \alpha) = - \cos \alpha$ , which means $\cos(\pi - \alpha) = \cos(\pi + \alpha)$
(To see this fact more clearly, imagine drawing horizontal lines in a y = cos x graph below the x-axis, and see that when the line intersects one part of the cosine graph, it intersects the opposite side, symmetrical to $\pi$ )

So that, $\cos^{-1}(\cos (\pi + \alpha)) = \cos^{-1} (\cos (\pi - \alpha))$

$\\ $But since$ \ 0 \leq \pi - \alpha \leq \pi \ $since$ \ \alpha \ $is acute$$

$\\ \therefore \ \cos^{-1}(\cos (\pi - \alpha)) = \pi - \alpha \ $by using$ \ (*)$

$\\ \therefore \ \cos^{-1}(\cos (\pi + \alpha)) = \pi - \alpha$

davidgoes4wce · Jul 6, 2015

Re: HSC 2015 3U Marathon

Sy123 said:
The first helpful thing to notice, is that the range of the cosine inverse function is between $0$ and $\pi$ . This means we can't do the old method of cancelling out the inverse cosine and the cosine because this would give us $\pi + \alpha$ which since $\alpha$ is acute would not be valid since it is outside the range.

However, we can still try to apply this by manipulating the expression. First, to give a proper definition of the trick, remember that:

$\cos^{-1}(\cos x) = x \ $for$ \ 0 \leq x \leq \pi \ (*)$

This is true precisely because of the definition of an inverse function, and precisely because that domain $0 \leq x \leq \pi$ is the domain of the original $y = \cos x$ function that we wish to 'invert'.

So proceeding from this, we want to manipulate the given expression into one in which we can use $(*)$ .

Remember that, $\cos (\pi + \alpha) = -\cos \alpha$ and $\cos (\pi - \alpha) = - \cos \alpha$ , which means $\cos(\pi - \alpha) = \cos(\pi + \alpha)$
(To see this fact more clearly, imagine drawing horizontal lines in a y = cos x graph below the x-axis, and see that when the line intersects one part of the cosine graph, it intersects the opposite side, symmetrical to $\pi$ )

So that, $\cos^{-1}(\cos (\pi + \alpha)) = \cos^{-1} (\cos (\pi - \alpha))$

$\\ $But since$ \ 0 \leq \pi - \alpha \leq \pi \ $since$ \ \alpha \ $is acute$$

$\\ \therefore \ \cos^{-1}(\cos (\pi - \alpha)) = \pi - \alpha \ $by using$ \ (*)$

$\\ \therefore \ \cos^{-1}(\cos (\pi + \alpha)) = \pi - \alpha$

Tjanks I didn't know that rule about cos (x +\alpha)) being equal to cos (x - \alpha)

davidgoes4wce · Jul 6, 2015

Re: HSC 2015 3U Marathon

I don't know if many of you guys have used or heard of 'Sami El Hosri' but I am pinching his questions from there. Any ideas about what you think about the standard of questions he sets ? For me I think its the real deal, about the right standard of HSC Extension 1 Papers.

davidgoes4wce · Jul 6, 2015

Re: HSC 2015 3U Marathon

If you guys are keen El Hosri, he has a MSC and DipEd

Generally speaking I have been throughly tested. I have spent 1 year doing Prelim and HSC Extension 1 maths questions. Make no mistake I do feel you need the full 2 years as there is alot of content to go through. I sometimes applaud the students that are doing Extension 1 and 2 together,the amount of time you have to put in , on top of the other subjects makes it a real challenge.

HSC 2015 MX1 Marathon (archive) (3 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Retired

Well-Known Member

Well-Known Member

Pew Pew

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

This too shall pass

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)