HSC 2015 MX1 Marathon (archive) (1 Viewer)

braintic · Aug 2, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Let $t=\tan{\frac{\theta}{2}}. \therefore \sin{\theta}=\frac{2t}{1+t^2}, \cos{\theta}=\frac{1-t^2}{1+t^2}. $ We can write the parametric forms as$: x=\sin{\theta}, y=\cos{\theta}. $ Since $\sin^2{\theta}+\cos^2{\theta}=1$, the cartesian form is$ :x^2+y^2=1.$

Any points missing from the circle?

leehuan · Aug 2, 2015

Re: HSC 2015 3U Marathon

pomsky said:
$$Q3 Solve the differential equation $ (1+x^2) \frac{dy}{dx} + xe^{-y} =0 $, given that the curve y= f(x) passes through (3,0).$

This seems rather demanding to be given to a 3U student.

3U also demands a substitution be given for integrals, but I will say Q1 is perfectly doable when it is given. Then again this is the marathon so this may not apply.

davidgoes4wce · Aug 2, 2015

Re: HSC 2015 3U Marathon

braintic said:
Any points missing from the circle?

The domain would be between

InteGrand · Aug 2, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
The domain would be between

$With the parameterisation $x(t)=\frac{2t}{1+t^2}$, we miss out on $x=0$ (but achieve every other value in $[-1,1]$), so we exclude the points $(0,\pm1)$ from the circle. (You can check that with the $y$ parameterisation, we attain all values in $[-1,1]$ EXCEPT $y=-1$, so every other point on the circle is obtained.)$

braintic · Aug 2, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$With the parameterisation $y(t)=\frac{1-t^2}{1+t^2}$, $y$ can take on any value in the range $-1<y\leq 1$, but it misses out on $y=1$, so the point $(0,-1)$ on the circle is excluded. Also, with $x(t)=\frac{2t}{1+t^2}$, we miss out on $x=0$ (but achieve every other value in $[-1,1]$), so we exclude the point $(0,1)$ from the circle. Overall, $(0,\pm 1)$ are the excluded points.$

Unless of course t=infinity is regarded as a "value" to be substituted for t.

braintic · Aug 2, 2015

Re: HSC 2015 3U Marathon

leehuan said:
This seems rather demanding to be given to a 3U student.

3U also demands a substitution be given for integrals, but I will say Q1 is perfectly doable when it is given. Then again this is the marathon so this may not apply.

It wouldn't be asked even in Ext 2 because
(i) the term "differential equation" is not in the syllabus (either Ext 1 or Ext 2)
(ii) although many students do use separation of variables, that concept is also not in the Ext 1 or Ext 2 course.

But it will be in the new syllabus, whenever that comes out.

InteGrand · Aug 2, 2015

Re: HSC 2015 3U Marathon

braintic said:
It wouldn't be asked even in Ext 2 because
(i) the term "differential equation" is not in the syllabus (either Ext 1 or Ext 2)
(ii) although many students do use separation of variables, that concept is also not in the Ext 1 or Ext 2 course.

But it will be in the new syllabus, whenever that comes out.

The words ''differential equation'' have been used in past HSC papers though. E.g. Q4(c) of the 2011 MX2 paper (see page 7 of the link) was a quite solid differential equations Q: http://www.boardofstudies.nsw.edu.a...ms/pdf_doc/2011-hsc-exam-mathematics-ext2.pdf

The 2/3 Unit has the words "differential equation" in it: http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths23u_syl.pdf

And when is the new syllabus expected to come out?

braintic · Aug 2, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
The words ''differential equation'' have been used in past HSC papers though. E.g. Q4(c) of the 2011 MX2 paper (see page 7 of the link) was a quite solid differential equations Q: http://www.boardofstudies.nsw.edu.a...ms/pdf_doc/2011-hsc-exam-mathematics-ext2.pdf

The 2/3 Unit has the words "differential equation" in it: http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths23u_syl.pdf

And when is the new syllabus expected to come out?

I stand corrected about the term.
But I maintain that separation of variables will not be asked.

And even though that question was in the Ext 2 HSC, it should not have been phrased like that because the concept of a 'solution' to a differential equation is not developed in the syllabus. I suspect many students would not have a clue what it was talking about.

davidgoes4wce · Aug 3, 2015

Re: HSC 2015 3U Marathon

The parabola x^2=8y has focus S. The chord of contact from the point T(2,-1) to the parabola cuts the parabola at A and B.

a) Find the equation of the chord of contact.
b) Show that the area of triangle ABS is 1/3 of the area of triangle ABT. (Hint: You do not need to find the length of AB.)

I found part (a) which is x-2y+2=0. I had an attempt at drawing part (b) on graph but not too sure how to solve part (b)

The way I read it is: A,B are two points on the parabola and S is the Focus which are all collinear points

davidgoes4wce · Aug 3, 2015

Re: HSC 2015 3U Marathon

Wait the Focus (0,a) should be (0,2) in this case. I think I got up too early and was not thinking.

davidgoes4wce · Aug 4, 2015

Re: HSC 2015 3U Marathon

It took me 3 pages of work and a night's sleep to get there but I got the answer.

davidgoes4wce · Aug 4, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce · Aug 4, 2015

Re: HSC 2015 3U Marathon

rand_althor · Aug 4, 2015

Re: HSC 2015 3U Marathon

A quicker way to do it is to find the perpendicular distance from S to AB and from T to AB. These distances would be the perpendicular heights of the triangles. Then using the area formula for a triangle you would find each area in terms of AB. When you find the ratio Area(ABS):Area(ABT), the AB cancels out and it simplifies to 1/3.

rand_althor · Aug 6, 2015

Re: HSC 2015 3U Marathon

$Prove by mathematical induction that for all positive integers $n$ that $a^{3n-1}-1$ is divisible by $a-1$ where $a$ is an integer.$

leehuan · Aug 6, 2015

Re: HSC 2015 3U Marathon

$\\ $Base case: Prove true for $n=1 \\ { a }^{ 3\left( 1 \right) -1 }-1={ a }^{ 2 }-1=\left( a+1 \right) \left( a-1 \right) =C\left( a-1 \right) \\ $noting $ a\in \mathbb Z $ hence divisible, thus true for $n=1. \\ $Inductive step: We assume that ${ a }^{ 3k-1 }-1=M\left( a-1 \right) ,\quad \\ M\in \mathbb Z , k\in \mathbb N \\ $and need to prove ${ a }^{ 3\left( k+1 \right) -1 }-1$ i.e. ${ a }^{ 3k+2 }-1$ is divisible by $\left( a-1 \right)$

$\\ { a }^{ 3k+2 }-1\\ ={ a }^{ 3 }\left( { a }^{ 3k-1 } \right) -1\\ ={ a }^{ 3 }\left( M\left( a-1 \right) +1 \right) -1 $ by inductive hypothesis$ \\ =M{ a }^{ 3 }\left( a-1 \right) +\left( { a }^{ 3 }-1 \right) \\ =\left( a-1 \right) \left( M{ a }^{ 3 }+\left( { a }^{ 2 }+a+1 \right) \right) \\ $which is divisible by $\left( a-1 \right) \\ $Thus because the statement is true for $n$=1, it follows that it is true for $n$=2, 3, 4, ... i.e. all positive integers $n$$

Drsoccerball · Aug 6, 2015

Re: HSC 2015 3U Marathon

leehuan said:
$\\ $Base case: Prove true for $n=1 \\ { a }^{ 3\left( 1 \right) -1 }-1={ a }^{ 2 }-1=\left( a+1 \right) \left( a-1 \right) =C\left( a-1 \right) \\ $noting $ a\in \mathbb Z $ hence divisible, thus true for $n=1. \\ $Inductive step: We assume that ${ a }^{ 3k-1 }-1=M\left( a-1 \right) ,\quad \\ M\in \mathbb Z , k\in \mathbb N \\ $and need to prove ${ a }^{ 3\left( k+1 \right) -1 }-1$ i.e. ${ a }^{ 3k+2 }-1$ is divisible by $\left( a-1 \right)$

$\\ { a }^{ 3k+2 }-1\\ ={ a }^{ 3 }\left( { a }^{ 3k-1 } \right) -1\\ ={ a }^{ 3 }\left( M\left( a-1 \right) +1 \right) -1 $ by inductive hypothesis$ \\ =M{ a }^{ 3 }\left( a-1 \right) +\left( { a }^{ 3 }-1 \right) \\ =\left( a-1 \right) \left( M{ a }^{ 3 }+\left( { a }^{ 2 }+a+1 \right) \right) \\ $which is divisible by $\left( a-1 \right) \\ $Thus because the statement is true for $n$=1, it follows that it is true for $n$=2, 3, 4, ... i.e. all positive integers $n$$

LOL my teacher forces us to say it like this :
The statement is true for n=1. If the statement is true for n=k, then it is also true for n=k+1, Hence by Mathematical Induction the statement is true for n>1 $\in \mathbb Z$

leehuan · Aug 6, 2015

Re: HSC 2015 3U Marathon

In class I think my teacher makes us say:

Hence we have satisfied the conditions for proof by induction. Thus the statement is true for all n <element> <positive natural>.

InteGrand · Aug 6, 2015

Re: HSC 2015 3U Marathon

You can just say something like 'Hence true by mathematical induction'.

rand_althor · Aug 6, 2015

Re: HSC 2015 3U Marathon

$$The letters of ENTERTAINMENT are arranged in a a row. Find the probability that:\\(a) The letters E are together. \\(b) Two Es are together and one is apart. \\(c) All the letters E are apart. \\(d) The word starts and ends with E. \\$

Answers: (a): 1/26 (b): 5/13 (c): 15/26 (d) 1/26

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