BOS MX2 Trials discussion thread. (1 Viewer)

[iBibah]

Hey carrot, what did you use to write this paper?

 

[asianese]

Also what the hell was the conics question? i) was just subbing in y into the hyperbola and showing that discriminant =0. then?? Anyone get it?

 

[zeebobDD]

Also what the hell was the conics question? i) was just subbing in y into the hyperbola and showing that discriminant =0. then?? Anyone get it?

don't you sub in y=mx+b into the equation of the hyperbola? then discriminant = 0

 

[deswa1]

Also what the hell was the conics question? i) was just subbing in y into the hyperbola and showing that discriminant =0. then?? Anyone get it?

Yeah, I derived it though. Basically the equation of the tangent will have the form y=mx+k. Sub this into the hyperbola and you end up with a quadratic in x. Now by letting the quadratic=0 (for tangent), you can find a value for k in terms of m,a,b (that thingo under the square root) and then you sub back in for k and you're done

 

[asianese]

Well yes but the question was kinda reversed. They gave you the line and then you had to show triangle =0, not ANY line y=mx+c into the hyperbola.

I meant the locus of T**

 

[Trebla]

(ii) is quite tricky because you need to derive the quadratic equation which has the roots y = mx ± c and convert into a quadratic equation in terms of m, which would solve out two values for the gradient. You can impose the fact that tangents are perpendicular to obtain the locus.

 

[Carrotsticks]

Bolded text below are my comments to his questions.

A lot of them are well known results in higher level mathematics. Also, some of those questions came from me. Namely:
Q8 A good question to test understanding of curve properties.
Q10 A better question, which simplifies down to a nice quadratic.
Q13b) Personally, I liked this question more than most others in the paper. I'm a bit fan of limits, limiting sums, infinite sums etc etc so I fell in love with this Q the moment I saw it.
Q15a) - though Carrot modified its original form (which had an induction component) and added in part iv) The Gamma Function is very sexy =)

The solution Q16 is actually not too bad when you think about it. I personally think it is not as difficult as some of the earlier questions.

Exactly, and this is quite often the case too in the actual HSC paper. The very last Q is actually fairly straightforward, but it's the intimidation that kills off students. They see a sigma or too and BAM, they skip it.

Haha effffff. I flicked to the very last part of Q16 just in case there was one of those limits questions or something that you could just read off but obviously there wasn't. If we had noticed those questions though...

If only...

I think the scare factor contributed to lots of the freak outs in this paper...haha

No guts. No glory.

Very much this.

Hey carrot, what did you use to write this paper?

Microsoft Word + Math Type.

Well yes but the question was kinda reversed. They gave you the line and then you had to show triangle =0, not ANY line y=mx+c into the hyperbola.

I meant the locus of T**

lol'd

 

[Beaconite]

holy fuck, hard ass exam

 

[Carrotsticks]

I'll start slowly releasing solutions as I make/double check them.

 

[math man]

I think overall a demanding test, a lot to do, but it is not overly difficult. There were only a few marks here and there.
I thought other parts were harder than all of q16. I'll say it is probably more difficult than most of the HSC papers in the
last 10 or so years, but yeh the earlier papers, especially in the 90s were still harder.

 

[zeebobDD]

I'll start slowly releasing solutions as I make/double check them.

you really love those sigma signs don't you?

 

[asianese]

The examiners love them too...

 

[Carrotsticks]

 

[Carrotsticks]

Oh and that was for the very first part of Q16. I totally forgot to name the part (sorry for the confusion).

 

[Sy123]

Am I right in saying that Q15 was one of the easier questions? I was able to get 8/15 for it without any 4U knowledge

 

[Carrotsticks]

Am I right in saying that Q15 was one of the easier questions? I was able to get 8/15 for it without any 4U knowledge

Q15 was a bit of a strange one. It was either too easy or hard. Not really 'in between'.

The ones that most people didn't seem to get were (a)(iv) and (b)(iv).

 

[tomp1612]

do you think the actual HSC will be this hard??

 

[deswa1]

Just did the Q16- that was very nice. What people said before was quite right imo- it not impossibly hard but it looks very intimidating haha.

 

[RealiseNothing]

Q15 was a bit of a strange one. It was either too easy or hard. Not really 'in between'.

The ones that most people didn't seem to get were (a)(iv) and (b)(iv).

I found the induction part of question 15 to be harder than those parts.

(a)(iv) you could find a recurrence formula for

(b)(iv) was something like the expression was actually and you take the limit of that which becomes or something.

 

[Carrotsticks]

I found the induction part of question 15 to be harder than those parts.

(a)(iv) you could find a recurrence formula for

(b)(iv) was something like the expression was actually and you take the limit of that which becomes or something.

You could find a recurrence formula for J_n, but that would be unnecessarily long. A simple substitution can get the answer quite quickly. Furthermore, using the substitution properly draws on the relationship between I_n and J_n. So for example, a substitution method would show that I_n = ... = J_n (not exactly equal, but you get the idea)

However, manually computing the recurrence formula for J_n would follow something more like I_n = ... = X, J_n = ... = X. Hence I_n = J_n.