Carrotsticks' 2014 Extension 1 HSC Solutions (1 Viewer)

[Kurosaki]

Would someone be able to upload the questions? We weren't allowed to take ours from the room!
Thanks.

http://community.boredofstudies.org/1125/mathematics-extension-1/330060/paper.html

 

[ZoeeR]

Thank you!

 

[garfield qiu]

Carrotsticks, I doubt your parametric solution for Q13 (c) (iii).
you said: This is the equation of a circle centred at the focus with radius a. We have the condition that
x =/= 0 because this can only occur if P lies precisely on the vertex of the parabola ie: t = 0 .
However, when this happens, then we cannot define Q because having a ratio of 0 :1 yields
an undefined point.
But please check out the definition of ratio, 0:1 means 0/1 which has meaning (P and Q have no distance); 1/0 is meaningless
Also, when you put t=0 back into (i), the coordinates of Q becomes (0,0) which is on the circle we got in (iii), and it completes the circle. That's why the question says it's a full circle, BOS is not that stupid as you think. So imo, you do need to separate the cases where x=0 or not.
When x=0, t=0, but at this moment t is not the slope of OQ because at this moment Q is on origin, then OQ doesn't have a slope, so (ii) has no problem.
Thus the whole question is good. It may become a reason of some people losing state rank no1

 

[RealiseNothing]

Carrotsticks, I doubt your parametric solution for Q13 (c) (iii).
you said: This is the equation of a circle centred at the focus with radius a. We have the condition that
x =/= 0 because this can only occur if P lies precisely on the vertex of the parabola ie: t = 0 .
However, when this happens, then we cannot define Q because having a ratio of 0 :1 yields
an undefined point.
But please check out the definition of ratio, 0:1 means 0/1 which has meaning (P and Q have no distance); 1/0 is meaningless
Also, when you put t=0 back into (i), the coordinates of Q becomes (0,0) which is on the circle we got in (iii), and it completes the circle. That's why the question says it's a full circle, BOS is not that stupid as you think. So imo, you do need to separate the cases where x=0 or not.
When x=0, t=0, but at this moment t is not the slope of OQ because at this moment Q is on origin, then OQ doesn't have a slope, so (ii) has no problem.
Thus the whole question is good. It may become a reason of some people losing state rank no1

The locus of Q is most definitely not a full circle.

The question also says that Q lies on a circle, this does not necessarily mean it maps out a full circle.

 

[RealiseNothing]

Also a cool way of doing the second part of the circle geo question:

We want to show that OP is a tangent to the circle through PQC. Let there be a point D that lies on both the arc PC and the line OP extended.

We know that angle PAO = angle APO by isosceles triangles, and angle APO = angle DPC by vertically opposite angles, and thus angle PAO = angle DPC.

However angle PAO = angle CQP from part (i). Thus angle DPC = angle CQP, and so CD=PC by equal angles subtend equal arcs.

If CD=PC then the point P and D coincide and so, other then P, the line OP does not pass through any points on the circle PQC. Hence OP is a tangent.

 

[Carrotsticks]

I am now looking a little more into the locus problem. Thanks to all who raised it.

 

[RealiseNothing]

Also a cool way of doing the second part of the circle geo question:

We want to show that OP is a tangent to the circle through PQC. Let there be a point D that lies on both the arc PC and the line OP extended.

We know that angle PAO = angle APO by isosceles triangles, and angle APO = angle DPC by vertically opposite angles, and thus angle PAO = angle DPC.

However angle PAO = angle CQP from part (i). Thus angle DPC = angle CQP, and so CD=PC by equal angles subtend equal arcs.

If CD=PC then the point P and D coincide and so, other then P, the line OP does not pass through any points on the circle PQC. Hence OP is a tangent.

 

[Carrotsticks]

Oh haha okay it didn't take long to see, just needed a minute of clear thinking to spot. I can confirm that my initial statement was erroneous, I will explain why below (it's been fixed now). Many thanks again to all who raised the problem.

When P slides along the parabola, Q slides along the arc of a circle and gets extremely close to (0,2a), but never reaches it.

Now our computations of the locus of Q to obtain the equation come from the assumption that x=/=0. HOWEVER, when we compute the case x=0 separately (I cannot believe that I did not do this, and that I disregarded it completely), it yields a point that 'fills the gap' and 'continues' the equation of the locus, thereby making the only 'hole' in the locus being the point (0,2a).

EDIT: 1:21AM grammar is not the strongest of grammars.

 

[RealiseNothing]

The probability question part (ii):







since

 

[Carrotsticks]

Carrotsticks, I doubt your parametric solution for Q13 (c) (iii).
you said: This is the equation of a circle centred at the focus with radius a. We have the condition that
x =/= 0 because this can only occur if P lies precisely on the vertex of the parabola ie: t = 0 .
However, when this happens, then we cannot define Q because having a ratio of 0 :1 yields
an undefined point.
But please check out the definition of ratio, 0:1 means 0/1 which has meaning (P and Q have no distance); 1/0 is meaningless
Also, when you put t=0 back into (i), the coordinates of Q becomes (0,0) which is on the circle we got in (iii), and it completes the circle. That's why the question says it's a full circle, BOS is not that stupid as you think. So imo, you do need to separate the cases where x=0 or not.
When x=0, t=0, but at this moment t is not the slope of OQ because at this moment Q is on origin, then OQ doesn't have a slope, so (ii) has no problem.
Thus the whole question is good. It may become a reason of some people losing state rank no1

Thank you for pointing out this erroneous statement. You are one of two students who have pointed this out. I am quite certain that the reason why I had made that mistake in the first place was because I had confused a ratio of 0:1 with the ratio 1:0 amongst the rush, without thinking about the bigger picture overall pointing out that the origin is most certainly a part of the locus equation.

Some things to note about your response:

1. I'm not too sure how I feel about your comment "BOS is not that stupid as you think", I would appreciate you not putting words in my mouth, thanks.

2. The equation of the locus is most certainly NOT a 'full circle'. It is excluding the point (0,2a) for reasons I have explained below.

3. I can see that you are talking about (ii) in a rather defensive manner, and I am not quite sure why you are doing so since nobody in the entire thread (or forum section for that matter) has had any problems with it.

 

[Carrotsticks]

The probability question part (ii):







since

Can you elaborate on your first line?

 

[RealiseNothing]

Can you elaborate on your first line?

If person 1 has a probability of A of eventually winning, we note that person 2 must also have a probability of A of eventually winning given that person 1 does not win on his first turn.

So we arrive at 3 outcomes:

Person 1 eventually wins - A

Person 2 eventually wins after person 1 does not win/lose on his first turn - rA

Person 2 wins due to person 1 losing on the first turn = q

This covers all outcomes and hence

 

[Carrotsticks]

That works very nicely, I did not see it myself.

 

[trapizi]

the probability question part (ii):







since

omg this is exactly what i did

 

[kimsungho]

hey carrot, would you need to show that there were 2 solutions not just one for the ski jump question to find max d? because i showed the pi/8 and justified with 2nd derivative test, but i didnt include the 2nd possible solution. will the marker take marks off for this?

 

[mreditor16]

hey carrot, would you need to show that there were 2 solutions not just one for the ski jump question to find max d? because i showed the pi/8 and justified with 2nd derivative test, but i didnt include the 2nd possible solution. will the marker take marks off for this?

Best to confirm with Carret, but that sounds fine to me

 

[mreditor16]

arghh so I showed that there were two stat points as 22.5 degrees (pi/8) and -67.5 degrees. I showed that the 22.5 degrees was a maxima by testing points around it, but just inspection of the other one in terms of the derivative told me that it was a minima so I didn't bother proving it (I wrote that it was a stationary point though).

Carrot do you think thats okay :/

it'll be fine, because the question specified that



As a result, you shouldn't have considered -67.5 to begin with.

 

[Chris100]

Yes, you should have calculated 2 different solutions but then you should have mentioned that because one of them isn't in the domain given by the question, only pi/8 can be considered
then u confirm that its a maximum value with a table or 2nd derivative or whatever

 

[Samir97]

Carrotsticks, for the parametric circle q, i didnt know how to use the previous part to find the curcle equation but I could immediately see that the centre was going to be S(0,a) so I just squared the X-coordinate of Q, subtracted a from the y co-ordinate (gave me [y-a]) and i squared that, then added both squares and the final term cancelled out to just a^2. Is that considered fudging? And would I still obtain full marks?

 

[Gumball]

Carrotsticks, for the parametric circle q, i didnt know how to use the previous part to find the curcle equation but I could immediately see that the centre was going to be S(0,a) so I just squared the X-coordinate of Q, subtracted a from the y co-ordinate (gave me [y-a]) and i squared that, then added both squares and the final term cancelled out to just a^2. Is that considered fudging? And would I still obtain full marks?

even though im not carret i believe it was fudging as there was a legit way to figure it out lol. you may be lucky to get even 1 but definitely not more than that. remember the test wasn't very hard so marking will be