Kurosaki
True Fail Kid
http://community.boredofstudies.org/1125/mathematics-extension-1/330060/paper.htmlWould 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.htmlWould someone be able to upload the questions? We weren't allowed to take ours from the room!
Thanks.
The locus of Q is most definitely not a full circle.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
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.
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.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
Can you elaborate on your first line?The probability question part (ii):
since
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.Can you elaborate on your first line?
omg this is exactly what i didthe probability question part (ii):
since
Best to confirm with Carret, but that sounds fine to mehey 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?
it'll be fine, because the question specified thatarghh 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 :/
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 beCarrotsticks, 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?