Specialist Mathematics Examination 2 discussion

[presto]

Did anyone label the asymptote at N=e^6 or N=403 for the N vs t graph in the phone question?
I didn't because I thought since the curve had a restricted domain (as it was specified by the question), the curve doesn't actually approach that value indefinitely.
But to be extra safe, I wrote an explanation for that beneath the graph, saying if it wasn't a restricted domain, there would be an asymptote at...etc.
Surely, I won't any lose marks for this?

[e^1]

I hope so!!!

This is true! Thank for keeping the hopes up!!

If anyone feels concerned about this, I also think they would accept this.

Anyway, I feel terrible. Hopefully all turns out well :/

[thecreeker]

what did you guys get for the area of the segment of the circle in question 1 SA
I hacked that out somehow to get  pi-2 sq units

[presto]

what did you guys get for the area of the segment of the circle in question 1 SA
I hacked that out somehow to get  pi-2 sq units

I got the same, I think that's right.

[Lejn]

what did you guys get for the area of the segment of the circle in question 1 SA
I hacked that out somehow to get  pi-2 sq units

Yeah same thing, sector minus a triangle.

[simba]

Anyone else find MC Q15 off? :/

[Uberjew]

what did you guys get for the area of the segment of the circle in question 1 SA
I hacked that out somehow to get  pi-2 sq units

Area of 1/4 circle - triangle
= pi/2/2 * 2^2 - 1/2 * root2 * 2root2
= pi - 2 sq units

the 1/4 circle is angle/2 * r^2 = pi/2/2 * r^2 = pi/4 * 4 = pi
triangle = 1/2 * base * height = 1/2 * 2root2 * root2 = root2^2 = 2

[xiayu123w]

I did really bad in exam one and didnt know how show the differential question in loge(N) question.

[Jeggz]

Anyone else find MC Q15 off? :/

Which was one was this?

[thecreeker]

Area of 1/4 circle - triangle
= pi/2/2 * 2^2 - 1/2 * root2 * 2root2
= pi - 2 sq units

the 1/4 circle is angle/2 * r^2 = pi/2/2 * r^2 = pi/4 * 4 = pi
triangle = 1/2 * base * height = 1/2 * 2root2 * root2 = root2^2 = 2

 I just found the square with sides 2root2, subtracted by the area of the circle and divided by 4. It seemed pretty dodgy when I did it, so hope you don't need a complicated approach to get that method mark

[simba]

Which was one was this?

It gave 3 vectors and asked which wasn't correct. I thought more than 1 were wrong (unless I misread the question..)

[Aelru]

transposed to obtain the circle's equation in terms of x solely, and then used area calculation multiplied by 2, irrespective of choosing either the top or bottom semicircle. ;;

[silverpixeli]

I just found the square with sides 2root2, subtracted by the area of the circle and divided by 4. It seemed pretty dodgy when I did it, so hope you don't need a complicated approach to get that method mark

should be fine, but guys the triangle was a right angle triangle on the side with 2 for the base and the height so you coulda all used 1/2 * 2 * 2 instead of all those root 2's

It gave 3 vectors and asked which wasn't correct. I thought more than 1 were wrong (unless I misread the question..)

I think i have E for this (u+w)dot(v)=12 or something because i was avoiding having to figure out C because it involved linear dependence, i got 6 for the dot product making 12 false

[Jeggz]

should be fine, but guys the triangle was a right angle triangle on the side with 2 for the base and the height so you coulda all used 1/2 * 2 * 2 instead of all those root 2's

I think i have E for this (u+w)dot(v)=12 or something because i was avoiding having to figure out C because it involved linear dependence, i got 6 for the dot product making 12 false

Thank god, that's what I got too!!

[~T]

I finished with half an hour to spare and spent 27 minutes checking over my Extended Response, so I'm really happy there. Then thought I should probably spend a bit of time on the MC section and found one that I'd stuffed up with 30 seconds to spare. So yeah, I walked out of the exam shaking like crazy, but I should be good for a 78-80. Very happy.