Specialist Mathematics Examination 2 discussion

[thecreeker]

Same here, dot product of E was 12

Had a mini heart attack when they didn't give the mass for the last question, only to realise it cancelled out anway. I know that got a few people in my cohort

[belsebob]

http://www.itute.com/board/viewtopic.php?f=3&t=2628&sid=fd1889ea5a219a92f264f4c51b91afda
Answers to multiple choice

[belsebob]

and has anyone got a copy of the exam

[lzxnl]

There was an error in the multiple choice section...
If z^3 has a principal argument in the second quadrant, there should be THREE different domains that identify possible values of the argument of z
(5pi/6,pi) , (pi/6,pi/3) and (-pi/2,-pi/3). Try each one of them.
AND VCAA dugs themselves an even bigger hole by asking in the question for "the complete set". Well please be as anal in your own questions before you criticise us for forgetting to put units in for area questions.

I thought the intersection of y=ax and arctan(bx) was nice. Anyone else get b<a<0?

Other than that...I felt the exam was relatively straightforward...there wasn't anything extreme or weird at all in this exam.
It would have been nice if VCAA didn't keep prodding us to find that the complex set required was (Re(z))^2=(Im(z))^2

Although who knows where I could have written something stupid like 1+1=3

[eddybaha]

i got a<b<0.....

[eddybaha]

Me too, I subbed in the values (1 for a, 2 for b) and it gave me 3 intersections.

yeah same, i tried for a>b but only got 1 solution

[Sentar]

Me too, I subbed in the values (1 for a, 2 for b) and it gave me 3 intersections.

But both b and a were less than 0. It was b<a<0 I'm fairly sure, I used -1 and -2.

[silverpixeli]

Nice pickup with the domain MC question nliu1995, I thought I was intelligent by immediately realising that there would be 2 sets
I got b<a<0 though, initially I thought a<b<0 because a had to be less than b, but thats only if both were positive, but the option says <0
make both negative and b<a<0 works

hey guys does anyone remember what question 4 was?

[Lejn]

I thought the intersection of y=ax and arctan(bx) was nice. Anyone else get b<a<0?

That was a very nice question, I enjoy how they reflected them. I guess it wouldn't have been hard in the end if you'd just check most on your calc.

[~T]

I thought the intersection of y=ax and arctan(bx) was nice. Anyone else get b<a<0?

Yup. That was the one that I mentioned above, changing in the last 30 seconds. I've never dropped a MC mark in a past exam, so that was unsettling.

[lzxnl]

The way I looked at it was this
For y=ax and y=arctan bx, you have a point of intersection at the origin
If the magnitude of the slope of y=ax was greater than the magnitude of the gradient of the tangent to y=arctan bx, then it will only end up intersecting the graph at the origin; hard to justify the logic but the gradient is always decreasing in magnitude, so if y=ax is initially steeper than y=arctan bx, it will always be.
The slope of the tangent to y=arctanb bx at the origin is simply b, so we have |b||>|a|
The only answer that matches this is b<a<0

Someone please check my working. It worked on the calculator.

[~T]

The way I looked at it was this
For y=ax and y=arctan bx, you have a point of intersection at the origin
If the magnitude of the slope of y=ax was greater than the magnitude of the gradient of the tangent to y=arctan bx, then it will only end up intersecting the graph at the origin; hard to justify the logic but the gradient is always decreasing in magnitude, so if y=ax is initially steeper than y=arctan bx, it will always be.
The slope of the tangent to y=arctanb bx at the origin is simply b, so we have |b||>|a|
The only answer that matches this is b<a<0

Someone please check my working. It worked on the calculator.

You're all good. I managed to check it on the calc in the last 30 seconds and change my answer, because in my working I'd written "|b|>|a|" and then I'd chosen the only option with b>a.
*facepalm*
BUT I changed it and that's all that matters.

[eddybaha]

The way I looked at it was this
For y=ax and y=arctan bx, you have a point of intersection at the origin
If the magnitude of the slope of y=ax was greater than the magnitude of the gradient of the tangent to y=arctan bx, then it will only end up intersecting the graph at the origin; hard to justify the logic but the gradient is always decreasing in magnitude, so if y=ax is initially steeper than y=arctan bx, it will always be.
The slope of the tangent to y=arctanb bx at the origin is simply b, so we have |b||>|a|
The only answer that matches this is b<a<0

Someone please check my working. It worked on the calculator.

obviously its right....like always

[Alwin]

There was an error in the multiple choice section...
If z^3 has a principal argument in the second quadrant, there should be THREE different domains that identify possible values of the argument of z
(5pi/6,pi) , (pi/6,pi/3) and (-pi/2,-pi/3). Try each one of them.
AND VCAA dugs themselves an even bigger hole by asking in the question for "the complete set". Well please be as anal in your own questions before you criticise us for forgetting to put units in for area questions.

I thought the intersection of y=ax and arctan(bx) was nice. Anyone else get b<a<0?

Yes... I noticed that complex q conundrum but thought I did the question wrong so chose some random answer XD


The "technical" way I went about doing that question arctan q for anyone interested was this:


When we sketch the graph, we see that the gradient of the line y=ax MUST be less that the gradient at x=0, or else there will not be three intersections.


But surprise surprise vcaa doesn't have this answer iirc. So, we times everything by -1

[AYNU]

any solutions up yet?