ATAR Notes: Forum
Archived Discussion => Mathematics/Science/Technology => 2015 Exam Discussion => Specialist Mathematics => Topic started by: Splash-Tackle-Flail on November 06, 2015, 10:27:59 am
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How did you guys find it?
Hard? easy?
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I went really bad
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Stupid mistake galore. Bye bye 40.
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Could anyone please upload a copy of the exam? Thanks
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yeah, answers to the exam would be great
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Hopefully full marked it.
Found it to be a pretty easy exam tbh.
Did way better in this than methods LOL
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For the cosec(2x)<cosec(x)
Is the answer (0,pi/3)?
And if for part a i wrote
Cos(x)=1/2
x=pi/6
Hence (0,pi/6), would I receive conseq. marks?
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For the cosec(2x)<cosec(x)
Is the answer (0,pi/3)?
i got (0,pi/3)U(pi/2,pi)
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I actually did pretty well on this. Probably some mistakes I didn't pick up on, but I'm happy with how I did. I honestly think I did better in this than in methods. I only didn't end up putting down an answer for the very last question. I think my workings will get me 1 mark for it though. Pretty happy overall.
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just wondering, does 1 mark exam 1 = 1 mark exam 2, or does each exam get taken and scaled individually?
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for the velocity question, did we have to explain why we chose our answer? i.e. since v=e, x=1 or something...
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i got (0,pi/3)U(pi/2,pi)
I got (0, pi/3), but not the other half.
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I got (0, pi/3), but not the other half.
SAME WTF?
which one is right?
also if you write 0+0i for the cartesian equation instead of just 0, is that right?
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i got the same as escobar, because 1/sinxcosx will always be negative in the second quadrant, whereas 1/sinx will always be positive
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I got (0, pi/3), but not the other half.
By the way, I managed to mess up the arctan integral question, but I fixed it up towards the end (discovered my error during checking). My mistake was I calculated the area I wanted to subtract from the rectangle, and my working was quite messy, and I had to quickly fix up the working to half make sense. If my working is a bit scattered and not in order, but it yields the correct result, what are my chances of retaining the method mark? (It was a 2 mark question, is that right?)
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i got the same as escobar, because 1/sinxcosx will always be negative in the second quadrant, whereas 1/sinx will always be positive
yep, same logic
i drew a graph and intepreted
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just wondering, does 1 mark exam 1 = 1 mark exam 2, or does each exam get taken and scaled individually?
I think in principle they're worth the same to your final score after scaling, that is i'm pretty sure they get scaled individually and then combined.
In practice exam 2 (tends to be) a little tougher and so a really good exam 2 score can turn around a poor exam 1 score (this is what happened to me! I stuffed up some small sections of exam 1 and got an A overall, but performed much better on exam 2)
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Anyone got the answers?
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can anyone post up a scaned copy by any chance?
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for question three where you had to find distance from origin, was the origin the starting point of j - 2k or is it (0,0)?
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for question three where you had to find distance from origin, was the origin the starting point of j - 2k or is it (0,0)?
i used (0,0)
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That was 13m right?
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i used (0,0)
do you get where i'm coming from though? it does seem a little ambiguous
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But the position vector is relative to an arbitrary, set origin O (maths terminology)
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But the position vector is relative to an arbitrary, set origin O (maths terminology)
Actually, now that I think about, I'm unsure as to what they were referring to as the origin. Was it the starting position or (0,0)?
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i've always seen the origin refer to (0,0) so idk
That was 13m right?
yeah
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what do you guys think the cut off will be for a+
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what do you guys think the cut off will be for a+
37 probably. More or less the same difficulty as last years, maybe slightly more difficult.
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Probably 36/37
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Lost 5 marks, 3 of which due to stupid arithmetic errors that could have been avoided with a cas.
What kind of score do I need in exam 2 for a 40?
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Really happy with today's exam, I thought it looked hard during reading time but it was actually alright.
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hahah fml I said -2i+2i=-4i
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I think (http://latex.codecogs.com/gif.latex?%5Cleft%20%28%200%2C%5Cfrac%7B%5Cpi%7D%7B3%7D%20%5Cright%20%29%5Ccup%5Cleft%20%28%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%5Cpi%20%5Cright%20%29) was correct for 7(b). When (http://latex.codecogs.com/gif.latex?x%5Cin%5Cleft%20%28%200%2C%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5Cright%20%29), you have (http://latex.codecogs.com/gif.latex?%5Ccsc%282x%29%3C%5Ccsc%28x%29) when (http://latex.codecogs.com/gif.latex?%5Csin%28x%29%3C%5Csin%282x%29), and from 7(a), this occurs when (http://latex.codecogs.com/gif.latex?x%20%5Cin%20%5Cleft%20%28%200%2C%5Cfrac%7B%5Cpi%7D%7B3%7D%20%5Cright%20%29). You also have this inequality when (http://latex.codecogs.com/gif.latex?x%20%5Cin%20%5Cleft%20%28%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%5Cpi%20%5Cright%20%29) because (http://latex.codecogs.com/gif.latex?%5Ccsc%282x%29%3C%200%20%3C%20%5Ccsc%28x%29) for these values. I think a common error would be writing something like (http://latex.codecogs.com/gif.latex?%5Ccsc%282x%29%20%3C%20%5Ccsc%28x%29%20%5CRightarrow%20%5Csin%28x%29%20%3C%20%5Csin%282x%29) and using graphs of the functions to work out the values. This doesn't work on the interval (http://latex.codecogs.com/gif.latex?%5Cleft%20%28%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%5Cpi%20%5Cright%20%29) because (http://latex.codecogs.com/gif.latex?%5Ccsc%282x%29%3C%200%20%3C%20%5Ccsc%28x%29) (you'd be dividing by a negative value, which changes the direction of the inequality).
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I think (http://latex.codecogs.com/gif.latex?%5Cleft%20%28%200%2C%5Cfrac%7B%5Cpi%7D%7B3%7D%20%5Cright%20%29%5Ccup%5Cleft%20%28%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%5Cpi%20%5Cright%20%29) was correct for 7(b). When (http://latex.codecogs.com/gif.latex?x%5Cin%5Cleft%20%28%200%2C%5Cfrac%7B%5Cpi%7D%7B3%7D%20%5Cright%20%29), you have (http://latex.codecogs.com/gif.latex?%5Ccsc%282x%29%3C%5Ccsc%28x%29) when (http://latex.codecogs.com/gif.latex?%5Csin%28x%29%3C%5Csin%282x%29). From 7(a), this occurs when (http://latex.codecogs.com/gif.latex?x%20%5Cin%20%5Cleft%20%28%200%2C%5Cfrac%7B%5Cpi%7D%7B3%7D%20%5Cright%20%29). You also have this inequality when (http://latex.codecogs.com/gif.latex?x%20%5Cin%20%5Cleft%20%28%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%5Cpi%20%5Cright%20%29) because (http://latex.codecogs.com/gif.latex?%5Ccsc%282x%29%3C%200%20%3C%20%5Ccsc%28x%29) for these values. I think a common error would be writing something like (http://latex.codecogs.com/gif.latex?%5Ccsc%282x%29%20%3C%20%5Ccsc%28x%29%20%5CRightarrow%20%5Csin%28x%29%20%3C%20%5Csin%282x%29) and using graphs of the functions to work out the values. This doesn't work on the interval (http://latex.codecogs.com/gif.latex?x%20%5Cin%20%5Cleft%20%28%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%5Cpi%20%5Cright%20%29) because (http://latex.codecogs.com/gif.latex?%5Ccsc%282x%29%3C%200%20%3C%20%5Ccsc%28x%29) because of signs (you'd be dividing by a negative value, which chances the direction of the inequality).
do you think 1 mark would be awarded if you assumed sin(2x) < sin(x) and solved correctly for x element of (0, pi/3)?
i know vcaa are pricks, but would you personally have awarded 1 mark out of 2 if this early error was made, but then the rest of the working out followed the domain etc?
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I think they'll probably end up giving 1 mark for (http://latex.codecogs.com/gif.latex?%5Cleft%20%28%200%2C%5Cfrac%7B%5Cpi%7D%7B3%7D%20%5Cright%20%29) (but you never know since we are dealing with VCAA).
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Clutching at straws here but oh well:
Would you get a mark for drawing the graphs but getting the equality sign the wrong way, hence putting x E (pi/3, pi/2) ?
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I'm 99% sure it's (0, pi/3) U (pi/2, pi). I graphed out cosec(2x) and cosec(x) on the same axes, then looked at it graphically. After x=pi/2, cosec(2x) goes all the way down to -infinity, while cosec(x) slowly returns to positive infinity.
Btw, I'm an idiot at Mechanics and for question 2 (box in a lift), all I wrote was F=ma. F = 20(1.2) = 24N. I did the exact same for the following question. Both questions were 2 marks. Do you guys reckon they'll give me 1 mark each or did I actually just fuck my chances of a low 40's score? :(
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I'm 99% sure it's (0, pi/3) U (pi/2, pi). I graphed out cosec(2x) and cosec(x) on the same axes, then looked at it graphically. After x=pi/2, cosec(2x) goes all the way down to -infinity, while cosec(x) slowly returns to positive infinity.
Was this in response to me? If so I never said it wasn't that I was just asking if I earned any marks, but thinking about it it's ludicrously unlikely I didnt
Btw, I'm a fucking idiot at Mechanics and for question 2 (box in a lift), all I wrote was F=ma. F = 20(1.2) = 24N. I did the exact same for the following question. Both questions were 2 marks. Do you guys reckon they'll give me 1 mark each or did I actually just fuck my chances of a low 40's score? :(
No way on the 1 mark, first mark for both would have probably been earnt by writing some variation of F(net) = R - W. But with a 36/40 you still have every chance to get a 40 if you kill it tomorrow.
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for the first question, would I get a mark for part be for writing A(dot)B=0 if perpendicular, don't ask me how I stuffed up that first question
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for the first question, would I get a mark for part be for writing A(dot)B=0 if perpendicular, don't ask me how I stuffed up that first question
You would get a mark.
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You would get a mark.
Wait we just needed to show the dot product was 0 right? like we didn't then have to go 0=cos(theta)->therefore theta is 90 degrees?
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Wait we just needed to show the dot product was 0 right? like we didn't then have to go 0=cos(theta)->therefore theta is 90 degrees?
Just needed to show the dot product = 0 because the dot product being = 0 can only occur as a result of theta = 90 degrees.
(Or if one of the vectors is the null vector but that's not related)