Specialist Exam 1

[wobblywobbly]

Really happy with today's exam, I thought it looked hard during reading time but it was actually alright.

[TiffanyF01]

hahah fml I said -2i+2i=-4i

[Conic]

I think was correct for 7(b). When , you have when , and from 7(a), this occurs when . You also have this inequality when because for these values. I think a common error would be writing something like and using graphs of the functions to work out the values. This doesn't work on the interval because (you'd be dividing by a negative value, which changes the direction of the inequality).

[cosine]

I think (Image removed from quote.) was correct for 7(b). When (Image removed from quote.), you have (Image removed from quote.) when (Image removed from quote.). From 7(a), this occurs when (Image removed from quote.). You also have this inequality when (Image removed from quote.) because (Image removed from quote.) for these values. I think a common error would be writing something like (Image removed from quote.) and using graphs of the functions to work out the values. This doesn't work on the interval (Image removed from quote.) because (Image removed from quote.) 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?

[Conic]

I think they'll probably end up giving 1 mark for (but you never know since we are dealing with VCAA).

[GeniDoi]

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) ?

[99.90 pls]

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?

[GeniDoi]

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.

[bobisnotmyname]

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

[GeniDoi]

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.

[Splash-Tackle-Flail]

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?

[GeniDoi]

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)