Post by: chuck981996 on November 06, 2014, 05:42:33 pm
Multiple choice answers thanks to jumcakes:
- A
- D
- D
- E
- C
- B
- E
- A
- A
- E
- E
- C
- C
- E
- B
- C
- B
- B
- C
- D
- D
- E
Also attached is an entirely BLANK copy of the ENTIRE exam.
5ci) d E [1,3)
5cii) d<1
5eii) u = 1/2(1 + sqrt(5)), v = 1/2(1 - sqrt(5))
The new file has been updated to include these errors.
UPDATED ERRATA:
2f,g) Units should be m/yr
N.B. The scan is randomly in colour on the last two pages because I'm at home and don't have access to my school's amazing super-scanner. (Clearly my scanner thinks "black & white" means greyscale) :'( I also couldn't include it in the original file and reattach because of the filesize limit - again, don't have access to a scanner capable of compressing and pixelating the files.
-Charlie
Post by: ivysan on November 06, 2014, 05:49:10 pm
Post by: RKTR on November 06, 2014, 06:03:28 pm
cii mine was d<1
fi u is the positive one while v is the negative one. since it says gradient at u is positive while gradient at v is negative
Post by: Blondie21 on November 06, 2014, 06:06:00 pm
Post by: speedy on November 06, 2014, 06:06:17 pm
5ci my answer was 1<= d<3
cii mine was d<1
fi u is the positive one while v is the negative one. since it says gradient at u is positive while gradient at v is negative
I got the same as you for cii and fi
Edit: Is 5b positive direction? Since we need to sub (x-1)
Post by: RKTR on November 06, 2014, 06:08:44 pm
I got the same as you for cii and fi
nice . how about ci? if d is 1, intercepts will be at (0,0) and (2,0) . so d can be 1?
Post by: speedy on November 06, 2014, 06:10:58 pm
nice . how about ci? if d is 1, intercepts will be at (0,0) and (2,0) . so d can be 1?
Mm, I didn't get the same as you but I think you're right. Since 0 isn't positive.
Post by: Aman476 on November 06, 2014, 06:14:35 pm
Post by: Aman476 on November 06, 2014, 06:18:41 pm
Post by: Mujteba on November 06, 2014, 06:20:48 pm
You need to reject the negative u solution and the positive v solution because the gradient at u is positive and is negative at v.
For 5ci
The boundary d=1 is included as if it were 1 then the intercept to the left of the positive one would be zero.
For 5cii
Should be d<1. You can see the result if you set up two inequalities for the two intercepts and set them both to >0. You end up with d<1, and d<5 so you take the intersection between both which is d<1
Also for 4e
Should n=14 because it is asking for min value of n, not all values of n for which blah blah blah...
Post by: chuck981996 on November 06, 2014, 06:21:11 pm
5ci my answer was 1<= d<3
cii mine was d<1
fi u is the positive one while v is the negative one. since it says gradient at u is positive while gradient at v is negative
You're right on all three counts. Sorry for the muck up!
Also for 4e
Should n=14 because it is asking for min value of n, not all values of n for which blah blah blah...
I think what I wrote is okay, because I'm stating n>=14 which implies n=14 is the minimum value, but I think VCAA would accept either :)
Is 5b positive direction? Since we need to sub (x-1)
We need to sub (x+1), so no.
Post by: jumcakes on November 06, 2014, 06:22:21 pm
- A
- D
- D
- E
- C
- B
- E
- A
- A
- E
- E
- C
- C
- E
- B
- C
- B
- B
- C
- D
- D
- E
Post by: GeniDoi on November 06, 2014, 06:38:22 pm
Post by: chuck981996 on November 06, 2014, 07:00:14 pm
e) ii) solution is partially incorrect, you needed to reject the negative solution for u (and hence the positive solution for v) as when u = (1-Sqrt(5))/2, g'(u) is negative and the question states that at g(u), the gradient is "m" where m is a positive number.
Thanks! RKTR already pointed this out - it's in the OP :)
Post by: Aman476 on November 06, 2014, 07:16:22 pm
You're fkn amazing man, you did like both exams in 20 min - glad i'm not competing with you in methods this year haha (but you're in spec - so for that :-[ :-[) hahaa
Post by: Brunette15 on November 06, 2014, 07:17:08 pm
Post by: GeniDoi on November 06, 2014, 07:17:25 pm
Thanks! RKTR already pointed this out - it's in the OP :)
No worries. Thanks for writing the solutions for these exams :D
Post by: NyaTaku on November 06, 2014, 07:18:21 pm
Does anyone have mc solutions with the actual questions written?
Would like these too.
Post by: outtaheresoon on November 06, 2014, 07:25:19 pm
Post by: chuck981996 on November 06, 2014, 07:38:27 pm
Chuck do you reckon you'll be able to keep being the sick kent you are and post the spec solutions tomorrow and monday? (I'm praying yes - seeing the solutions completely gets rid of that uncertainty/anxiety post exam) haha
You're fkn amazing man, you did like both exams in 20 min - glad i'm not competing with you in methods this year haha (but you're in spec - so for that :-[ :-[) hahaa
Nah I won't do spesh - I'll be too busy worrying about my own solutions!!
Post by: keltingmeith on November 06, 2014, 07:42:10 pm
Post by: speedy on November 06, 2014, 07:43:13 pm
Nah I won't do spesh - I'll be too busy worrying about my own solutions!!
Could you explain why 5b is in the negative direction? Maybe I'm going full retard here or something but I don't get it. Shouldn't it be in the positive direction?
Post by: jumcakes on November 06, 2014, 07:50:01 pm
Could you explain why 5b is in the negative direction? Maybe I'm going full retard here or something but I don't get it. Shouldn't it be in the positive direction?
Probably would be better on the main discussion board - but we want the transformation that maps f(x) to g(x) - that is the transformation that maps the quartic f(x) with intercepts at (1,0) and (3,0) to the quartic g(x) with intercepts (0,0) and (2,0) - so we move one left.
Additionally, (x-3)(x-1)(x^2+3) becomes ((x+1)-3)((x+1)-1)((x+1)^2+3) = (x-2)x((x+1)^2+3)= g(x) which is what we want!
Post by: chuck981996 on November 06, 2014, 07:52:40 pm
Could you explain why 5b is in the negative direction? Maybe I'm going full retard here or something but I don't get it. Shouldn't it be in the positive direction?
Because the transformation maps x->x+1.
To see very clearly, have a look at the graphs I did on Wolfram Alpha (attached). See how, to get g(x) from f(x) we must translate it by 1 in the negative direction?
Post by: speedy on November 06, 2014, 07:58:03 pm
Thought it was g(x) to f(x) lolz, sorry guys
Post by: S61778 on November 06, 2014, 08:05:11 pm
Post by: vizard12 on November 06, 2014, 08:11:50 pm
I then calculated standerd devition using squareroot(intergral from 0 to 50 of (x-25)^2 * h(x) dx).
then i solved z=(x-25)/standard deviation just caluated for x.
i got 13.7227 for x, have i completely overcomplicated this and will get 0 marks?
cheers
Post by: swagsxcboi on November 06, 2014, 08:19:19 pm
Isn't q17 on multi choice A?nah, A, when a=3 gives you infinitely many solutions iirc
Post by: Yuki N. on November 06, 2014, 08:22:33 pm
Post by: althepal on November 06, 2014, 09:10:43 pm
Post by: swagsxcboi on November 06, 2014, 09:17:23 pm
if I didn't put units in the majority of my answers, will I loose one mark per answernah you'll be fine, you don't need units.
Post by: NyaTaku on November 06, 2014, 09:18:54 pm
For 2f. I used 10 instead of -10 and ended up with a positive answer. How many marks will I lose for that?
1 mark if they're mean, 2 if they're brutal.
nah you'll be fine, you don't need units.
Can't tell if serious
Post by: swagsxcboi on November 06, 2014, 09:21:41 pm
1 mark if they're mean, 2 if they're brutal.
Can't tell if serious
In methods, you NEVER are required to write units. Ever. There are no exceptions. Every question can be answered to full marks without writing units.check previous assessment reports
The only time you write units is if YOU have changed the units of the answer you are stating, in which case you have changed the magnitude of the answer and must justify it with the appropriate unit of measurement. Like the question "Find the height in cm of the plant to the nearest mm" or whatever it was, unless you converted your answer to mm, you can simply leave it unitless.
Post by: speedy on November 06, 2014, 09:30:37 pm
check previous assessment reports
In Euler's stupid question thread, someone apparently emailed VCAA and they said to include them.
Edit: I am on the side of not including them though - however I did just in case :S
Post by: swagsxcboi on November 06, 2014, 09:32:28 pm
apologies for messy working, just ask for clarification
didn't do the last two MC because I have no idea how to
Post by: chuck981996 on November 06, 2014, 09:42:31 pm
Hey for question 4d i calculated the z value using .15 as area which gives z=-1.03643.
I then calculated standerd devition using squareroot(intergral from 0 to 50 of (x-25)^2 * h(x) dx).
then i solved z=(x-25)/standard deviation just caluated for x.
i got 13.7227 for x, have i completely overcomplicated this and will get 0 marks?
cheers
The height isn't normally distributed, I'm afraid. It's governed by the PDF given on page 19. :( I'm sorry to say it but I don't think that answer can be awarded any marks.
if I didn't put units in the majority of my answers, will I loose one mark per answer
I think you may have been given misinformation by other people here, or confusing information, so I will tell you what I think:
If you don't have an SI unit like m or m^3 you will lose a mark. If you don't have a unit like "wombats" you will probably lose a mark - but it depends what other people did. I'm so sorry to tell you this.
Can someone explain why q2f is m/hr and not m/yr for units?
Ugh omfg because I'm stupid. You're right: it should be m/yr. Sorry about that.
Edit: updated it in the OP.
Post by: jumcakes on November 06, 2014, 09:55:44 pm
On the other hand, 1)b) might need units (maximum is 1600 wombats, minimum 800 wombats, since the question is asking for the maximum populations rather than the range of n. I'm under the impression that if they asked for the maximum/minimum of n(t), 1600 and 800 without units would be acceptable.
Another example would be that it's fine to say the period is 6, instead of 6 hours, but again depends on the nature of the question.
Post by: Aman476 on November 06, 2014, 10:01:53 pm
|z+congjug(z)| = |z-conjug(z)|
i ended up with x^2 = y^2 and so that y = sqrt(x^2) = |x|
but if you think about what it's saying - the distance between z and the negative of its conjugate being equal to distance between z and its conjugate being equal it is any point along y = x and y=-x (this is what vcaa say as well) my question is why this doesn't come out in the algebra?
Post by: speedy on November 06, 2014, 10:04:44 pm
I know this should be in the spec from but given chuck is here haha, has anyone seen last years answer to exam 2 question 2 b they ask you to graph the locus points such that:
|z+congjug(z)| = |z-conjug(z)|
i ended up with x^2 = y^2 and so that y = sqrt(x^2) = |x|
but if you think about what it's saying - the distance between z and the negative of its conjugate being equal to distance between z and its conjugate being equal it is any point along y = x and y=-x (this is what vcaa say as well) my question is why this doesn't come out in the algebra?
I don't understand this? y = |x| = +/- x
Post by: Aman476 on November 06, 2014, 10:07:12 pm
I don't understand this? y = |x| = +/- x
|x| = +x
|-x| = x
you only get the parts of the lines above the x axis, but the answer is the lines unrestricted (so i drew like a modulus, but the answer is just two lines going on forever, -x and +x
Post by: Conic on November 06, 2014, 10:15:02 pm
x^2 = y^2 and so that y = sqrt(x^2) = |x|It should be
Post by: Aman476 on November 06, 2014, 10:18:16 pm
It should berather than
, since you took the square root of both sides, and
and
. You could also do this using another method:
OMGGGG thankyouuuuuu haha
Post by: Mayesy on November 06, 2014, 10:22:48 pm
Post by: simondaaanng on November 06, 2014, 10:23:31 pm
Post by: NyaTaku on November 06, 2014, 10:25:09 pm
Q4e. "Find the minimum value of n". Wouldn't it be n=14 instead of n>=14?
n=14, pretty sure it's been said already.
Post by: Mayesy on November 06, 2014, 10:35:02 pm
Post by: chuck981996 on November 06, 2014, 10:48:02 pm
Q4e. "Find the minimum value of n". Wouldn't it be n=14 instead of n>=14?
I think my answer's good too though - VCAA would accept either.
Post by: jumcakes on November 06, 2014, 10:56:48 pm
Also, why is A for question 3d 4, I had 37e/25
Nearest integer :(
Post by: Mayesy on November 06, 2014, 11:27:44 pm
Can someone please explain why question 9 in mc is A and not E?Still waiting for an explanation anyone?
Post by: NyaTaku on November 06, 2014, 11:30:03 pm
Still waiting for an explanation anyone?
Range of f(x) = Domain of f inverse.
ran f= (4,inf) therefore the domain of the inverse would be (4,inf)
Post by: Mayesy on November 06, 2014, 11:54:29 pm
Range of f(x) = Domain of f inverse.Ahh right I forgot that it approaches 4 from above, and when I graphed it on my calculator is appeared to begin at 4, my mistake, thanks.
ran f= (4,inf) therefore the domain of the inverse would be (4,inf)
Post by: hisoka on November 07, 2014, 07:02:43 pm
Credits: Dr He and Big Garry Zhu
Post by: keltingmeith on November 07, 2014, 07:05:55 pm
Question 5.c.i, should be [1,3) since 0 isn't a positive number, and if it were translated 1 unit to the left, one of the intersections would be x=0.
Credits: Dr He and Big Garry Zhu
This has been fixed, if you check the updated question 5 upload.