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AEA June 2011

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I know, they should do it like those music exams where you have pass, merit and distinuction. I completely agree with you!

How annoying if someone was just one mark off a merit, but that other person completely failed that paper, and they both got U. Not fair, isn't it!

Yeah it's not fair!!! Sorry I acciently negged you D: will pos later!!

Reply 201

Agent Fiasco

Original post by nick_the_big

got the same . hopefully is correct

me too XD

the only question i did quite well was the circle one with the square

wat did u guys gat as the centre and the radius and was was ur max area and were u able to show that its a max???

Reply 202

RK92

Original post by implus

This is how I did question 1:

$let \ \theta +35^{\circ} = A \\ and \ let \ \theta -53^{\circ} = B \\tanA=CotB \\ \frac{SinA}{CosA}=\frac{CosB}{SinB} \\ SinASinB=CosACosB \\ CosACosB-SinASinB=0 \\ Cos(A+B)=0 \\ Cos(2\theta -18^{\circ})=0 \\ 2\theta -18^{\circ}= 90^{\circ}, 270^{\circ} \\ \theta = 54^{\circ},144^{\circ}$

another way:

tan[ (x-9) - 44 ] * tan[ (x-9) + 44 } = 1

use tan double angle, you get difference of two square on top and bottom of the fractions and you can tidy up to:
tan²(x-9) . [1+tan²44] = [1+tan²44]
tan(x-9) = +/- 1
so in the range 0<x<180
tan(x-9) = 1
and tan(x-9) = -1

giving 54 and 144

(of course, i forgot to deal with the +/- and so didnt mention the fact that tan(135)=-1 :rolleyes:

)

(edited 12 years ago)

Reply 203

Piecewise

Original post by RK92

even simpler:

tan[ (x-9) - 44 ] * tan[ (x-9) + 44 } = 1

use tan double angle, you get difference of two square on top and bottom of the fractions and you can tidy up to:
tan²(x-9) . [1+tan²44] = [1+tan²44]
tan(x-9) = +/- 1
so in the range 0<x<180
tan(x-9) = 1
and tan(x-9) = -1

giving 54 and 144

(of course, i forgot to check deal with the +/- and forgot that tan(135)=-1 :rolleyes:

)

Far simpler: as

\cot(x) = \tan(90^{\circ}-x)

, we have

\cot(\theta-53^{\circ}) = \tan(143^{\circ}-\theta)

; hence what we were asked to solve is

\tan(\theta+35^{\circ}) = \tan~(143^{\circ}-\theta)

from which

\theta = 54^{\circ}+90^{\circ}n

. Obviously

n = 0

, and

n = 1

give values of

\theta

in the desired range! :tongue:

Reply 204

ali25

I could not get the first part of the vector question, which is annoying because I could have done the rest of it, |I just couldn't because I couldn't do part i.

Reply 205

lawwainam

is the answer for part C of Q3 be 70?

Reply 206

lawwainam

Original post by ali25

I could not get the first part of the vector question, which is annoying because I could have done the rest of it, |I just couldn't because I couldn't do part i.

Me too. I don't know how to do the reflection of the 3D vector :frown:

Reply 207

M1K3

Original post by Agent Fiasco

me too XD

the only question i did quite well was the circle one with the square

wat did u guys gat as the centre and the radius and was was ur max area and were u able to show that its a max???

(0,1/2) mate, n i think i got the radius to be 1/2, I hope i got that right as it was the only question which seemed to go well :smile:

As well, the value of alpha i got to be pi/6, can't remember what i got for area.

I decided to leave determining whether it was max or not, i thought i wud leave it till the end as I already spent too much time (on Q1 mainly), and only had about an hour for Q5,6,7 :angry:

so yeah, couldn't look at Q7 at all, only parts of Q6.

Reply 208

OneMomentPlease

Original post by lawwainam

is the answer for part C of Q3 be 70?

yeah I got 70,

how do yo show that a^2 etc... is what they gave, I could'nt get it :mad:

Reply 209

SmileyGurl13

So who else thinks they got a U?

Reply 210

lawwainam

Original post by OneMomentPlease

yeah I got 70,

how do yo show that a^2 etc... is what they gave, I could'nt get it :mad:

Do you mean Q5 (c) (i)?
I couldn't get it too. :frown:

, sorry
By the way, is q=6, p=3 for Q.7?

Reply 211

OneMomentPlease

Original post by lawwainam

Do you mean Q5 (c) (i)?
I couldn't get it too. :frown:

, sorry
By the way, is q=6, p=3 for Q.7?

Ah ok, I really don't know why I couldnt get it, seemed simple enough...

And yeah I think so, well thats what I remember putting :smile:

Reply 212

superkinetic

Original post by Webbykun

Oh dude stop this! xD Okay I'm just gonna assume now I haven't got a distinction and prepare for my insurance uni :smile:

What's yr firm? Are you majoring in math in future?

Original post by lawwainam

Do you mean Q5 (c) (i)?
I couldn't get it too. :frown:

, sorry
By the way, is q=6, p=3 for Q.7?

Yeah that is correct.

Reply 213

AnonyMatt

Original post by superkinetic

Yeah you're to use dot product, to prove

\vec{AP} \cdot \vec{AB} = 0

, as cos90=0. But then you got to find B first which depends on the P' from (a). That is the only method I can think of.

I didn't use vectors at all.

Since you were given 120, the opposite angle is 60 degrees.

Thus the necessary angle is 180 - 60 - 30 = 90.

The reason I had to do it this way is because I didn't know how to find P', or B, or anything.

How did people find the inverse function of m at the end? :s-smilie:

I couldn't find 'q' either. That was annoying.

I didn't do question 1 either.

I got 60 for the sum to infinity... :frown:

I couldn't show the last parts of the circle questions - I said that the radius of the circle was the normal to the curve, and that equals y - 1/2 - but I couldn't get the algebra out. FML.

Definitely not got a distinction. Ugh. Might not be going to uni. D:
*fingers crossed I got a 2 SOMEWHERE on STEP*

Reply 214

lawwainam

Original post by AnonyMatt

I couldn't find 'q' either. That was annoying.

I didn't do question 1 either.

I got 60 for the sum to infinity... :frown:

I used the quadratic formula to find the inverse function.
Am I right?
And for Q4, alpha=pi/6, right?

(edited 12 years ago)

Reply 215

jumping_homeless

Hey guys, a couple of things.

Firstly, what did people get for 7 c) (10 marks)?

Part i) was to find m^(-1), can't remember the rest of the question though :frown:

Also what are your thoughts about what the grade boundaries are going to be this year?

I'd say this is the hardest paper I've seen and 2006/7 had distinction grades at 67 and 68 respectively so maybe 65 ish?

The only thing I'm clinging onto was the fact I got all of the vector question, struggled with 3 b, c, 5 c and 7 b & c.

Thanks.

Reply 216

kenyan1231

Original post by SmileyGurl13

So who else thinks they got a U?

ME!

Reply 217

lawwainam

Original post by jumping_homeless

Hey guys, a couple of things.

Firstly, what did people get for 7 c) (10 marks)?

Part i) was to find m^(-1), can't remember the rest of the question though :frown:

The grade boundaries of 2010 scared me.

Reply 218

DFranklin

Original post by lawwainam

The grade boundaries of 2010 scared me.

2010 was an aberration - very easy paper.

Reply 219

superkinetic

Original post by AnonyMatt

But that's not true that opposite angles in a kite add up to 180 deg; it's coincidental that both opposite angles add up to 180 deg, as one of the angle is 90 deg.

Assume there are 2 similar angles of 100 deg in a kite. Do the remaining opposite angles add up to 180 deg?

Anw, do u require AEA for yr offer?