General thoughts- Extension 1 maths (1 Viewer)

[MonsterCookie]

Wow...that was actually easier than i expected it to be...i think it is probably my least hated exam so far...which is nice. I actually kinda enjoyed doing it.
I was so glad that they didn't have on of those really hard binomial questions. I just can't seem to get my head around them.
And i wasn't wishing for more time, because i git do do each question to the best of my ability...and i answered alot more questions than i anticipated i would.
And my theory worked (better than expected too)...i decided that if i went into the exam with no confidence whatsoever...then when i saw the paper...and realised how bloody hard it was...i would have no confidence to be shattered, and hence i couldn't freak out and screw up...the great thing was that when i say that the paper was easier than i expected...i got confidence!!! YAY!!!
Overall...i am very pleased with the exam, and i think that i will get a good mark...and a much better one than i anticipated during the exam.

I hope everyone else found the paper ok, and good luck with your last few exams all

 

[woho]

i found it harder than 4 unit too- there's so much room for error in the 3 unit exam, while 4 unit, you got the proofs, you're looking at a great scaled mark.

Ok then. Personally I struggle with 4u and usually pwn 3unit, so that's prob y I found it easier.

But yay for scaling

 

[MonsterCookie]

Blah. It started of very easy, and then some harder questions poped up and screwed me over =/

Mainly the circle geo question pissed me off like hell. I am very good at circle geo, and every past ext1 paper I've done, I got the answer in like 10 seconds, and here I was sitting for like 5-10 mins trying to figure it out and couldn't.
That's what really really pissed me off >_<

The circle geometry question was Q4c) right...that one was easy.

 

[Starcraftmazter]

Ok, for 5c, I got x=1/root 2 and y=root 3/2.

Those are the only two answers, people whom think there's 4 are thinking of it the other way around, if you had x and y and had to find the angle.

But there are only 2 answes for 5c.

 

[Legham]

In relation to question 5c...damnit I forgot how I did it, gime a sec.

angle BCX = angle BDA = theta

angle DAX = 180 - 90 - theta

angle AXP = 180 - 90 - (90 - theta) = theta

angle QXC = angle AXP vert. opposite

.: angle qxc = angle QCX = theta

angle pxd = angle axd - (180 - 90 - theta) = theta

angle bxq = angle pxd = theta

angle bxq + angle qxc = 90 = 2 theta

.: theta = 45

in triangle xbc angle xbc = 180 - 3 theta = 45

,: angle bxq = angle xbq = theta = 45



I think thats different that how i did it in the exam :S uh oh!

 

[Vasc0]

Oh thats reassuring then Someone was talking about limits and shit and i didnt use anything like that :/ Just subbed t = 10 into the equation for x.

I got 283 for that...didn't you guys integrate then find the constant? C = -250...so when you get 513 or whatever you subtract 250 from that.

Also for 1c) what was the answer? I'm not sure if I got that one

 

[Shmegel]

i barely studies for maths ext, so its not like i expected much anyway...

but this test was quite simple, but i scrwed up bad...ran out of time, couldnt even touch question 7 and most of 6...:rofl: thank god this doesnt have to count

 

[mattmaug]

angle BCX = angle BDA = theta

angle DAX = 180 - 90 - theta

angle AXP = 180 - 90 - (90 - theta) = theta

angle QXC = angle AXP vert. opposite

.: angle qxc = angle QCX = theta

angle pxd = angle axd - (180 - 90 - theta) = theta

angle bxq = angle pxd = theta

angle bxq + angle qxc = 90 = 2 theta

.: theta = 45

in triangle xbc angle xbc = 180 - 3 theta = 45

,: angle bxq = angle xbq = theta = 45



I think thats different that how i did it in the exam :S uh oh!

i proved triangle XQB was isosceles, so QX=BQ
then proved XQC was also isosceles so QX=QC
therefore BQ=QC so it bisected it

 

[Forbidden.]

i barely studies for maths ext, so its not like i expected much anyway...

but this test was quite simple, but i scrwed up bad...ran out of time, couldnt even touch question 7 and most of 6...:rofl: thank god this doesnt have to count

Same, but I managed to solve the Cartesian equation question but left the rest of Question 7 out.

 

[t.j.howlett]

hey guys the exam wasnt too bad i made a few dumb mistakes which i was pissed about :burn: for the probability with the people sitting (5b) did u have to do (4!x2!)/6! ?

 

[Legham]

i proved triangle XQB was isosceles, so QX=BQ
then proved XQC was also isosceles so QX=QC
therefore BQ=QC so it bisected it

Yeah, that was part (ii).. mine was a solution to (i)

 

[anniea89]

I found it hard. I had to make a half arsed attmept at some questions

 

[MonsterCookie]

Q4c)

(i) let angle QXB = theta

angle PXD = angle QXB (vert. opp. angles)

angle XDP = 180-(90+theta) = 90 - theta (angle sum triangle)

angle ADB = angle ACB = 90 - theta (angles in same segment)

angle XBC = 180 - (90+90-theta) = theta (angle sum triangle BXC)

therefore angle QXB = angle QBX = theta (equals to equals)


(ii)

Then you prove that triangle BXQ is isosceles (base angles equal as proved above)
therefore BQ = XQ (equal sides isos triangle)

angle BXC = 90 (AC perpendicular to BD)
and angle BXC = angle BXQ + angle QXC angles)

therefore angle QXC = 90 - theta (ajacent complimentry angles)

therefore triangle QXC is isos (base angles equal)
therefore QX = QC (equal sides isos triangle)

therefore BQ = CQ (equals to equals)
therefore Q bisects BC (BQ = CQ)

There you go...i hope that explains it.

 

[joshuajspence]

hey guys the exam wasnt too bad i made a few dumb mistakes which i was pissed about :burn: for the probability with the people sitting (5b) did u have to do (4!x2!)/6! ?

I had 4! * 4 / 6!

Because the parents could sit at either end (2) and they could switch seats (2)

therefore 2*2*4! / 6!

 

[victorheaven]

it's 4!*3!/6!

 

[MonsterCookie]

I had 4! * 4 / 6!

Because the parents could sit at either end (2) and they could switch seats (2)

therefore 2*2*4! / 6!

I had 4!x3! for the no. of ways the kids could sit next to each other, and then 6! for the total number of ways the family could be arranged...
therefore the probabliity was 4!x3!/6!
which was 1/5

what did everybody else get???

 

[mrzeidan1]

That exam was alright i thought

i got 283 for the parachuting lady

and 4!x3!/6! for the probability

missed a few marks, 1ST PARAMETRICS WHAT THE HELL , and also 6b.iii and 7aii.

finished the projectile hopefully its right

hoping for a 75+/84

 

[gaoOO]

I had 4!x3! for the no. of ways the kids could sit next to each other, and then 6! for the total number of ways the family could be arranged...
therefore the probabliity was 4!x3!/6!
which was 1/5

what did everybody else get???

mm, i got 1/5 as well.

but i realised i did the last part of question 7 b) wrong.

they wanted the lengths, i gave them the range

 

[fishy89sg]

OK, 2 marks gone for that question........

lets keep on revealin my mistakes guys! lol

 

[lilfrenchi]

I had 4! * 4 / 6!

Because the parents could sit at either end (2) and they could switch seats (2)

therefore 2*2*4! / 6!

i counted the 4 children as one since they are seatin together then multiplied it by the way they can be arranged together. then divided that by 6!
therefore 3!*4!/6!
duno if im right tho