Trial Revision Thread (1 Viewer)

[Riviet]

With my 4u trial coming up this Monday (CSSA), I thought a thread for help with some of the harder questions from past papers, mainly catholic ones, would be useful in preparation for the exam.

I've had a look at the '05 catholic paper (can be found here*) and it definitely doesn't look easy.

Feel free to ask questions or contribute by helping others with questions from past trials.

*courtesy of buchanan

 

[YBK]

cool, i'll be back with questions

 

[Riviet]

Here are the main questions that I'm having trouble with in the '05 catholic:
1 b)
2 e)
3 b) ii)
5 b) i)
6 b) ii)
7 a) iii) iv) and b) ii) iii)
8 c)

edit: I got the worked solutions for the '05 catholic from Sober.

 

[Sober]

1 b)

P(x) = (x-a)² * Q(x)
P'(x) = 2(a-x) * Q(x) + (x-a)² * Q'(x)
= (x-a) * (2Q(x) + (x-a)Q'(x))
= (x-a) * R(x)

 

[hyparzero]

Solution to 8c)i)

Yea, during MSN convo with Riviet:

 

[Sober]

I found the answer sheet to this exam (MX2 CSSA 2005) amongst my notes and have uploaded it to a server, PM me for the URL.

 

[Riviet]

Thanks again hyparzero and Sober, really appreciate it.

 

[vafa]

I have finished this paper and I actually was going to publish the answeres but unfortunately the system does not allow me to upload my scans. can anyone email me and then i will send my solutions to him/her and then we can share them.


Vafa11@aol.com

 

[hyparzero]

I have finished this paper and I actually was going to publish the answeres but unfortunately the system does not allow me to upload my scans. can anyone email me and then i will send my solutions to him/her and then we can share them.


Vafa11@aol.com

could u send me the solutions: ye.zero@gmail.com,
also, for question 8c)ii), i get the feeling it wants me to use the chain rule, since there are more than 1 variable (radius and height) if A is constant.

 

[vafa]

I was successful to compress the files. by i can only send them one by one...!View attachment 13463

 

[YBK]

for question 1b) i

why cant we just do this:
(x-a)^2 = P(x)

and then differentiate


then sub x = a into the derrivative and we get 0...

part ii, I got right... but vafa sent the solution to that

 

[bboyelement]

for question 1b) i

why cant we just do this:
(x-a)^2 = P(x)

and then differentiate


then sub x = a into the derrivative and we get 0...

part ii, I got right... but vafa sent the solution to that

because we dont know what P(x) is ... what you did just then was making P(x) = x^2-2ax+a^2 ... but they didnt say that so you cant assume ... while all polynomials have the form of (x-a)^n*Q(x)+R(x)

as they said that P(x) = 0 with double root at a then sub in the equation and you would get R(x) = 0 or you could just say that a root has no remainder hence thats why Sober didnt put R(x)

 

[YBK]

3 b) ii)

after drawing the parallelogram label the angle that is made in the bottom right corner.

the angle would be 2pi/3 since the side OC bisects BOA, which is pi/3 (parallelogram rule)

after you've got the bottom right angle, you can use the cosine rule.

OC^2 = 4^2 + 2^2 - 2*4*2cos(2pi/r)
Therefore OC = 2 root 7

OC is |z1 + z2|
therefore |z1 + z2| = 2 root 7

 

[YBK]

because we dont know what P(x) is ... what you did just then was making P(x) = x^2-2ax+a^2 ... but they didnt say that so you cant assume ... while all polynomials have the form of (x-a)^n*Q(x)+R(x)

as they said that P(x) = 0 with double root at a then sub in the equation and you would get R(x) = 0 or you could just say that a root has no remainder hence thats why Sober didnt put R(x)

ah okiez, thanks..

 

[Riviet]

Has anyone had a look at the '04 catholic? I did it today and the first few seemed alright, but 6, 7 and 8 got pretty nasty.

 

[YBK]

Has anyone had a look at the '04 catholic? I did it today and the first few seemed alright, but 6, 7 and 8 got pretty nasty. I'll post up the ones I'm having difficulty with later.

P.S Anyone got solutions for it?

yup...

i do i do...

i've done up to half of question 5
lol

 

[bboyelement]

question 1b(ii) where it askes show that mx³ - 3x² + 1 = 0 has a double root ... how would you show that its a double root ... i mean you solved it simutaneously with the equation of the tangent and the curve to get this new equation hence if the tangent touches the curve then it would touch at two identical points and hence the new equation would have a double root ... but how would you go about 'showing' mathematically...
off cssa 2005 btw

 

[YBK]

b) ii)

this one's pretty simple once you get the little trick:

Let roots be a,b,c
you know that roots a + b + c = 0

therefore a + b = - c
b + c = - a
and c + a = -b

now we can substitute into the expressions that they've given us.

(a + b) / c^2 = -c/c^2 = -1/c

and so on.

now x = -1/c
therefore c = -1/x

now sub -1/x into the original equation

(-1/x)^3 + 2(-1/x) + 1 = 0

therefore

x^3 - 2x^2 - 1 = 0

 

[YBK]

question 1b(ii) where it askes show mx³ - 3x² + 1 = 0 how would you show that its a double root ... i mean you solved it simutaneously with the equation of the tangent and the curve to get this new equation hence if the tangent meets the curve then it would meet at a double root ... but how would you go about 'showing' mathematically...

you don't have to... all you need to do is say that it's a tangent where they meet and that means that there's a double root.

 

[Riviet]

Cool, thanks for that YBK. =)

In the '04 catholic, can anyone do Q6 a) ii) where you have to show the speed is Vsin@?

Also having trouble with Q7 a) b) and Q8 b).