cssa 2005 (1 Viewer)

[kony]

hey there,

does anyone have the solutions to the ext2 2005 CSSA trials? this is one tricky paper.

thanks

 

[ssglain]

I wouldn't say it's unexpected, but certainly a nice surprise to see you here. And to answer your question, no I don't have the solutions you're after.

I must say I'm rather amused that people have started with trial papers so early on - or maybe I'm being unrealistically optimistic. Guess I should be waiting to shoot my own foot any second now.

 

[Trebla]

I wouldn't say it's a very tricky paper. In my opinion the 2006 paper was a lot more trickier especially with the last question.
Yes, I do have the solutions to them. I'm not going to bother scanning it, because it's a lot of pages. So, just name a question you're genuinely stuck on, and I'll provide assistance.

 

[kony]

thanks trebla. i say it's tricky because i've done with no queries every other cssa paper in the 2 topics i'm being tested on in the upcoming exams, polynomials and conics.

for this paper:

q1, b, (ii) "the line y=mx is a tangent to the curve y = 3 - 1/x². Show that the equation mx³ - 3x² + 1 = 0 has a double root and hence find any values of m."

i've tried every method i know to no avail.

q7, b The equation x² + x + 1 = 0 has roots a, b. Tn = a^n + b^n, n = 1,2,3,...

(ii) show that Tn = -T(n-1) - T(n-2)

note that n, n-1, n-2 are all subscripts, indicating a series of some sort.

over to you, trebla. or anyone else who might give this a go

 

[Yip]

1.(b)(ii) mx=3 - 1/x²
Solutions mx³ - 3x² + 1 = 0 represent x-coordinates of interesection between the line and the curve. Since we know the line is tangential to the curve, there will be a double root that represents this point of tangency.

let f(x)=mx³ - 3x² + 1
f'(x)=3mx^2 -6x=0
x=0 is obviously not the double root [f(0)=1]
thus double root is x=2/m
f(2/m)=(-4/m^2)+1=0
m=2,-2

7.(b)(i)
Substitute a and b into quadratic
a² + a + 1 = 0
b² + b + 1 = 0

Multiply both equations by a^n-2 and b^n-2 respectively,
a^n + a^(n-1) + a^(n-2) = 0
b^n + b^(n-1) + b^(n-2) = 0

Adding the above equations,
a^n+b^n + a^(n-1)+b^(n-1) + a^(n-2)+b^(n-2) = 0
T[n]=-T[n-1]-T[n-2]

 

[kony]

thank you yip!!

only, in the first question, it asks you to show that mx³ - 3x² + 1 = 0 has a double root, not assume it.

any ideas?

and how did you work out the method to do the second question? i mean, what you did is obviously right, but how would you know this was the way to do it before you did it? i'm specifically referring to the multiplying both equations by a^(n-2) thing.

 

[kony]

aaah okay how could i have forgotten.

hehe i'm glad you're not doing the hsc this year.

 

[Trebla]

thank you yip!!

only, in the first question, it asks you to show that mx³ - 3x² + 1 = 0 has a double root, not assume it.

any ideas?

The question was worded very badly here. In the solutions its states that "This equation has a double root if the line is tangent to the curve". Note that m is a variable so it would be impossible to actually prove that the more general equation mx³ - 3x² + 1 = 0 has a double root if m is not found and is considered to be variable.

and how did you work out the method to do the second question? i mean, what you did is obviously right, but how would you know this was the way to do it before you did it? i'm specifically referring to the multiplying both equations by a^(n-2) thing.

The trick is to look ahead in your working before plunging into the algebra. Most people who do well in Extension 2 will be able to look ahead and mentally plan out their method of solving it. It all involves a bit of deductive logical thinking from the smallest of clues.
A good way to start is to list what you know. Given that a and b are roots of the equation x² + x + 1 = 0, then therefore
a² + a + 1 = 0 (1)
b² + b + 1 = 0 (2)
Now notice the result you are trying to prove: T_n = - T_{n - 1} - T_{n - 2}
Rearrange it a bit and you get T_n + T_{n - 1} + T_{n - 2} = 0
Look at the original equation x² + x + 1 = 0. Did you notice that the powers of x decreased by 1 in each progressive term? Notice how similar that looks to the expression we are trying to prove.
Now all we have to do is make everything in terms of n. Hence we multiply (1) by a^{n - 2} and (2) by b^{n - 2} to obtain the nth degree as well as the terms with powers of (n - 1) and (n - 2) respectively. A bit of simple substitution would then eventuate into the required result.

An interesting thing to note is that the suggested solutions from the CSSA is completely different and much more brief comapred to the one shown here. It just says something like:

For n = 3, 4, 5, ......

aⁿ + bⁿ = (a + b)(a^{n - 1} + b^{n - 1}) - ab(a^{n - 2} + b^{n - 2})

a + b = -1
ab = 1

.: aⁿ + bⁿ = - (a^{n - 1} + b^{n - 1}) - (a^{n - 2} + b^{n - 2})

.: T_n = - T_{n - 1} - T_{n - 2}

doubt many would have thought of that though lol....


That's the beauty of Extension 2 Maths, finding those extremely concise clues and turning what looks like a hard problem into a very simple one. That mentally planning ahead thing is a skill that comes with plenty of practice.