STEP I, II, III 2000 solutions

SimonM

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(Updated as far as #62) SimonM - 24.03.2009

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Hmm, maybe I picked the wrong questions the last time I tried this paper - 1 and 2 didn't take nearly quite as much time as 5, 8, or especially 7 (see previous thread for solution). Note to self: work on getting rid of paranoid fear of geometry.

For the geometry problems there might be a more elegant approach I missed - if someone could point out a less tedious method, that would be greatly appreciated. Especially for question 5 - I was approaching that more as a linear algebra problem than geometry...

Attached files

III-2000-Q01.pdf (29.6 KB)
III-2000-Q02.pdf (45.9 KB)
III-2000-Q03.pdf (44.7 KB)
III-2000-Q05.pdf (44.3 KB)
III-2000-Q08.pdf (45.1 KB)

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Zhen Lin

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I tried to do 4 and 6 today but failed to complete the problems. However I suppose I did not try hard enough. For question 4, I was unable to show even the first result (somehow I got the result that there must be an intersection if $| a | \ge \frac{1}{2}$ instead of the required $| a | \ge 1$ ); for question 6 I'm not sure how to approach the last part using the previous results. Any hints?

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DFranklin

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For Q4, try to directly solve x+c = f(x). You end up with a quadratic involving c and a. Consider the discriminant D of that quadratic.

If there's to be an intersection regardless of c, then we must have D>=0 for all c. But D is a quadratic function of c, so for that to be true, the discriminant of D (considered as a function of c) must be >=0. This gives a quadratic inequality in a, which eventually gives the desired result.

For Q6, try a substitution x=y-a for a suitable choice of a.

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Zhen Lin

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Ah. Do the discriminant inequality twice... of course. I did it once, and then for some reason or other decided to minimise the discriminant by varying c in order to eliminate c as a variable. But since the first inequality requires $D_1(c) \ge 0$ , shouldn't the second inequality be $D_2(a) \le 0$ ?

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DFranklin

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(Original post by Zhen Lin)
Ah. Do the discriminant inequality twice... of course. I did it once, and then for some reason or other decided to minimise the discriminant by varying c in order to eliminate c as a variable. But since the first inequality requires $D_1(c) \ge 0$ , shouldn't the second inequality be $D_2(a) \le 0$ ?

Probably. I'm getting tangled up trying not to give any of the actual equations - but I'm sure you can work it out!

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DFranklin

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Re Q5: I think the idea of the D(a,b)c+D(c, a)b + D(b, c)a=0 identity was to show you can write c as a combination of b and a. So you can find the image of c under the transformation, and check if this equals r. I haven't actually done any working to see if it's noticably easier than what you did, but my feeling is that it should be.

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DFranklin

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Re: Q8. When it comes to solving the recurrence relation, I think you've approached it the wrong way around. Essentially, you've derived a formula for the solution, but you haven't shown it's the only possible solution.

In fact, to answer the question you don't need to derive it. Consider

$\displaystyle c_n = \frac{\alpha^{2n-1} + \alpha^{-(2n-1)}}{\sqrt 5}$

All you need is to show

, and $c_n = 3c_{n-1}-c_{n-2}$ . Then write

; you have

and $b_n = 3b_{n-1}-b_{n-2}$ , so trivially b_n = 0 for all n by induction.

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Zhen Lin

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(Original post by DFranklin)
Re Q5: I think the idea of the D(a,b)c+D(c, a)b + D(b, c)a=0 identity was to show you can write c as a combination of b and a.

Hmm, but is it really a linear combination? Since the coordinates of c are being used on the other side of the resulting equation:

$\displaystyle \mathbf{c} = \frac{\Delta(\mathbf{b}, \mathbf{c}) \mathbf{a} + \Delta(\mathbf{c}, \mathbf{a}) \mathbf{b}}{\Delta(\mathbf{a}, \mathbf{b})}$

But I suppose it could be considered one since a, b and c are known constants.

(Original post by DFranklin)
Re: Q8. When it comes to solving the recurrence relation, I think you've approached it the wrong way around. Essentially, you've derived a formula for the solution, but you haven't shown it's the only possible solution.

Gaa. But I do not see how your method shoes that it is the only possible solution either - it certainly shows that it is a solution, and in a very simple manner indeed.

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DFranklin

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(Original post by Zhen Lin)
Hmm, but is it really a linear combination? Since the coordinates of c are being used on the other side of the resulting equation:

$\displaystyle \mathbf{c} = \frac{\Delta(\mathbf{b}, \mathbf{c}) \mathbf{a} + \Delta(\mathbf{c}, \mathbf{a}) \mathbf{b}}{\Delta(\mathbf{a}, \mathbf{b})}$

Yes, it's a linear combination. In the sense that we can find scalars $\lambda, \mu$ with ${\bf c} = \lambda {\bf a} +\mu {\bf b}$ . The fact that $\lambda, \mu$ are functions of c doesn't matter.

Gaa. But I do not see how your method shoes that it is the only possible solution either - it certainly shows that it is a solution, and in a very simple manner indeed.

I think it does show it's the only solution. Because if a_n is a solution and c_n is a solution, then b_n = a_n-c_n is zero for all n, so a_n = c_n for all n.

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Zhen Lin

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Hmm. I think I must have misunderstood what you meant by "only solution". :-) I thought it meant that there were no other functions which would give the same sequence for all n... that probably can't be proven. Or rather, is easily disproven... after all, $a_n + sin(n\pi) = a_n \, \forall n \in \mathbb{N}$ .

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generalebriety

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(Original post by Zhen Lin)
Hmm. I think I must have misunderstood what you meant by "only solution". :-) I thought it meant that there were no other functions which would give the same sequence for all n... that probably can't be proven. Or rather, is easily disproven... after all, $a_n + sin(n\pi) = a_n \, \forall n \in \mathbb{N}$ .

Surely, if two functions give the same outputs for the same inputs, they're the same function?

If c_n = a_n + sin(n.pi), then c_n - a_n = 0 for all n, so c_n = a_n. sin(n.pi) is always zero, and you wouldn't call a_n + 0 something different from a_n, would you?

A function is defined by its inputs and respective outputs, not the actual way you write the function. For example, |x| and sqrt(x^2) are the same function, even though they're written differently; they have the same domain, and for each x, we have that |x| = sqrt(x^2) holds. So they're the same.

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Zhen Lin

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Ah, but it's not all inputs, now is it? I mean, $a_n + \sin(n\pi) \ne a_n \, \forall n \notin \mathbb{N}$ , and $| x | \ne \sqrt{x^2} \, \forall x \notin \mathbb{R}$ . But yes, I suppose within the domain in question it is the same function.

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generalebriety

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(Original post by Zhen Lin)
Ah, but it's not all inputs, now is it? I mean, $a_n + \sin(n\pi) \ne a_n \, \forall n \notin \mathbb{N}$ , and $| x | \ne \sqrt{x^2} \, \forall x \notin \mathbb{R}$ . But yes, I suppose within the domain in question it is the same function.

Functions are defined over domains, though. The domain is a crucial part of the definition of a given function. And I assume here that the domain of the function is the integers or something (without bothering to look back at the question), right?

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squeezebox

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I know this is old, but regarding question 8, is it expected that I know how to solve recurrence relations in that way?
Would proving the result by induction suffice?

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DFranklin

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(Original post by squeezebox)
I know this is old, but regarding question 8, is it expected that I know how to solve recurrence relations in that way?

The classic 'examiners' answer' would be "you only need to know the official STEP syllabus". The pragmatic answer is that it's probably worthwhile knowing it. (The basic concept is very simply: "guess" that there are solutions of the form $x^\lambda$ and use that to get a quadratic in $\lambda$ . It's very similar to how you solve 2nd order diff equations).

It's probably worth knowing that the explicit formula for the Fibonacci series as well. (e.g. http://www.mcs.surrey.ac.uk/Personal...a.html#formula)

Would proving the result by induction suffice?

Yes, and in this case it's the better solution, as I explained to Zhen.

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squeezebox

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(Original post by DFranklin)
The classic 'examiners' answer' would be "you only need to know the official STEP syllabus". The pragmatic answer is that it's probably worthwhile knowing it. (The basic concept is very simply: "guess" that there are solutions of the form $x^\lambda$ and use that to get a quadratic in $\lambda$ . It's very similar to how you solve 2nd order diff equations).

It's probably worth knowing that the explicit formula for the Fibonacci series as well. (e.g. http://www.mcs.surrey.ac.uk/Personal...a.html#formula)

Yes, and in this case it's the better solution, as I explained to Zhen

Thats a relief, I would never have come up with that!

Since the method is similar to solving 2nd order diff equations, if there was a repeated solution, would it be like $(C + Dn)x^\lambda$ ?

Thanks

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DFranklin

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[QUOTE=squeezebox]Thats a relief, I would never have come up with that!

Since the method is similar to solving 2nd order diff equations, if there was a repeated solution, would it be like $(C + Dn)x^\lambda$ Yes (that's a very inspired guess - I'm impressed!)

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Trevor 12345

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I know this thread is a little old now, however I think the op's solution to Q1 isn't quite what the examiner is looking for. They want you to find that the point m is at (-b,a), rather than show that (-b,a) fits the equation, with a little thought and care it is a fairly straightforwards result to find.

I also feel that the final "proof" needs a little more too in order to gain the marks.

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DFranklin

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(Original post by Mrm.)
I know this thread is a little old now, however I think the op's solution to Q1 isn't quite what the examiner is looking for. They want you to find that the point m is at (-b,a), rather than show that (-b,a) fits the equation, with a little thought and care it is a fairly straightforwards result to find.

It might not be exactly what they want, but I think it very unlikely you'd lose any marks. It's obvious there's only one intersection, so if you show (-b,a) is on both lines, then the point M must be (-b,a).

I also feel that the final "proof" needs a little more too in order to gain the marks.

Again, I think you'd be OK.

Without disregarding your comments, in both cases I think the answer falls into the "If it's not what the examiners wanted, they should have worded the question better". I really doubt the examiners will take marks off in such cases - they very much tend to give the benefit of the doubt.

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Trevor 12345

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(Original post by DFranklin)
It might not be exactly what they want, but I think it very unlikely you'd lose any marks. It's obvious there's only one intersection, so if you show (-b,a) is on both lines, then the point M must be (-b,a).

Again, I think you'd be OK.

Without disregarding your comments, in both cases I think the answer falls into the "If it's not what the examiners wanted, they should have worded the question better". I really doubt the examiners will take marks off in such cases - they very much tend to give the benefit of the doubt.

Hi Dave

in hindsight you are probably correct. I just had the feeling that enough information is given to actually find the co-ordinates of m, as opposed o simply substituting in the values into the expression. More often than not in A-levels and other school based examinations a question so worded implies that the solution given is what the candidate should arrive at, as opposed to using the solution in itself to prove that it satisfies a given condition. That said I do indeed accept the points that you made.

With regards to the proof I just felt a slightly fuller and more rigorous explanation is possible, Perhaps something along the lines of : consider the matrix R = (row 1 = 0 1) (row 2 = -1 0) acting on the point (a,b)...det R = 1 so distance from the centre of rotation is preserved, rotation is of 90 degrees ... etc...)

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