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STEP I, II, III 2000 solutions

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Sorry if this is unecessary but I found probably a slightly easier start to III Q2.

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Reply 81
Avatar for rnd
rnd
6
DeanK22
II / 1

Guess a solution to 1N=1a+1b \displaystyle \frac{1}{N} = \frac{1}{a} + \frac{1}{b} you are provided with 12=13+16 \displaystyle \frac{1}{2} = \frac{1}{3} + \frac{1}{6} and also 13=14+112 \displaystyle \frac{1}{3} = \frac{1}{4} + \frac{1}{12}

The solution is clear - 1N=1N+1+1N2+N \displaystyle \frac{1}{N} = \frac{1}{N + 1} + \frac{1}{N^2 + N}

To prove this is so; 1N≡1N+1+1N2+N=N+1N(N+1)=1N \displaystyle \frac{1}{N} \equiv \frac{1}{N + 1} + \frac{1}{N^2 + N} = \frac{N+1}{N(N+1)} = \frac{1}{N}

Looking at (a−N)(b−N)=N2 \displaystyle (a-N)(b-N) = N^2 for N is a prime it is apparent that the only factors of N^2 are N2=N×N=N2×1 \displaystyle N^2 = N \times N = N^2 \times 1

This would imply that if distinct natural numbers (a,b) are sought for the equation 1N=1a+1b \frac{1}{N} = \frac{1}{a} + \frac{1}{b} for N prime then there is only one pair of solutions.

wip for 2N \displaystyle \frac{2}{N}


wip?

Is it OK just to say

If 1N=1N+1+1N(N+1)\frac{1}{N}=\frac{1}{N+1}+\frac{1}{N(N+1)}

2N=2N+1+2N(N+1)\frac{2}{N}=\frac{2}{N+1}+\frac{2}{N(N+1)}

2N=1(N+1)/2+1N(N+1)/2\frac{2}{N}=\frac{1}{(N+1)/2}+\frac{1}{N(N+1)/2}

and these denominators are distinct integers as N is odd and the uniqueness follows from part 1.

?
Reply 82
Avatar for rnd
rnd
6
Step II q 2 last part (alternative version) Don't suppose it's what they had in mind but is it OK?

(x−a)4(x-a)^4 is a factor of x6+4x5−5x4−40x3−40x2+32x+kx^6+4x^5-5x^4-40x^3-40x^2+32x+k

So factorising...(actually I divided)

x6+4x5−5x4−40x3−40x2+32x+k=(x−a)4(x2+4(1+a)x+10a2+16a−5)x^6+4x^5-5x^4-40x^3-40x^2+32x+k=(x-a)^4(x^2+4(1+a)x+10a^2+16a-5)

Expanding and comparing coefficients of x^3

20a3+40a2−20a−40=020a^3+40a^2-20a-40=0

a3+2a2−a−2=0a^3+2a^2-a-2=0

a3−a+2a2−2=0a^3-a+2a^2-2=0

a(a2−1)+2(a2−1)=0a(a^2-1)+2(a^2-1)=0

(a+2)(a2−1)=0(a+2)(a^2-1)=0

a=-2 turns out to be the right one and then k=a4(10a2+16a−5)=48k=a^4(10a^2+16a-5)=48
I would like a little more justification than "a=-2 turns out to be the right one" (although I haven't got the question to hand - there may be an obvious reason).

Only other thing is that in STEP, if they say "hence", you won't get marks for an alternative method. (Again, I don't know if they do say hence).
Reply 84
Avatar for rnd
rnd
6
Thanks for the feedback.

The first part of the question was to show that if (x-a)^2 is a factor of p(x) then p'(a)=0 and then the same thing for (x-a)^4.

As for justifying a=-2 rather than -1 or 1 is it OK to say something to the effect that with a=-2 all the other coefficients come out right?

There was no hence this time but I wasn't happy about not using the first part of the quesiton.
rnd
Thanks for the feedback.

The first part of the question was to show that if (x-a)^2 is a factor of p(x) then p'(a)=0 and then the same thing for (x-a)^4.

As for justifying a=-2 rather than -1 or 1 is it OK to say something to the effect that with a=-2 all the other coefficients come out right?Not really. Two reasons: firstly "saying isn't showing". That is, if you say "with a=-2 all the other coefficients come out right", I'm left thinking you probably haven't even checked that, but are simply hoping that it's true. You need to actually show the calculations.

Secondly, it's not enough to show "it works for a = -2"; that doesn't rule out a = 1 or a = -1. So you need to show it doesn't work for those cases.
Reply 86
Avatar for rnd
rnd
6
DFranklin
Not really. Two reasons: firstly "saying isn't showing". That is, if you say "with a=-2 all the other coefficients come out right", I'm left thinking you probably haven't even checked that, but are simply hoping that it's true. You need to actually show the calculations.

Secondly, it's not enough to show "it works for a = -2"; that doesn't rule out a = 1 or a = -1. So you need to show it doesn't work for those cases.


Thanks again. Well, :smile:, when I checked and saw I had the same answer as Dean I got lazy. If I ever took a STEP exam I would be more careful. I've checked now and it all works out. What I meant though was, even given all that checking, is this a valid solution?
As long as you've shown "it works for a = -2, and it doesn't work for a = 1 or a = -1", then yes. It's still not enough to have simply shown "it works for a = -2".

(And I do appreciate why you may want to take short-cuts here that you wouldn't in an actual exam. But if you ask me "is this OK?" in an "exam question thread", I'm going to tell you whether I think it's OK to write in an exam).
Reply 88
Avatar for rnd
rnd
6
DFranklin
...I'm going to tell you whether I think it's OK to write in an exam...


and I'm glad that you do. I appreciate your help and your attention to detail.

Edit: I just read posts 82 and 83 above. Food for thought there for me perhaps.
Reply 89
Avatar for rnd
rnd
6
Full post here:http://www.thestudentroom.co.uk/showpost.php?p=17773188&postcount=60

I expanded tan⁡(x+π/4)\tan(x+\pi/4) to get tan⁡(x+π/4)=1+2x+...\tan(x+\pi/4)=1+2x+... which leads to the required result.

As ever, corrections and comments are welcome.
Reply 90
Avatar for rnd
rnd
6

By sheer observation the complex number β=(1+i)(4−i)3(−20+i) \displaystyle \beta = (1+i)(4-i)^3(-20+i) shows the fact; 3arctan(14)+arctan(120)+arctan(11985)=π4 \displaystyle 3arctan(\frac{1}{4}) + arctan(\frac{1}{20}) + arctan(\frac{1}{1985}) = \frac{\pi}{4}


Full post here http://www.thestudentroom.co.uk/showpost.php?p=17174111&postcount=54

Well, I can't do "sheer observation" so I went for sheer calculation.

Rearranging the required result to make calculation easier then establishing that

arg⁡((4+i)3(20+i)(1−i))=arg⁡((52+47i)(21−19i))=arg⁡(1985−i)\arg((4+i)^3(20+i)(1-i))=\arg((52+47i)(21-19i))=\arg(1985-i)

This can be rearranged to give the requred result.
Reply 91
Avatar for amoled
amoled
3
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 D1(c)≥0D_1(c) \ge 0, shouldn't the second inequality be D2(a)≤0D_2(a) \le 0?


Ok i know this is like 3 years old but Im stuck on this question and am wondering why the discriminant of the discriminant must be negative. It seems to give the required result that way but i dont really understand why.
Thanks for your help. I hope this is the right place to ask this.

amoled.
Reply 92
Avatar for Nazgul855
Nazgul855
There's an easier way to do the second part of question 2 paper 1. Notice that:
(x^2 -1)(x^4+ x^2 +1)=x^6 -1
Then write out a gereral term for each side and you dont have to do any multiplying out. Message me if you want me to explain this further.
(edited 12 years ago)
Reply 93
Avatar for Zuzuzu
Zuzuzu
12
Second part to STEP II question 1 which was left as WIP:

1N=1N+1+1N(N+1)⇒2N=2N+1+2N(N+1) \displaystyle \frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)} \Rightarrow \frac{2}{N} = \frac{2}{N+1} + \frac{2}{N(N+1)} .

We know N is odd, as it is a prime > 2. Therefore, let N=2k+1,k∈Z \displaystyle N = 2k + 1, k \in \mathbb{Z},
so 2N=2N+1+2N(N+1) \displaystyle \frac{2}{N} = \frac{2}{N+1} + \frac{2}{N(N+1)}
[INDENT]=22k+2+2(2k+1)(2k+2)\displaystyle = \frac{2}{2k+2} + \frac{2}{(2k+1)(2k+2)}[/INDENT]
[INDENT]=1k+1+1(k+1)(2k+1)\displaystyle= \frac{1}{k+1} + \frac{1}{(k+1)(2k+1)}[/INDENT]

From which it follows using the working from the previous part of the question, that 2N\displaystyle \frac{2}{N} can be expressed uniquely as the sum of two unit fractions.
(edited 12 years ago)
I don't follow Brian everit's solution to III Q6 in the bit about showing that there is always a non-negative root. He says that the product of the roots is non-negative. How does he know that? where did he find the other roots? All we know is that a^2 is a root, right?
Original post by ben-smith
I don't follow Brian everit's solution to III Q6 in the bit about showing that there is always a non-negative root. He says that the product of the roots is non-negative. How does he know that? where did he find the other roots? All we know is that a^2 is a root, right?
Suppose the roots are α,β,γ\alpha, \beta, \gamma. Then

u3+2pu2+(p2−4r)u−q2=(u−α)(u−β)(u−γ)u^3+2pu^2+(p^2-4r)u-q^2 = (u-\alpha)(u-\beta)(u-\gamma).

By considering the constant term, αβγ=q2\alpha \beta \gamma = q^2.
Original post by DFranklin
Suppose the roots are α,β,γ\alpha, \beta, \gamma. Then

u3+2pu2+(p2−4r)u−q2=(u−α)(u−β)(u−γ)u^3+2pu^2+(p^2-4r)u-q^2 = (u-\alpha)(u-\beta)(u-\gamma).

By considering the constant term, αβγ=q2\alpha \beta \gamma = q^2.


Cheers. I should have seen that.
why can't you just say a^2 is a root and it is non-negative so the cubic always has a non negative root?
Seems to trivialise the question which I guess is a bad sign...
Reply 97
Avatar for nuodai
nuodai
17
Original post by ben-smith
Cheers. I should have seen that.
why can't you just say a^2 is a root and it is non-negative so the cubic always has a non negative root?
Seems to trivialise the question which I guess is a bad sign...


I think the word "always" comes into play here; i.e. for any values of p,q,r, rather than just the ones in this instance that depend upon a,b,c.
The question isn't brilliantly worded, but I think the way it's supposed to flow is:

Suppose this quartic identity holds. Using the various relationships between p,q,r and a,b,c, show that a^2 is a root of {cubic}.

Now, suppose you have a cubic of the form {cubic}. Show that this cubic always has a non-negative root.

In particular, in the 2nd case you should not assume that we can find a, b, c such that the given quartic identity holds.

I doubt it was many marks though - I think the main intent is to get you to look at a certain cubic that may help for the last bit (and to explciitly point out a root of that cubic).

I think what Brian did diverges from the examiners plan for the next part, incidentally.
For what the examiners were probably expecting, have a look at Square's solution at http://www.thestudentroom.co.uk/showpost.php?p=12420496&postcount=47 (and also my comment at #49).

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